Optimal. Leaf size=357 \[ \frac {e x \left (B \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )+A c e (3 c d-b e)\right )}{c^3}+\frac {\log \left (a+b x+c x^2\right ) \left (A c e \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )+B \left (-3 c^2 d e (a e+b d)+b c e^2 (2 a e+3 b d)-b^3 e^3+c^3 d^3\right )\right )}{2 c^4}-\frac {\tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (b^2 c e \left (-4 a B e^2+3 A c d e+3 B c d^2\right )-b c^2 \left (-3 a A e^3-9 a B d e^2+3 A c d^2 e+B c d^3\right )+2 c^2 \left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right )-b^3 c e^2 (A e+3 B d)+b^4 B e^3\right )}{c^4 \sqrt {b^2-4 a c}}+\frac {e^2 x^2 (A c e-b B e+3 B c d)}{2 c^2}+\frac {B e^3 x^3}{3 c} \]
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Rubi [A] time = 0.65, antiderivative size = 357, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {800, 634, 618, 206, 628} \begin {gather*} \frac {\log \left (a+b x+c x^2\right ) \left (A c e \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )+B \left (-3 c^2 d e (a e+b d)+b c e^2 (2 a e+3 b d)-b^3 e^3+c^3 d^3\right )\right )}{2 c^4}+\frac {e x \left (B \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )+A c e (3 c d-b e)\right )}{c^3}-\frac {\tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (b^2 c e \left (-4 a B e^2+3 A c d e+3 B c d^2\right )-b c^2 \left (-3 a A e^3-9 a B d e^2+3 A c d^2 e+B c d^3\right )+2 c^2 \left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right )-b^3 c e^2 (A e+3 B d)+b^4 B e^3\right )}{c^4 \sqrt {b^2-4 a c}}+\frac {e^2 x^2 (A c e-b B e+3 B c d)}{2 c^2}+\frac {B e^3 x^3}{3 c} \end {gather*}
Antiderivative was successfully verified.
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Rule 206
Rule 618
Rule 628
Rule 634
Rule 800
Rubi steps
\begin {align*} \int \frac {(A+B x) (d+e x)^3}{a+b x+c x^2} \, dx &=\int \left (\frac {e \left (A c e (3 c d-b e)+B \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )\right )}{c^3}+\frac {e^2 (3 B c d-b B e+A c e) x}{c^2}+\frac {B e^3 x^2}{c}+\frac {A c \left (c^2 d^3-3 a c d e^2+a b e^3\right )-a B e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )+\left (A c e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )+B \left (c^3 d^3-b^3 e^3-3 c^2 d e (b d+a e)+b c e^2 (3 b d+2 a e)\right )\right ) x}{c^3 \left (a+b x+c x^2\right )}\right ) \, dx\\ &=\frac {e \left (A c e (3 c d-b e)+B \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )\right ) x}{c^3}+\frac {e^2 (3 B c d-b B e+A c e) x^2}{2 c^2}+\frac {B e^3 x^3}{3 c}+\frac {\int \frac {A c \left (c^2 d^3-3 a c d e^2+a b e^3\right )-a B e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )+\left (A c e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )+B \left (c^3 d^3-b^3 e^3-3 c^2 d e (b d+a e)+b c e^2 (3 b d+2 a e)\right )\right ) x}{a+b x+c x^2} \, dx}{c^3}\\ &=\frac {e \left (A c e (3 c d-b e)+B \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )\right ) x}{c^3}+\frac {e^2 (3 B c d-b B e+A c e) x^2}{2 c^2}+\frac {B e^3 x^3}{3 c}+\frac {\left (A c e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )+B \left (c^3 d^3-b^3 e^3-3 c^2 d e (b d+a e)+b c e^2 (3 b d+2 a e)\right )\right ) \int \frac {b+2 c x}{a+b x+c x^2} \, dx}{2 c^4}+\frac {\left (b^4 B e^3-b^3 c e^2 (3 B d+A e)+b^2 c e \left (3 B c d^2+3 A c d e-4 a B e^2\right )-b c^2 \left (B c d^3+3 A c d^2 e-9 a B d e^2-3 a A e^3\right )+2 c^2 \left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right )\right ) \int \frac {1}{a+b