Optimal. Leaf size=542 \[ \frac {\left (B (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+A f \left (b^2 f^2-2 c f (b e-a f)+c^2 \left (e^2-d f\right )\right )\right ) x}{f^4}-\frac {\left (A c f (c e-2 b f)-B \left (b^2 f^2-2 c f (b e-a f)+c^2 \left (e^2-d f\right )\right )\right ) x^2}{2 f^3}-\frac {c (B c e-2 b B f-A c f) x^3}{3 f^2}+\frac {B c^2 x^4}{4 f}-\frac {\left (A f \left (c^2 \left (e^4-4 d e^2 f+2 d^2 f^2\right )-f^2 \left (2 a b e f-2 a^2 f^2-b^2 \left (e^2-2 d f\right )\right )+2 c f \left (a f \left (e^2-2 d f\right )-b \left (e^3-3 d e f\right )\right )\right )-B \left (c^2 \left (e^5-5 d e^3 f+5 d^2 e f^2\right )+f^2 \left (a^2 e f^2-2 a b f \left (e^2-2 d f\right )+b^2 \left (e^3-3 d e f\right )\right )+2 c f \left (a e f \left (e^2-3 d f\right )-b \left (e^4-4 d e^2 f+2 d^2 f^2\right )\right )\right )\right ) \tanh ^{-1}\left (\frac {e+2 f x}{\sqrt {e^2-4 d f}}\right )}{f^5 \sqrt {e^2-4 d f}}+\frac {\left (A f (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+B \left (c^2 \left (e^4-3 d e^2 f+d^2 f^2\right )-f^2 \left (2 a b e f-a^2 f^2-b^2 \left (e^2-d f\right )\right )+2 c f \left (a f \left (e^2-d f\right )-b \left (e^3-2 d e f\right )\right )\right )\right ) \log \left (d+e x+f x^2\right )}{2 f^5} \]
[Out]
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Rubi [A]
time = 0.68, antiderivative size = 542, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1025, 648, 632,
212, 642} \begin {gather*} \frac {\log \left (d+e x+f x^2\right ) \left (B \left (-f^2 \left (-a^2 f^2+2 a b e f-\left (b^2 \left (e^2-d f\right )\right )\right )+2 c f \left (a f \left (e^2-d f\right )-b \left (e^3-2 d e f\right )\right )+c^2 \left (d^2 f^2-3 d e^2 f+e^4\right )\right )+A f (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )\right )}{2 f^5}-\frac {\tanh ^{-1}\left (\frac {e+2 f x}{\sqrt {e^2-4 d f}}\right ) \left (A f \left (-f^2 \left (-2 a^2 f^2+2 a b e f-\left (b^2 \left (e^2-2 d f\right )\right )\right )+2 c f \left (a f \left (e^2-2 d f\right )-b \left (e^3-3 d e f\right )\right )+c^2 \left (2 d^2 f^2-4 d e^2 f+e^4\right )\right )-B \left (f^2 \left (a^2 e f^2-2 a b f \left (e^2-2 d f\right )+b^2 \left (e^3-3 d e f\right )\right )+2 c f \left (a e f \left (e^2-3 d f\right )-b \left (2 d^2 f^2-4 d e^2 f+e^4\right )\right )+c^2 \left (5 d^2 e f^2-5 d e^3 f+e^5\right )\right )\right )}{f^5 \sqrt {e^2-4 d f}}+\frac {x \left (A f \left (-2 c f (b e-a f)+b^2 f^2+c^2 \left (e^2-d f\right )\right )+B (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )\right )}{f^4}-\frac {x^2 \left (A c f (c e-2 b f)-B \left (-2 c f (b e-a f)+b^2 f^2+c^2 \left (e^2-d f\right )\right )\right )}{2 f^3}-\frac {c x^3 (-A c f-2 b B f+B c e)}{3 f^2}+\frac {B c^2 x^4}{4 f} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
Rule 212
Rule 632
Rule 642
Rule 648
Rule 1025
Rubi steps
