Optimal. Leaf size=186 \[ \frac{d \text{PolyLog}\left (2,\frac{c (-a-b x+1)}{-a c+b d+c}\right )}{2 c^2}-\frac{d \text{PolyLog}\left (2,\frac{c (a+b x+1)}{a c-b d+c}\right )}{2 c^2}+\frac{d \log (-a-b x+1) \log \left (\frac{b (c x+d)}{-a c+b d+c}\right )}{2 c^2}-\frac{d \log (a+b x+1) \log \left (-\frac{b (c x+d)}{a c-b d+c}\right )}{2 c^2}+\frac{(-a-b x+1) \log (-a-b x+1)}{2 b c}+\frac{(a+b x+1) \log (a+b x+1)}{2 b c} \]
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Rubi [A] time = 0.237145, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {6115, 2409, 2389, 2295, 2394, 2393, 2391} \[ \frac{d \text{PolyLog}\left (2,\frac{c (-a-b x+1)}{-a c+b d+c}\right )}{2 c^2}-\frac{d \text{PolyLog}\left (2,\frac{c (a+b x+1)}{a c-b d+c}\right )}{2 c^2}+\frac{d \log (-a-b x+1) \log \left (\frac{b (c x+d)}{-a c+b d+c}\right )}{2 c^2}-\frac{d \log (a+b x+1) \log \left (-\frac{b (c x+d)}{a c-b d+c}\right )}{2 c^2}+\frac{(-a-b x+1) \log (-a-b x+1)}{2 b c}+\frac{(a+b x+1) \log (a+b x+1)}{2 b c} \]
Antiderivative was successfully verified.
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Rule 6115
Rule 2409
Rule 2389
Rule 2295
Rule 2394
Rule 2393
Rule 2391
Rubi steps
\begin{align*} \int \frac{\tanh ^{-1}(a+b x)}{c+\frac{d}{x}} \, dx &=-\left (\frac{1}{2} \int \frac{\log (1-a-b x)}{c+\frac{d}{x}} \, dx\right )+\frac{1}{2} \int \frac{\log (1+a+b x)}{c+\frac{d}{x}} \, dx\\ &=-\left (\frac{1}{2} \int \left (\frac{\log (1-a-b x)}{c}-\frac{d \log (1-a-b x)}{c (d+c x)}\right ) \, dx\right )+\frac{1}{2} \int \left (\frac{\log (1+a+b x)}{c}-\frac{d \log (1+a+b x)}{c (d+c x)}\right ) \, dx\\ &=-\frac{\int \log (1-a-b x) \, dx}{2 c}+\frac{\int \log (1+a+b x) \, dx}{2 c}+\frac{d \int \frac{\log (1-a-b x)}{d+c x} \, dx}{2 c}-\frac{d \int \frac{\log (1+a+b x)}{d+c x} \, dx}{2 c}\\ &=-\frac{d \log (1+a+b x) \log \left (-\frac{b (d+c x)}{c+a c-b d}\right )}{2 c^2}+\frac{d \log (1-a-b x) \log \left (\frac{b (d+c x)}{c-a c+b d}\right )}{2 c^2}+\frac{\operatorname{Subst}(\int \log (x) \, dx,x,1-a-b x)}{2 b c}+\frac{\operatorname{Subst}(\int \log (x) \, dx,x,1+a+b x)}{2 b c}+\frac{(b d) \int \frac{\log \left (-\frac{b (d+c x)}{-(1-a) c-b d}\right )}{1-a-b x} \, dx}{2 c^2}+\frac{(b d) \int \frac{\log \left (\frac{b (d+c x)}{-(1+a) c+b d}\right )}{1+a+b x} \, dx}{2 c^2}\\ &=\frac{(1-a-b x) \log (1-a-b x)}{2 b c}+\frac{(1+a+b x) \log (1+a+b x)}{2 b c}-\frac{d \log (1+a+b x) \log \left (-\frac{b (d+c x)}{c+a c-b d}\right )}{2 c^2}+\frac{d \log (1-a-b x) \log \left (\frac{b (d+c x)}{c-a c+b d}\right )}{2 c^2}-\frac{d \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{c