Optimal. Leaf size=277 \[ \frac{12 \sqrt [4]{a} b^{5/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{5 \sqrt{a x+b x^3}}+\frac{24 b^{3/2} x \left (a+b x^2\right )}{5 \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{a x+b x^3}}-\frac{24 \sqrt [4]{a} b^{5/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a x+b x^3}}-\frac{12 b \sqrt{a x+b x^3}}{5 x}-\frac{2 \left (a x+b x^3\right )^{3/2}}{5 x^4} \]
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Rubi [A] time = 0.260832, antiderivative size = 277, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {2020, 2032, 329, 305, 220, 1196} \[ \frac{24 b^{3/2} x \left (a+b x^2\right )}{5 \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{a x+b x^3}}+\frac{12 \sqrt [4]{a} b^{5/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a x+b x^3}}-\frac{24 \sqrt [4]{a} b^{5/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a x+b x^3}}-\frac{12 b \sqrt{a x+b x^3}}{5 x}-\frac{2 \left (a x+b x^3\right )^{3/2}}{5 x^4} \]
Antiderivative was successfully verified.
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Rule 2020
Rule 2032
Rule 329
Rule 305
Rule 220
Rule 1196
Rubi steps
\begin{align*} \int \frac{\left (a x+b x^3\right )^{3/2}}{x^5} \, dx &=-\frac{2 \left (a x+b x^3\right )^{3/2}}{5 x^4}+\frac{1}{5} (6 b) \int \frac{\sqrt{a x+b x^3}}{x^2} \, dx\\ &=-\frac{12 b \sqrt{a x+b x^3}}{5 x}-\frac{2 \left (a x+b x^3\right )^{3/2}}{5 x^4}+\frac{1}{5} \left (12 b^2\right ) \int \frac{x}{\sqrt{a x+b x^3}} \, dx\\ &=-\frac{12 b \sqrt{a x+b x^3}}{5 x}-\frac{2 \left (a x+b x^3\right )^{3/2}}{5 x^4}+\frac{\left (12 b^2 \sqrt{x} \sqrt{a+b x^2}\right ) \int \frac{\sqrt{x}}{\sqrt{a+b x^2}} \, dx}{5 \sqrt{a x+b x^3}}\\ &=-\frac{12 b \sqrt{a x+b x^3}}{5 x}-\frac{2 \left (a x+b x^3\right )^{3/2}}{5 x^4}+\frac{\left (24 b^2 \sqrt{x} \sqrt{a+b x^2}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+b x^4}} \, dx,x,\sqrt{x}\right )}{5 \sqrt{a x+b x^3}}\\ &=-\frac{12 b \sqrt{a x+b x^3}}{5 x}-\frac{2 \left (a x+b x^3\right )^{3/2}}{5 x^4}+\frac{\left (24 \sqrt{a} b^{3/2} \sqrt{x} \sqrt{a+b x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^4}} \, dx,x,\sqrt{x}\right )}{5 \sqrt{a x+b x^3}}-\frac{\left (24 \sqrt{a} b^{3/2} \sqrt{x} \sqrt{a+b x^2}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{b} x^2}{\sqrt{a}}}{\sqrt{a+b x^4}} \, dx,x,\sqrt{x}\right )}{5 \sqrt{a x+b x^3}}\\ &=\frac{24 b^{3/2} x \left (a+b x^2\right )}{5 \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{a x+b x^3}}-\frac{12 b \sqrt{a x+b x^3}}{5 x}-\frac{2 \left (a x+b x^3\right )^{3/2}}{5 x^4}-\frac{24 \sqrt [4]{a} b^{5/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a x+b x^3}}+\frac{12 \sqrt [4]{a} b^{5/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a x+b x^3}}\\ \end{align*}
Mathematica [C] time = 0.013007, size = 54, normalized size = 0.19 \[ -\frac{2 a \sqrt{x \left (a+b x^2\right )} \, _2F_1\left (-\frac{3}{2},-\frac{5}{4};-\frac{1}{4};-\frac{b x^2}{a}\right )}{5 x^3 \sqrt{\frac{b x^2}{a}+1}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.016, size = 196, normalized size = 0.7 \begin{align*} -{\frac{2\,a}{5\,{x}^{3}}\sqrt{b{x}^{3}+ax}}-{\frac{ \left ( 14\,b{x}^{2}+14\,a \right ) b}{5}{\frac{1}{\sqrt{x \left ( b{x}^{2}+a \right ) }}}}+{\frac{12\,b}{5}\sqrt{-ab}\sqrt{{b \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{-2\,{\frac{b}{\sqrt{-ab}} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }}\sqrt{-{bx{\frac{1}{\sqrt{-ab}}}}} \left ( -2\,{\frac{\sqrt{-ab}}{b}{\it EllipticE} \left ( \sqrt{{\frac{b}{\sqrt{-ab}} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }},1/2\,\sqrt{2} \right ) }+{\frac{1}{b}\sqrt{-ab}{\it EllipticF} \left ( \sqrt{{b \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ){\frac{1}{\sqrt{-ab}}}}},{\frac{\sqrt{2}}{2}} \right ) } \right ){\frac{1}{\sqrt{b{x}^{3}+ax}}}} \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 \frac{{\left (b x^{3} + a x\right )}^{\frac{3}{2}}}{x^{5}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{3} + a x}{\left (b x^{2} + a\right )}}{x^{4}}, x\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{\left (x \left (a + b x^{2}\right )\right )^{\frac{3}{2}}}{x^{5}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{3} + a x\right )}^{\frac{3}{2}}}{x^{5}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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