Optimal. Leaf size=53 \[ -\frac {x}{a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {\text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{2 a^2}+\frac {\text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{2 a^2} \]
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Rubi [A]
time = 0.17, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6179, 6181,
5556, 3382, 6115, 3393} \begin {gather*} \frac {\text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{2 a^2}+\frac {\text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{2 a^2}-\frac {x}{a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \end {gather*}
Antiderivative was successfully verified.
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Rule 3382
Rule 3393
Rule 5556
Rule 6115
Rule 6179
Rule 6181
Rubi steps
\begin {align*} \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx &=-\frac {x}{a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {\int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx}{a}+(3 a) \int \frac {x^2}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx\\ &=-\frac {x}{a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {\text {Subst}\left (\int \frac {\cosh ^4(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}+\frac {3 \text {Subst}\left (\int \frac {\cosh ^2(x) \sinh ^2(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=-\frac {x}{a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {\text {Subst}\left (\int \left (\frac {3}{8 x}+\frac {\cosh (2 x)}{2 x}+\frac {\cosh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}+\frac {3 \text {Subst}\left (\int \left (-\frac {1}{8 x}+\frac {\cosh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=-\frac {x}{a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {\text {Subst}\left (\int \frac {\cosh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{8 a^2}+\frac {3 \text {Subst}\left (\int \frac {\cosh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{8 a^2}+\frac {\text {Subst}\left (\int \frac {\cosh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{2 a^2}\\ &=-\frac {x}{a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {\text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{2 a^2}+\frac {\text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{2 a^2}\\ \end {align*}
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Mathematica [A]
time = 0.06, size = 75, normalized size = 1.42 \begin {gather*} \frac {-2 a x+\left (-1+a^2 x^2\right )^2 \tanh ^{-1}(a x) \text {Chi}\left (2 \tanh ^{-1}(a x)\right )+\left (-1+a^2 x^2\right )^2 \tanh ^{-1}(a x) \text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{2 a^2 \left (-1+a^2 x^2\right )^2 \tanh ^{-1}(a x)} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 3.66, size = 54, normalized size = 1.02
method | result | size |
derivativedivides | \(\frac {-\frac {\sinh \left (4 \arctanh \left (a x \right )\right )}{8 \arctanh \left (a x \right )}+\frac {\hyperbolicCosineIntegral \left (4 \arctanh \left (a x \right )\right )}{2}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{4 \arctanh \left (a x \right )}+\frac {\hyperbolicCosineIntegral \left (2 \arctanh \left (a x \right )\right )}{2}}{a^{2}}\) | \(54\) |
default | \(\frac {-\frac {\sinh \left (4 \arctanh \left (a x \right )\right )}{8 \arctanh \left (a x \right )}+\frac {\hyperbolicCosineIntegral \left (4 \arctanh \left (a x \right )\right )}{2}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{4 \arctanh \left (a x \right )}+\frac {\hyperbolicCosineIntegral \left (2 \arctanh \left (a x \right )\right )}{2}}{a^{2}}\) | \(54\) |
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 225 vs.
\(2 (48) = 96\).
time = 0.38, size = 225, normalized size = 4.25 \begin {gather*} -\frac {8 \, a x - {\left ({\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) + {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) + {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) + {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{4 \, {\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )} \end {gather*}
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 \begin {gather*} - \int \frac {x}{a^{6} x^{6} \operatorname {atanh}^{2}{\left (a x \right )} - 3 a^{4} x^{4} \operatorname {atanh}^{2}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname {atanh}^{2}{\left (a x \right )} - \operatorname {atanh}^{2}{\left (a x \right )}}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} -\int \frac {x}{{\mathrm {atanh}\left (a\,x\right )}^2\,{\left (a^2\,x^2-1\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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