Optimal. Leaf size=108 \[ -\frac {\tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (2 c (A c d-a B e)-b c (A e+B d)+b^2 B e\right )}{c^2 \sqrt {b^2-4 a c}}+\frac {\log \left (a+b x+c x^2\right ) (A c e-b B e+B c d)}{2 c^2}+\frac {B e x}{c} \]
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Rubi [A] time = 0.13, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {773, 634, 618, 206, 628} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (2 c (A c d-a B e)-b c (A e+B d)+b^2 B e\right )}{c^2 \sqrt {b^2-4 a c}}+\frac {\log \left (a+b x+c x^2\right ) (A c e-b B e+B c d)}{2 c^2}+\frac {B e x}{c} \end {gather*}
Antiderivative was successfully verified.
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Rule 206
Rule 618
Rule 628
Rule 634
Rule 773
Rubi steps
\begin {align*} \int \frac {(A+B x) (d+e x)}{a+b x+c x^2} \, dx &=\frac {B e x}{c}+\frac {\int \frac {A c d-a B e+(B c d-b B e+A c e) x}{a+b x+c x^2} \, dx}{c}\\ &=\frac {B e x}{c}+\frac {(B c d-b B e+A c e) \int \frac {b+2 c x}{a+b x+c x^2} \, dx}{2 c^2}+\frac {\left (b^2 B e-b c (B d+A e)+2 c (A c d-a B e)\right ) \int \frac {1}{a+b x+c x^2} \, dx}{2 c^2}\\ &=\frac {B e x}{c}+\frac {(B c d-b B e+A c e) \log \left (a+b x+c x^2\right )}{2 c^2}-\frac {\left (b^2 B e-b c (B d+A e)+2 c (A c d-a B e)\right ) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{c^2}\\ &=\frac {B e x}{c}-\frac {\left (b^2 B e-b c (B d+A e)+2 c (A c d-a B e)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c}}+\frac {(B c d-b B e+A c e) \log \left (a+b x+c x^2\right )}{2 c^2}\\ \end {align*}
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Mathematica [A] time = 0.09, size = 108, normalized size = 1.00 \begin {gather*} \frac {\frac {2 \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right ) \left (2 c (A c d-a B e)-b c (A e+B d)+b^2 B e\right )}{\sqrt {4 a c-b^2}}+\log (a+x (b+c x)) (A c e-b B e+B c d)+2 B c e x}{2 c^2} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(A+B x) (d+e x)}{a+b x+c x^2} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 0.42, size = 369, normalized size = 3.42 \begin {gather*} \left [\frac {2 \, {\left (B b^{2} c - 4 \, B a c^{2}\right )} e x + \sqrt {b^{2} - 4 \, a c} {\left ({\left (B b c - 2 \, A c^{2}\right )} d - {\left (B b^{2} - {\left (2 \, B a + A b\right )} c\right )} e\right )} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + {\left ({\left (B b^{2} c - 4 \, B a c^{2}\right )} d - {\left (B b^{3} + 4 \, A a c^{2} - {\left (4 \, B a b + A b^{2}\right )} c\right )} e\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}, \frac {2 \, {\left (B b^{2} c - 4 \, B a c^{2}\right )} e x + 2 \, \sqrt {-b^{2} + 4 \, a c} {\left ({\left (B b c - 2 \, A c^{2}\right )} d - {\left (B b^{2} - {\left (2 \, B a + A b\right )} c\right )} e\right )} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + {\left ({\left (B b^{2} c - 4 \, B a c^{2}\right )} d - {\left (B b^{3} + 4 \, A a c^{2} - {\left (4 \, B a b + A b^{2}\right )} c\right )} e\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.16, size = 112, normalized size = 1.04 \begin {gather*} \frac {B x e}{c} + \frac {{\left (B c d - B b e + A c e\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{2}} - \frac {{\left (B b c d - 2 \, A c^{2} d - B b^{2} e + 2 \, B a c e + A b c e\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} c^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.05, size = 261, normalized size = 2.42 \begin {gather*} -\frac {A b e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}+\frac {2 A d \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}-\frac {2 B a e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}+\frac {B \,b^{2} e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{2}}-\frac {B b d \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}+\frac {A e \ln \left (c \,x^{2}+b x +a \right )}{2 c}-\frac {B b e \ln \left (c \,x^{2}+b x +a \right )}{2 c^{2}}+\frac {B d \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {B e x}{c} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.70, size = 163, normalized size = 1.51 \begin {gather*} \frac {\ln \left (c\,x^2+b\,x+a\right )\,\left (B\,b^3\,e+4\,A\,a\,c^2\,e+4\,B\,a\,c^2\,d-A\,b^2\,c\,e-B\,b^2\,c\,d-4\,B\,a\,b\,c\,e\right )}{2\,\left (4\,a\,c^3-b^2\,c^2\right )}-\frac {\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )\,\left (A\,b\,c\,e-B\,b^2\,e-2\,A\,c^2\,d+2\,B\,a\,c\,e+B\,b\,c\,d\right )}{c^2\,\sqrt {4\,a\,c-b^2}}+\frac {B\,e\,x}{c} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 3.44, size = 677, normalized size = 6.27 \begin {gather*} \frac {B e x}{c} + \left (- \frac {\sqrt {- 4 a c + b^{2}} \left (A b c e - 2 A c^{2} d + 2 B a c e - B b^{2} e + B b c d\right )}{2 c^{2} \left (4 a c - b^{2}\right )} - \frac {- A c e + B b e - B c d}{2 c^{2}}\right ) \log {\left (x + \frac {2 A a c e - A b c d - B a b e + 2 B a c d - 4 a c^{2} \left (- \frac {\sqrt {- 4 a c + b^{2}} \left (A b c e - 2 A c^{2} d + 2 B a c e - B b^{2} e + B b c d\right )}{2 c^{2} \left (4 a c - b^{2}\right )} - \frac {- A c e + B b e - B c d}{2 c^{2}}\right ) + b^{2} c \left (- \frac {\sqrt {- 4 a c + b^{2}} \left (A b c e - 2 A c^{2} d + 2 B a c e - B b^{2} e + B b c d\right )}{2 c^{2} \left (4 a c - b^{2}\right )} - \frac {- A c e + B b e - B c d}{2 c^{2}}\right )}{A b c e - 2 A c^{2} d + 2 B a c e - B b^{2} e + B b c d} \right )} + \left (\frac {\sqrt {- 4 a c + b^{2}} \left (A b c e - 2 A c^{2} d + 2 B a c e - B b^{2} e + B b c d\right )}{2 c^{2} \left (4 a c - b^{2}\right )} - \frac {- A c e + B b e - B c d}{2 c^{2}}\right ) \log {\left (x + \frac {2 A a c e - A b c d - B a b e + 2 B a c d - 4 a c^{2} \left (\frac {\sqrt {- 4 a c + b^{2}} \left (A b c e - 2 A c^{2} d + 2 B a c e - B b^{2} e + B b c d\right )}{2 c^{2} \left (4 a c - b^{2}\right )} - \frac {- A c e + B b e - B c d}{2 c^{2}}\right ) + b^{2} c \left (\frac {\sqrt {- 4 a c + b^{2}} \left (A b c e - 2 A c^{2} d + 2 B a c e - B b^{2} e + B b c d\right )}{2 c^{2} \left (4 a c - b^{2}\right )} - \frac {- A c e + B b e - B c d}{2 c^{2}}\right )}{A b c e - 2 A c^{2} d + 2 B a c e - B b^{2} e + B b c d} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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