Optimal. Leaf size=211 \[ -\frac {1}{7 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^7}-\frac {x}{21 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^6}-\frac {1+a^2 x^2}{105 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {x}{105 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {1+a^2 x^2}{315 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {2 x}{315 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {2 \left (1+a^2 x^2\right )}{315 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {4 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{315 a} \]
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Rubi [A]
time = 0.28, antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {6113, 6143,
6181, 5556, 12, 3379} \begin {gather*} -\frac {2 x}{315 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {x}{105 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {x}{21 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^6}-\frac {2 \left (a^2 x^2+1\right )}{315 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {a^2 x^2+1}{315 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {a^2 x^2+1}{105 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {1}{7 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^7}+\frac {4 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{315 a} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 3379
Rule 5556
Rule 6113
Rule 6143
Rule 6181
Rubi steps
\begin {align*} \int \frac {1}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^8} \, dx &=-\frac {1}{7 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^7}+\frac {1}{7} (2 a) \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^7} \, dx\\ &=-\frac {1}{7 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^7}-\frac {x}{21 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^6}-\frac {1+a^2 x^2}{105 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}+\frac {1}{105} (4 a) \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5} \, dx\\ &=-\frac {1}{7 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^7}-\frac {x}{21 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^6}-\frac {1+a^2 x^2}{105 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {x}{105 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {1+a^2 x^2}{315 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}+\frac {1}{315} (4 a) \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3} \, dx\\ &=-\frac {1}{7 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^7}-\frac {x}{21 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^6}-\frac {1+a^2 x^2}{105 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {x}{105 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {1+a^2 x^2}{315 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {2 x}{315 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {2 \left (1+a^2 x^2\right )}{315 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {1}{315} (8 a) \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx\\ &=-\frac {1}{7 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^7}-\frac {x}{21 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^6}-\frac {1+a^2 x^2}{105 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {x}{105 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {1+a^2 x^2}{315 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {2 x}{315 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {2 \left (1+a^2 x^2\right )}{315 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {8 \text {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{315 a}\\ &=-\frac {1}{7 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^7}-\frac {x}{21 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^6}-\frac {1+a^2 x^2}{105 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {x}{105 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {1+a^2 x^2}{315 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {2 x}{315 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {2 \left (1+a^2 x^2\right )}{315 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {8 \text {Subst}\left (\int \frac {\sinh (2 x)}{2 x} \, dx,x,\tanh ^{-1}(a x)\right )}{315 a}\\ &=-\frac {1}{7 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^7}-\frac {x}{21 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^6}-\frac {1+a^2 x^2}{105 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {x}{105 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {1+a^2 x^2}{315 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {2 x}{315 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {2 \left (1+a^2 x^2\right )}{315 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {4 \text {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{315 a}\\ &=-\frac {1}{7 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^7}-\frac {x}{21 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^6}-\frac {1+a^2 x^2}{105 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {x}{105 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {1+a^2 x^2}{315 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {2 x}{315 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {2 \left (1+a^2 x^2\right )}{315 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {4 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{315 a}\\ \end {align*}
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Mathematica [A]
time = 0.06, size = 128, normalized size = 0.61 \begin {gather*} \frac {45+15 a x \tanh ^{-1}(a x)+3 \left (1+a^2 x^2\right ) \tanh ^{-1}(a x)^2+3 a x \tanh ^{-1}(a x)^3+\left (1+a^2 x^2\right ) \tanh ^{-1}(a x)^4+2 a x \tanh ^{-1}(a x)^5+2 \left (1+a^2 x^2\right ) \tanh ^{-1}(a x)^6+4 \left (-1+a^2 x^2\right ) \tanh ^{-1}(a x)^7 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{315 a \left (-1+a^2 x^2\right ) \tanh ^{-1}(a x)^7} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 5.89, size = 128, normalized size = 0.61
method | result | size |
derivativedivides | \(\frac {-\frac {1}{14 \arctanh \left (a x \right )^{7}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{14 \arctanh \left (a x \right )^{7}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{42 \arctanh \left (a x \right )^{6}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{105 \arctanh \left (a x \right )^{5}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{210 \arctanh \left (a x \right )^{4}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{315 \arctanh \left (a x \right )^{3}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{315 \arctanh \left (a x \right )^{2}}-\frac {2 \cosh \left (2 \arctanh \left (a x \right )\right )}{315 \arctanh \left (a x \right )}+\frac {4 \hyperbolicSineIntegral \left (2 \arctanh \left (a x \right )\right )}{315}}{a}\) | \(128\) |
default | \(\frac {-\frac {1}{14 \arctanh \left (a x \right )^{7}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{14 \arctanh \left (a x \right )^{7}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{42 \arctanh \left (a x \right )^{6}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{105 \arctanh \left (a x \right )^{5}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{210 \arctanh \left (a x \right )^{4}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{315 \arctanh \left (a x \right )^{3}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{315 \arctanh \left (a x \right )^{2}}-\frac {2 \cosh \left (2 \arctanh \left (a x \right )\right )}{315 \arctanh \left (a x \right )}+\frac {4 \hyperbolicSineIntegral \left (2 \arctanh \left (a x \right )\right )}{315}}{a}\) | \(128\) |
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 [A]
time = 0.37, size = 249, normalized size = 1.18 \begin {gather*} \frac {2 \, {\left ({\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 )^{7} + 4 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right )^{5} + 2 \, {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{6} + 24 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + 4 \, {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{4} + 480 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) + 48 \, {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 2880\right )}}{315 \, {\left (a^{3} x^{2} - a\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{7}} \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 {1}{\left (a x - 1\right )^{2} \left (a x + 1\right )^{2} \operatorname {atanh}^{8}{\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 = -0.00 \begin {gather*} \int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^8\,{\left (a^2\,x^2-1\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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