Optimal. Leaf size=257 \[ \frac {x}{5 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {8 x}{15 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}+\frac {a^2 x^2+1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {8}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {32}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {4}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}+\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{15 a}+\frac {16 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{15 a} \]
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Rubi [A] time = 1.35, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 49, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {5966, 6032, 6028, 5996, 6034, 5448, 12, 3298} \[ \frac {x}{5 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {8 x}{15 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}+\frac {a^2 x^2+1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {8}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {32}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {4}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}+\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{15 a}+\frac {16 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{15 a} \]
Antiderivative was successfully verified.
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Rule 12
Rule 3298
Rule 5448
Rule 5966
Rule 5996
Rule 6028
Rule 6032
Rule 6034
Rubi steps
\begin {align*} \int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^6} \, dx &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}+\frac {1}{5} (4 a) \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^5} \, dx\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}+\frac {1}{5} \int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^4} \, dx+\frac {1}{5} \left (3 a^2\right ) \int \frac {x^2}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^4} \, dx\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}-\frac {1}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {3}{5} \int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^4} \, dx-\frac {3}{5} \int \frac {1}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4} \, dx+\frac {1}{15} (4 a) \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^3} \, dx\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}-\frac {4}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {2 x}{15 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {2}{15} \int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx-\frac {1}{5} (2 a) \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3} \, dx+\frac {1}{5} (4 a) \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^3} \, dx+\frac {1}{5} \left (2 a^2\right ) \int \frac {x^2}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}-\frac {4}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {8 x}{15 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {x}{5 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {2}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {1+a^2 x^2}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+2 \left (\frac {2}{5} \int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx\right )-\frac {2}{5} \int \frac {1}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2} \, dx+\frac {1}{15} (8 a) \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx-\frac {1}{5} (4 a) \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx+\frac {1}{5} \left (6 a^2\right ) \int \frac {x^2}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}-\frac {4}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {8 x}{15 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {x}{5 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {2}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {2}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {1+a^2 x^2}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {6}{5} \int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx-\frac {6}{5} \int \frac {1}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2} \, dx+\frac {8 \operatorname {Subst}\left (\int \frac {\cosh ^3(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{15 a}-\frac {4 \operatorname {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}-\frac {1}{5} (4 a) \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx+2 \left (-\frac {2}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {1}{5} (8 a) \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx\right )\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}-\frac {4}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {8 x}{15 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {x}{5 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {4}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {8}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {1+a^2 x^2}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {8 \operatorname {Subst}\left (\int \left (\frac {\sinh (2 x)}{4 x}+\frac {\sinh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{15 a}-\frac {4 \operatorname {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}-\frac {4 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{2 x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}+2 \left (-\frac {2}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {8 \operatorname {Subst}\left (\int \frac {\cosh ^3(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}\right )-\frac {1}{5} (12 a) \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx+\frac {1}{5} (24 a) \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}-\frac {4}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {8 x}{15 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {x}{5 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {4}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {8}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {1+a^2 x^2}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {\operatorname {Subst}\left (\int \frac {\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{15 a}+\frac {2 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{15 a}-\frac {2 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}-\frac {4 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{2 x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}+2 \left (-\frac {2}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {8 \operatorname {Subst}\left (\int \left (\frac {\sinh (2 x)}{4 x}+\frac {\sinh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}\right )-\frac {12 \operatorname {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}+\frac {24 \operatorname {Subst}\left (\int \frac {\cosh ^3(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}-\frac {4}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {8 x}{15 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {x}{5 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {4}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {8}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {1+a^2 x^2}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {4 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{15 a}+\frac {\text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{15 