Optimal. Leaf size=203 \[ \frac {1}{4} \text {Li}_2\left (\frac {-a-b x+1}{-a-b+1}\right )-\frac {1}{4} \text {Li}_2\left (\frac {-a-b x+1}{-a+b+1}\right )+\frac {1}{4} \text {Li}_2\left (\frac {a+b x+1}{a-b+1}\right )-\frac {1}{4} \text {Li}_2\left (\frac {a+b x+1}{a+b+1}\right )+\frac {1}{4} \log \left (-\frac {b (1-x)}{-a-b+1}\right ) \log (-a-b x+1)-\frac {1}{4} \log \left (\frac {b (x+1)}{-a+b+1}\right ) \log (-a-b x+1)-\frac {1}{4} \log \left (\frac {b (1-x)}{a+b+1}\right ) \log (a+b x+1)+\frac {1}{4} \log \left (-\frac {b (x+1)}{a-b+1}\right ) \log (a+b x+1) \]
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Rubi [A] time = 0.26, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {6115, 2409, 2394, 2393, 2391} \[ \frac {1}{4} \text {PolyLog}\left (2,\frac {-a-b x+1}{-a-b+1}\right )-\frac {1}{4} \text {PolyLog}\left (2,\frac {-a-b x+1}{-a+b+1}\right )+\frac {1}{4} \text {PolyLog}\left (2,\frac {a+b x+1}{a-b+1}\right )-\frac {1}{4} \text {PolyLog}\left (2,\frac {a+b x+1}{a+b+1}\right )+\frac {1}{4} \log \left (-\frac {b (1-x)}{-a-b+1}\right ) \log (-a-b x+1)-\frac {1}{4} \log \left (\frac {b (x+1)}{-a+b+1}\right ) \log (-a-b x+1)-\frac {1}{4} \log \left (\frac {b (1-x)}{a+b+1}\right ) \log (a+b x+1)+\frac {1}{4} \log \left (-\frac {b (x+1)}{a-b+1}\right ) \log (a+b x+1) \]
Antiderivative was successfully verified.
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Rule 2391
Rule 2393
Rule 2394
Rule 2409
Rule 6115
Rubi steps
\begin {align*} \int \frac {\tanh ^{-1}(a+b x)}{1-x^2} \, dx &=-\left (\frac {1}{2} \int \frac {\log (1-a-b x)}{1-x^2} \, dx\right )+\frac {1}{2} \int \frac {\log (1+a+b x)}{1-x^2} \, dx\\ &=-\left (\frac {1}{2} \int \left (\frac {\log (1-a-b x)}{2 (1-x)}+\frac {\log (1-a-b x)}{2 (1+x)}\right ) \, dx\right )+\frac {1}{2} \int \left (\frac {\log (1+a+b x)}{2 (1-x)}+\frac {\log (1+a+b x)}{2 (1+x)}\right ) \, dx\\ &=-\left (\frac {1}{4} \int \frac {\log (1-a-b x)}{1-x} \, dx\right )-\frac {1}{4} \int \frac {\log (1-a-b x)}{1+x} \, dx+\frac {1}{4} \int \frac {\log (1+a+b x)}{1-x} \, dx+\frac {1}{4} \int \frac {\log (1+a+b x)}{1+x} \, dx\\ &=\frac {1}{4} \log \left (-\frac {b (1-x)}{1-a-b}\right ) \log (1-a-b x)-\frac {1}{4} \log \left (\frac {b (1+x)}{1-a+b}\right ) \log (1-a-b x)-\frac {1}{4} \log \left (\frac {b (1-x)}{1+a+b}\right ) \log (1+a+b x)+\frac {1}{4} \log \left (-\frac {b (1+x)}{1+a-b}\right ) \log (1+a+b x)+\frac {1}{4} b \int \frac {\log \left (-\frac {b (1-x)}{1-a-b}\right )}{1-a-b x} \, dx+\frac {1}{4} b \int \frac {\log \left (\frac {b (1-x)}{1+a+b}\right )}{1+a+b x} \, dx-\frac {1}{4} b \int \frac {\log \left (-\frac {b (1+x)}{-1+a-b}\right )}{1-a-b x} \, dx-\frac {1}{4} b \int \frac {\log \left (\frac {b (1+x)}{-1-a+b}\right )}{1+a+b x} \, dx\\ &=\frac {1}{4} \log \left (-\frac {b (1-x)}{1-a-b}\right ) \log (1-a-b x)-\frac {1}{4} \log \left (\frac {b (1+x)}{1-a+b}\right ) \log (1-a-b x)-\frac {1}{4} \log \left (\frac {b (1-x)}{1+a+b}\right ) \log (1+a+b x)+\frac {1}{4} \log \left (-\frac {b (1+x)}{1+a-b}\right ) \log (1+a+b x)-\frac {1}{4} \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{1-a-b}\right )}{x} \, dx,x,1-a-b x\right )+\frac {1}{4} \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{-1+a-b}\right )}{x} \, dx,x,1-a-b x\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{-1-a+b}\right )}{x} \, dx,x,1+a+b x\right )+\frac {1}{4} \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{1+a+b}\right )}{x} \, dx,x,1+a+b