Optimal. Leaf size=176 \[ \frac{3}{4 b^2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}+\frac{5}{4 b x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac{1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}-\frac{15 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac{15}{4 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]
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Rubi [A] time = 0.128015, antiderivative size = 176, 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, 2163, 2162} \[ \frac{3}{4 b^2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}+\frac{5}{4 b x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac{1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}-\frac{15 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac{15}{4 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]
Antiderivative was successfully verified.
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Rule 2168
Rule 2163
Rule 2162
Rubi steps
\begin{align*} \int \frac{1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))^3} \, dx &=-\frac{1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}-\frac{3 \int \frac{1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2} \, dx}{4 b}\\ &=-\frac{1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}+\frac{15 \int \frac{1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b^2}\\ &=\frac{3}{4 b^2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}-\frac{15 \int \frac{1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{5}{4 b x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{3}{4 b^2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}+\frac{15 \int \frac{1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac{15}{4 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{5}{4 b x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{3}{4 b^2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}+\frac{(15 b) \int \frac{1}{\sqrt{x} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac{15 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac{15}{4 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{5}{4 b x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{3}{4 b^2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}\\ \end{align*}
Mathematica [A] time = 0.166529, size = 141, normalized size = 0.8 \[ -\frac{b \sqrt{x}}{2 \tanh ^{-1}(\tanh (a+b x))^2 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}-\frac{7 b \sqrt{x}}{4 \tanh ^{-1}(\tanh (a+b x)) \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3}-\frac{2}{\sqrt{x} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3}-\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{4 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{7/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.142, size = 181, normalized size = 1. \begin{align*} -{\frac{7\,{b}^{2}}{4\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{x}^{{\frac{3}{2}}}}-{\frac{9\,ab}{4\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}\sqrt{x}}-{\frac{9\,b \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{4\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}\sqrt{x}}-{\frac{15\,b}{4\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b}}}}-2\,{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}\sqrt{x}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.13912, size = 466, normalized size = 2.65 \begin{align*} \left [\frac{15 \,{\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{b x - 2 \, a \sqrt{x} \sqrt{-\frac{b}{a}} - a}{b x + a}\right ) - 2 \,{\left (15 \, b^{2} x^{2} + 25 \, a b x + 8 \, a^{2}\right )} \sqrt{x}}{8 \,{\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}}, \frac{15 \,{\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{\frac{b}{a}}}{b \sqrt{x}}\right ) -{\left (15 \, b^{2} x^{2} + 25 \, a b x + 8 \, a^{2}\right )} \sqrt{x}}{4 \,{\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}}\right ] \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{1}{x^{\frac{3}{2}} \operatorname{atanh}^{3}{\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 = 1.10973, size = 80, normalized size = 0.45 \begin{align*} -\frac{15 \, b \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} a^{3}} - \frac{2}{a^{3} \sqrt{x}} - \frac{7 \, b^{2} x^{\frac{3}{2}} + 9 \, a b \sqrt{x}}{4 \,{\left (b x + a\right )}^{2} a^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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