Optimal. Leaf size=104 \[ -2 a e x-2 b e x \tanh ^{-1}(c x)+\frac {e \left (a+b \tanh ^{-1}(c x)\right )^2}{b c}-\frac {b e \log \left (1-c^2 x^2\right )}{c}+x \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \left (d+e \log \left (1-c^2 x^2\right )\right )^2}{4 c e} \]
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Rubi [A]
time = 0.14, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6220, 2525,
2437, 2338, 6127, 6021, 266, 6095} \begin {gather*} x \left (a+b \tanh ^{-1}(c x)\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )+\frac {e \left (a+b \tanh ^{-1}(c x)\right )^2}{b c}-2 a e x+\frac {b \left (e \log \left (1-c^2 x^2\right )+d\right )^2}{4 c e}-\frac {b e \log \left (1-c^2 x^2\right )}{c}-2 b e x \tanh ^{-1}(c x) \end {gather*}
Antiderivative was successfully verified.
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Rule 266
Rule 2338
Rule 2437
Rule 2525
Rule 6021
Rule 6095
Rule 6127
Rule 6220
Rubi steps
\begin {align*} \int \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx &=x \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )-(b c) \int \frac {x \left (d+e \log \left (1-c^2 x^2\right )\right )}{1-c^2 x^2} \, dx+\left (2 c^2 e\right ) \int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=x \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )-\frac {1}{2} (b c) \text {Subst}\left (\int \frac {d+e \log \left (1-c^2 x\right )}{1-c^2 x} \, dx,x,x^2\right )-(2 e) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx+(2 e) \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx\\ &=-2 a e x+\frac {e \left (a+b \tanh ^{-1}(c x)\right )^2}{b c}+x \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \text {Subst}\left (\int \frac {d+e \log (x)}{x} \, dx,x,1-c^2 x^2\right )}{2 c}-(2 b e) \int \tanh ^{-1}(c x) \, dx\\ &=-2 a e x-2 b e x \tanh ^{-1}(c x)+\frac {e \left (a+b \tanh ^{-1}(c x)\right )^2}{b c}+x \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \left (d+e \log \left (1-c^2 x^2\right )\right )^2}{4 c e}+(2 b c e) \int \frac {x}{1-c^2 x^2} \, dx\\ &=-2 a e x-2 b e x \tanh ^{-1}(c x)+\frac {e \left (a+b \tanh ^{-1}(c x)\right )^2}{b c}-\frac {b e \log \left (1-c^2 x^2\right )}{c}+x \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \left (d+e \log \left (1-c^2 x^2\right )\right )^2}{4 c e}\\ \end {align*}
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Mathematica [A]
time = 0.02, size = 144, normalized size = 1.38 \begin {gather*} a d x-2 a e x+\frac {2 a e \tanh ^{-1}(c x)}{c}+b d x \tanh ^{-1}(c x)-2 b e x \tanh ^{-1}(c x)+\frac {b e \tanh ^{-1}(c x)^2}{c}+\frac {b d \log \left (1-c^2 x^2\right )}{2 c}-\frac {b e \log \left (1-c^2 x^2\right )}{c}+a e x \log \left (1-c^2 x^2\right )+b e x \tanh ^{-1}(c x) \log \left (1-c^2 x^2\right )+\frac {b e \log ^2\left (1-c^2 x^2\right )}{4 c} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
3.
time = 4.36, size = 2529, normalized size = 24.32
method | result | size |
default | \(\text {Expression too large to display}\) | \(2529\) |
risch | \(\text {Expression too large to display}\) | \(3552\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [C] Result contains complex when optimal does not.