x}{-(1-a) c-b d}\right )}{x} \, dx,x,1-a-b x\right )}{2 c^2}+\frac{d \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{c x}{-(1+a) c+b d}\right )}{x} \, dx,x,1+a+b x\right )}{2 c^2}\\ &=\frac{(1-a-b x) \log (1-a-b x)}{2 b c}+\frac{(1+a+b x) \log (1+a+b x)}{2 b c}-\frac{d \log (1+a+b x) \log \left (-\frac{b (d+c x)}{c+a c-b d}\right )}{2 c^2}+\frac{d \log (1-a-b x) \log \left (\frac{b (d+c x)}{c-a c+b d}\right )}{2 c^2}+\frac{d \text{Li}_2\left (\frac{c (1-a-b x)}{c-a c+b d}\right )}{2 c^2}-\frac{d \text{Li}_2\left (\frac{c (1+a+b x)}{c+a c-b d}\right )}{2 c^2}\\ \end{align*}
Mathematica [C] time = 3.92718, size = 759, normalized size = 4.08 \[ \frac{b d (b d-a c) \text{PolyLog}\left (2,\exp \left (2 \left (\tanh ^{-1}\left (a-\frac{b d}{c}\right )-\tanh ^{-1}(a+b x)\right )\right )\right )+b d (a c-b d) \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a+b x)}\right )+b c d \sqrt{-a^2+\frac{2 a b d}{c}-\frac{b^2 d^2}{c^2}+1} \tanh ^{-1}(a+b x)^2 e^{\tanh ^{-1}\left (a-\frac{b d}{c}\right )}-2 a^2 c^2 \tanh ^{-1}(a+b x)+2 b^2 d^2 \tanh ^{-1}\left (a-\frac{b d}{c}\right ) \log \left (1-\exp \left (2 \left (\tanh ^{-1}\left (a-\frac{b d}{c}\right )-\tanh ^{-1}(a+b x)\right )\right )\right )-2 b^2 d^2 \tanh ^{-1}(a+b x) \log \left (1-\exp \left (2 \left (\tanh ^{-1}\left (a-\frac{b d}{c}\right )-\tanh ^{-1}(a+b x)\right )\right )\right )+2 b^2 d^2 \tanh ^{-1}(a+b x) \tanh ^{-1}\left (a-\frac{b d}{c}\right )-2 b^2 d^2 \tanh ^{-1}\left (a-\frac{b d}{c}\right ) \log \left (-i \sinh \left (\tanh ^{-1}\left (a-\frac{b d}{c}\right )-\tanh ^{-1}(a+b x)\right )\right )+2 b^2 c d x \tanh ^{-1}(a+b x)-i \pi b^2 d^2 \log \left (\frac{1}{\sqrt{1-(a+b x)^2}}\right )+b^2 d^2 \tanh ^{-1}(a+b x)^2-i \pi b^2 d^2 \tanh ^{-1}(a+b x)+2 b^2 d^2 \tanh ^{-1}(a+b x) \log \left (e^{-2 \tanh ^{-1}(a+b x)}+1\right )+i \pi b^2 d^2 \log \left (e^{2 \tanh ^{-1}(a+b x)}+1\right )+2 a c^2 \log \left (\frac{1}{\sqrt{1-(a+b x)^2}}\right )-2 a b c^2 x \tanh ^{-1}(a+b x)-2 a b c d \tanh ^{-1}\left (a-\frac{b d}{c}\right ) \log \left (1-\exp \left (2 \left (\tanh ^{-1}\left (a-\frac{b d}{c}\right )-\tanh ^{-1}(a+b x)\right )\right )\right )+2 a b c d \tanh ^{-1}(a+b x) \log \left (1-\exp \left (2 \left (\tanh ^{-1}\left (a-\frac{b d}{c}\right )-\tanh ^{-1}(a+b x)\right )\right )\right )-2 b c d \log \left (\frac{1}{\sqrt{1-(a+b x)^2}}\right )+i \pi a b c d \log \left (\frac{1}{\sqrt{1-(a+b x)^2}}\right )-a b c d \tanh ^{-1}(a+b x)^2-b c d \tanh ^{-1}(a+b x)^2+2 a b c d \tanh ^{-1}(a+b x)-2 a b c d \tanh ^{-1}(a+b x) \tanh ^{-1}\left (a-\frac{b d}{c}\right )+i \pi a b c