Optimal. Leaf size=618 \[ \frac{e \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{a^{3/2} d^2}-\frac{2 c x}{a^2 d \sqrt{a+c x^2}}+\frac{c d x \left (a f^2+c \left (e^2-d f\right )\right )+a e \left (a f^2+c \left (e^2-2 d f\right )\right )}{a d^2 \sqrt{a+c x^2} \left ((c d-a f)^2+a c e^2\right )}+\frac{f \left (e \left (e-\sqrt{e^2-4 d f}\right ) \left (a f^2+c \left (e^2-2 d f\right )\right )-2 \left (a f^2 \left (e^2-d f\right )+c \left (d^2 f^2-3 d e^2 f+e^4\right )\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{f \left (e \left (\sqrt{e^2-4 d f}+e\right ) \left (a f^2+c \left (e^2-2 d f\right )\right )-2 \left (a f^2 \left (e^2-d f\right )+c \left (d^2 f^2-3 d e^2 f+e^4\right )\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{e}{a d^2 \sqrt{a+c x^2}}-\frac{1}{a d x \sqrt{a+c x^2}} \]
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Rubi [A] time = 2.2803, antiderivative size = 618, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.407, Rules used = {6728, 271, 191, 266, 51, 63, 208, 1017, 1034, 725, 206} \[ \frac{e \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{a^{3/2} d^2}-\frac{2 c x}{a^2 d \sqrt{a+c x^2}}+\frac{c d x \left (a f^2+c \left (e^2-d f\right )\right )+a e \left (a f^2+c \left (e^2-2 d f\right )\right )}{a d^2 \sqrt{a+c x^2} \left ((c d-a f)^2+a c e^2\right )}+\frac{f \left (e \left (e-\sqrt{e^2-4 d f}\right ) \left (a f^2+c \left (e^2-2 d f\right )\right )-2 \left (a f^2 \left (e^2-d f\right )+c \left (d^2 f^2-3 d e^2 f+e^4\right )\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{f \left (e \left (\sqrt{e^2-4 d f}+e\right ) \left (a f^2+c \left (e^2-2 d f\right )\right )-2 \left (a f^2 \left (e^2-d f\right )+c \left (d^2 f^2-3 d e^2 f+e^4\right )\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{e}{a d^2 \sqrt{a+c x^2}}-\frac{1}{a d x \sqrt{a+c x^2}} \]
Antiderivative was successfully verified.
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Rule 6728
Rule 271
Rule 191
Rule 266
Rule 51
Rule 63
Rule 208
Rule 1017
Rule 1034
Rule 725
Rule 206
Rubi steps
\begin{align*} \int \frac{1}{x^2 \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx &=\int \left (\frac{1}{d x^2 \left (a+c x^2\right )^{3/2}}-\frac{e}{d^2 x \left (a+c x^2\right )^{3/2}}+\frac{e^2-d f+e f x}{d^2 \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )}\right ) \, dx\\ &=\frac{\int \frac{e^2-d f+e f x}{\left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx}{d^2}+\frac{\int \frac{1}{x^2 \left (a+c x^2\right )^{3/2}} \, dx}{d}-\frac{e \int \frac{1}{x \left (a+c x^2\right )^{3/2}} \, dx}{d^2}\\ &=-\frac{1}{a d x \sqrt{a+c x^2}}+\frac{a e \left (a f^2+c \left (e^2-2 d f\right )\right )+c d \left (a f^2+c \left (e^2-d f\right )\right ) x}{a d^2 \left (a c e^2+(c d-a f)^2\right ) \sqrt{a+c x^2}}-\frac{(2 c) \int \frac{1}{\left (a+c x^2\right )^{3/2}} \, dx}{a d}-\frac{e \operatorname{Subst}\left (\int \frac{1}{x (a+c x)^{3/2}} \, dx,x,x^2\right )}{2 d^2}+\frac{\int \frac{2 a c \left (a f^2 \left (e^2-d f\right )+c \left (e^4-3 d e^2 f+d^2 f^2\right )\right )+2 a c e f \left (a f^2+c \left (e^2-2 d f\right )\right ) x}{\sqrt{a+c x^2} \left (d+e x+f x^2\right )} \, dx}{2 a c d^2 \left (a c e^2+(c d-a f)^2\right )}\\ &=-\frac{e}{a d^2 \sqrt{a+c x^2}}-\frac{1}{a d x \sqrt{a+c x^2}}-\frac{2 c x}{a^2 d \sqrt{a+c x^2}}+\frac{a e \left (a f^2+c \left (e^2-2 d f\right )\right )+c d \left (a f^2+c \left (e^2-d f\right )\right ) x}{a d^2 \left (a c e^2+(c d-a f)^2\right ) \sqrt{a+c x^2}}-\frac{e \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+c x}} \, dx,x,x^2\right )}{2 a d^2}-\frac{\left (f \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 \left (a f^2 \left (e^2-d f\right )+c \left (e^4-3 d e^2 f+d^2 f^2\right )\right )\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} \left (a c e^2+(c d-a f)^2\right )}+\frac{\left (f \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 \left (a f^2 \left (e^2-d f\right )+c \left (e^4-3 d e^2 f+d^2 f^2\right )\right )\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} \left (a c e^2+(c d-a f)^2\right )}\\ &=-\frac{e}{a d^2 \sqrt{a+c x^2}}-\frac{1}{a d x \sqrt{a+c x^2}}-\frac{2 c x}{a^2 d \sqrt{a+c x^2}}+\frac{a e \left (a f^2+c \left (e^2-2 d f\right )\right )+c d \left (a f^2+c \left (e^2-d f\right )\right ) x}{a d^2 \left (a c e^2+(c d-a f)^2\right ) \sqrt{a+c x^2}}-\frac{e \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{c}+\frac{x^2}{c}} \, dx,x,\sqrt{a+c x^2}\right )}{a c d^2}+\frac{\left (f \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 \left (a f^2 \left (e^2-d f\right )+c \left (e^4-3 d e^2 f+d^2 f^2\right )\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 )}{d^2 \sqrt{e^2-4 d f} \left (a c e^2+(c d-a f)^2\right )}-\frac{\left (f \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 \left (a f^2 \left (e^2-d f\right )+c \left (e^4-3 d e^2 f+d^2 f^2\right )\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 )}{d^2 \sqrt{e^2-4 d f} \left (a c e^2+(c d-a f)^2\right )}\\ &=-\frac{e}{a d^2 \sqrt{a+c x^2}}-\frac{1}{a d x \sqrt{a+c x^2}}-\frac{2 c x}{a^2 d \sqrt{a+c x^2}}+\frac{a e \left (a f^2+c \left (e^2-2 d f\right )\right )+c d \left (a f^2+c \left (e^2-d f\right )\right ) x}{a d^2 \left (a c e^2+(c d-a f)^2\right ) \sqrt{a+c x^2}}+\frac{f \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 \left (a f^2 \left (e^2-d f\right )+c \left (e^4-3 d e^2 f+d^2 f^2\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} d^2 \sqrt{e^2-4 d f} \left (a c e^2+(c d-a f)^2\right ) \sqrt{2 a f^2+c \left (e^2-2 d f-e \sqrt{e^2-4 d f}\right )}}-\frac{f \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 \left (a f^2 \left (e^2-d f\right )+c \left (e^4-3 d e^2 f+d^2 f^2\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} d^2 \sqrt{e^2-4 d f} \left (a c e^2+(c d-a f)^2\right ) \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 )}{a^{3/2} d^2}\\ \end{align*}
Mathematica [C] time = 4.9239, size = 557, normalized size = 0.9 \[ -\frac{\frac{d \left (a+2 c x^2\right )}{a^2 x \sqrt{a+c x^2}}-\frac{f \left (\frac{e^2-2 d f}{\sqrt{e^2-4 d f}}+e\right ) \left (2 a f+c x \left (e-\sqrt{e^2-4 d f}\right )\right )}{a \sqrt{a+c x^2} \left (4 a f^2+c \left (e-\sqrt{e^2-4 d f}\right )^2\right )}-\frac{f \left (\frac{2 d f-e^2}{\sqrt{e^2-4 d f}}+e\right ) \left (2 a f+c x \left (\sqrt{e^2-4 d f}+e\right )\right )}{a \sqrt{a+c x^2} \left (4 a f^2+c \left (\sqrt{e^2-4 d f}+e\right )^2\right )}+\frac{\sqrt{2} f^3 \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac{2 a f+c x \left (\sqrt{e^2-4 d f}-e\right )}{\sqrt{a+c x^2} \sqrt{4 a f^2-2 c \left (e \sqrt{e^2-4 d f}+2 d f-e^2\right )}}\right )}{\sqrt{e^2-4 d f} \left (2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )\right )^{3/2}}+\frac{\sqrt{2} f^3 \left (e \sqrt{e^2-4 d f}+2 d f-e^2\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{a+c x^2} \sqrt{4 a f^2+2 c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{e^2-4 d f} \left (2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )\right )^{3/2}}+\frac{e \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c x^2}{a}+1\right )}{a \sqrt{a+c x^2}}}{d^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.273, size = 2046, normalized size = 3.3 \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 \frac{1}{{\left (c x^{2} + a\right )}^{\frac{3}{2}}{\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} \left (a + c x^{2}\right )^{\frac{3}{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] time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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