Optimal. Leaf size=200 \[ \frac{b e \text{PolyLog}(2,-c x)}{2 d^2}-\frac{b e \text{PolyLog}(2,c x)}{2 d^2}+\frac{b e \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right )}{2 d^2}-\frac{b e \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 d^2}-\frac{e \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}+\frac{e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d^2}-\frac{a+b \tanh ^{-1}(c x)}{d x}-\frac{a e \log (x)}{d^2}-\frac{b c \log \left (1-c^2 x^2\right )}{2 d}+\frac{b c \log (x)}{d} \]
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Rubi [A] time = 0.20372, antiderivative size = 200, normalized size of antiderivative = 1., 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 5940
Rule 5916
Rule 266
Rule 36
Rule 29
Rule 31
Rule 5912
Rule 5920
Rule 2402
Rule 2315
Rule 2447
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*}
Mathematica [C] time = 3.22214, size = 360, normalized size = 1.8 \[ -\frac{-b d e \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )+b d e \text{PolyLog}\left (2,e^{-2 \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )+\frac{2 a d^2}{x}+2 a d e \log (x)-2 a d e \log (d+e x)+\frac{b e^2 \sqrt{1-\frac{c^2 d^2}{e^2}} \tanh ^{-1}(c x)^2 e^{-\tanh ^{-1}\left (\frac{c d}{e}\right )}}{c}-2 b c d^2 \log \left (\frac{c x}{\sqrt{1-c^2 x^2}}\right )+\frac{1}{2} i \pi b d e \log \left (1-c^2 x^2\right )+\frac{2 b d^2 \tanh ^{-1}(c x)}{x}+b d e \tanh ^{-1}(c x)^2-2 b d e \tanh ^{-1}(c x) \tanh ^{-1}\left (\frac{c d}{e}\right )-i \pi b d e \tanh ^{-1}(c x)+2 b d e \tanh ^{-1}(c x) \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )+i \pi b d e \log \left (e^{2 \tanh ^{-1}(c x)}+1\right )-2 b d e \tanh ^{-1}\left (\frac{c d}{e}\right ) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-2 b d e \tanh ^{-1}(c x) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )+2 b d e \tanh ^{-1}\left (\frac{c d}{e}\right ) \log \left (i \sinh \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )\right )-\frac{b e^2 \tanh ^{-1}(c x)^2}{c}}{2 d^3} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.134, size = 279, normalized size = 1.4 \begin{align*}{\frac{ae\ln \left ( cxe+cd \right ) }{{d}^{2}}}-{\frac{a}{dx}}-{\frac{ae\ln \left ( cx \right ) }{{d}^{2}}}+{\frac{b{\it Artanh} \left ( cx \right ) e\ln \left ( cxe+cd \right ) }{{d}^{2}}}-{\frac{b{\it Artanh} \left ( cx \right ) }{dx}}-{\frac{b{\it Artanh} \left ( cx \right ) e\ln \left ( cx \right ) }{{d}^{2}}}-{\frac{bc\ln \left ( cx-1 \right ) }{2\,d}}+{\frac{bc\ln \left ( cx \right ) }{d}}-{\frac{bc\ln \left ( cx+1 \right ) }{2\,d}}+{\frac{be{\it dilog} \left ( cx \right ) }{2\,{d}^{2}}}+{\frac{be{\it dilog} \left ( cx+1 \right ) }{2\,{d}^{2}}}+{\frac{be\ln \left ( cx \right ) \ln \left ( cx+1 \right ) }{2\,{d}^{2}}}-{\frac{b\ln \left ( cxe+cd \right ) e}{2\,{d}^{2}}\ln \left ({\frac{cxe+e}{-cd+e}} \right ) }-{\frac{be}{2\,{d}^{2}}{\it dilog} \left ({\frac{cxe+e}{-cd+e}} \right ) }+{\frac{b\ln \left ( cxe+cd \right ) e}{2\,{d}^{2}}\ln \left ({\frac{cxe-e}{-cd-e}} \right ) }+{\frac{be}{2\,{d}^{2}}{\it dilog} \left ({\frac{cxe-e}{-cd-e}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} a{\left (\frac{e \log \left (e x + d\right )}{d^{2}} - \frac{e \log \left (x\right )}{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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \operatorname{artanh}\left (c x\right ) + a}{e x^{3} + d x^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{artanh}\left (c x\right ) + a}{{\left (e x + d\right )} x^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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