Optimal. Leaf size=120 \[ -\frac {x}{3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {x}{6 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {a^2 x^2+1}{12 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {1}{4 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}+\frac {\text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{3 a} \]
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Rubi [A] time = 0.29, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {5966, 5996, 6032, 6034, 3312, 3301, 5968} \[ -\frac {x}{3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {x}{6 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {a^2 x^2+1}{12 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {1}{4 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}+\frac {\text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{3 a} \]
Antiderivative was successfully verified.
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Rule 3301
Rule 3312
Rule 5966
Rule 5968
Rule 5996
Rule 6032
Rule 6034
Rubi steps
\begin {align*} \int \frac {1}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5} \, dx &=-\frac {1}{4 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}+\frac {1}{2} a \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4} \, dx\\ &=-\frac {1}{4 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {x}{6 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {1+a^2 x^2}{12 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}+\frac {1}{3} a \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2} \, dx\\ &=-\frac {1}{4 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {x}{6 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {1+a^2 x^2}{12 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {x}{3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {1}{3} \int \frac {1}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx+\frac {1}{3} a^2 \int \frac {x^2}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx\\ &=-\frac {1}{4 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {x}{6 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {1+a^2 x^2}{12 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {x}{3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {\operatorname {Subst}\left (\int \frac {\cosh ^2(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{3 a}+\frac {\operatorname {Subst}\left (\int \frac {\sinh ^2(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{3 a}\\ &=-\frac {1}{4 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {x}{6 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {1+a^2 x^2}{12 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {x}{3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {\operatorname {Subst}\left (\int \left (\frac {1}{2 x}-\frac {\cosh (2 x)}{2 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{3 a}+\frac {\operatorname {Subst}\left (\int \left (\frac {1}{2 x}+\frac {\cosh (2 x)}{2 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{3 a}\\ &=-\frac {1}{4 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {x}{6 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {1+a^2 x^2}{12 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {x}{3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+2 \frac {\operatorname {Subst}\left (\int \frac {\cosh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{6 a}\\ &=-\frac {1}{4 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {x}{6 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {1+a^2 x^2}{12 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {x}{3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {\text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{3 a}\\ \end {align*}
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Mathematica [A] time = 0.10, size = 84, normalized size = 0.70 \[ \frac {4 \left (a^2 x^2-1\right ) \tanh ^{-1}(a x)^4 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )+\left (a^2 x^2+1\right ) \tanh ^{-1}(a x)^2+4 a x \tanh ^{-1}(a x)^3+2 a x \tanh ^{-1}(a x)+3}{12 a \left (a^2 x^2-1\right ) \tanh ^{-1}(a x)^4} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.70, size = 171, normalized size = 1.42 \[ \frac {4 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + {\left ({\left (a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) + {\left (a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{4} + 8 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) + 2 \, {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 24}{6 \, {\left (a^{3} x^{2} - a\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{4}} \]
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 \[ \int \frac {1}{{\left (a^{2} x^{2} - 1\right )}^{2} \operatorname {artanh}\left (a x\right )^{5}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.19, size = 83, normalized size = 0.69 \[ \frac {-\frac {1}{8 \arctanh \left (a x \right )^{4}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{8 \arctanh \left (a x \right )^{4}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{12 \arctanh \left (a x \right )^{3}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{12 \arctanh \left (a x \right )^{2}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{6 \arctanh \left (a x \right )}+\frac {\Chi \left (2 \arctanh \left (a x \right )\right )}{3}}{a} \]
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 \[ \frac {2 \, a x \log \left (a x + 1\right )^{3} - 2 \, a x \log \left (-a x + 1\right )^{3} + 4 \, a x \log \left (a x + 1\right ) + {\left (a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} + {\left (a^{2} x^{2} + 6 \, a x \log \left (a x + 1\right ) + 1\right )} \log \left (-a x + 1\right )^{2} - 2 \, {\left (3 \, a x \log \left (a x + 1\right )^{2} + 2 \, a x + {\left (a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right ) + 12}{3 \, {\left ({\left (a^{3} x^{2} - a\right )} \log \left (a x + 1\right )^{4} - 4 \, {\left (a^{3} x^{2} - a\right )} \log \left (a x + 1\right )^{3} \log \left (-a x + 1\right ) + 6 \, {\left (a^{3} x^{2} - a\right )} \log \left (a x + 1\right )^{2} \log \left (-a x + 1\right )^{2} - 4 \, {\left (a^{3} x^{2} - a\right )} \log \left (a x + 1\right ) \log \left (-a x + 1\right )^{3} + {\left (a^{3} x^{2} - a\right )} \log \left (-a x + 1\right )^{4}\right )}} - \int -\frac {2 \, {\left (a^{2} x^{2} + 1\right )}}{3 \, {\left ({\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) - {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (-a x + 1\right )\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^5\,{\left (a^2\,x^2-1\right )}^2} \,d x \]
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 \[ \int \frac {1}{\left (a x - 1\right )^{2} \left (a x + 1\right )^{2} \operatorname {atanh}^{5}{\left (a x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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