Optimal. Leaf size=279 \[ -\frac{21 a^{5/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{10 b^{11/4} \sqrt{a x+b x^3}}+\frac{21 a^{5/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 b^{11/4} \sqrt{a x+b x^3}}+\frac{7 x \sqrt{a x+b x^3}}{5 b^2}-\frac{21 a x \left (a+b x^2\right )}{5 b^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{a x+b x^3}}-\frac{x^4}{b \sqrt{a x+b x^3}} \]
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Rubi [A] time = 0.268844, antiderivative size = 279, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.412, Rules used = {2022, 2024, 2032, 329, 305, 220, 1196} \[ -\frac{21 a^{5/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{10 b^{11/4} \sqrt{a x+b x^3}}+\frac{21 a^{5/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 b^{11/4} \sqrt{a x+b x^3}}+\frac{7 x \sqrt{a x+b x^3}}{5 b^2}-\frac{21 a x \left (a+b x^2\right )}{5 b^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{a x+b x^3}}-\frac{x^4}{b \sqrt{a x+b x^3}} \]
Antiderivative was successfully verified.
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Rule 2022
Rule 2024
Rule 2032
Rule 329
Rule 305
Rule 220
Rule 1196
Rubi steps
\begin{align*} \int \frac{x^6}{\left (a x+b x^3\right )^{3/2}} \, dx &=-\frac{x^4}{b \sqrt{a x+b x^3}}+\frac{7 \int \frac{x^3}{\sqrt{a x+b x^3}} \, dx}{2 b}\\ &=-\frac{x^4}{b \sqrt{a x+b x^3}}+\frac{7 x \sqrt{a x+b x^3}}{5 b^2}-\frac{(21 a) \int \frac{x}{\sqrt{a x+b x^3}} \, dx}{10 b^2}\\ &=-\frac{x^4}{b \sqrt{a x+b x^3}}+\frac{7 x \sqrt{a x+b x^3}}{5 b^2}-\frac{\left (21 a \sqrt{x} \sqrt{a+b x^2}\right ) \int \frac{\sqrt{x}}{\sqrt{a+b x^2}} \, dx}{10 b^2 \sqrt{a x+b x^3}}\\ &=-\frac{x^4}{b \sqrt{a x+b x^3}}+\frac{7 x \sqrt{a x+b x^3}}{5 b^2}-\frac{\left (21 a \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 b^2 \sqrt{a x+b x^3}}\\ &=-\frac{x^4}{b \sqrt{a x+b x^3}}+\frac{7 x \sqrt{a x+b x^3}}{5 b^2}-\frac{\left (21 a^{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 b^{5/2} \sqrt{a x+b x^3}}+\frac{\left (21 a^{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 b^{5/2} \sqrt{a x+b x^3}}\\ &=-\frac{x^4}{b \sqrt{a x+b x^3}}-\frac{21 a x \left (a+b x^2\right )}{5 b^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{a x+b x^3}}+\frac{7 x \sqrt{a x+b x^3}}{5 b^2}+\frac{21 a^{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 b^{11/4} \sqrt{a x+b x^3}}-\frac{21 a^{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 )}{10 b^{11/4} \sqrt{a x+b x^3}}\\ \end{align*}
Mathematica [C] time = 0.0262226, size = 68, normalized size = 0.24 \[ \frac{2 x^2 \left (7 a \sqrt{\frac{b x^2}{a}+1} \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\frac{b x^2}{a}\right )-7 a+b x^2\right )}{5 b^2 \sqrt{x \left (a+b x^2\right )}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.015, size = 200, normalized size = 0.7 \begin{align*}{\frac{a{x}^{2}}{{b}^{2}}{\frac{1}{\sqrt{ \left ({\frac{a}{b}}+{x}^{2} \right ) bx}}}}+{\frac{2\,x}{5\,{b}^{2}}\sqrt{b{x}^{3}+ax}}-{\frac{21\,a}{10\,{b}^{3}}\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{x^{6}}{{\left (b x^{3} + a x\right )}^{\frac{3}{2}}}\,{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} x^{4}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}, 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{x^{6}}{\left (x \left (a + b x^{2}\right )\right )^{\frac{3}{2}}}\, 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{x^{6}}{{\left (b x^{3} + a x\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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