time = 0.27, size = 181, normalized size = 1.74 \begin {gather*} -{\left (c^{2} {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )} - x \log \left (-c^{2} x^{2} + 1\right )\right )} b \operatorname {artanh}\left (c x\right ) e + a d x - {\left (c^{2} {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )} - x \log \left (-c^{2} x^{2} + 1\right )\right )} a e + \frac {{\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d}{2 \, c} + \frac {{\left ({\left (i \, \pi + 2 \, \log \left (c x - 1\right ) - 2\right )} \log \left (c x + 1\right ) + {\left (i \, \pi - 2\right )} \log \left (c x - 1\right )\right )} b e}{2 \, c} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.44, size = 189, normalized size = 1.82 \begin {gather*} \frac {4 \, a c d x - 8 \, a c x \cosh \left (1\right ) - 8 \, a c x \sinh \left (1\right ) + {\left (b \cosh \left (1\right ) + b \sinh \left (1\right )\right )} \log \left (-c^{2} x^{2} + 1\right )^{2} + {\left (b \cosh \left (1\right ) + b \sinh \left (1\right )\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )^{2} + 2 \, {\left (b d + 2 \, {\left (a c x - b\right )} \cosh \left (1\right ) + 2 \, {\left (a c x - b\right )} \sinh \left (1\right )\right )} \log \left (-c^{2} x^{2} + 1\right ) + 2 \, {\left (b c d x - 2 \, {\left (b c x - a\right )} \cosh \left (1\right ) + {\left (b c x \cosh \left (1\right ) + b c x \sinh \left (1\right )\right )} \log \left (-c^{2} x^{2} + 1\right ) - 2 \, {\left (b c x - a\right )} \sinh \left (1\right )\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{4 \, c} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.67, size = 148, normalized size = 1.42 \begin {gather*} \begin {cases} a d x + a e x \log {\left (- c^{2} x^{2} + 1 \right )} - 2 a e x + \frac {2 a e \operatorname {atanh}{\left (c x \right )}}{c} + b d x \operatorname {atanh}{\left (c x \right )} + b e x \log {\left (- c^{2} x^{2} + 1 \right )} \operatorname {atanh}{\left (c x \right )} - 2 b e x \operatorname {atanh}{\left (c x \right )} + \frac {b d \log {\left (- c^{2} x^{2} + 1 \right )}}{2 c} + \frac {b e \log {\left (- c^{2} x^{2} + 1 \right )}^{2}}{4 c} - \frac {b e \log {\left (- c^{2} x^{2} + 1 \right )}}{c} + \frac {b e \operatorname {atanh}^{2}{\left (c x \right )}}{c} & \text {for}\: c \neq 0 \\a d x & \text {otherwise} \end {cases} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.46, size = 165, normalized size = 1.59 \begin {gather*} -\frac {1}{2} \, b e x \log \left (-c x + 1\right )^{2} + \frac {1}{2} \, {\left (b d + 2 \, a e - 2 \, b e\right )} x \log \left (c x + 1\right ) + \frac {1}{2} \, {\left (b e x + \frac {b e}{c}\right )} \log \left (c x + 1\right )^{2} - \frac {b e \log \left (c x - 1\right )^{2}}{2 \, c} + {\left (a d - 2 \, a e\right )} x - \frac {1}{2} \, {\left ({\left (b d - 2 \, a e - 2 \, b e\right )} x - \frac {2 \, b e \log \left (c x - 1\right )}{c}\right )} \log \left (-c x + 1\right ) + \frac {{\left (b d + 2 \, a e - 2 \, b e\right )} \log \left (c x + 1\right )}{2 \, c} + \frac {{\left (b d - 2 \, a e - 2 \, b e\right )} \log \left (c x - 1\right )}{2 \, c} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.79, size = 385, normalized size = 3.70 \begin {gather*} a\,d\,x-2\,a\,e\,x+\frac {b\,e\,{\ln \left (c\,x+1\right )}^2}{2\,c}+\frac {b\,e\,{\ln \left (1-c\,x\right )}^2}{2\,c}+a\,e\,x\,\ln \left (1-c^2\,x^2\right )+\frac {b\,d\,x\,\ln \left (c\,x+1\right )}{2}-\frac {b\,d\,x\,\ln \left (1-c\,x\right )}{2}-b\,e\,x\,\ln \left (c\,x+1\right )+b\,e\,x\,\ln \left (1-c\,x\right )-\frac {a\,e\,\ln \left (c\,x-1\right )}{c}+\frac {a\,e\,\ln \left (c\,x+1\right )}{c}+\frac {b\,d\,\ln \left (c\,x-1\right )}{2\,c}+\frac {b\,d\,\ln \left (c\,x+1\right )}{2\,c}-\frac {b\,e\,\ln \left (c\,x-1\right )}{c}-\frac {b\,e\,\ln \left (c\,x+1\right )}{c}+\frac {b\,e\,x\,\ln \left (c\,x+1\right )\,\ln \left (1-c^2\,x^2\right )}{2}-\frac {b\,e\,x\,\ln \left (1-c\,x\right )\,\ln \left (1-c^2\,x^2\right )}{2}+\frac {b\,e\,\ln \left (1-c^2\,x^2\right )\,\ln \left (-2\,a\,e-2\,a\,c\,e\,x\right )}{2\,c}+\frac {b\,e\,\ln \left (1-c^2\,x^2\right )\,\ln \left (2\,a\,c\,e\,x-2\,a\,e\right )}{2\,c}-\frac {b\,e\,\ln \left (c\,x+1\right )\,\ln \left (-2\,a\,e-2\,a\,c\,e\,x\right )}{2\,c}-\frac {b\,e\,\ln \left (c\,x+1\right )\,\ln \left (2\,a\,c\,e\,x-2\,a\,e\right )}{2\,c}-\frac {b\,e\,\ln \left (1-c\,x\right )\,\ln \left (-2\,a\,e-2\,a\,c\,e\,x\right )}{2\,c}-\frac {b\,e\,\ln \left (1-c\,x\right )\,\ln \left (2\,a\,c\,e\,x-2\,a\,e\right )}{2\,c} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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