Optimal. Leaf size=543 \[ \frac{b \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{2 a^{3/2} d}-\frac{f \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (e-\sqrt{e^2-4 d f}\right )\right )-b \left (e-\sqrt{e^2-4 d f}\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \sqrt{f \left (2 a f-b \left (e-\sqrt{e^2-4 d f}\right )\right )+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{f \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (\sqrt{e^2-4 d f}+e\right )\right )-b \left (\sqrt{e^2-4 d f}+e\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}+\frac{e \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{\sqrt{a} d^2}-\frac{\sqrt{a+b x+c x^2}}{a d x} \]
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Rubi [A] time = 4.59176, antiderivative size = 543, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6728, 730, 724, 206, 1032} \[ \frac{b \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{2 a^{3/2} d}-\frac{f \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (e-\sqrt{e^2-4 d f}\right )\right )-b \left (e-\sqrt{e^2-4 d f}\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2-\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \sqrt{f \left (2 a f-b \left (e-\sqrt{e^2-4 d f}\right )\right )+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{f \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac{4 a f+2 x \left (b f-c \left (\sqrt{e^2-4 d f}+e\right )\right )-b \left (\sqrt{e^2-4 d f}+e\right )}{2 \sqrt{2} \sqrt{a+b x+c x^2} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+\sqrt{e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}+\frac{e \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{\sqrt{a} d^2}-\frac{\sqrt{a+b x+c x^2}}{a d x} \]
Antiderivative was successfully verified.
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Rule 6728
Rule 730
Rule 724
Rule 206
Rule 1032
Rubi steps
\begin{align*} \int \frac{1}{x^2 \sqrt{a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx &=\int \left (\frac{1}{d x^2 \sqrt{a+b x+c x^2}}-\frac{e}{d^2 x \sqrt{a+b x+c x^2}}+\frac{e^2-d f+e f x}{d^2 \sqrt{a+b x+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+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{d^2}+\frac{\int \frac{1}{x^2 \sqrt{a+b x+c x^2}} \, dx}{d}-\frac{e \int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx}{d^2}\\ &=-\frac{\sqrt{a+b x+c x^2}}{a d x}-\frac{b \int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx}{2 a d}+\frac{(2 e) \operatorname{Subst}\left (\int \frac{1}{4 a-x^2} \, dx,x,\frac{2 a+b x}{\sqrt{a+b x+c x^2}}\right )}{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+b x+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+b x+c x^2}} \, dx}{d^2 \sqrt{e^2-4 d f}}\\ &=-\frac{\sqrt{a+b x+c x^2}}{a d x}+\frac{e \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{\sqrt{a} d^2}+\frac{b \operatorname{Subst}\left (\int \frac{1}{4 a-x^2} \, dx,x,\frac{2 a+b x}{\sqrt{a+b x+c x^2}}\right )}{a d}+\frac{\left (2 f \left (e^2-2 d f-e \sqrt{e^2-4 d f}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{16 a f^2-8 b f \left (e+\sqrt{e^2-4 d f}\right )+4 c \left (e+\sqrt{e^2-4 d f}\right )^2-x^2} \, dx,x,\frac{4 a f-b \left (e+\sqrt{e^2-4 d f}\right )-\left (-2 b f+2 c \left (e+\sqrt{e^2-4 d f}\right )\right ) x}{\sqrt{a+b x+c x^2}}\right )}{d^2 \sqrt{e^2-4 d f}}-\frac{\left (2 f \left (e^2-2 d f+e \sqrt{e^2-4 d f}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{16 a f^2-8 b f \left (e-\sqrt{e^2-4 d f}\right )+4 c \left (e-\sqrt{e^2-4 d f}\right )^2-x^2} \, dx,x,\frac{4 a f-b \left (e-\sqrt{e^2-4 d f}\right )-\left (-2 b f+2 c \left (e-\sqrt{e^2-4 d f}\right )\right ) x}{\sqrt{a+b x+c x^2}}\right )}{d^2 \sqrt{e^2-4 d f}}\\ &=-\frac{\sqrt{a+b x+c x^2}}{a d x}+\frac{b \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{2 a^{3/2} d}+\frac{e \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{\sqrt{a} d^2}-\frac{f \left (e^2-2 d f+e \sqrt{e^2-4 d f}\right ) \tanh ^{-1}\left (\frac{4 a f-b \left (e-\sqrt{e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt{e^2-4 d f}\right )\right ) x}{2 \sqrt{2} \sqrt{c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt{e^2-4 d f}} \sqrt{a+b x+c x^2}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \sqrt{c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt{e^2-4 d f}}}+\frac{f \left (e^2-2 d f-e \sqrt{e^2-4 d f}\right ) \tanh ^{-1}\left (\frac{4 a f-b \left (e+\sqrt{e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt{e^2-4 d f}\right )\right ) x}{2 \sqrt{2} \sqrt{c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt{e^2-4 d f}} \sqrt{a+b x+c x^2}}\right )}{\sqrt{2} d^2 \sqrt{e^2-4 d f} \sqrt{c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt{e^2-4 d f}}}\\ \end{align*}
Mathematica [A] time = 1.59828, size = 533, normalized size = 0.98 \[ -\frac{-\frac{b d \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )}{a^{3/2}}+\frac{\sqrt{2} f \left (e \sqrt{e^2-4 d f}+2 d f-e^2\right ) \tanh ^{-1}\left (\frac{4 a f-b \left (\sqrt{e^2-4 d f}+e-2 f x\right )-2 c x \left (\sqrt{e^2-4 d f}+e\right )}{2 \sqrt{2} \sqrt{a+x (b+c x)} \sqrt{f \left (2 a f-b \left (\sqrt{e^2-4 d f}+e\right )\right )+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{e^2-4 d f} \sqrt{f \left (2 a f-b \left (\sqrt{e^2-4 d f}+e\right )\right )+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{\sqrt{2} f \left (\frac{e^2-2 d f}{\sqrt{e^2-4 d f}}+e\right ) \tanh ^{-1}\left (\frac{4 a f+b \left (\sqrt{e^2-4 d f}-e+2 f x\right )+2 c x \left (\sqrt{e^2-4 d f}-e\right )}{2 \sqrt{2} \sqrt{a+x (b+c x)} \sqrt{f \left (2 a f+b \left (\sqrt{e^2-4 d f}-e\right )\right )+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{f \left (2 a f+b \left (\sqrt{e^2-4 d f}-e\right )\right )+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{2 d \sqrt{a+x (b+c x)}}{a x}-\frac{2 e \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )}{\sqrt{a}}}{2 d^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.344, size = 983, normalized size = 1.8 \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} + b x + 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(-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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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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