\begin {align*} \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{d+e x+f x^2} \, dx &=\int \left (\frac {B (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+A f \left (b^2 f^2-2 c f (b e-a f)+c^2 \left (e^2-d f\right )\right )}{f^4}-\frac {\left (A c f (c e-2 b f)-B \left (b^2 f^2-2 c f (b e-a f)+c^2 \left (e^2-d f\right )\right )\right ) x}{f^3}-\frac {c (B c e-2 b B f-A c f) x^2}{f^2}+\frac {B c^2 x^3}{f}+\frac {-B d (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+A f \left (2 c d f (b e-a f)-f^2 \left (b^2 d-a^2 f\right )-c^2 d \left (e^2-d f\right )\right )+\left (A f (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+B \left (c^2 \left (e^4-3 d e^2 f+d^2 f^2\right )-f^2 \left (2 a b e f-a^2 f^2-b^2 \left (e^2-d f\right )\right )+2 c f \left (a f \left (e^2-d f\right )-b \left (e^3-2 d e f\right )\right )\right )\right ) x}{f^4 \left (d+e x+f x^2\right )}\right ) \, dx\\ &=\frac {\left (B (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+A f \left (b^2 f^2-2 c f (b e-a f)+c^2 \left (e^2-d f\right )\right )\right ) x}{f^4}-\frac {\left (A c f (c e-2 b f)-B \left (b^2 f^2-2 c f (b e-a f)+c^2 \left (e^2-d f\right )\right )\right ) x^2}{2 f^3}-\frac {c (B c e-2 b B f-A c f) x^3}{3 f^2}+\frac {B c^2 x^4}{4 f}+\frac {\int \frac {-B d (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+A f \left (2 c d f (b e-a f)-f^2 \left (b^2 d-a^2 f\right )-c^2 d \left (e^2-d f\right )\right )+\left (A f (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+B \left (c^2 \left (e^4-3 d e^2 f+d^2 f^2\right )-f^2 \left (2 a b e f-a^2 f^2-b^2 \left (e^2-d f\right )\right )+2 c f \left (a f \left (e^2-d f\right )-b \left (e^3-2 d e f\right )\right )\right )\right ) x}{d+e x+f x^2} \, dx}{f^4}\\ &=\frac {\left (B (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+A f \left (b^2 f^2-2 c f (b e-a f)+c^2 \left (e^2-d f\right )\right )\right ) x}{f^4}-\frac {\left (A c f (c e-2 b f)-B \left (b^2 f^2-2 c f (b e-a f)+c^2 \left (e^2-d f\right )\right )\right ) x^2}{2 f^3}-\frac {c (B c e-2 b B f-A c f) x^3}{3 f^2}+\frac {B c^2 x^4}{4 f}+\frac {\left (A f (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+B \left (c^2 \left (e^4-3 d e^2 f+d^2 f^2\right )-f^2 \left (2 a b e f-a^2 f^2-b^2 \left (e^2-d f\right )\right )+2 c f \left (a f \left (e^2-d f\right )-b \left (e^3-2 d e f\right )\right )\right )\right ) \int \frac {e+2 f x}{d+e x+f x^2} \, dx}{2 f^5}+\frac {\left (A f \left (c^2 \left (e^4-4 d e^2 f+2 d^2 f^2\right )-f^2 \left (2 a b e f-2 a^2 f^2-b^2 \left (e^2-2 d f\right )\right )+2 c f \left (a f \left (e^2-2 d f\right )-b \left (e^3-3 d e f\right )\right )\right )-B \left (c^2 \left (e^5-5 d e^3 f+5 d^2 e f^2\right )+f^2 \left (a^2 e f^2-2 a b f \left (e^2-2 d f\right )+b^2 \left (e^3-3 d e f\right )\right )+2 c f \left (a e f \left (e^2-3 d f\right )-b \left (e^4-4 d e^2 f+2 d^2 f^2\right )\right )\right )\right ) \int \frac {1}{d+e x+f x^2} \, dx}{2 f^5}\\ &=\frac {\left (B (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+A f \left (b^2 f^2-2 c f (b e-a f)+c^2 \left (e^2-d f\right )\right )\right ) x}{f^4}-\frac {\left (A c f (c e-2 b f)-B \left (b^2 f^2-2 c f (b e-a f)+c^2 \left (e^2-d f\right )\right )\right ) x^2}{2 f^3}-\frac {c (B c e-2 b B f-A c f) x^3}{3 f^2}+\frac {B c^2 x^4}{4 f}+\frac {\left (A f (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+B \left (c^2 \left (e^4-3 d e^2 f+d^2 f^2\right )-f^2 \left (2 a b e f-a^2 f^2-b^2 \left (e^2-d f\right )\right )+2 c f \left (a f \left (e^2-d f\right )-b \left (e^3-2 d e f\right )\right )\right )\right ) \log \left (d+e x+f x^2\right )}{2 f^5}-\frac {\left (A f \left (c^2 \left (e^4-4 d e^2 f+2 d^2 f^2\right )-f^2 \left (2 a b e f-2 a^2 f^2-b^2 \left (e^2-2 d f\right )\right )+2 c f \left (a f \left (e^2-2 d f\right )-b \left (e^3-3 d e f\right )\right )\right )-B \left (c^2 \left (e^5-5 d e^3 f+5 d^2 e f^2\right )+f^2 \left (a^2 e f^2-2 a b f \left (e^2-2 d f\right )+b^2 \left (e^3-3 d e f\right )\right )+2 c f \left (a e f \left (e^2-3 d f\right )-b \left (e^4-4 d e^2 f+2 d^2 f^2\right )\right )\right )\right ) \text {Subst}\left (\int \frac {1}{e^2-4 d f-x^2} \, dx,x,e+2 f x\right )}{f^5}\\ &=\frac {\left (B (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+A f \left (b^2 f^2-2 c f (b e-a f)+c^2 \left (e^2-d f\right )\right )\right ) x}{f^4}-\frac {\left (A c f (c e-2 b f)-B \left (b^2 f^2-2 c f (b e-a f)+c^2 \left (e^2-d f\right )\right )\right ) x^2}{2 f^3}-\frac {c (B c e-2 b B f-A c f) x^3}{3 f^2}+\frac {B c^2 x^4}{4 f}-\frac {\left (A f \left (c^2 \left (e^4-4 d e^2 f+2 d^2 f^2\right )-f^2 \left (2 a b e f-2 a^2 f^2-b^2 \left (e^2-2 d f\right )\right )+2 c f \left (a f \left (e^2-2 d f\right )-b \left (e^3-3 d e f\right )\right )\right )-B \left (c^2 \left (e^5-5 d e^3 f+5 d^2 e f^2\right )+f^2 \left (a^2 e f^2-2 a b f \left (e^2-2 d f\right )+b^2 \left (e^3-3 d e f\right )\right )+2 c f \left (a e f \left (e^2-3 d f\right )-b \left (e^4-4 d e^2 f+2 d^2 f^2\right )\right )\right )\right ) \tanh ^{-1}\left (\frac {e+2 f x}{\sqrt {e^2-4 d f}}\right )}{f^5 \sqrt {e^2-4 d f}}+\frac {\left (A f (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+B \left (c^2 \left (e^4-3 d e^2 f+d^2 f^2\right )-f^2 \left (2 a b e f-a^2 f^2-b^2 \left (e^2-d f\right )\right )+2 c f \left (a f \left (e^2-d f\right )-b \left (e^3-2 d e f\right )\right )\right )\right ) \log \left (d+e x+f x^2\right )}{2 f^5}\\ \end {align*}
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Mathematica [A]
time = 0.41, size = 535, normalized size = 0.99 \begin {gather*} \frac {12 f \left (-B (c e-b f) \left (f (-b e+2 a f)+c \left (e^2-2 d f\right )\right )+A f \left (b^2 f^2+2 c f (-b e+a f)+c^2 \left (e^2-d f\right )\right )\right ) x+6 f^2 \left (A c f (-c e+2 b f)+B \left (b^2 f^2+2 c f (-b e+a f)+c^2 \left (e^2-d f\right )\right )\right ) x^2+4 c f^3 (-B c e+2 b B f+A c f) x^3+3 B c^2 f^4 x^4-\frac {12 \left (-A f \left (c^2 \left (e^4-4 d e^2 f+2 d^2 f^2\right )+f^2 \left (-2 a b e f+2 a^2 f^2+b^2 \left (e^2-2 d f\right )\right )+2 c f \left (a f \left (e^2-2 d