x\right )\\ &=\frac {1}{4} \log \left (-\frac {b (1-x)}{1-a-b}\right ) \log (1-a-b x)-\frac {1}{4} \log \left (\frac {b (1+x)}{1-a+b}\right ) \log (1-a-b x)-\frac {1}{4} \log \left (\frac {b (1-x)}{1+a+b}\right ) \log (1+a+b x)+\frac {1}{4} \log \left (-\frac {b (1+x)}{1+a-b}\right ) \log (1+a+b x)+\frac {1}{4} \text {Li}_2\left (\frac {1-a-b x}{1-a-b}\right )-\frac {1}{4} \text {Li}_2\left (\frac {1-a-b x}{1-a+b}\right )+\frac {1}{4} \text {Li}_2\left (\frac {1+a+b x}{1+a-b}\right )-\frac {1}{4} \text {Li}_2\left (\frac {1+a+b x}{1+a+b}\right )\\ \end {align*}
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Mathematica [A] time = 0.04, size = 203, normalized size = 1.00 \[ \frac {1}{4} \text {Li}_2\left (\frac {-a-b x+1}{-a-b+1}\right )-\frac {1}{4} \text {Li}_2\left (\frac {-a-b x+1}{-a+b+1}\right )+\frac {1}{4} \text {Li}_2\left (\frac {a+b x+1}{a-b+1}\right )-\frac {1}{4} \text {Li}_2\left (\frac {a+b x+1}{a+b+1}\right )+\frac {1}{4} \log \left (-\frac {b (1-x)}{-a-b+1}\right ) \log (-a-b x+1)-\frac {1}{4} \log \left (\frac {b (x+1)}{-a+b+1}\right ) \log (-a-b x+1)-\frac {1}{4} \log \left (\frac {b (1-x)}{a+b+1}\right ) \log (a+b x+1)+\frac {1}{4} \log \left (-\frac {b (x+1)}{a-b+1}\right ) \log (a+b x+1) \]
Antiderivative was successfully verified.
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fricas [F] time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\operatorname {artanh}\left (b x + a\right )}{x^{2} - 1}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 196, normalized size = 0.97 \[ \frac {\arctanh \left (b x +a \right ) \ln \left (b x +b \right )}{2}-\frac {\arctanh \left (b x +a \right ) \ln \left (b x -b \right )}{2}+\frac {\dilog \left (\frac {b x +a -1}{-b -1+a}\right )}{4}+\frac {\ln \left (b x +b \right ) \ln \left (\frac {b x +a -1}{-b -1+a}\right )}{4}-\frac {\dilog \left (\frac {b x +a +1}{1+a -b}\right )}{4}-\frac {\ln \left (b x +b \right ) \ln \left (\frac {b x +a +1}{1+a -b}\right )}{4}+\frac {\dilog \left (\frac {b x +a +1}{1+a +b}\right )}{4}+\frac {\ln \left (b x -b \right ) \ln \left (\frac {b x +a +1}{1+a +b}\right )}{4}-\frac {\dilog \left (\frac {b x +a -1}{b -1+a}\right )}{4}-\frac {\ln \left (b x -b \right ) \ln \left (\frac {b x +a -1}{b -1+a}\right )}{4} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.32, size = 198, normalized size = 0.98 \[ \frac {1}{4} \, b {\left (\frac {\log \left (x - 1\right ) \log \left (\frac {b x - b}{a + b + 1} + 1\right ) + {\rm Li}_2\left (-\frac {b x - b}{a + b + 1}\right )}{b} - \frac {\log \left (x - 1\right ) \log \left (\frac {b x - b}{a + b - 1} + 1\right ) + {\rm Li}_2\left (-\frac {b x - b}{a + b - 1}\right )}{b} - \frac {\log \left (x + 1\right ) \log \left (\frac {b x + b}{a - b + 1} + 1\right ) + {\rm Li}_2\left (-\frac {b x + b}{a - b + 1}\right )}{b} + \frac {\log \left (x + 1\right ) \log \left (\frac {b x + b}{a - b - 1} + 1\right ) + {\rm Li}_2\left (-\frac {b x + b}{a - b - 1}\right )}{b}\right )} + \frac {1}{2} \, {\left (\log \left (x + 1\right ) - \log \left (x - 1\right )\right )} \operatorname {artanh}\left (b x + a\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ -\int \frac {\mathrm {atanh}\left (a+b\,x\right )}{x^2-1} \,d x \]
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 \[ - \int \frac {\operatorname {atanh}{\left (a + b x \right )}}{x^{2} - 1}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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