Optimal. Leaf size=193 \[ \frac{\left (2 a^2 c d-a b (b d+3 c e)+b^3 e\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{c^2 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac{\left (a b d+a c e+b^2 (-e)\right ) \log \left (a x^2+b x+c\right )}{2 c^2 \left (a d^2-e (b d-c e)\right )}+\frac{e^3 \log (d+e x)}{d^2 \left (a d^2-e (b d-c e)\right )}-\frac{\log (x) (b d+c e)}{c^2 d^2}-\frac{1}{c d x} \]
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Rubi [A] time = 0.343094, antiderivative size = 193, 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, 893, 634, 618, 206, 628} \[ \frac{\left (2 a^2 c d-a b (b d+3 c e)+b^3 e\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{c^2 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac{\left (a b d+a c e+b^2 (-e)\right ) \log \left (a x^2+b x+c\right )}{2 c^2 \left (a d^2-e (b d-c e)\right )}+\frac{e^3 \log (d+e x)}{d^2 \left (a d^2-e (b d-c e)\right )}-\frac{\log (x) (b d+c e)}{c^2 d^2}-\frac{1}{c d x} \]
Antiderivative was successfully verified.
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Rule 1569
Rule 893
Rule 634
Rule 618
Rule 206
Rule 628
Rubi steps
\begin{align*} \int \frac{1}{\left (a+\frac{c}{x^2}+\frac{b}{x}\right ) x^4 (d+e x)} \, dx &=\int \frac{1}{x^2 (d+e x) \left (c+b x+a x^2\right )} \, dx\\ &=\int \left (\frac{1}{c d x^2}+\frac{-b d-c e}{c^2 d^2 x}+\frac{e^4}{d^2 \left (a d^2-e (b d-c e)\right ) (d+e x)}+\frac{-a^2 c d-b^3 e+a b (b d+2 c e)+a \left (a b d-b^2 e+a c e\right ) x}{c^2 \left (a d^2-e (b d-c e)\right ) \left (c+b x+a x^2\right )}\right ) \, dx\\ &=-\frac{1}{c d x}-\frac{(b d+c e) \log (x)}{c^2 d^2}+\frac{e^3 \log (d+e x)}{d^2 \left (a d^2-e (b d-c e)\right )}+\frac{\int \frac{-a^2 c d-b^3 e+a b (b d+2 c e)+a \left (a b d-b^2 e+a c e\right ) x}{c+b x+a x^2} \, dx}{c^2 \left (a d^2-e (b d-c e)\right )}\\ &=-\frac{1}{c d x}-\frac{(b d+c e) \log (x)}{c^2 d^2}+\frac{e^3 \log (d+e x)}{d^2 \left (a d^2-e (b d-c e)\right )}+\frac{\left (a b d-b^2 e+a c e\right ) \int \frac{b+2 a x}{c+b x+a x^2} \, dx}{2 c^2 \left (a d^2-e (b d-c e)\right )}-\frac{\left (2 a^2 c d+b^3 e-a b (b d+3 c e)\right ) \int \frac{1}{c+b x+a x^2} \, dx}{2 c^2 \left (a d^2-e (b d-c e)\right )}\\ &=-\frac{1}{c d x}-\frac{(b d+c e) \log (x)}{c^2 d^2}+\frac{e^3 \log (d+e x)}{d^2 \left (a d^2-e (b d-c e)\right )}+\frac{\left (a b d-b^2 e+a c e\right ) \log \left (c+b x+a x^2\right )}{2 c^2 \left (a d^2-e (b d-c e)\right )}+\frac{\left (2 a^2 c d+b^3 e-a b (b d+3 c e)\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 a x\right )}{c^2 \left (a d^2-e (b d-c e)\right )}\\ &=-\frac{1}{c d x}+\frac{\left (2 a^2 c d+b^3 e-a b (b d+3 c e)\right ) \tanh ^{-1}\left (\frac{b+2 a x}{\sqrt{b^2-4 a c}}\right )}{c^2 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )}-\frac{(b d+c e) \log (x)}{c^2 d^2}+\frac{e^3 \log (d+e x)}{d^2 \left (a d^2-e (b d-c e)\right )}+\frac{\left (a b d-b^2 e+a c e\right ) \log \left (c+b x+a x^2\right )}{2 c^2 \left (a d^2-e (b d-c e)\right )}\\ \end{align*}
Mathematica [A] time = 0.177771, size = 194, normalized size = 1.01 \[ \frac{\left (2 a^2 c d-a b (b d+3 c e)+b^3 e\right ) \tan ^{-1}\left (\frac{2 a x+b}{\sqrt{4 a c-b^2}}\right )}{c^2 \sqrt{4 a c-b^2} \left (e (b d-c e)-a d^2\right )}+\frac{\left (a b d+a c e+b^2 (-e)\right ) \log (x (a x+b)+c)}{2 c^2 \left (a d^2+e (c e-b d)\right )}+\frac{e^3 \log (d+e x)}{a d^4+d^2 e (c e-b d)}-\frac{\log (x) (b d+c e)}{c^2 d^2}-\frac{1}{c d x} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.011, size = 412, normalized size = 2.1 \begin{align*}{\frac{{e}^{3}\ln \left ( ex+d \right ) }{{d}^{2} \left ( a{d}^{2}-bde+{e}^{2}c \right ) }}+{\frac{a\ln \left ( a{x}^{2}+bx+c \right ) bd}{ \left ( 2\,a{d}^{2}-2\,bde+2\,{e}^{2}c \right ){c}^{2}}}+{\frac{a\ln \left ( a{x}^{2}+bx+c \right ) e}{ \left ( 2\,a{d}^{2}-2\,bde+2\,{e}^{2}c \right ) c}}-{\frac{\ln \left ( a{x}^{2}+bx+c \right ){b}^{2}e}{ \left ( 2\,a{d}^{2}-2\,bde+2\,{e}^{2}c \right ){c}^{2}}}-2\,{\frac{{a}^{2}d}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) c\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,ax+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+{\frac{a{b}^{2}d}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ){c}^{2}}\arctan \left ({(2\,ax+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+3\,{\frac{abe}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) c\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,ax+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-{\frac{{b}^{3}e}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ){c}^{2}}\arctan \left ({(2\,ax+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}-{\frac{1}{cdx}}-{\frac{b\ln \left ( x \right ) }{{c}^{2}d}}-{\frac{\ln \left ( x \right ) e}{c{d}^{2}}} \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 = 1.10823, size = 284, normalized size = 1.47 \begin{align*} \frac{{\left (a b d - b^{2} e + a c e\right )} \log \left (a x^{2} + b x + c\right )}{2 \,{\left (a c^{2} d^{2} - b c^{2} d e + c^{3} e^{2}\right )}} + \frac{e^{4} \log \left ({\left | x e + d \right |}\right )}{a d^{4} e - b d^{3} e^{2} + c d^{2} e^{3}} + \frac{{\left (a b^{2} d - 2 \, a^{2} c d - b^{3} e + 3 \, a b c e\right )} \arctan \left (\frac{2 \, a x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (a c^{2} d^{2} - b c^{2} d e + c^{3} e^{2}\right )} \sqrt{-b^{2} + 4 \, a c}} - \frac{{\left (b d + c e\right )} \log \left ({\left | x \right |}\right )}{c^{2} d^{2}} - \frac{1}{c d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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