Optimal. Leaf size=452 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right ) \left (a f^2+2 c \left (e^2-d f\right )\right )}{2 \sqrt{c} f^3}-\frac{\left (e \left (e-\sqrt{e^2-4 d f}\right ) \left (a f^2+c \left (e^2-2 d f\right )\right )-2 d f \left (a f^2+c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} f^3 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{\left (e \left (\sqrt{e^2-4 d f}+e\right ) \left (a f^2+c \left (e^2-2 d f\right )\right )-2 d f \left (a f^2+c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} f^3 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{\sqrt{a+c x^2} (2 e-f x)}{2 f^2} \]
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Rubi [A] time = 1.96739, antiderivative size = 452, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {1069, 1080, 217, 206, 1034, 725} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right ) \left (a f^2+2 c \left (e^2-d f\right )\right )}{2 \sqrt{c} f^3}-\frac{\left (e \left (e-\sqrt{e^2-4 d f}\right ) \left (a f^2+c \left (e^2-2 d f\right )\right )-2 d f \left (a f^2+c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} f^3 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{\left (e \left (\sqrt{e^2-4 d f}+e\right ) \left (a f^2+c \left (e^2-2 d f\right )\right )-2 d f \left (a f^2+c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} f^3 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{\sqrt{a+c x^2} (2 e-f x)}{2 f^2} \]
Antiderivative was successfully verified.
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Rule 1069
Rule 1080
Rule 217
Rule 206
Rule 1034
Rule 725
Rubi steps
\begin{align*} \int \frac{x^2 \sqrt{a+c x^2}}{d+e x+f x^2} \, dx &=-\frac{(2 e-f x) \sqrt{a+c x^2}}{2 f^2}-\frac{\int \frac{a c d f-c e (2 c d-a f) x-c \left (a f^2+2 c \left (e^2-d f\right )\right ) x^2}{\sqrt{a+c x^2} \left (d+e x+f x^2\right )} \, dx}{2 c f^2}\\ &=-\frac{(2 e-f x) \sqrt{a+c x^2}}{2 f^2}-\frac{\int \frac{a c d f^2+c d \left (a f^2+2 c \left (e^2-d f\right )\right )+\left (-c e f (2 c d-a f)+c e \left (a f^2+2 c \left (e^2-d f\right )\right )\right ) x}{\sqrt{a+c x^2} \left (d+e x+f x^2\right )} \, dx}{2 c f^3}+\frac{\left (a f^2+2 c \left (e^2-d f\right )\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx}{2 f^3}\\ &=-\frac{(2 e-f x) \sqrt{a+c x^2}}{2 f^2}+\frac{\left (a f^2+2 c \left (e^2-d f\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )}{2 f^3}+\frac{\left (e \left (e-\sqrt{e^2-4 d f}\right ) \left (a f^2+c \left (e^2-2 d f\right )\right )-2 d f \left (a f^2+c \left (e^2-d f\right )\right )\right ) \int \frac{1}{\left (e-\sqrt{e^2-4 d f}+2 f x\right ) \sqrt{a+c x^2}} \, dx}{f^3 \sqrt{e^2-4 d f}}-\frac{\left (e \left (e+\sqrt{e^2-4 d f}\right ) \left (a f^2+c \left (e^2-2 d f\right )\right )-2 d f \left (a f^2+c \left (e^2-d f\right )\right )\right ) \int \frac{1}{\left (e+\sqrt{e^2-4 d f}+2 f x\right ) \sqrt{a+c x^2}} \, dx}{f^3 \sqrt{e^2-4 d f}}\\ &=-\frac{(2 e-f x) \sqrt{a+c x^2}}{2 f^2}+\frac{\left (a f^2+2 c \left (e^2-d f\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{2 \sqrt{c} f^3}-\frac{\left (e \left (e-\sqrt{e^2-4 d f}\right ) \left (a f^2+c \left (e^2-2 d f\right )\right )-2 d f \left (a f^2+c \left (e^2-d f\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 a f^2+c \left (e-\sqrt{e^2-4 d f}\right )^2-x^2} \, dx,x,\frac{2 a f-c \left (e-\sqrt{e^2-4 d f}\right ) x}{\sqrt{a+c x^2}}\right )}{f^3 \sqrt{e^2-4 d f}}+\frac{\left (e \left (e+\sqrt{e^2-4 d f}\right ) \left (a f^2+c \left (e^2-2 d f\right )\right )-2 d f \left (a f^2+c \left (e^2-d f\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 a f^2+c \left (e+\sqrt{e^2-4 d f}\right )^2-x^2} \, dx,x,\frac{2 a f-c \left (e+\sqrt{e^2-4 d f}\right ) x}{\sqrt{a+c x^2}}\right )}{f^3 \sqrt{e^2-4 d f}}\\ &=-\frac{(2 e-f x) \sqrt{a+c x^2}}{2 f^2}+\frac{\left (a f^2+2 c \left (e^2-d f\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{2 \sqrt{c} f^3}-\frac{\left (e \left (e-\sqrt{e^2-4 d f}\right ) \left (a f^2+c \left (e^2-2 d f\right )\right )-2 d f \left (a f^2+c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c \left (e-\sqrt{e^2-4 d f}\right ) x}{\sqrt{2} \sqrt{2 a f^2+c \left (e^2-2 d f-e \sqrt{e^2-4 d f}\right )} \sqrt{a+c x^2}}\right )}{\sqrt{2} f^3 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e^2-2 d f-e \sqrt{e^2-4 d f}\right )}}+\frac{\left (e \left (e+\sqrt{e^2-4 d f}\right ) \left (a f^2+c \left (e^2-2 d f\right )\right )-2 d f \left (a f^2+c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c \left (e+\sqrt{e^2-4 d f}\right ) x}{\sqrt{2} \sqrt{2 a f^2+c \left (e^2-2 d f+e \sqrt{e^2-4 d f}\right )} \sqrt{a+c x^2}}\right )}{\sqrt{2} f^3 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e^2-2 d f+e \sqrt{e^2-4 d f}\right )}}\\ \end{align*}
Mathematica [A] time = 2.59567, size = 516, normalized size = 1.14 \[ \frac{\frac{2 f \left (\frac{a^{3/2} \sqrt{\frac{c x^2}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{c}}+a x+c x^3\right )}{\sqrt{a+c x^2}}+\frac{\left (\frac{2 d f-e^2}{\sqrt{e^2-4 d f}}+e\right ) \left (\sqrt{4 a f^2-2 c \left (e \sqrt{e^2-4 d f}+2 d f-e^2\right )} \tanh ^{-1}\left (\frac{2 a f+c x \left (\sqrt{e^2-4 d f}-e\right )}{\sqrt{a+c x^2} \sqrt{4 a f^2-2 c \left (e \sqrt{e^2-4 d f}+2 d f-e^2\right )}}\right )-\sqrt{c} \left (\sqrt{e^2-4 d f}-e\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )\right )}{f}+\frac{\left (e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \left (\sqrt{4 a f^2+2 c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )} \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{a+c x^2} \sqrt{4 a f^2+2 c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )+\sqrt{c} \left (\sqrt{e^2-4 d f}+e\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )\right )}{f \sqrt{e^2-4 d f}}-2 \sqrt{a+c x^2} \left (\frac{e^2-2 d f}{\sqrt{e^2-4 d f}}+e\right )-2 \sqrt{a+c x^2} \left (\frac{2 d f-e^2}{\sqrt{e^2-4 d f}}+e\right )}{4 f^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.319, size = 7739, normalized size = 17.1 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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^{2} \sqrt{a + c x^{2}}}{d + e x + f x^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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