BUG: Timedelta created by to_timedelta does not add correctly to datetime in Pandas 2

[Flix6x]

Pandas version checks

• I have checked that this issue has not already been reported.

• I have confirmed this bug exists on the latest version of pandas.

• I have confirmed this bug exists on the main branch of pandas.

Reproducible Example

from datetime import datetime
import pandas as pd

# Create a datetime object (midnight)
dt = datetime(2015, 1, 1, 0)

# Create a pd.Timedelta object (1 hour)
delta_1 = pd.Timedelta("1H")

# Create a pd.Timedelta object from a DateTimeIndex frequency (1 hour)
delta_2 = pd.to_timedelta(pd.date_range('2015-01-01 00', '2015-01-01 01', freq="1H").freq)

# First check that the pd.Timedelta objects are equal
assert(delta_1 == delta_2)  # True

# Then check whether adding each pd.Timedelta to the datetime object moves the datetime by one hour: this does not behave correctly since Pandas 2.0
assert(dt + delta_1 == dt + delta_2)  # False -> AssertionError

# Left-hand side looks right: the hour was added
dt + delta_1  # datetime.datetime(2015, 1, 1, 1, 0)

# Right-hand side appears to be wrong: the hour was not added
dt + delta_2  # datetime.datetime(2015, 1, 1, 0, 0)

Issue Description

It seems that the pd.Timedelta object produced by calling pd.to_timedelta on a pd.DateTimeIndex frequency can no longer be added to a datetime.datetime object since pandas>1.5.3.

It's not evident to me why the two pd.Timedelta objects (appearing to be equal) are behaving differently.

Expected Behavior

Running the above example in pandas==1.5.3 (and also 1.4.4) works as expected.

Installed Versions

INSTALLED VERSIONS

commit : 4149f32
python : 3.9.9.final.0
python-bits : 64
OS : Linux
OS-release : 5.15.11-76051511-generic
Version : #202112220937164018548121.04~b3a2c21-Ubuntu SMP Mon Jan 3 16:5
machine : x86_64
processor :
byteorder : little
LC_ALL : None
LANG : en_US.UTF-8
LOCALE : en_US.UTF-8
pandas : 2.1.0.dev0+956.g4149f32388
numpy : 1.22.4
pytz : 2022.2.1
dateutil : 2.8.2
setuptools : 65.2.0
pip : 23.1.2
Cython : 0.29.30
pytest : 7.0.1
hypothesis : None
sphinx : 4.4.0
blosc : None
feather : None
xlsxwriter : None
lxml.etree : None
html5lib : 1.1
pymysql : None
psycopg2 : 2.9.3
jinja2 : 3.1.2
IPython : None
pandas_datareader: None
bs4 : 4.11.1
bottleneck : None
brotli : None
fastparquet : None
fsspec : None
gcsfs : None
matplotlib : 3.5.1
numba : 0.56.0
numexpr : None
odfpy : None
openpyxl : 3.0.10
pandas_gbq : None
pyarrow : None
pyreadstat : None
pyxlsb : None
s3fs : None
scipy : 1.8.0
snappy : None
sqlalchemy : 1.4.31
tables : None
tabulate : 0.8.9
xarray : None
xlrd : 2.0.1
zstandard : None
tzdata : 2022.5
qtpy : None
pyqt5 : None

[jbrockmendel]

I have an explanation but not a solution.

delta_2 here has unit="s". Timedeltas with unit="s" can go outside the range supported by datetime.timedelta, so the way they are initialized passes seconds=0 to the parent class (datetime.timedelta) constructor. So the pytimedelta struct (which pandas ignores) looks like timedelta(0). When you do dt + delta_2, that calls the pydatetime's __add__ method, which treats delta_2 as a regular pytimedelta, not a pd.Timedelta. So it uses the 0, giving the incorrect result.

Reversing the operation and doing delta_2 + dt gives the correct result, as does pd.Timestamp(dt) + delta_2.

A solution here would need to be in the stdlib datetime.__add__ returning NotImplemented