Optimal. Leaf size=99 \[ -\frac {\text {Int}\left (\frac {1}{x^2 \tanh ^{-1}(a x)},x\right )}{a}-\frac {a x}{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {a x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {3}{2} \text {Chi}\left (2 \tanh ^{-1}(a x)\right )+\frac {1}{2} \text {Chi}\left (4 \tanh ^{-1}(a x)\right )-\frac {1}{a x \tanh ^{-1}(a x)} \]
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Rubi [A] time = 0.65, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {1}{x \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx \]
Verification is Not applicable to the result.
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Rubi steps
\begin {align*} \int \frac {1}{x \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx &=a^2 \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx+\int \frac {1}{x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2} \, dx\\ &=-\frac {a x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+a \int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx+a^2 \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2} \, dx+\left (3 a^3\right ) \int \frac {x^2}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx+\int \frac {1}{x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2} \, dx\\ &=-\frac {1}{a x \tanh ^{-1}(a x)}-\frac {a x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}-\frac {a x}{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+3 \operatorname {Subst}\left (\int \frac {\cosh ^2(x) \sinh ^2(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )-\frac {\int \frac {1}{x^2 \tanh ^{-1}(a x)} \, dx}{a}+a \int \frac {1}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx+a^3 \int \frac {x^2}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx+\operatorname {Subst}\left (\int \frac {\cosh ^4(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )\\ &=-\frac {1}{a x \tanh ^{-1}(a x)}-\frac {a x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}-\frac {a x}{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+3 \operatorname {Subst}\left (\int \left (-\frac {1}{8 x}+\frac {\cosh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )-\frac {\int \frac {1}{x^2 \tanh ^{-1}(a x)} \, dx}{a}+\operatorname {Subst}\left (\int \frac {\cosh ^2(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )+\operatorname {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 )+\operatorname {Subst}\left (\int \frac {\sinh ^2(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )\\ &=-\frac {1}{a x \tanh ^{-1}(a x)}-\frac {a x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}-\frac {a x}{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {1}{8} \operatorname {Subst}\left (\int \frac {\cosh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )+\frac {3}{8} \operatorname {Subst}\left (\int \frac {\cosh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {\cosh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )-\frac {\int \frac {1}{x^2 \tanh ^{-1}(a x)} \, dx}{a}-\operatorname {Subst}\left (\int \left (\frac {1}{2 x}-\frac {\cosh (2 x)}{2 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )+\operatorname {Subst}\left (\int \left (\frac {1}{2 x}+\frac {\cosh (2 x)}{2 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )\\ &=-\frac {1}{a x \tanh ^{-1}(a x)}-\frac {a x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}-\frac {a x}{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {1}{2} \text {Chi}\left (2 \tanh ^{-1}(a x)\right )+\frac {1}{2} \text {Chi}\left (4 \tanh ^{-1}(a x)\right )+2 \left (\frac {1}{2} \operatorname {Subst}\left (\int \frac {\cosh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )\right )-\frac {\int \frac {1}{x^2 \tanh ^{-1}(a x)} \, dx}{a}\\ &=-\frac {1}{a x \tanh ^{-1}(a x)}-\frac {a x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}-\frac {a x}{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {3}{2} \text {Chi}\left (2 \tanh ^{-1}(a x)\right )+\frac {1}{2} \text {Chi}\left (4 \tanh ^{-1}(a x)\right )-\frac {\int \frac {1}{x^2 \tanh ^{-1}(a x)} \, dx}{a}\\ \end {align*}
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Mathematica [A] time = 4.69, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx \]
Verification is Not applicable to the result.
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fricas [A] time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {1}{{\left (a^{6} x^{7} - 3 \, a^{4} x^{5} + 3 \, a^{2} x^{3} - x\right )} \operatorname {artanh}\left (a x\right )^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {1}{{\left (a^{2} x^{2} - 1\right )}^{3} x \operatorname {artanh}\left (a x\right )^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.64, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \left (-a^{2} x^{2}+1\right )^{3} \arctanh \left (a x \right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {2}{{\left (a^{5} x^{5} - 2 \, a^{3} x^{3} + a x\right )} \log \left (a x + 1\right ) - {\left (a^{5} x^{5} - 2 \, a^{3} x^{3} + a x\right )} \log \left (-a x + 1\right )} + \int -\frac {2 \, {\left (5 \, a^{2} x^{2} - 1\right )}}{{\left (a^{7} x^{8} - 3 \, a^{5} x^{6} + 3 \, a^{3} x^{4} - a x^{2}\right )} \log \left (a x + 1\right ) - {\left (a^{7} x^{8} - 3 \, a^{5} x^{6} + 3 \, a^{3} x^{4} - a x^{2}\right )} \log \left (-a x + 1\right )}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [A] time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {1}{x\,{\mathrm {atanh}\left (a\,x\right )}^2\,{\left (a^2\,x^2-1\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {1}{a^{6} x^{7} \operatorname {atanh}^{2}{\left (a x \right )} - 3 a^{4} x^{5} \operatorname {atanh}^{2}{\left (a x \right )} + 3 a^{2} x^{3} \operatorname {atanh}^{2}{\left (a x \right )} - x \operatorname {atanh}^{2}{\left (a x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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