Optimal. Leaf size=59 \[ \frac{16 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{693 b^3}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2}+\frac{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b} \]
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Rubi [A] time = 0.02927, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2157, 30} \[ \frac{16 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{693 b^3}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2}+\frac{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b} \]
Antiderivative was successfully verified.
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Rule 2168
Rule 2157
Rule 30
Rubi steps
\begin{align*} \int x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx &=\frac{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac{4 \int x \tanh ^{-1}(\tanh (a+b x))^{7/2} \, dx}{7 b}\\ &=\frac{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2}+\frac{8 \int \tanh ^{-1}(\tanh (a+b x))^{9/2} \, dx}{63 b^2}\\ &=\frac{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2}+\frac{8 \operatorname{Subst}\left (\int x^{9/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{63 b^3}\\ &=\frac{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{63 b^2}+\frac{16 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{693 b^3}\\ \end{align*}
Mathematica [A] time = 0.0312256, size = 49, normalized size = 0.83 \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{7/2} \left (-44 b x \tanh ^{-1}(\tanh (a+b x))+8 \tanh ^{-1}(\tanh (a+b x))^2+99 b^2 x^2\right )}{693 b^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.033, size = 69, normalized size = 1.2 \begin{align*} 2\,{\frac{1/11\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{11/2}+1/9\, \left ( -2\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +2\,bx \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{9/2}+1/7\, \left ( bx-{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{7/2}}{{b}^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.80105, size = 57, normalized size = 0.97 \begin{align*} \frac{2 \,{\left (63 \, b^{3} x^{3} + 35 \, a b^{2} x^{2} - 20 \, a^{2} b x + 8 \, a^{3}\right )}{\left (b x + a\right )}^{\frac{5}{2}}}{693 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.98195, size = 146, normalized size = 2.47 \begin{align*} \frac{2 \,{\left (63 \, b^{5} x^{5} + 161 \, a b^{4} x^{4} + 113 \, a^{2} b^{3} x^{3} + 3 \, a^{3} b^{2} x^{2} - 4 \, a^{4} b x + 8 \, a^{5}\right )} \sqrt{b x + a}}{693 \, b^{3}} \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 [B] time = 1.18087, size = 227, normalized size = 3.85 \begin{align*} \frac{\sqrt{2}{\left (\frac{33 \, \sqrt{2}{\left (15 \,{\left (b x + a\right )}^{\frac{7}{2}} - 42 \,{\left (b x + a\right )}^{\frac{5}{2}} a + 35 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{2}\right )} a^{2}}{b^{2}} + \frac{22 \, \sqrt{2}{\left (35 \,{\left (b x + a\right )}^{\frac{9}{2}} - 135 \,{\left (b x + a\right )}^{\frac{7}{2}} a + 189 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{2} - 105 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{3}\right )} a}{b^{2}} + \frac{\sqrt{2}{\left (315 \,{\left (b x + a\right )}^{\frac{11}{2}} - 1540 \,{\left (b x + a\right )}^{\frac{9}{2}} a + 2970 \,{\left (b x + a\right )}^{\frac{7}{2}} a^{2} - 2772 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{3} + 1155 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{4}\right )}}{b^{2}}\right )}}{3465 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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