Optimal. Leaf size=210 \[ \frac{\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )^2}-\frac{\log (d+e x) \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right )}{\left (a e^2-b d e+c d^2\right )^2}+\frac{e \sqrt{b^2-4 a c} (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (a e^2-b d e+c d^2\right )^2}+\frac{2 c d-b e}{(d+e x) \left (a e^2-b d e+c d^2\right )} \]
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Rubi [A] time = 0.299542, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {800, 634, 618, 206, 628} \[ \frac{\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )^2}-\frac{\log (d+e x) \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right )}{\left (a e^2-b d e+c d^2\right )^2}+\frac{e \sqrt{b^2-4 a c} (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (a e^2-b d e+c d^2\right )^2}+\frac{2 c d-b e}{(d+e x) \left (a e^2-b d e+c d^2\right )} \]
Antiderivative was successfully verified.
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Rule 800
Rule 634
Rule 618
Rule 206
Rule 628
Rubi steps
\begin{align*} \int \frac{b+2 c x}{(d+e x)^2 \left (a+b x+c x^2\right )} \, dx &=\int \left (\frac{e (-2 c d+b e)}{\left (c d^2-b d e+a e^2\right ) (d+e x)^2}+\frac{e \left (-2 c^2 d^2-b^2 e^2+2 c e (b d+a e)\right )}{\left (c d^2-b d e+a e^2\right )^2 (d+e x)}+\frac{-2 b^2 c d e+4 a c^2 d e+b^3 e^2+b c \left (c d^2-3 a e^2\right )+c \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) x}{\left (c d^2-b d e+a e^2\right )^2 \left (a+b x+c x^2\right )}\right ) \, dx\\ &=\frac{2 c d-b e}{\left (c d^2-b d e+a e^2\right ) (d+e x)}-\frac{\left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^2}+\frac{\int \frac{-2 b^2 c d e+4 a c^2 d e+b^3 e^2+b c \left (c d^2-3 a e^2\right )+c \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) x}{a+b x+c x^2} \, dx}{\left (c d^2-b d e+a e^2\right )^2}\\ &=\frac{2 c d-b e}{\left (c d^2-b d e+a e^2\right ) (d+e x)}-\frac{\left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^2}-\frac{\left (\left (b^2-4 a c\right ) e (2 c d-b e)\right ) \int \frac{1}{a+b x+c x^2} \, dx}{2 \left (c d^2-b d e+a e^2\right )^2}+\frac{\left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \int \frac{b+2 c x}{a+b x+c x^2} \, dx}{2 \left (c d^2-b d e+a e^2\right )^2}\\ &=\frac{2 c d-b e}{\left (c d^2-b d e+a e^2\right ) (d+e x)}-\frac{\left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^2}+\frac{\left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \log \left (a+b x+c x^2\right )}{2 \left (c d^2-b d e+a e^2\right )^2}+\frac{\left (\left (b^2-4 a c\right ) e (2 c d-b e)\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{\left (c d^2-b d e+a e^2\right )^2}\\ &=\frac{2 c d-b e}{\left (c d^2-b d e+a e^2\right ) (d+e x)}+\frac{\sqrt{b^2-4 a c} e (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{\left (c d^2-b d e+a e^2\right )^2}-\frac{\left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^2}+\frac{\left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \log \left (a+b x+c x^2\right )}{2 \left (c d^2-b d e+a e^2\right )^2}\\ \end{align*}
Mathematica [A] time = 0.323495, size = 177, normalized size = 0.84 \[ \frac{\log (d+e x) \left (4 c e (a e+b d)-2 b^2 e^2-4 c^2 d^2\right )+\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log (a+x (b+c x))-2 e \sqrt{4 a c-b^2} (b e-2 c d) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )+\frac{2 (2 c d-b e) \left (e (a e-b d)+c d^2\right )}{d+e x}}{2 \left (e (a e-b d)+c d^2\right )^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.01, size = 560, normalized size = 2.7 \begin{align*} 2\,{\frac{\ln \left ( ex+d \right ) ac{e}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}-{\frac{\ln \left ( ex+d \right ){b}^{2}{e}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}+2\,{\frac{\ln \left ( ex+d \right ) bcde}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}-2\,{\frac{\ln \left ( ex+d \right ){c}^{2}{d}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}-{\frac{be}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) \left ( ex+d \right ) }}+2\,{\frac{cd}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) \left ( ex+d \right ) }}-{\frac{c\ln \left ( c{x}^{2}+bx+a \right ) a{e}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}+{\frac{\ln \left ( c{x}^{2}+bx+a \right ){b}^{2}{e}^{2}}{2\, \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}-{\frac{c\ln \left ( c{x}^{2}+bx+a \right ) bde}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}+{\frac{{c}^{2}\ln \left ( c{x}^{2}+bx+a \right ){d}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}}-4\,{\frac{c{e}^{2}ab}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+8\,{\frac{a{c}^{2}de}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+{\frac{{b}^{3}{e}^{2}}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}}\arctan \left ({(2\,cx+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}-2\,{\frac{{b}^{2}cde}{ \left ( a{e}^{2}-bde+c{d}^{2} \right ) ^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) } \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 [A] time = 11.6136, size = 1634, normalized size = 7.78 \begin{align*} \text{result too large to display} \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 = 1.15526, size = 494, normalized size = 2.35 \begin{align*} \frac{{\left (2 \, b^{2} c d e^{3} - 8 \, a c^{2} d e^{3} - b^{3} e^{4} + 4 \, a b c e^{4}\right )} \arctan \left (-\frac{{\left (2 \, c d - \frac{2 \, c d^{2}}{x e + d} - b e + \frac{2 \, b d e}{x e + d} - \frac{2 \, a e^{2}}{x e + d}\right )} e^{\left (-1\right )}}{\sqrt{-b^{2} + 4 \, a c}}\right ) e^{\left (-2\right )}}{{\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} \sqrt{-b^{2} + 4 \, a c}} + \frac{{\left (2 \, c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2} - 2 \, a c e^{2}\right )} \log \left (-c + \frac{2 \, c d}{x e + d} - \frac{c 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{a e^{2}}{{\left (x e + d\right )}^{2}}\right )}{2 \,{\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )}} + \frac{\frac{2 \, c d e^{2}}{x e + d} - \frac{b e^{3}}{x e + d}}{c d^{2} e^{2} - b d e^{3} + a e^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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