Optimal. Leaf size=290 \[ \frac{\left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+A b) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right ),\frac{1}{2}\right )}{a^{3/4} b^{3/4} e^{3/2} \sqrt{a+b x^2}}-\frac{2 \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+A b) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{a^{3/4} b^{3/4} e^{3/2} \sqrt{a+b x^2}}+\frac{2 \sqrt{e x} \sqrt{a+b x^2} (a B+A b)}{a \sqrt{b} e^2 \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{2 A \sqrt{a+b x^2}}{a e \sqrt{e x}} \]
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Rubi [A] time = 0.222748, antiderivative size = 290, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {453, 329, 305, 220, 1196} \[ \frac{\left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+A b) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{a^{3/4} b^{3/4} e^{3/2} \sqrt{a+b x^2}}-\frac{2 \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+A b) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{a^{3/4} b^{3/4} e^{3/2} \sqrt{a+b x^2}}+\frac{2 \sqrt{e x} \sqrt{a+b x^2} (a B+A b)}{a \sqrt{b} e^2 \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{2 A \sqrt{a+b x^2}}{a e \sqrt{e x}} \]
Antiderivative was successfully verified.
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Rule 453
Rule 329
Rule 305
Rule 220
Rule 1196
Rubi steps
\begin{align*} \int \frac{A+B x^2}{(e x)^{3/2} \sqrt{a+b x^2}} \, dx &=-\frac{2 A \sqrt{a+b x^2}}{a e \sqrt{e x}}+\frac{(A b+a B) \int \frac{\sqrt{e x}}{\sqrt{a+b x^2}} \, dx}{a e^2}\\ &=-\frac{2 A \sqrt{a+b x^2}}{a e \sqrt{e x}}+\frac{(2 (A b+a B)) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{a e^3}\\ &=-\frac{2 A \sqrt{a+b x^2}}{a e \sqrt{e x}}+\frac{(2 (A b+a B)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{\sqrt{a} \sqrt{b} e^2}-\frac{(2 (A b+a B)) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{b} x^2}{\sqrt{a} e}}{\sqrt{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{\sqrt{a} \sqrt{b} e^2}\\ &=-\frac{2 A \sqrt{a+b x^2}}{a e \sqrt{e x}}+\frac{2 (A b+a B) \sqrt{e x} \sqrt{a+b x^2}}{a \sqrt{b} e^2 \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{2 (A b+a B) \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{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{a^{3/4} b^{3/4} e^{3/2} \sqrt{a+b x^2}}+\frac{(A b+a B) \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{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{a^{3/4} b^{3/4} e^{3/2} \sqrt{a+b x^2}}\\ \end{align*}
Mathematica [C] time = 0.0423765, size = 82, normalized size = 0.28 \[ \frac{x \left (2 x^2 \sqrt{\frac{b x^2}{a}+1} (a B+A b) \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{b x^2}{a}\right )-6 A \left (a+b x^2\right )\right )}{3 a (e x)^{3/2} \sqrt{a+b x^2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.017, size = 378, normalized size = 1.3 \begin{align*}{\frac{1}{aeb} \left ( 2\,A\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticE} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) ab-A\sqrt{{ \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{2}\sqrt{{ \left ( -bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{-{bx{\frac{1}{\sqrt{-ab}}}}}{\it EllipticF} \left ( \sqrt{{ \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}},{\frac{\sqrt{2}}{2}} \right ) ab+2\,B\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticE} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ){a}^{2}-B\sqrt{{ \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{2}\sqrt{{ \left ( -bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{-{bx{\frac{1}{\sqrt{-ab}}}}}{\it EllipticF} \left ( \sqrt{{ \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}},{\frac{\sqrt{2}}{2}} \right ){a}^{2}-2\,A{x}^{2}{b}^{2}-2\,Aab \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}{\frac{1}{\sqrt{ex}}}} \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{B x^{2} + A}{\sqrt{b x^{2} + a} \left (e 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{{\left (B x^{2} + A\right )} \sqrt{b x^{2} + a} \sqrt{e x}}{b e^{2} x^{4} + a e^{2} x^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [C] time = 4.85839, size = 97, normalized size = 0.33 \begin{align*} \frac{A \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{4}, \frac{1}{2} \\ \frac{3}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{a} e^{\frac{3}{2}} \sqrt{x} \Gamma \left (\frac{3}{4}\right )} + \frac{B x^{\frac{3}{2}} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt{a} e^{\frac{3}{2}} \Gamma \left (\frac{7}{4}\right )} \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{B x^{2} + A}{\sqrt{b x^{2} + a} \left (e x\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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