Optimal. Leaf size=356 \[ -\frac{\sqrt{2} \sqrt{d x^2+2} (-10 a d f+3 b d+2 b f) \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{3}}\right ),1-\frac{3 d}{2 f}\right )}{5 d f^{3/2} \sqrt{f x^2+3} \sqrt{\frac{d x^2+2}{f x^2+3}}}+\frac{x \sqrt{d x^2+2} \left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right )}{15 d^2 f \sqrt{f x^2+3}}-\frac{\sqrt{2} \sqrt{d x^2+2} \left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{3}}\right )|1-\frac{3 d}{2 f}\right )}{15 d^2 f^{3/2} \sqrt{f x^2+3} \sqrt{\frac{d x^2+2}{f x^2+3}}}+\frac{x \sqrt{d x^2+2} \sqrt{f x^2+3} (5 a d f+3 b d-4 b f)}{15 d f}+\frac{b x \left (d x^2+2\right )^{3/2} \sqrt{f x^2+3}}{5 d} \]
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Rubi [A] time = 0.322451, antiderivative size = 356, 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} \[ \frac{x \sqrt{d x^2+2} \left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right )}{15 d^2 f \sqrt{f x^2+3}}-\frac{\sqrt{2} \sqrt{d x^2+2} \left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{3}}\right )|1-\frac{3 d}{2 f}\right )}{15 d^2 f^{3/2} \sqrt{f x^2+3} \sqrt{\frac{d x^2+2}{f x^2+3}}}-\frac{\sqrt{2} \sqrt{d x^2+2} (-10 a d f+3 b d+2 b f) F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{3}}\right )|1-\frac{3 d}{2 f}\right )}{5 d f^{3/2} \sqrt{f x^2+3} \sqrt{\frac{d x^2+2}{f x^2+3}}}+\frac{x \sqrt{d x^2+2} \sqrt{f x^2+3} (5 a d f+3 b d-4 b f)}{15 d f}+\frac{b x \left (d x^2+2\right )^{3/2} \sqrt{f x^2+3}}{5 d} \]
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{2+d x^2} \sqrt{3+f x^2} \, dx &=\frac{b x \left (2+d x^2\right )^{3/2} \sqrt{3+f x^2}}{5 d}+\frac{\int \frac{\sqrt{2+d x^2} \left (-3 (2 b-5 a d)+(3 b d-4 b f+5 a d f) x^2\right )}{\sqrt{3+f x^2}} \, dx}{5 d}\\ &=\frac{(3 b d-4 b f+5 a d f) x \sqrt{2+d x^2} \sqrt{3+f x^2}}{15 d f}+\frac{b x \left (2+d x^2\right )^{3/2} \sqrt{3+f x^2}}{5 d}+\frac{\int \frac{-6 (3 b d+2 b f-10 a d f)+\left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) x^2}{\sqrt{2+d x^2} \sqrt{3+f x^2}} \, dx}{15 d f}\\ &=\frac{(3 b d-4 b f+5 a d f) x \sqrt{2+d x^2} \sqrt{3+f x^2}}{15 d f}+\frac{b x \left (2+d x^2\right )^{3/2} \sqrt{3+f x^2}}{5 d}-\frac{(2 (3 b d+2 b f-10 a d f)) \int \frac{1}{\sqrt{2+d x^2} \sqrt{3+f x^2}} \, dx}{5 d f}+\frac{\left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) \int \frac{x^2}{\sqrt{2+d x^2} \sqrt{3+f x^2}} \, dx}{15 d f}\\ &=\frac{\left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) x \sqrt{2+d x^2}}{15 d^2 f \sqrt{3+f x^2}}+\frac{(3 b d-4 b f+5 a d f) x \sqrt{2+d x^2} \sqrt{3+f x^2}}{15 d f}+\frac{b x \left (2+d x^2\right )^{3/2} \sqrt{3+f x^2}}{5 d}-\frac{\sqrt{2} (3 b d+2 b f-10 a d f) \sqrt{2+d x^2} F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{3}}\right )|1-\frac{3 d}{2 f}\right )}{5 d f^{3/2} \sqrt{\frac{2+d x^2}{3+f x^2}} \sqrt{3+f x^2}}-\frac{\left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) \int \frac{\sqrt{2+d x^2}}{\left (3+f x^2\right )^{3/2}} \, dx}{5 d^2 f}\\ &=\frac{\left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) x \sqrt{2+d x^2}}{15 d^2 f \sqrt{3+f x^2}}+\frac{(3 b d-4 b f+5 a d f) x \sqrt{2+d x^2} \sqrt{3+f x^2}}{15 d f}+\frac{b x \left (2+d x^2\right )^{3/2} \sqrt{3+f x^2}}{5 d}-\frac{\sqrt{2} \left (5 a d f (3 d+2 f)-2 b \left (9 d^2-6 d f+4 f^2\right )\right ) \sqrt{2+d x^2} E\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{3}}\right )|1-\frac{3 d}{2 f}\right )}{15 d^2 f^{3/2} \sqrt{\frac{2+d x^2}{3+f x^2}} \sqrt{3+f x^2}}-\frac{\sqrt{2} (3 b d+2 b f-10 a d f) \sqrt{2+d x^2} F\left (\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{3}}\right )|1-\frac{3 d}{2 f}\right )}{5 d f^{3/2} \sqrt{\frac{2+d x^2}{3+f x^2}} \sqrt{3+f x^2}}\\ \end{align*}
Mathematica [C] time = 0.323886, size = 186, normalized size = 0.52 \[ \frac{i \sqrt{3} (3 d-2 f) (5 a d f-6 b d+2 b f) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{2}}\right ),\frac{2 f}{3 d}\right )+i \sqrt{3} \left (2 b \left (9 d^2-6 d f+4 f^2\right )-5 a d f (3 d+2 f)\right ) E\left (i \sinh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{2}}\right )|\frac{2 f}{3 d}\right )+\sqrt{d} f x \sqrt{d x^2+2} \sqrt{f x^2+3} \left (5 a d f+3 b d \left (f x^2+1\right )+2 b f\right )}{15 d^{3/2} f^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.023, size = 775, normalized size = 2.2 \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} + 2} \sqrt{f x^{2} + 3}\,{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} + 2} \sqrt{f x^{2} + 3}, 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{d x^{2} + 2} \sqrt{f x^{2} + 3}\, 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} + 2} \sqrt{f x^{2} + 3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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