Optimal. Leaf size=114 \[ -\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{e}+\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e}+\frac {b \text {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{2 e}-\frac {b \text {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e} \]
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Rubi [A]
time = 0.05, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6057, 2449,
2352, 2497} \begin {gather*} \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e}-\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e}-\frac {b \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{2 e}+\frac {b \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{2 e} \end {gather*}
Antiderivative was successfully verified.
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Rule 2352
Rule 2449
Rule 2497
Rule 6057
Rubi steps
\begin {align*} \int \frac {a+b \tanh ^{-1}(c x)}{d+e x} \, dx &=-\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{e}+\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e}+\frac {(b c) \int \frac {\log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{e}-\frac {(b c) \int \frac {\log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{1-c^2 x^2} \, dx}{e}\\ &=-\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{e}+\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e}-\frac {b \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e}+\frac {b \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+c x}\right )}{e}\\ &=-\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{e}+\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e}+\frac {b \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{2 e}-\frac {b \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e}\\ \end {align*}
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Mathematica [C] Result contains complex when optimal does not.
time = 0.18, size = 244, normalized size = 2.14 \begin {gather*} \frac {a \log (d+e x)}{e}+\frac {b \left (-\frac {1}{4} \left (\pi -2 i \tanh ^{-1}(c x)\right )^2+\left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )^2-i \left (\pi -2 i \tanh ^{-1}(c x)\right ) \log \left (1+e^{2 \tanh ^{-1}(c x)}\right )+2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right ) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )+\left (i \pi +2 \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{\sqrt {1-c^2 x^2}}\right )+\tanh ^{-1}(c x) \left (\log \left (1-c^2 x^2\right )+2 \log \left (i \sinh \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )\right )\right )-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right ) \log \left (2 i \sinh \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )\right )-\text {PolyLog}\left (2,-e^{2 \tanh ^{-1}(c x)}\right )-\text {PolyLog}\left (2,e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )\right )}{2 e} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.19, size = 158, normalized size = 1.39
method | result | size |
derivativedivides | \(\frac {\frac {a c \ln \left (c e x +d c \right )}{e}+\frac {b c \ln \left (c e x +d c \right ) \arctanh \left (c x \right )}{e}+\frac {b c \ln \left (c e x +d c \right ) \ln \left (\frac {c e x -e}{-d c -e}\right )}{2 e}+\frac {b c \dilog \left (\frac {c e x -e}{-d c -e}\right )}{2 e}-\frac {b c \ln \left (c e x +d c \right ) \ln \left (\frac {c e x +e}{-d c +e}\right )}{2 e}-\frac {b c \dilog \left (\frac {c e x +e}{-d c +e}\right )}{2 e}}{c}\) | \(158\) |
default | \(\frac {\frac {a c \ln \left (c e x +d c \right )}{e}+\frac {b c \ln \left (c e x +d c \right ) \arctanh \left (c x \right )}{e}+\frac {b c \ln \left (c e x +d c \right ) \ln \left (\frac {c e x -e}{-d c -e}\right )}{2 e}+\frac {b c \dilog \left (\frac {c e x -e}{-d c -e}\right )}{2 e}-\frac {b c \ln \left (c e x +d c \right ) \ln \left (\frac {c e x +e}{-d c +e}\right )}{2 e}-\frac {b c \dilog \left (\frac {c e x +e}{-d c +e}\right )}{2 e}}{c}\) | \(158\) |
risch | \(-\frac {b \dilog \left (\frac {\left (-c x +1\right ) e -d c -e}{-d c -e}\right )}{2 e}-\frac {b \ln \left (-c x +1\right ) \ln \left (\frac {\left (-c x +1\right ) e -d c -e}{-d c -e}\right )}{2 e}+\frac {a \ln \left (\left (-c x +1\right ) e -d c -e \right )}{e}+\frac {b \dilog \left (\frac {\left (c x +1\right ) e +d c -e}{d c -e}\right )}{2 e}+\frac {b \ln \left (c x +1\right ) \ln \left (\frac {\left (c x +1\right ) e +d c -e}{d c -e}\right )}{2 e}\) | \(167\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \operatorname {atanh}{\left (c x \right )}}{d + e x}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\mathrm {atanh}\left (c\,x\right )}{d+e\,x} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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