x+c x^2} \, dx}{2 c^4}\\ &=\frac {e \left (A c e (3 c d-b e)+B \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )\right ) x}{c^3}+\frac {e^2 (3 B c d-b B e+A c e) x^2}{2 c^2}+\frac {B e^3 x^3}{3 c}+\frac {\left (A c e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )+B \left (c^3 d^3-b^3 e^3-3 c^2 d e (b d+a e)+b c e^2 (3 b d+2 a e)\right )\right ) \log \left (a+b x+c x^2\right )}{2 c^4}-\frac {\left (b^4 B e^3-b^3 c e^2 (3 B d+A e)+b^2 c e \left (3 B c d^2+3 A c d e-4 a B e^2\right )-b c^2 \left (B c d^3+3 A c d^2 e-9 a B d e^2-3 a A e^3\right )+2 c^2 \left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{c^4}\\ &=\frac {e \left (A c e (3 c d-b e)+B \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )\right ) x}{c^3}+\frac {e^2 (3 B c d-b B e+A c e) x^2}{2 c^2}+\frac {B e^3 x^3}{3 c}-\frac {\left (b^4 B e^3-b^3 c e^2 (3 B d+A e)+b^2 c e \left (3 B c d^2+3 A c d e-4 a B e^2\right )-b c^2 \left (B c d^3+3 A c d^2 e-9 a B d e^2-3 a A e^3\right )+2 c^2 \left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^4 \sqrt {b^2-4 a c}}+\frac {\left (A c e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )+B \left (c^3 d^3-b^3 e^3-3 c^2 d e (b d+a e)+b c e^2 (3 b d+2 a e)\right )\right ) \log \left (a+b x+c x^2\right )}{2 c^4}\\ \end {align*}
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Mathematica [A] time = 0.29, size = 352, normalized size = 0.99 \begin {gather*} \frac {6 c e x \left (B \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )+A c e (3 c d-b e)\right )+3 \log (a+x (b+c x)) \left (A c e \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )+B \left (-3 c^2 d e (a e+b d)+b c e^2 (2 a e+3 b d)-b^3 e^3+c^3 d^3\right )\right )+\frac {6 \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right ) \left (b^2 c e \left (-4 a B e^2+3 A c d e+3 B c d^2\right )+b c^2 \left (3 a A e^3+9 a B d e^2-3 A c d^2 e-B c d^3\right )+2 c^2 \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (a e^2-3 c d^2\right )\right )-b^3 c e^2 (A e+3 B d)+b^4 B e^3\right )}{\sqrt {4 a c-b^2}}+3 c^2 e^2 x^2 (A c e-b B e+3 B c d)+2 B c^3 e^3 x^3}{6 c^4} \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)^3}{a+b x+c x^2} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 0.54, size = 1139, normalized size = 3.19
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.18, size = 402, normalized size = 1.13 \begin {gather*} \frac {2 \, B c^{2} x^{3} e^{3} + 9 \, B c^{2} d x^{2} e^{2} + 18 \, B c^{2} d^{2} x e - 3 \, B b c x^{2} e^{3} + 3 \, A c^{2} x^{2} e^{3} - 18 \, B b c d x e^{2} + 18 \, A c^{2} d x e^{2} + 6 \, B b^{2} x e^{3} - 6 \, B a c x e^{3} - 6 \, A b c x e^{3}}{6 \, c^{3}} + \frac {{\left (B c^{3} d^{3} - 3 \, B b c^{2} d^{2} e + 3 \, A c^{3} d^{2} e + 3 \, B b^{2} c d e^{2} - 3 \, B a c^{2} d e^{2} - 3 \, A b c^{2} d e^{2} - B b^{3} e^{3} + 2 \, B a b c e^{3} + A b^{2} c e^{3} - A a c^{2} e^{3}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{4}} - \frac {{\left (B b c^{3} d^{3} - 2 \, A c^{4} d^{3} - 3 \, B b^{2} c^{2} d^{2} e + 6 \, B a c^{3} d^{2} e + 3 \, A b c^{3} d^{2} e + 3 \, B b^{3} c d e^{2} - 9 \, B a b c^{2} d e^{2} - 3 \, A b^{2} c^{2} d e^{2} + 6 \, A a c^{3} d e^{2} - B b^{4} e^{3} + 4 \, B a b^{2} c e^{3} + A b^{3} c e^{3} - 2 \, B a^{2} c^{2} e^{3} - 3 \, A a b c^{2} e^{3}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} c^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.05, size = 946, normalized size = 2.65 \begin {gather*} \frac {B \,e^{3} x^{3}}{3 