Optimal. Leaf size=274 \[ -\frac{\left (-b^2 c \left (4 a d^2-c e^2\right )+6 a b c^2 d e+2 a c^2 \left (a d^2-c e^2\right )-2 b^3 c d e+b^4 d^2\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{a^2 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}-\frac{(b d-c e) \left (-2 a c d+b^2 d-b c e\right ) \log \left (a x^2+b x+c\right )}{2 a^2 \left (a d^2-e (b d-c e)\right )^2}-\frac{d^4}{e^3 (d+e x) \left (a d^2-e (b d-c e)\right )}-\frac{d^3 \log (d+e x) \left (2 a d^2-e (3 b d-4 c e)\right )}{e^3 \left (a d^2-e (b d-c e)\right )^2}+\frac{x}{a e^2} \]
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Rubi [A] time = 0.56307, antiderivative size = 274, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {1569, 1628, 634, 618, 206, 628} \[ -\frac{\left (-b^2 c \left (4 a d^2-c e^2\right )+6 a b c^2 d e+2 a c^2 \left (a d^2-c e^2\right )-2 b^3 c d e+b^4 d^2\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{a^2 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}-\frac{(b d-c e) \left (-2 a c d+b^2 d-b c e\right ) \log \left (a x^2+b x+c\right )}{2 a^2 \left (a d^2-e (b d-c e)\right )^2}-\frac{d^4}{e^3 (d+e x) \left (a d^2-e (b d-c e)\right )}-\frac{d^3 \log (d+e x) \left (2 a d^2-e (3 b d-4 c e)\right )}{e^3 \left (a d^2-e (b d-c e)\right )^2}+\frac{x}{a e^2} \]
Antiderivative was successfully verified.
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Rule 1569
Rule 1628
Rule 634
Rule 618
Rule 206
Rule 628
Rubi steps
\begin{align*} \int \frac{x^2}{\left (a+\frac{c}{x^2}+\frac{b}{x}\right ) (d+e x)^2} \, dx &=\int \frac{x^4}{(d+e x)^2 \left (c+b x+a x^2\right )} \, dx\\ &=\int \left (\frac{1}{a e^2}+\frac{d^4}{e^2 \left (a d^2-e (b d-c e)\right ) (d+e x)^2}+\frac{d^3 \left (-2 a d^2+e (3 b d-4 c e)\right )}{e^2 \left (a d^2-e (b d-c e)\right )^2 (d+e x)}+\frac{-c \left (b^2 d^2-2 b c d e-c \left (a d^2-c e^2\right )\right )-(b d-c e) \left (b^2 d-2 a c d-b c e\right ) x}{a \left (a d^2-e (b d-c e)\right )^2 \left (c+b x+a x^2\right )}\right ) \, dx\\ &=\frac{x}{a e^2}-\frac{d^4}{e^3 \left (a d^2-e (b d-c e)\right ) (d+e x)}-\frac{d^3 \left (2 a d^2-e (3 b d-4 c e)\right ) \log (d+e x)}{e^3 \left (a d^2-e (b d-c e)\right )^2}+\frac{\int \frac{-c \left (b^2 d^2-2 b c d e-c \left (a d^2-c e^2\right )\right )-(b d-c e) \left (b^2 d-2 a c d-b c e\right ) x}{c+b x+a x^2} \, dx}{a \left (a d^2-e (b d-c e)\right )^2}\\ &=\frac{x}{a e^2}-\frac{d^4}{e^3 \left (a d^2-e (b d-c e)\right ) (d+e x)}-\frac{d^3 \left (2 a d^2-e (3 b d-4 c e)\right ) \log (d+e x)}{e^3 \left (a d^2-e (b d-c e)\right )^2}-\frac{\left ((b d-c e) \left (b^2 d-2 a c d-b c e\right )\right ) \int \frac{b+2 a x}{c+b x+a x^2} \, dx}{2 a^2 \left (a d^2-e (b d-c e)\right )^2}+\frac{\left (b^4 d^2-2 b^3 c d e+6 a b c^2 d e+2 a c^2 \left (a d^2-c e^2\right )-b^2 c \left (4 a d^2-c e^2\right )\right ) \int \frac{1}{c+b x+a x^2} \, dx}{2 a^2 \left (a d^2-e (b d-c e)\right )^2}\\ &=\frac{x}{a e^2}-\frac{d^4}{e^3 \left (a d^2-e (b d-c e)\right ) (d+e x)}-\frac{d^3 \left (2 a d^2-e (3 b d-4 c e)\right ) \log (d+e x)}{e^3 \left (a d^2-e (b d-c e)\right )^2}-\frac{(b d-c e) \left (b^2 d-2 a c d-b c e\right ) \log \left (c+b x+a x^2\right )}{2 a^2 \left (a d^2-e (b d-c e)\right )^2}-\frac{\left (b^4 d^2-2 b^3 c d e+6 a b c^2 d e+2 a c^2 \left (a d^2-c e^2\right )-b^2 c \left (4 a d^2-c e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 a x\right )}{a^2 \left (a d^2-e (b d-c e)\right )^2}\\ &=\frac{x}{a e^2}-\frac{d^4}{e^3 \left (a d^2-e (b d-c e)\right ) (d+e x)}-\frac{\left (b^4 d^2-2 b^3 c d e+6 a b c^2 d e+2 a c^2 \left (a d^2-c e^2\right )-b^2 c \left (4 a d^2-c e^2\right )\right ) \tanh ^{-1}\left (\frac{b+2 a x}{\sqrt{b^2-4 a c}}\right )}{a^2 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}-\frac{d^3 \left (2 a d^2-e (3 b d-4 c e)\right ) \log (d+e x)}{e^3 \left (a d^2-e (b d-c e)\right )^2}-\frac{(b d-c e) \left (b^2 d-2 a c d-b c e\right ) \log \left (c+b x+a x^2\right )}{2 a^2 \left (a d^2-e (b d-c e)\right )^2}\\ \end{align*}
Mathematica [A] time = 0.315942, size = 269, normalized size = 0.98 \[ \frac{\left (b^2 c \left (c e^2-4 a d^2\right )+6 a b c^2 d e+2 a c^2 \left (a d^2-c e^2\right )-2 b^3 c d e+b^4 d^2\right ) \tan ^{-1}\left (\frac{2 a x+b}{\sqrt{4 a c-b^2}}\right )}{a^2 \sqrt{4 a c-b^2} \left (a d^2+e (c e-b d)\right )^2}+\frac{(b d-c e) \left (2 a c d+b^2 (-d)+b c e\right ) \log (x (a x+b)+c)}{2 a^2 \left (a d^2+e (c e-b d)\right )^2}-\frac{d^4}{e^3 (d+e x) \left (a d^2+e (c e-b d)\right )}-\frac{\log (d+e x) \left (2 a d^5+d^3 e (4 c e-3 b d)\right )}{e^3 \left (a d^2+e (c e-b d)\right )^2}+\frac{x}{a e^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.012, size = 765, normalized size = 2.8 \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(-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(-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 [A] time = 1.13066, size = 643, normalized size = 2.35 \begin{align*} -\frac{d^{4} e^{3}}{{\left (a d^{2} e^{6} - b d e^{7} + c e^{8}\right )}{\left (x e + d\right )}} - \frac{{\left (b^{4} d^{2} e^{2} - 4 \, a b^{2} c d^{2} e^{2} + 2 \, a^{2} c^{2} d^{2} e^{2} - 2 \, b^{3} c d e^{3} + 6 \, a b c^{2} d e^{3} + b^{2} c^{2} e^{4} - 2 \, a c^{3} e^{4}\right )} \arctan \left (-\frac{{\left (2 \, a d - \frac{2 \, a d^{2}}{x e + d} - b e + \frac{2 \, b d e}{x e + d} - \frac{2 \, c e^{2}}{x e + d}\right )} e^{\left (-1\right )}}{\sqrt{-b^{2} + 4 \, a c}}\right ) e^{\left (-2\right )}}{{\left (a^{4} d^{4} - 2 \, a^{3} b d^{3} e + a^{2} b^{2} d^{2} e^{2} + 2 \, a^{3} c d^{2} e^{2} - 2 \, a^{2} b c d e^{3} + a^{2} c^{2} e^{4}\right )} \sqrt{-b^{2} + 4 \, a c}} + \frac{{\left (x e + d\right )} e^{\left (-3\right )}}{a} - \frac{{\left (b^{3} d^{2} - 2 \, a b c d^{2} - 2 \, b^{2} c d e + 2 \, a c^{2} d e + b c^{2} e^{2}\right )} \log \left (-a + \frac{2 \, a d}{x e + d} - \frac{a d^{2}}{{\left (x e + d\right )}^{2}} - \frac{b e}{x e + d} + \frac{b d e}{{\left (x e + d\right )}^{2}} - \frac{c e^{2}}{{\left (x e + d\right )}^{2}}\right )}{2 \,{\left (a^{4} d^{4} - 2 \, a^{3} b d^{3} e + a^{2} b^{2} d^{2} e^{2} + 2 \, a^{3} c d^{2} e^{2} - 2 \, a^{2} b c d e^{3} + a^{2} c^{2} e^{4}\right )}} + \frac{{\left (2 \, a d + b e\right )} e^{\left (-3\right )} \log \left (\frac{{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right )}{a^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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