a}-\frac {2 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}+2 \left (-\frac {2}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {\operatorname {Subst}\left (\int \frac {\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}+\frac {2 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}\right )-\frac {12 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{2 x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}+\frac {24 \operatorname {Subst}\left (\int \left (\frac {\sinh (2 x)}{4 x}+\frac {\sinh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}-\frac {4}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {8 x}{15 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {x}{5 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {4}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {8}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {1+a^2 x^2}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{3 a}+\frac {\text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{15 a}+2 \left (-\frac {2}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{5 a}+\frac {\text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{5 a}\right )+\frac {3 \operatorname {Subst}\left (\int \frac {\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}-\frac {4}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {8 x}{15 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {x}{5 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {4}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {8}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {1+a^2 x^2}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{3 a}+\frac {2 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{3 a}+2 \left (-\frac {2}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{5 a}+\frac {\text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{5 a}\right )\\ \end {align*}
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Mathematica [A] time = 0.33, size = 166, normalized size = 0.65 \[ -\frac {3 a^4 x^4 \tanh ^{-1}(a x)^4+3 a^3 x^3 \tanh ^{-1}(a x)^3-2 \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^5 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )-16 \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^5 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )+24 a^2 x^2 \tanh ^{-1}(a x)^4+3 a^2 x^2 \tanh ^{-1}(a x)^2+5 \tanh ^{-1}(a x)^4+5 a x \tanh ^{-1}(a x)^3+\tanh ^{-1}(a x)^2+3 a x \tanh ^{-1}(a x)+3}{15 a \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^5} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.52, size = 341, normalized size = 1.33 \[ \frac {{\left (8 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) - 8 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) + {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) - {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{5} - 2 \, {\left (3 \, a^{4} x^{4} + 24 \, a^{2} x^{2} + 5\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{4} - 4 \, {\left (3 \, a^{3} x^{3} + 5 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} - 48 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) - 8 \, {\left (3 \, a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 96}{15 \, {\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{5}} \]
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 \[ \int -\frac {1}{{\left (a^{2} x^{2} - 1\right )}^{3} \operatorname {artanh}\left (a x\right )^{6}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.23, size = 182, normalized size = 0.71 \[ \frac {-\frac {3}{40 \arctanh \left (a x \right )^{5}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{10 \arctanh \left (a x \right )^{5}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{20 \arctanh \left (a x \right )^{4}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{30 \arctanh \left (a x \right )^{3}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{30 \arctanh \left (a x \right )^{2}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{15 \arctanh \left (a x \right )}+\frac {2 \Shi \left (2 \arctanh \left (a x \right )\right )}{15}-\frac {\cosh \left (4 \arctanh \left (a x \right )\right )}{40 \arctanh \left (a x \right )^{5}}-\frac {\sinh \left (4 \arctanh \left (a x \right )\right )}{40 \arctanh \left (a x \right )^{4}}-\frac {\cosh \left (4 \arctanh \left (a x \right )\right )}{30 \arctanh \left (a x \right )^{3}}-\frac {\sinh \left (4 \arctanh \left (a x \right )\right )}{15 \arctanh \left (a x \right )^{2}}-\frac {4 \cosh \left (4 \arctanh \left (a x \right )\right )}{15 \arctanh \left (a x \right )}+\frac {16 \Shi \left (4 \arctanh \left (a x \right )\right )}{15}}{a} \]
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 \, {\left ({\left (3 \, a^{4} x^{4} + 24 \, a^{2} x^{2} + 5\right )} \log \left (a x + 1\right )^{4} + {\left (3 \, a^{4} x^{4} + 24 \, a^{2} x^{2} + 5\right )} \log \left (-a x + 1\right )^{4} + 2 \, {\left (3 \, a^{3} x^{3} + 5 \, a x\right )} \log \left (a x + 1\right )^{3} - 2 \, {\left (3 \, a^{3} x^{3} + 5 \, a x + 2 \, {\left (3 \, a^{4} x^{4} + 24 \, a^{2} x^{2} + 5\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )^{3} + 24 \, a x \log \left (a x + 1\right ) + 4 \, {\left (3 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} + 2 \, {\left (6 \, a^{2} x^{2} + 3 \, {\left (3 \, a^{4} x^{4} + 24 \, a^{2} x^{2} + 5\right )} \log \left (a x + 1\right )^{2} + 3 \, {\left (3 \, a^{3} x^{3} + 5 \, a x\right )} \log \left (a x + 1\right ) + 2\right )} \log \left (-a x + 1\right )^{2} - 2 \, {\left (2 \, {\left (3 \, a^{4} x^{4} + 24 \, a^{2} x^{2} + 5\right )} \log \left (a x + 1\right )^{3} + 3 \, {\left (3 \, a^{3} x^{3} + 5 \, a x\right )} \log \left (a x + 1\right )^{2} + 12 \, a x + 4 \, {\left (3 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right ) + 48\right )}}{15 \, {\left ({\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (a x + 1\right )^{5} - 5 \, {\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (a x + 1\right )^{4} \log \left (-a x + 1\right ) + 10 \, {\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (a x + 1\right )^{3} \log \left (-a x + 1\right )^{2} - 10 \, {\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (a x + 1\right )^{2} \log \left (-a x + 1\right )^{3} + 5 \, {\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (a x + 1\right ) \log \left (-a x + 1\right )^{4} - {\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (-a x + 1\right )^{5}\right )}} + \int -\frac {8 \, {\left (15 \, a^{3} x^{3} + 17 \, a x\right )}}{15 \, {\left ({\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) - {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, 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 = -0.00 \[ -\int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^6\,{\left (a^2\,x^2-1\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {1}{a^{6} x^{6} \operatorname {atanh}^{6}{\left (a x \right )} - 3 a^{4} x^{4} \operatorname {atanh}^{6}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname {atanh}^{6}{\left (a x \right )} - \operatorname {atanh}^{6}{\left (a x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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