Optimal. Leaf size=60 \[ a c d x+a d \log (x)+\frac {1}{2} b d \log \left (1-c^2 x^2\right )-\frac {1}{2} b d \text {Li}_2(-c x)+\frac {1}{2} b d \text {Li}_2(c x)+b c d x \tanh ^{-1}(c x) \]
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Rubi [A] time = 0.07, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5940, 5910, 260, 5912} \[ -\frac {1}{2} b d \text {PolyLog}(2,-c x)+\frac {1}{2} b d \text {PolyLog}(2,c x)+a c d x+a d \log (x)+\frac {1}{2} b d \log \left (1-c^2 x^2\right )+b c d x \tanh ^{-1}(c x) \]
Antiderivative was successfully verified.
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Rule 260
Rule 5910
Rule 5912
Rule 5940
Rubi steps
\begin {align*} \int \frac {(d+c d x) \left (a+b \tanh ^{-1}(c x)\right )}{x} \, dx &=\int \left (c d \left (a+b \tanh ^{-1}(c x)\right )+\frac {d \left (a+b \tanh ^{-1}(c x)\right )}{x}\right ) \, dx\\ &=d \int \frac {a+b \tanh ^{-1}(c x)}{x} \, dx+(c d) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx\\ &=a c d x+a d \log (x)-\frac {1}{2} b d \text {Li}_2(-c x)+\frac {1}{2} b d \text {Li}_2(c x)+(b c d) \int \tanh ^{-1}(c x) \, dx\\ &=a c d x+b c d x \tanh ^{-1}(c x)+a d \log (x)-\frac {1}{2} b d \text {Li}_2(-c x)+\frac {1}{2} b d \text {Li}_2(c x)-\left (b c^2 d\right ) \int \frac {x}{1-c^2 x^2} \, dx\\ &=a c d x+b c d x \tanh ^{-1}(c x)+a d \log (x)+\frac {1}{2} b d \log \left (1-c^2 x^2\right )-\frac {1}{2} b d \text {Li}_2(-c x)+\frac {1}{2} b d \text {Li}_2(c x)\\ \end {align*}
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Mathematica [A] time = 0.07, size = 54, normalized size = 0.90 \[ \frac {1}{2} d \left (2 a c x+2 a \log (x)+b \log \left (1-c^2 x^2\right )-b \text {Li}_2(-c x)+b \text {Li}_2(c x)+2 b c x \tanh ^{-1}(c x)\right ) \]
Antiderivative was successfully verified.
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fricas [F] time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a c d x + a d + {\left (b c d x + b d\right )} \operatorname {artanh}\left (c x\right )}{x}, x\right ) \]
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 \[ \int \frac {{\left (c d x + d\right )} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 86, normalized size = 1.43 \[ d a \ln \left (c x \right )+a c d x +d b \arctanh \left (c x \right ) \ln \left (c x \right )+b c d x \arctanh \left (c x \right )+\frac {d b \ln \left (c x -1\right )}{2}+\frac {d b \ln \left (c x +1\right )}{2}-\frac {d b \dilog \left (c x \right )}{2}-\frac {d b \dilog \left (c x +1\right )}{2}-\frac {d b \ln \left (c x \right ) \ln \left (c x +1\right )}{2} \]
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 \[ a c d x + \frac {1}{2} \, {\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d + \frac {1}{2} \, b d \int \frac {\log \left (c x + 1\right ) - \log \left (-c x + 1\right )}{x}\,{d x} + a d \log \relax (x) \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,\left (d+c\,d\,x\right )}{x} \,d x \]
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 \[ d \left (\int a c\, dx + \int \frac {a}{x}\, dx + \int b c \operatorname {atanh}{\left (c x \right )}\, dx + \int \frac {b \operatorname {atanh}{\left (c x \right )}}{x}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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