Optimal. Leaf size=184 \[ -\frac{\log \left (d+e x+f x^2\right ) \left (A f (c e-b f)-B \left (a f^2-b e f-c d f+c e^2\right )\right )}{2 f^3}-\frac{\tanh ^{-1}\left (\frac{e+2 f x}{\sqrt{e^2-4 d f}}\right ) \left (A f \left (2 a f^2-b e f-2 c d f+c e^2\right )+B \left (f \left (-a e f-2 b d f+b e^2\right )-c \left (e^3-3 d e f\right )\right )\right )}{f^3 \sqrt{e^2-4 d f}}-\frac{x (-A c f-b B f+B c e)}{f^2}+\frac{B c x^2}{2 f} \]
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Rubi [A] time = 0.345593, antiderivative size = 182, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {1628, 634, 618, 206, 628} \[ -\frac{\log \left (d+e x+f x^2\right ) \left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-d f\right )\right )}{2 f^3}-\frac{\tanh ^{-1}\left (\frac{e+2 f x}{\sqrt{e^2-4 d f}}\right ) \left (A f \left (2 a f^2-b e f-2 c d f+c e^2\right )+B f \left (-a e f-2 b d f+b e^2\right )-B c \left (e^3-3 d e f\right )\right )}{f^3 \sqrt{e^2-4 d f}}-\frac{x (-A c f-b B f+B c e)}{f^2}+\frac{B c x^2}{2 f} \]
Antiderivative was successfully verified.
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Rule 1628
Rule 634
Rule 618
Rule 206
Rule 628
Rubi steps
\begin{align*} \int \frac{(A+B x) \left (a+b x+c x^2\right )}{d+e x+f x^2} \, dx &=\int \left (-\frac{B c e-b B f-A c f}{f^2}+\frac{B c x}{f}+\frac{-A f (c d-a f)+B d (c e-b f)-\left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-d f\right )\right ) x}{f^2 \left (d+e x+f x^2\right )}\right ) \, dx\\ &=-\frac{(B c e-b B f-A c f) x}{f^2}+\frac{B c x^2}{2 f}+\frac{\int \frac{-A f (c d-a f)+B d (c e-b f)-\left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-d f\right )\right ) x}{d+e x+f x^2} \, dx}{f^2}\\ &=-\frac{(B c e-b B f-A c f) x}{f^2}+\frac{B c x^2}{2 f}+\frac{\left (-B f (b e-a f)-A f (c e-b f)+B c \left (e^2-d f\right )\right ) \int \frac{e+2 f x}{d+e x+f x^2} \, dx}{2 f^3}+\frac{\left (B f \left (b e^2-2 b d f-a e f\right )-B c \left (e^3-3 d e f\right )+A f \left (c e^2-2 c d f-b e f+2 a f^2\right )\right ) \int \frac{1}{d+e x+f x^2} \, dx}{2 f^3}\\ &=-\frac{(B c e-b B f-A c f) x}{f^2}+\frac{B c x^2}{2 f}-\frac{\left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-d f\right )\right ) \log \left (d+e x+f x^2\right )}{2 f^3}-\frac{\left (B f \left (b e^2-2 b d f-a e f\right )-B c \left (e^3-3 d e f\right )+A f \left (c e^2-2 c d f-b e f+2 a f^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{e^2-4 d f-x^2} \, dx,x,e+2 f x\right )}{f^3}\\ &=-\frac{(B c e-b B f-A c f) x}{f^2}+\frac{B c x^2}{2 f}-\frac{\left (B f \left (b e^2-2 b d f-a e f\right )-B c \left (e^3-3 d e f\right )+A f \left (c e^2-2 c d f-b e f+2 a f^2\right )\right ) \tanh ^{-1}\left (\frac{e+2 f x}{\sqrt{e^2-4 d f}}\right )}{f^3 \sqrt{e^2-4 d f}}-\frac{\left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-d f\right )\right ) \log \left (d+e x+f x^2\right )}{2 f^3}\\ \end{align*}
Mathematica [A] time = 0.202927, size = 175, normalized size = 0.95 \[ \frac{-\frac{2 \tan ^{-1}\left (\frac{e+2 f x}{\sqrt{4 d f-e^2}}\right ) \left (A f \left (-2 a f^2+b e f+2 c d f-c e^2\right )+B f \left (a e f+2 b d f-b e^2\right )+B c \left (e^3-3 d e f\right )\right )}{\sqrt{4 d f-e^2}}+\log (d+x (e+f x)) \left (B f (a f-b e)+A f (b f-c e)+B c \left (e^2-d f\right )\right )+2 f x (A c f+b B f-B c e)+B c f^2 x^2}{2 f^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.175, size = 510, normalized size = 2.8 \begin{align*}{\frac{Bc{x}^{2}}{2\,f}}+{\frac{Acx}{f}}+{\frac{Bbx}{f}}-{\frac{Bcex}{{f}^{2}}}+{\frac{\ln \left ( f{x}^{2}+ex+d \right ) Ab}{2\,f}}-{\frac{\ln \left ( f{x}^{2}+ex+d \right ) Ace}{2\,{f}^{2}}}+{\frac{\ln \left ( f{x}^{2}+ex+d \right ) Ba}{2\,f}}-{\frac{\ln \left ( f{x}^{2}+ex+d \right ) Bbe}{2\,{f}^{2}}}-{\frac{\ln \left ( f{x}^{2}+ex+d \right ) Bcd}{2\,{f}^{2}}}+{\frac{\ln \left ( f{x}^{2}+ex+d \right ) Bc{e}^{2}}{2\,{f}^{3}}}+2\,{\frac{Aa}{\sqrt{4\,df-{e}^{2}}}\arctan \left ({\frac{2\,fx+e}{\sqrt{4\,df-{e}^{2}}}} \right ) }-2\,{\frac{Acd}{f\sqrt{4\,df-{e}^{2}}}\arctan \left ({\frac{2\,fx+e}{\sqrt{4\,df-{e}^{2}}}} \right ) }-2\,{\frac{Bbd}{f\sqrt{4\,df-{e}^{2}}}\arctan \left ({\frac{2\,fx+e}{\sqrt{4\,df-{e}^{2}}}} \right ) }+3\,{\frac{Bcde}{{f}^{2}\sqrt{4\,df-{e}^{2}}}\arctan \left ({\frac{2\,fx+e}{\sqrt{4\,df-{e}^{2}}}} \right ) }-{\frac{Abe}{f}\arctan \left ({(2\,fx+e){\frac{1}{\sqrt{4\,df-{e}^{2}}}}} \right ){\frac{1}{\sqrt{4\,df-{e}^{2}}}}}+{\frac{Ac{e}^{2}}{{f}^{2}}\arctan \left ({(2\,fx+e){\frac{1}{\sqrt{4\,df-{e}^{2}}}}} \right ){\frac{1}{\sqrt{4\,df-{e}^{2}}}}}-{\frac{aBe}{f}\arctan \left ({(2\,fx+e){\frac{1}{\sqrt{4\,df-{e}^{2}}}}} \right ){\frac{1}{\sqrt{4\,df-{e}^{2}}}}}+{\frac{bB{e}^{2}}{{f}^{2}}\arctan \left ({(2\,fx+e){\frac{1}{\sqrt{4\,df-{e}^{2}}}}} \right ){\frac{1}{\sqrt{4\,df-{e}^{2}}}}}-{\frac{Bc{e}^{3}}{{f}^{3}}\arctan \left ({(2\,fx+e){\frac{1}{\sqrt{4\,df-{e}^{2}}}}} \right ){\frac{1}{\sqrt{4\,df-{e}^{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 [A] time = 1.79901, size = 1280, normalized size = 6.96 \begin{align*} \left [\frac{{\left (B c e^{2} f^{2} - 4 \, B c d f^{3}\right )} x^{2} -{\left (B c e^{3} - 2 \, A a f^{3} +{\left (2 \,{\left (B b + A c\right )} d +{\left (B a + A b\right )} e\right )} f^{2} -{\left (3 \, B c d e +{\left (B b + A c\right )} e^{2}\right )} f\right )} \sqrt{e^{2} - 4 \, d f} \log \left (\frac{2 \, f^{2} x^{2} + 2 \, e f x + e^{2} - 2 \, d f - \sqrt{e^{2} - 4 \, d f}{\left (2 \, f x + e\right )}}{f x^{2} + e x + d}\right ) - 2 \,{\left (B c e^{3} f + 4 \,{\left (B b + A c\right )} d f^{3} -{\left (4 \, B c d e +{\left (B b + A c\right )} e^{2}\right )} f^{2}\right )} x +{\left (B c e^{4} - 4 \,{\left (B a + A b\right )} d f^{3} +{\left (4 \, B c d^{2} + 4 \,{\left (B b + A c\right )} d e +{\left (B a + A b\right )} e^{2}\right )} f^{2} -{\left (5 \, B c d e^{2} +{\left (B b + A c\right )} e^{3}\right )} f\right )} \log \left (f x^{2} + e x + d\right )}{2 \,{\left (e^{2} f^{3} - 4 \, d f^{4}\right )}}, \frac{{\left (B c e^{2} f^{2} - 4 \, B c d f^{3}\right )} x^{2} + 2 \,{\left (B c e^{3} - 2 \, A a f^{3} +{\left (2 \,{\left (B b + A c\right )} d +{\left (B a + A b\right )} e\right )} f^{2} -{\left (3 \, B c d e +{\left (B b + A c\right )} e^{2}\right )} f\right )} \sqrt{-e^{2} + 4 \, d f} \arctan \left (-\frac{\sqrt{-e^{2} + 4 \, d f}{\left (2 \, f x + e\right )}}{e^{2} - 4 \, d f}\right ) - 2 \,{\left (B c e^{3} f + 4 \,{\left (B b + A c\right )} d f^{3} -{\left (4 \, B c d e +{\left (B b + A c\right )} e^{2}\right )} f^{2}\right )} x +{\left (B c e^{4} - 4 \,{\left (B a + A b\right )} d f^{3} +{\left (4 \, B c d^{2} + 4 \,{\left (B b + A c\right )} d e +{\left (B a + A b\right )} e^{2}\right )} f^{2} -{\left (5 \, B c d e^{2} +{\left (B b + A c\right )} e^{3}\right )} f\right )} \log \left (f x^{2} + e x + d\right )}{2 \,{\left (e^{2} f^{3} - 4 \, d f^{4}\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [B] time = 25.2751, size = 1260, normalized size = 6.85 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.14854, size = 258, normalized size = 1.4 \begin{align*} \frac{B c f x^{2} + 2 \, B b f x + 2 \, A c f x - 2 \, B c x e}{2 \, f^{2}} - \frac{{\left (B c d f - B a f^{2} - A b f^{2} + B b f e + A c f e - B c e^{2}\right )} \log \left (f x^{2} + x e + d\right )}{2 \, f^{3}} - \frac{{\left (2 \, B b d f^{2} + 2 \, A c d f^{2} - 2 \, A a f^{3} - 3 \, B c d f e + B a f^{2} e + A b f^{2} e - B b f e^{2} - A c f e^{2} + B c e^{3}\right )} \arctan \left (\frac{2 \, f x + e}{\sqrt{4 \, d f - e^{2}}}\right )}{\sqrt{4 \, d f - e^{2}} f^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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