Optimal. Leaf size=101 \[ \frac{32 x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{105 b^3}-\frac{16 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac{256 \tanh ^{-1}(\tanh (a+b x))^{13/2}}{15015 b^5}-\frac{128 x \tanh ^{-1}(\tanh (a+b x))^{11/2}}{1155 b^4}+\frac{2 x^4 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b} \]
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Rubi [A] time = 0.0672489, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2157, 30} \[ \frac{32 x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{105 b^3}-\frac{16 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac{256 \tanh ^{-1}(\tanh (a+b x))^{13/2}}{15015 b^5}-\frac{128 x \tanh ^{-1}(\tanh (a+b x))^{11/2}}{1155 b^4}+\frac{2 x^4 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b} \]
Antiderivative was successfully verified.
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Rule 2168
Rule 2157
Rule 30
Rubi steps
\begin{align*} \int x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx &=\frac{2 x^4 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac{8 \int x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx}{5 b}\\ &=\frac{2 x^4 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac{16 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac{48 \int x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2} \, dx}{35 b^2}\\ &=\frac{2 x^4 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac{16 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac{32 x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{105 b^3}-\frac{64 \int x \tanh ^{-1}(\tanh (a+b x))^{9/2} \, dx}{105 b^3}\\ &=\frac{2 x^4 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac{16 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac{32 x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{105 b^3}-\frac{128 x \tanh ^{-1}(\tanh (a+b x))^{11/2}}{1155 b^4}+\frac{128 \int \tanh ^{-1}(\tanh (a+b x))^{11/2} \, dx}{1155 b^4}\\ &=\frac{2 x^4 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac{16 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac{32 x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{105 b^3}-\frac{128 x \tanh ^{-1}(\tanh (a+b x))^{11/2}}{1155 b^4}+\frac{128 \operatorname{Subst}\left (\int x^{11/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{1155 b^5}\\ &=\frac{2 x^4 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac{16 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac{32 x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{105 b^3}-\frac{128 x \tanh ^{-1}(\tanh (a+b x))^{11/2}}{1155 b^4}+\frac{256 \tanh ^{-1}(\tanh (a+b x))^{13/2}}{15015 b^5}\\ \end{align*}
Mathematica [A] time = 0.0371079, size = 83, normalized size = 0.82 \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2} \left (-3432 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+2288 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-832 b x \tanh ^{-1}(\tanh (a+b x))^3+128 \tanh ^{-1}(\tanh (a+b x))^4+3003 b^4 x^4\right )}{15015 b^5} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.033, size = 154, normalized size = 1.5 \begin{align*} 2\,{\frac{1/13\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{13/2}+1/11\, \left ( -4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +4\,bx \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{11/2}+1/9\, \left ( 2\, \left ( bx-{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}+ \left ( -2\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +2\,bx \right ) ^{2} \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{9/2}+2/7\, \left ( bx-{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2} \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 ) ^{7/2}+1/5\, \left ( bx-{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{4} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{5/2}}{{b}^{5}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.78044, size = 86, normalized size = 0.85 \begin{align*} \frac{2 \,{\left (1155 \, b^{5} x^{5} + 315 \, a b^{4} x^{4} - 280 \, a^{2} b^{3} x^{3} + 240 \, a^{3} b^{2} x^{2} - 192 \, a^{4} b x + 128 \, a^{5}\right )}{\left (b x + a\right )}^{\frac{3}{2}}}{15015 \, b^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.68228, size = 180, normalized size = 1.78 \begin{align*} \frac{2 \,{\left (1155 \, b^{6} x^{6} + 1470 \, a b^{5} x^{5} + 35 \, a^{2} b^{4} x^{4} - 40 \, a^{3} b^{3} x^{3} + 48 \, a^{4} b^{2} x^{2} - 64 \, a^{5} b x + 128 \, a^{6}\right )} \sqrt{b x + a}}{15015 \, b^{5}} \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 = 1.17114, size = 203, normalized size = 2.01 \begin{align*} \frac{\sqrt{2}{\left (\frac{13 \, \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 )} a}{b^{4}} + \frac{5 \, \sqrt{2}{\left (693 \,{\left (b x + a\right )}^{\frac{13}{2}} - 4095 \,{\left (b x + a\right )}^{\frac{11}{2}} a + 10010 \,{\left (b x + a\right )}^{\frac{9}{2}} a^{2} - 12870 \,{\left (b x + a\right )}^{\frac{7}{2}} a^{3} + 9009 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{4} - 3003 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{5}\right )}}{b^{4}}\right )}}{45045 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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