Optimal. Leaf size=186 \[ \frac {d \text {Li}_2\left (\frac {c (-a-b x+1)}{-a c+c+b d}\right )}{2 c^2}-\frac {d \text {Li}_2\left (\frac {c (a+b x+1)}{a c+c-b d}\right )}{2 c^2}+\frac {d \log (-a-b x+1) \log \left (\frac {b (c x+d)}{-a c+b d+c}\right )}{2 c^2}-\frac {d \log (a+b x+1) \log \left (-\frac {b (c x+d)}{a c-b d+c}\right )}{2 c^2}+\frac {(-a-b x+1) \log (-a-b x+1)}{2 b c}+\frac {(a+b x+1) \log (a+b x+1)}{2 b c} \]
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Rubi [A] time = 0.24, antiderivative size = 186, normalized size of antiderivative = 1.00, 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 2295
Rule 2389
Rule 2391
Rule 2393
Rule 2394
Rule 2409
Rule 6115
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*}
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Mathematica [C] time = 4.15, size = 759, normalized size = 4.08 \[ \frac {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)+b d (b d-a c) \text {Li}_2\left (\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}\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 )+b d (a c-b d) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(a+b x)}\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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fricas [F] time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x \operatorname {artanh}\left (b x + a\right )}{c x + d}, x\right ) \]
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 \[ \int \frac {\operatorname {artanh}\left (b x + a\right )}{c + \frac {d}{x}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.12, size = 297, normalized size = 1.60 \[ \frac {\arctanh \left (b x +a \right ) x}{c}+\frac {\arctanh \left (b x +a \right ) a}{b c}-\frac {\arctanh \left (b x +a \right ) d \ln \left (c \left (b x +a \right )-a c +b d \right )}{c^{2}}+\frac {d \ln \left (c \left (b x +a \right )-a c +b d \right ) \ln \left (\frac {c \left (b x +a \right )+c}{a c -b d +c}\right )}{2 c^{2}}+\frac {d \dilog \left (\frac {c \left (b x +a \right )+c}{a c -b d +c}\right )}{2 c^{2}}-\frac {d \ln \left (c \left (b x +a \right )-a c +b d \right ) \ln \left (\frac {c \left (b x +a \right )-c}{a c -b d -c}\right )}{2 c^{2}}-\frac {d \dilog \left (\frac {c \left (b x +a \right )-c}{a c -b d -c}\right )}{2 c^{2}}+\frac {\ln \left (a^{2} c^{2}-2 a b c d +b^{2} d^{2}+2 \left (c \left (b x +a \right )-a c +b d \right ) a c -2 \left (c \left (b x +a \right )-a c +b d \right ) b d +\left (c \left (b x +a \right )-a c +b d \right )^{2}-c^{2}\right )}{2 b c} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.32, size = 192, normalized size = 1.03 \[ \frac {1}{2} \, b {\left (\frac {{\left (\log \left (c x + d\right ) \log \left (\frac {b c x + b d}{a c - b d + c} + 1\right ) + {\rm Li}_2\left (-\frac {b c x + b d}{a c - b d + c}\right )\right )} d}{b c^{2}} - \frac {{\left (\log \left (c x + d\right ) \log \left (\frac {b c x + b d}{a c - b d - c} + 1\right ) + {\rm Li}_2\left (-\frac {b c x + b d}{a c - b d - c}\right )\right )} d}{b c^{2}} + \frac {{\left (a + 1\right )} \log \left (b x + a + 1\right )}{b^{2} c} - \frac {{\left (a - 1\right )} \log \left (b x + a - 1\right )}{b^{2} c}\right )} + {\left (\frac {x}{c} - \frac {d \log \left (c x + d\right )}{c^{2}}\right )} \operatorname {artanh}\left (b x + a\right ) \]
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 \[ \int \frac {\mathrm {atanh}\left (a+b\,x\right )}{c+\frac {d}{x}} \,d x \]
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 \[ \int \frac {x \operatorname {atanh}{\left (a + b x \right )}}{c x + d}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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