f\right )-b \left (e^3-3 d e f\right )\right )\right )+B \left (c^2 \left (e^5-5 d e^3 f+5 d^2 e f^2\right )+f^2 \left (a^2 e f^2+2 a b f \left (-e^2+2 d f\right )+b^2 \left (e^3-3 d e f\right )\right )-2 c f \left (-a e f \left (e^2-3 d f\right )+b \left (e^4-4 d e^2 f+2 d^2 f^2\right )\right )\right )\right ) \tan ^{-1}\left (\frac {e+2 f x}{\sqrt {-e^2+4 d f}}\right )}{\sqrt {-e^2+4 d f}}+6 \left (A f (-c e+b f) \left (f (-b e+2 a f)+c \left (e^2-2 d f\right )\right )+B \left (c^2 \left (e^4-3 d e^2 f+d^2 f^2\right )+f^2 \left (-2 a b e f+a^2 f^2+b^2 \left (e^2-d f\right )\right )-2 c f \left (a f \left (-e^2+d f\right )+b \left (e^3-2 d e f\right )\right )\right )\right ) \log (d+x (e+f x))}{12 f^5} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.27, size = 800, normalized size = 1.48
method | result | size |
default | \(\frac {\frac {1}{4} B \,c^{2} x^{4} f^{3}+\frac {1}{3} A \,c^{2} f^{3} x^{3}+\frac {2}{3} B b c \,f^{3} x^{3}-\frac {1}{3} B \,c^{2} e \,f^{2} x^{3}+A b c \,f^{3} x^{2}-\frac {1}{2} A \,c^{2} e \,f^{2} x^{2}+B a c \,f^{3} x^{2}+\frac {1}{2} B \,b^{2} f^{3} x^{2}-B b c e \,f^{2} x^{2}-\frac {1}{2} B \,c^{2} d \,f^{2} x^{2}+\frac {1}{2} B \,c^{2} e^{2} f \,x^{2}+2 A a c \,f^{3} x +A \,b^{2} f^{3} x -2 A b c e \,f^{2} x -A \,c^{2} d \,f^{2} x +A \,c^{2} e^{2} f x +2 B a b \,f^{3} x -2 B a c e \,f^{2} x -B \,b^{2} e \,f^{2} x -2 B b c d \,f^{2} x +2 B b c \,e^{2} f x +2 B \,c^{2} d e f x -B \,c^{2} e^{3} x}{f^{4}}+\frac {\frac {\left (2 A a b \,f^{4}-2 A a c e \,f^{3}-A \,b^{2} e \,f^{3}-2 A b c d \,f^{3}+2 A b c \,e^{2} f^{2}+2 A \,c^{2} d e \,f^{2}-A \,c^{2} e^{3} f +B \,a^{2} f^{4}-2 B a b e \,f^{3}-2 B a c d \,f^{3}+2 B a c \,e^{2} f^{2}-B \,b^{2} d \,f^{3}+B \,b^{2} e^{2} f^{2}+4 B b c d e \,f^{2}-2 B b c \,e^{3} f +B \,c^{2} d^{2} f^{2}-3 B \,c^{2} d \,e^{2} f +B \,c^{2} e^{4}\right ) \ln \left (f \,x^{2}+e x +d \right )}{2 f}+\frac {2 \left (A \,a^{2} f^{4}-2 A a c d \,f^{3}-A \,b^{2} d \,f^{3}+2 A b c d e \,f^{2}+A \,c^{2} d^{2} f^{2}-A \,c^{2} d \,e^{2} f -2 B a b d \,f^{3}+2 B a c d e \,f^{2}+B \,b^{2} d e \,f^{2}+2 B b c \,d^{2} f^{2}-2 B b c d \,e^{2} f -2 B \,c^{2} d^{2} e f +B \,c^{2} d \,e^{3}-\frac {\left (2 A a b \,f^{4}-2 A a c e \,f^{3}-A \,b^{2} e \,f^{3}-2 A b c d \,f^{3}+2 A b c \,e^{2} f^{2}+2 A \,c^{2} d e \,f^{2}-A \,c^{2} e^{3} f +B \,a^{2} f^{4}-2 B a b e \,f^{3}-2 B a c d \,f^{3}+2 B a c \,e^{2} f^{2}-B \,b^{2} d \,f^{3}+B \,b^{2} e^{2} f^{2}+4 B b c d e \,f^{2}-2 B b c \,e^{3} f +B \,c^{2} d^{2} f^{2}-3 B \,c^{2} d \,e^{2} f +B \,c^{2} e^{4}\right ) e}{2 f}\right ) \arctan \left (\frac {2 f x +e}{\sqrt {4 d f -e^{2}}}\right )}{\sqrt {4 d f -e^{2}}}}{f^{4}}\) | \(800\) |
risch | \(\text {Expression too large to display}\) | \(60299\) |
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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Fricas [A]
time = 0.51, size = 1921, normalized size = 3.54 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 4663 vs.
