Optimal. Leaf size=110 \[ \frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{693 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{8 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{99 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]
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Rubi [A] time = 0.057208, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2171, 2167} \[ \frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{693 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{8 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{99 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]
Antiderivative was successfully verified.
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Rule 2171
Rule 2167
Rubi steps
\begin{align*} \int \frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{13/2}} \, dx &=\frac{2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{(4 b) \int \frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{11/2}} \, dx}{11 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{8 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{99 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{\left (8 b^2\right ) \int \frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{9/2}} \, dx}{99 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{693 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{8 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{99 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end{align*}
Mathematica [A] time = 0.0450775, size = 66, normalized size = 0.6 \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{7/2} \left (-154 b x \tanh ^{-1}(\tanh (a+b x))+63 \tanh ^{-1}(\tanh (a+b x))^2+99 b^2 x^2\right )}{693 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.163, size = 105, normalized size = 1. \begin{align*} -{\frac{2}{11\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -11\,bx} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{7}{2}}}{x}^{-{\frac{11}{2}}}}-{\frac{8\,b}{11\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -11\,bx} \left ( -{\frac{1}{9\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -9\,bx} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{7}{2}}}{x}^{-{\frac{9}{2}}}}+{\frac{2\,b}{63\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{7}{2}}}{x}^{-{\frac{7}{2}}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.47315, size = 61, normalized size = 0.55 \begin{align*} -\frac{2 \,{\left (8 \, b^{3} x^{3} - 20 \, a b^{2} x^{2} + 35 \, a^{2} b x + 63 \, a^{3}\right )}{\left (b x + a\right )}^{\frac{5}{2}}}{693 \, a^{3} x^{\frac{11}{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.04776, size = 162, normalized size = 1.47 \begin{align*} -\frac{2 \,{\left (8 \, b^{5} x^{5} - 4 \, a b^{4} x^{4} + 3 \, a^{2} b^{3} x^{3} + 113 \, a^{3} b^{2} x^{2} + 161 \, a^{4} b x + 63 \, a^{5}\right )} \sqrt{b x + a}}{693 \, a^{3} x^{\frac{11}{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.19404, size = 105, normalized size = 0.95 \begin{align*} -\frac{\sqrt{2}{\left (\frac{99 \, \sqrt{2} b^{11}}{a} + 4 \,{\left (\frac{2 \, \sqrt{2}{\left (b x + a\right )} b^{11}}{a^{3}} - \frac{11 \, \sqrt{2} b^{11}}{a^{2}}\right )}{\left (b x + a\right )}\right )}{\left (b x + a\right )}^{\frac{7}{2}} b}{693 \,{\left ({\left (b x + a\right )} b - a b\right )}^{\frac{11}{2}}{\left | b \right |}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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