Optimal. Leaf size=200 \[ -\frac {e \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}+\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d^2}-\frac {a+b \tanh ^{-1}(c x)}{d x}-\frac {a e \log (x)}{d^2}-\frac {b c \log \left (1-c^2 x^2\right )}{2 d}+\frac {b e \text {Li}_2(-c x)}{2 d^2}-\frac {b e \text {Li}_2(c x)}{2 d^2}+\frac {b e \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{2 d^2}-\frac {b e \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{2 d^2}+\frac {b c \log (x)}{d} \]
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Rubi [A] time = 0.20, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.579, Rules used = {5940, 5916, 266, 36, 29, 31, 5912, 5920, 2402, 2315, 2447} \[ \frac {b e \text {PolyLog}(2,-c x)}{2 d^2}-\frac {b e \text {PolyLog}(2,c x)}{2 d^2}+\frac {b e \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{2 d^2}-\frac {b e \text {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 d^2}-\frac {e \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}+\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d^2}-\frac {a+b \tanh ^{-1}(c x)}{d x}-\frac {a e \log (x)}{d^2}-\frac {b c \log \left (1-c^2 x^2\right )}{2 d}+\frac {b c \log (x)}{d} \]
Antiderivative was successfully verified.
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Rule 29
Rule 31
Rule 36
Rule 266
Rule 2315
Rule 2402
Rule 2447
Rule 5912
Rule 5916
Rule 5920
Rule 5940
Rubi steps
\begin {align*} \int \frac {a+b \tanh ^{-1}(c x)}{x^2 (d+e x)} \, dx &=\int \left (\frac {a+b \tanh ^{-1}(c x)}{d x^2}-\frac {e \left (a+b \tanh ^{-1}(c x)\right )}{d^2 x}+\frac {e^2 \left (a+b \tanh ^{-1}(c x)\right )}{d^2 (d+e x)}\right ) \, dx\\ &=\frac {\int \frac {a+b \tanh ^{-1}(c x)}{x^2} \, dx}{d}-\frac {e \int \frac {a+b \tanh ^{-1}(c x)}{x} \, dx}{d^2}+\frac {e^2 \int \frac {a+b \tanh ^{-1}(c x)}{d+e x} \, dx}{d^2}\\ &=-\frac {a+b \tanh ^{-1}(c x)}{d x}-\frac {a e \log (x)}{d^2}-\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^2}+\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d^2}+\frac {b e \text {Li}_2(-c x)}{2 d^2}-\frac {b e \text {Li}_2(c x)}{2 d^2}+\frac {(b c) \int \frac {1}{x \left (1-c^2 x^2\right )} \, dx}{d}+\frac {(b c e) \int \frac {\log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d^2}-\frac {(b c e) \int \frac {\log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{1-c^2 x^2} \, dx}{d^2}\\ &=-\frac {a+b \tanh ^{-1}(c x)}{d x}-\frac {a e \log (x)}{d^2}-\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^2}+\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d^2}+\frac {b e \text {Li}_2(-c x)}{2 d^2}-\frac {b e \text {Li}_2(c x)}{2 d^2}-\frac {b e \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d^2}+\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )}{2 d}+\frac {(b e) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+c x}\right )}{d^2}\\ &=-\frac {a+b \tanh ^{-1}(c x)}{d x}-\frac {a e \log (x)}{d^2}-\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^2}+\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d^2}+\frac {b e \text {Li}_2(-c x)}{2 d^2}-\frac {b e \text {Li}_2(c x)}{2 d^2}+\frac {b e \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{2 d^2}-\frac {b e \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d^2}+\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 d}+\frac {\left (b c^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-c^2 x} \, dx,x,x^2\right )}{2 d}\\ &=-\frac {a+b \tanh ^{-1}(c x)}{d x}+\frac {b c \log (x)}{d}-\frac {a e \log (x)}{d^2}-\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^2}+\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d^2}-\frac {b c \log \left (1-c^2 x^2\right )}{2 d}+\frac {b e \text {Li}_2(-c x)}{2 d^2}-\frac {b e \text {Li}_2(c x)}{2 d^2}+\frac {b e \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{2 d^2}-\frac {b e \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d^2}\\ \end {align*}
