Optimal. Leaf size=57 \[ \frac{16 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^3}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac{2 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b} \]
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Rubi [A] time = 0.0307876, antiderivative size = 57, 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} \[ \frac{16 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^3}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac{2 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b} \]
Antiderivative was successfully verified.
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Rule 2168
Rule 2157
Rule 30
Rubi steps
\begin{align*} \int \frac{x^2}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx &=\frac{2 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac{4 \int x \sqrt{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b}\\ &=\frac{2 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac{8 \int \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx}{3 b^2}\\ &=\frac{2 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac{8 \operatorname{Subst}\left (\int x^{3/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^3}\\ &=\frac{2 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac{16 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^3}\\ \end{align*}
Mathematica [A] time = 0.0337379, size = 49, normalized size = 0.86 \[ \frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (-20 b x \tanh ^{-1}(\tanh (a+b x))+8 \tanh ^{-1}(\tanh (a+b x))^2+15 b^2 x^2\right )}{15 b^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.06, size = 68, normalized size = 1.2 \begin{align*} 2\,{\frac{1/5\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{5/2}+1/3\, \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 ) ^{3/2}+ \left ( bx-{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{{b}^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.80668, size = 57, normalized size = 1. \begin{align*} \frac{2 \,{\left (3 \, b^{3} x^{3} - a b^{2} x^{2} + 4 \, a^{2} b x + 8 \, a^{3}\right )}}{15 \, \sqrt{b x + a} b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.37504, size = 73, normalized size = 1.28 \begin{align*} \frac{2 \,{\left (3 \, b^{2} x^{2} - 4 \, a b x + 8 \, a^{2}\right )} \sqrt{b x + a}}{15 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\sqrt{\operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.18468, size = 50, normalized size = 0.88 \begin{align*} \frac{2 \,{\left (3 \,{\left (b x + a\right )}^{\frac{5}{2}} - 10 \,{\left (b x + a\right )}^{\frac{3}{2}} a + 15 \, \sqrt{b x + a} a^{2}\right )}}{15 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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