d \tanh ^{-1}(a+b x)-2 a b c d \tanh ^{-1}(a+b x) \log \left (e^{-2 \tanh ^{-1}(a+b x)}+1\right )-i \pi a b c d \log \left (e^{2 \tanh ^{-1}(a+b x)}+1\right )+2 a b c d \tanh ^{-1}\left (a-\frac{b d}{c}\right ) \log \left (-i \sinh \left (\tanh ^{-1}\left (a-\frac{b d}{c}\right )-\tanh ^{-1}(a+b x)\right )\right )}{2 b c^2 (b d-a c)} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.197, size = 297, normalized size = 1.6 \begin{align*}{\frac{{\it Artanh} \left ( bx+a \right ) x}{c}}+{\frac{{\it Artanh} \left ( bx+a \right ) a}{bc}}-{\frac{{\it Artanh} \left ( bx+a \right ) d\ln \left ( c \left ( bx+a \right ) -ac+bd \right ) }{{c}^{2}}}-{\frac{d\ln \left ( c \left ( bx+a \right ) -ac+bd \right ) }{2\,{c}^{2}}\ln \left ({\frac{c \left ( bx+a \right ) -c}{ac-bd-c}} \right ) }-{\frac{d}{2\,{c}^{2}}{\it dilog} \left ({\frac{c \left ( bx+a \right ) -c}{ac-bd-c}} \right ) }+{\frac{d\ln \left ( c \left ( bx+a \right ) -ac+bd \right ) }{2\,{c}^{2}}\ln \left ({\frac{c \left ( bx+a \right ) +c}{ac-bd+c}} \right ) }+{\frac{d}{2\,{c}^{2}}{\it dilog} \left ({\frac{c \left ( bx+a \right ) +c}{ac-bd+c}} \right ) }+{\frac{\ln \left ({a}^{2}{c}^{2}-2\,abcd+{b}^{2}{d}^{2}+2\, \left ( c \left ( bx+a \right ) -ac+bd \right ) ac-2\, \left ( c \left ( bx+a \right ) -ac+bd \right ) bd+ \left ( c \left ( bx+a \right ) -ac+bd \right ) ^{2}-{c}^{2} \right ) }{2\,bc}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.961282, size = 259, normalized size = 1.39 \begin{align*} \frac{1}{2} \, b{\left (\frac{{\left (\log \left (c x + d\right ) \log \left (\frac{b c x + b d}{a c - b d + c} + 1\right ) +{\rm Li}_2\left (-\frac{b c x + b d}{a c - b d + c}\right )\right )} d}{b c^{2}} - \frac{{\left (\log \left (c x + d\right ) \log \left (\frac{b c x + b d}{a c - b d - c} + 1\right ) +{\rm Li}_2\left (-\frac{b c x + b d}{a c - b d - c}\right )\right )} d}{b c^{2}} + \frac{{\left (a + 1\right )} \log \left (b x + a + 1\right )}{b^{2} c} - \frac{{\left (a - 1\right )} \log \left (b x + a - 1\right )}{b^{2} c}\right )} +{\left (\frac{x}{c} - \frac{d \log \left (c x + d\right )}{c^{2}}\right )} \operatorname{artanh}\left (b x + a\right ) \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{x \operatorname{artanh}\left (b x + a\right )}{c x + d}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \operatorname{atanh}{\left (a + b x \right )}}{c x + d}\, dx \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{\operatorname{artanh}\left (b x + a\right )}{c + \frac{d}{x}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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