Optimal. Leaf size=367 \[ -\frac{f \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{f \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{e \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a} d^2}-\frac{\sqrt{a+c x^2}}{a d x} \]
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Rubi [A] time = 1.19924, antiderivative size = 367, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {6728, 264, 266, 63, 208, 1034, 725, 206} \[ -\frac{f \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{f \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{e \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a} d^2}-\frac{\sqrt{a+c x^2}}{a d x} \]
Antiderivative was successfully verified.
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Rule 6728
Rule 264
Rule 266
Rule 63
Rule 208
Rule 1034
Rule 725
Rule 206
Rubi steps
\begin{align*} \int \frac{1}{x^2 \sqrt{a+c x^2} \left (d+e x+f x^2\right )} \, dx &=\int \left (\frac{1}{d x^2 \sqrt{a+c x^2}}-\frac{e}{d^2 x \sqrt{a+c x^2}}+\frac{e^2-d f+e f x}{d^2 \sqrt{a+c x^2} \left (d+e x+f x^2\right )}\right ) \, dx\\ &=\frac{\int \frac{e^2-d f+e f x}{\sqrt{a+c x^2} \left (d+e x+f x^2\right )} \, dx}{d^2}+\frac{\int \frac{1}{x^2 \sqrt{a+c x^2}} \, dx}{d}-\frac{e \int \frac{1}{x \sqrt{a+c x^2}} \, dx}{d^2}\\ &=-\frac{\sqrt{a+c x^2}}{a d x}-\frac{e \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+c x}} \, dx,x,x^2\right )}{2 d^2}-\frac{\left (f \left (e^2-2 d f-e \sqrt{e^2-4 d f}\right )\right ) \int \frac{1}{\left (e+\sqrt{e^2-4 d f}+2 f x\right ) \sqrt{a+c x^2}} \, dx}{d^2 \sqrt{e^2-4 d f}}+\frac{\left (f \left (e^2-2 d f+e \sqrt{e^2-4 d f}\right )\right ) \int \frac{1}{\left (e-\sqrt{e^2-4 d f}+2 f x\right ) \sqrt{a+c x^2}} \, dx}{d^2 \sqrt{e^2-4 d f}}\\ &=-\frac{\sqrt{a+c x^2}}{a d x}-\frac{e \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{c}+\frac{x^2}{c}} \, dx,x,\sqrt{a+c x^2}\right )}{c d^2}+\frac{\left (f \left (e^2-2 d f-e \sqrt{e^2-4 d f}\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 )}{d^2 \sqrt{e^2-4 d f}}-\frac{\left (f \left (e^2-2 d f+e \sqrt{e^2-4 d f}\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 )}{d^2 \sqrt{e^2-4 d f}}\\ &=-\frac{\sqrt{a+c x^2}}{a d x}-\frac{f \left (e^2-2 d f+e \sqrt{e^2-4 d f}\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} d^2 \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{f \left (e^2-2 d f-e \sqrt{e^2-4 d f}\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} d^2 \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{e \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a} d^2}\\ \end{align*}
Mathematica [A] time = 0.902893, size = 356, normalized size = 0.97 \[ -\frac{\frac{\sqrt{2} f \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac{2 a f+c x \left (\sqrt{e^2-4 d f}-e\right )}{\sqrt{a+c x^2} \sqrt{4 a f^2-2 c \left (e \sqrt{e^2-4 d f}+2 d f-e^2\right )}}\right )}{\sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{\sqrt{2} f \left (e \sqrt{e^2-4 d f}+2 d f-e^2\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{a+c x^2} \sqrt{4 a f^2+2 c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{2 d \sqrt{a+c x^2}}{a x}-\frac{2 e \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a}}}{2 d^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.273, size = 736, normalized size = 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 \frac{1}{\sqrt{c x^{2} + a}{\left (f x^{2} + e x + d\right )} x^{2}}\,{d x} \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{1}{x^{2} \sqrt{a + c x^{2}} \left (d + e x + f x^{2}\right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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