Optimal. Leaf size=59 \[ \frac{16 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{105 b^3}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b} \]
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Rubi [A] time = 0.0297581, 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} \[ \frac{16 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{105 b^3}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b} \]
Antiderivative was successfully verified.
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Rule 2168
Rule 2157
Rule 30
Rubi steps
\begin{align*} \int x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac{4 \int x \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx}{3 b}\\ &=\frac{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac{8 \int \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx}{15 b^2}\\ &=\frac{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac{8 \operatorname{Subst}\left (\int x^{5/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{15 b^3}\\ &=\frac{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac{16 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{105 b^3}\\ \end{align*}
Mathematica [A] time = 0.0274241, size = 49, normalized size = 0.83 \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (-28 b x \tanh ^{-1}(\tanh (a+b x))+8 \tanh ^{-1}(\tanh (a+b x))^2+35 b^2 x^2\right )}{105 b^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.046, size = 69, normalized size = 1.2 \begin{align*} 2\,{\frac{1/7\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{7/2}+1/5\, \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 ) ^{5/2}+1/3\, \left ( bx-{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}}{{b}^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.79401, size = 57, normalized size = 0.97 \begin{align*} \frac{2 \,{\left (15 \, b^{3} x^{3} + 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt{b x + a}}{105 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.542, size = 97, normalized size = 1.64 \begin{align*} \frac{2 \,{\left (15 \, b^{3} x^{3} + 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt{b x + a}}{105 \, 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 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 = 1.17487, size = 50, normalized size = 0.85 \begin{align*} \frac{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 )}}{105 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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