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Mathematica [C] time = 3.22, size = 360, normalized size = 1.80 \[ -\frac {\frac {2 a d^2}{x}+2 a d e \log (x)-2 a d e \log (d+e x)+\frac {b e^2 \sqrt {1-\frac {c^2 d^2}{e^2}} \tanh ^{-1}(c x)^2 e^{-\tanh ^{-1}\left (\frac {c d}{e}\right )}}{c}-2 b c d^2 \log \left (\frac {c x}{\sqrt {1-c^2 x^2}}\right )+\frac {1}{2} i \pi b d e \log \left (1-c^2 x^2\right )+\frac {2 b d^2 \tanh ^{-1}(c x)}{x}-b d e \text {Li}_2\left (e^{-2 \tanh ^{-1}(c x)}\right )+b d e \text {Li}_2\left (e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )+b d e \tanh ^{-1}(c x)^2-2 b d e \tanh ^{-1}(c x) \tanh ^{-1}\left (\frac {c d}{e}\right )-i \pi b d e \tanh ^{-1}(c x)+2 b d e \tanh ^{-1}(c x) \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )+i \pi b d e \log \left (e^{2 \tanh ^{-1}(c x)}+1\right )-2 b d e \tanh ^{-1}\left (\frac {c d}{e}\right ) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-2 b d e \tanh ^{-1}(c x) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )+2 b d e \tanh ^{-1}\left (\frac {c d}{e}\right ) \log \left (i \sinh \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )\right )-\frac {b e^2 \tanh ^{-1}(c x)^2}{c}}{2 d^3} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {artanh}\left (c x\right ) + a}{e x^{3} + d x^{2}}, x\right ) \]
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 \[ \int \frac {b \operatorname {artanh}\left (c x\right ) + a}{{\left (e x + d\right )} x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 279, normalized size = 1.40 \[ -\frac {a}{d x}-\frac {a e \ln \left (c x \right )}{d^{2}}+\frac {a e \ln \left (c x e +c d \right )}{d^{2}}-\frac {b \arctanh \left (c x \right )}{d x}-\frac {b \arctanh \left (c x \right ) e \ln \left (c x \right )}{d^{2}}+\frac {b \arctanh \left (c x \right ) e \ln \left (c x e +c d \right )}{d^{2}}+\frac {c b \ln \left (c x \right )}{d}-\frac {c b \ln \left (c x -1\right )}{2 d}-\frac {c b \ln \left (c x +1\right )}{2 d}+\frac {b e \dilog \left (c x \right )}{2 d^{2}}+\frac {b e \dilog \left (c x +1\right )}{2 d^{2}}+\frac {b e \ln \left (c x \right ) \ln \left (c x +1\right )}{2 d^{2}}-\frac {b \ln \left (c x e +c d \right ) \ln \left (\frac {c x e +e}{-c d +e}\right ) e}{2 d^{2}}-\frac {b \dilog \left (\frac {c x e +e}{-c d +e}\right ) e}{2 d^{2}}+\frac {b \ln \left (c x e +c d \right ) \ln \left (\frac {c x e -e}{-c d -e}\right ) e}{2 d^{2}}+\frac {b \dilog \left (\frac {c x e -e}{-c d -e}\right ) e}{2 d^{2}} \]
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 \[ a {\left (\frac {e \log \left (e x + d\right )}{d^{2}} - \frac {e \log \relax (x)}{d^{2}} - \frac {1}{d x}\right )} + \frac {1}{2} \, b \int \frac {\log \left (c x + 1\right ) - \log \left (-c x + 1\right )}{e x^{3} + d x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {atanh}\left (c\,x\right )}{x^2\,\left (d+e\,x\right )} \,d x \]
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 \[ \int \frac {a + b \operatorname {atanh}{\left (c x \right )}}{x^{2} \left (d + e x\right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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