c}+\frac {3 A a b \,e^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{2}}-\frac {6 A a d \,e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}-\frac {A \,b^{3} e^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{3}}+\frac {3 A \,b^{2} d \,e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{2}}-\frac {3 A b \,d^{2} e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}+\frac {A \,e^{3} x^{2}}{2 c}+\frac {2 A \,d^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}+\frac {2 B \,a^{2} e^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{2}}-\frac {4 B a \,b^{2} e^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{3}}+\frac {9 B a b d \,e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{2}}-\frac {6 B a \,d^{2} 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^{4} e^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{4}}-\frac {3 B \,b^{3} d \,e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{3}}+\frac {3 B \,b^{2} d^{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^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}-\frac {B b \,e^{3} x^{2}}{2 c^{2}}+\frac {3 B d \,e^{2} x^{2}}{2 c}-\frac {A a \,e^{3} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{2}}+\frac {A \,b^{2} e^{3} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{3}}-\frac {3 A b d \,e^{2} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{2}}-\frac {A b \,e^{3} x}{c^{2}}+\frac {3 A \,d^{2} e \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {3 A d \,e^{2} x}{c}+\frac {B a b \,e^{3} \ln \left (c \,x^{2}+b x +a \right )}{c^{3}}-\frac {3 B a d \,e^{2} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{2}}-\frac {B a \,e^{3} x}{c^{2}}-\frac {B \,b^{3} e^{3} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{4}}+\frac {3 B \,b^{2} d \,e^{2} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{3}}+\frac {B \,b^{2} e^{3} x}{c^{3}}-\frac {3 B b \,d^{2} e \ln \left (c \,x^{2}+b x +a \right )}{2 c^{2}}-\frac {3 B b d \,e^{2} x}{c^{2}}+\frac {B \,d^{3} \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {3 B \,d^{2} 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 = 3.11, size = 539, normalized size = 1.51 \begin {gather*} x^2\,\left (\frac {A\,e^3+3\,B\,d\,e^2}{2\,c}-\frac {B\,b\,e^3}{2\,c^2}\right )-x\,\left (\frac {b\,\left (\frac {A\,e^3+3\,B\,d\,e^2}{c}-\frac {B\,b\,e^3}{c^2}\right )}{c}-\frac {3\,d\,e\,\left (A\,e+B\,d\right )}{c}+\frac {B\,a\,e^3}{c^2}\right )-\frac {\ln \left (c\,x^2+b\,x+a\right )\,\left (-8\,B\,a^2\,b\,c^2\,e^3+12\,B\,a^2\,c^3\,d\,e^2+4\,A\,a^2\,c^3\,e^3+6\,B\,a\,b^3\,c\,e^3-15\,B\,a\,b^2\,c^2\,d\,e^2-5\,A\,a\,b^2\,c^2\,e^3+12\,B\,a\,b\,c^3\,d^2\,e+12\,A\,a\,b\,c^3\,d\,e^2-4\,B\,a\,c^4\,d^3-12\,A\,a\,c^4\,d^2\,e-B\,b^5\,e^3+3\,B\,b^4\,c\,d\,e^2+A\,b^4\,c\,e^3-3\,B\,b^3\,c^2\,d^2\,e-3\,A\,b^3\,c^2\,d\,e^2+B\,b^2\,c^3\,d^3+3\,A\,b^2\,c^3\,d^2\,e\right )}{2\,\left (4\,a\,c^5-b^2\,c^4\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 (2\,B\,a^2\,c^2\,e^3-4\,B\,a\,b^2\,c\,e^3+9\,B\,a\,b\,c^2\,d\,e^2+3\,A\,a\,b\,c^2\,e^3-6\,B\,a\,c^3\,d^2\,e-6\,A\,a\,c^3\,d\,e^2+B\,b^4\,e^3-3\,B\,b^3\,c\,d\,e^2-A\,b^3\,c\,e^3+3\,B\,b^2\,c^2\,d^2\,e+3\,A\,b^2\,c^2\,d\,e^2-B\,b\,c^3\,d^3-3\,A\,b\,c^3\,d^2\,e+2\,A\,c^4\,d^3\right )}{c^4\,\sqrt {4\,a\,c-b^2}}+\frac {B\,e^3\,x^3}{3\,c} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 25.81, size = 2759, normalized size = 7.73
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
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