Optimal. Leaf size=177 \[ \frac{3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}{64 b^{5/2}}+\frac{3 \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{64 b^2}+\frac{x^{3/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{32 b}-\frac{1}{8} x^{5/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac{1}{4} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{3/2} \]
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Rubi [A] time = 0.103711, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2169, 2165} \[ \frac{3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}{64 b^{5/2}}+\frac{3 \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{64 b^2}+\frac{x^{3/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{32 b}-\frac{1}{8} x^{5/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac{1}{4} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{3/2} \]
Antiderivative was successfully verified.
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Rule 2169
Rule 2165
Rubi steps
\begin{align*} \int x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx &=\frac{1}{4} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}-\frac{1}{8} \left (3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int x^{3/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \, dx\\ &=-\frac{1}{8} x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}+\frac{1}{4} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac{1}{16} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \int \frac{x^{3/2}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=-\frac{1}{8} x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}+\frac{x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{32 b}+\frac{1}{4} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}-\frac{\left (3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac{\sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{64 b}\\ &=-\frac{1}{8} x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}+\frac{x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{32 b}+\frac{3 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{64 b^2}+\frac{1}{4} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac{\left (3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac{1}{\sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{128 b^2}\\ &=\frac{3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}{64 b^{5/2}}-\frac{1}{8} x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}+\frac{x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{32 b}+\frac{3 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{64 b^2}+\frac{1}{4} x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}\\ \end{align*}
Mathematica [A] time = 0.0891484, size = 122, normalized size = 0.69 \[ \frac{\sqrt{b} \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (11 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+11 b x \tanh ^{-1}(\tanh (a+b x))^2-3 \tanh ^{-1}(\tanh (a+b x))^3-3 b^3 x^3\right )+3 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4 \log \left (\sqrt{b} \sqrt{\tanh ^{-1}(\tanh (a+b x))}+b \sqrt{x}\right )}{64 b^{5/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.115, size = 471, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{\frac{3}{2}} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{\frac{3}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.30879, size = 412, normalized size = 2.33 \begin{align*} \left [\frac{3 \, a^{4} \sqrt{b} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) + 2 \,{\left (16 \, b^{4} x^{3} + 24 \, a b^{3} x^{2} + 2 \, a^{2} b^{2} x - 3 \, a^{3} b\right )} \sqrt{b x + a} \sqrt{x}}{128 \, b^{3}}, -\frac{3 \, a^{4} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) -{\left (16 \, b^{4} x^{3} + 24 \, a b^{3} x^{2} + 2 \, a^{2} b^{2} x - 3 \, a^{3} b\right )} \sqrt{b x + a} \sqrt{x}}{64 \, b^{3}}\right ] \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 = 1.20397, size = 198, normalized size = 1.12 \begin{align*} \frac{1}{384} \, \sqrt{2}{\left (8 \, \sqrt{2}{\left (\sqrt{b x + a}{\left (2 \,{\left (4 \, x + \frac{a}{b}\right )} x - \frac{3 \, a^{2}}{b^{2}}\right )} \sqrt{x} - \frac{3 \, a^{3} \log \left ({\left | -\sqrt{b} \sqrt{x} + \sqrt{b x + a} \right |}\right )}{b^{\frac{5}{2}}}\right )} a + \sqrt{2}{\left ({\left (2 \,{\left (4 \,{\left (6 \, x + \frac{a}{b}\right )} x - \frac{5 \, a^{2}}{b^{2}}\right )} x + \frac{15 \, a^{3}}{b^{3}}\right )} \sqrt{b x + a} \sqrt{x} + \frac{15 \, a^{4} \log \left ({\left | -\sqrt{b} \sqrt{x} + \sqrt{b x + a} \right |}\right )}{b^{\frac{7}{2}}}\right )} b\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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