Optimal. Leaf size=381 \[ -\frac{e^{3/2} \sqrt{c+d x^2} (-10 a d f+b c f+b d e) \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ),1-\frac{d e}{c f}\right )}{15 d f^{3/2} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{x \sqrt{c+d x^2} \left (5 a d f (c f+d e)-2 b \left (c^2 f^2-c d e f+d^2 e^2\right )\right )}{15 d^2 f \sqrt{e+f x^2}}-\frac{\sqrt{e} \sqrt{c+d x^2} \left (5 a d f (c f+d e)-2 b \left (c^2 f^2-c d e f+d^2 e^2\right )\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{15 d^2 f^{3/2} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{x \sqrt{c+d x^2} \sqrt{e+f x^2} (5 a d f-2 b c f+b d e)}{15 d f}+\frac{b x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}{5 d} \]
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Rubi [A] time = 0.377796, antiderivative size = 381, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {528, 531, 418, 492, 411} \[ \frac{x \sqrt{c+d x^2} \left (5 a d f (c f+d e)-2 b \left (c^2 f^2-c d e f+d^2 e^2\right )\right )}{15 d^2 f \sqrt{e+f x^2}}-\frac{\sqrt{e} \sqrt{c+d x^2} \left (5 a d f (c f+d e)-2 b \left (c^2 f^2-c d e f+d^2 e^2\right )\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{15 d^2 f^{3/2} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}-\frac{e^{3/2} \sqrt{c+d x^2} (-10 a d f+b c f+b d e) F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{15 d f^{3/2} \sqrt{e+f x^2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac{x \sqrt{c+d x^2} \sqrt{e+f x^2} (5 a d f-2 b c f+b d e)}{15 d f}+\frac{b x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}{5 d} \]
Antiderivative was successfully verified.
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Rule 528
Rule 531
Rule 418
Rule 492
Rule 411
Rubi steps
\begin{align*} \int \left (a+b x^2\right ) \sqrt{c+d x^2} \sqrt{e+f x^2} \, dx &=\frac{b x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}{5 d}+\frac{\int \frac{\sqrt{c+d x^2} \left (-(b c-5 a d) e+(b d e-2 b c f+5 a d f) x^2\right )}{\sqrt{e+f x^2}} \, dx}{5 d}\\ &=\frac{(b d e-2 b c f+5 a d f) x \sqrt{c+d x^2} \sqrt{e+f x^2}}{15 d f}+\frac{b x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}{5 d}+\frac{\int \frac{-c e (b d e+b c f-10 a d f)+\left (5 a d f (d e+c f)-2 b \left (d^2 e^2-c d e f+c^2 f^2\right )\right ) x^2}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{15 d f}\\ &=\frac{(b d e-2 b c f+5 a d f) x \sqrt{c+d x^2} \sqrt{e+f x^2}}{15 d f}+\frac{b x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}{5 d}-\frac{(c e (b d e+b c f-10 a d f)) \int \frac{1}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{15 d f}+\frac{\left (5 a d f (d e+c f)-2 b \left (d^2 e^2-c d e f+c^2 f^2\right )\right ) \int \frac{x^2}{\sqrt{c+d x^2} \sqrt{e+f x^2}} \, dx}{15 d f}\\ &=\frac{\left (5 a d f (d e+c f)-2 b \left (d^2 e^2-c d e f+c^2 f^2\right )\right ) x \sqrt{c+d x^2}}{15 d^2 f \sqrt{e+f x^2}}+\frac{(b d e-2 b c f+5 a d f) x \sqrt{c+d x^2} \sqrt{e+f x^2}}{15 d f}+\frac{b x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}{5 d}-\frac{e^{3/2} (b d e+b c f-10 a d f) \sqrt{c+d x^2} F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{15 d f^{3/2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt{e+f x^2}}-\frac{\left (e \left (5 a d f (d e+c f)-2 b \left (d^2 e^2-c d e f+c^2 f^2\right )\right )\right ) \int \frac{\sqrt{c+d x^2}}{\left (e+f x^2\right )^{3/2}} \, dx}{15 d^2 f}\\ &=\frac{\left (5 a d f (d e+c f)-2 b \left (d^2 e^2-c d e f+c^2 f^2\right )\right ) x \sqrt{c+d x^2}}{15 d^2 f \sqrt{e+f x^2}}+\frac{(b d e-2 b c f+5 a d f) x \sqrt{c+d x^2} \sqrt{e+f x^2}}{15 d f}+\frac{b x \left (c+d x^2\right )^{3/2} \sqrt{e+f x^2}}{5 d}-\frac{\sqrt{e} \left (5 a d f (d e+c f)-2 b \left (d^2 e^2-c d e f+c^2 f^2\right )\right ) \sqrt{c+d x^2} E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{15 d^2 f^{3/2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt{e+f x^2}}-\frac{e^{3/2} (b d e+b c f-10 a d f) \sqrt{c+d x^2} F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )|1-\frac{d e}{c f}\right )}{15 d f^{3/2} \sqrt{\frac{e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt{e+f x^2}}\\ \end{align*}
Mathematica [C] time = 0.75238, size = 267, normalized size = 0.7 \[ \frac{-i e \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{f x^2}{e}+1} (c f-d e) (5 a d f+b c f-2 b d e) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{d}{c}}\right ),\frac{c f}{d e}\right )+i e \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{f x^2}{e}+1} \left (2 b \left (c^2 f^2-c d e f+d^2 e^2\right )-5 a d f (c f+d e)\right ) E\left (i \sinh ^{-1}\left (\sqrt{\frac{d}{c}} x\right )|\frac{c f}{d e}\right )+f x \sqrt{\frac{d}{c}} \left (c+d x^2\right ) \left (e+f x^2\right ) \left (5 a d f+b c f+b d \left (e+3 f x^2\right )\right )}{15 d f^2 \sqrt{\frac{d}{c}} \sqrt{c+d x^2} \sqrt{e+f x^2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.015, size = 865, normalized size = 2.3 \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{\left (b x^{2} + a\right )} \sqrt{d x^{2} + c} \sqrt{f x^{2} + e}\,{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 ({\left (b x^{2} + a\right )} \sqrt{d x^{2} + c} \sqrt{f x^{2} + e}, 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 \left (a + b x^{2}\right ) \sqrt{c + d x^{2}} \sqrt{e + f x^{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{\left (b x^{2} + a\right )} \sqrt{d x^{2} + c} \sqrt{f x^{2} + e}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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