\(2 (520) = 1040\).
time = 91.26, size = 4663, normalized size = 8.60 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 3.08, size = 738, normalized size = 1.36 \begin {gather*} \frac {3 \, B c^{2} f^{3} x^{4} + 8 \, B b c f^{3} x^{3} + 4 \, A c^{2} f^{3} x^{3} - 4 \, B c^{2} f^{2} x^{3} e - 6 \, B c^{2} d f^{2} x^{2} + 6 \, B b^{2} f^{3} x^{2} + 12 \, B a c f^{3} x^{2} + 12 \, A b c f^{3} x^{2} - 12 \, B b c f^{2} x^{2} e - 6 \, A c^{2} f^{2} x^{2} e - 24 \, B b c d f^{2} x - 12 \, A c^{2} d f^{2} x + 24 \, B a b f^{3} x + 12 \, A b^{2} f^{3} x + 24 \, A a c f^{3} x + 6 \, B c^{2} f x^{2} e^{2} + 24 \, B c^{2} d f x e - 12 \, B b^{2} f^{2} x e - 24 \, B a c f^{2} x e - 24 \, A b c f^{2} x e + 24 \, B b c f x e^{2} + 12 \, A c^{2} f x e^{2} - 12 \, B c^{2} x e^{3}}{12 \, f^{4}} + \frac {{\left (B c^{2} d^{2} f^{2} - B b^{2} d f^{3} - 2 \, B a c d f^{3} - 2 \, A b c d f^{3} + B a^{2} f^{4} + 2 \, A a b f^{4} + 4 \, B b c d f^{2} e + 2 \, A c^{2} d f^{2} e - 2 \, B a b f^{3} e - A b^{2} f^{3} e - 2 \, A a c f^{3} e - 3 \, B c^{2} d f e^{2} + B b^{2} f^{2} e^{2} + 2 \, B a c f^{2} e^{2} + 2 \, A b c f^{2} e^{2} - 2 \, B b c f e^{3} - A c^{2} f e^{3} + B c^{2} e^{4}\right )} \log \left (f x^{2} + x e + d\right )}{2 \, f^{5}} + \frac {{\left (4 \, B b c d^{2} f^{3} + 2 \, A c^{2} d^{2} f^{3} - 4 \, B a b d f^{4} - 2 \, A b^{2} d f^{4} - 4 \, A a c d f^{4} + 2 \, A a^{2} f^{5} - 5 \, B c^{2} d^{2} f^{2} e + 3 \, B b^{2} d f^{3} e + 6 \, B a c d f^{3} e + 6 \, A b c d f^{3} e - B a^{2} f^{4} e - 2 \, A a b f^{4} e - 8 \, B b c d f^{2} e^{2} - 4 \, A c^{2} d f^{2} e^{2} + 2 \, B a b f^{3} e^{2} + A b^{2} f^{3} e^{2} + 2 \, A a c f^{3} e^{2} + 5 \, B c^{2} d f e^{3} - B b^{2} f^{2} e^{3} - 2 \, B a c f^{2} e^{3} - 2 \, A b c f^{2} e^{3} + 2 \, B b c f e^{4} + A c^{2} f e^{4} - B c^{2} e^{5}\right )} \arctan \left (\frac {2 \, f x + e}{\sqrt {4 \, d f - e^{2}}}\right )}{\sqrt {4 \, d f - e^{2}} f^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 4.85, size = 893, normalized size = 1.65 \begin {gather*} x^3\,\left (\frac {A\,c^2+2\,B\,b\,c}{3\,f}-\frac {B\,c^2\,e}{3\,f^2}\right )+x\,\left (\frac {A\,b^2+2\,B\,a\,b+2\,A\,a\,c}{f}-\frac {d\,\left (\frac {A\,c^2+2\,B\,b\,c}{f}-\frac {B\,c^2\,e}{f^2}\right )}{f}+\frac {e\,\left (\frac {e\,\left (\frac {A\,c^2+2\,B\,b\,c}{f}-\frac {B\,c^2\,e}{f^2}\right )}{f}-\frac {B\,b^2+2\,A\,c\,b+2\,B\,a\,c}{f}+\frac {B\,c^2\,d}{f^2}\right )}{f}\right )-x^2\,\left (\frac {e\,\left (\frac {A\,c^2+2\,B\,b\,c}{f}-\frac {B\,c^2\,e}{f^2}\right )}{2\,f}-\frac {B\,b^2+2\,A\,c\,b+2\,B\,a\,c}{2\,f}+\frac {B\,c^2\,d}{2\,f^2}\right )-\frac {\ln \left (f\,x^2+e\,x+d\right )\,\left (-4\,B\,a^2\,d\,f^5+B\,a^2\,e^2\,f^4+8\,B\,a\,b\,d\,e\,f^4-8\,A\,a\,b\,d\,f^5-2\,B\,a\,b\,e^3\,f^3+2\,A\,a\,b\,e^2\,f^4+8\,B\,a\,c\,d^2\,f^4-10\,B\,a\,c\,d\,e^2\,f^3+8\,A\,a\,c\,d\,e\,f^4+2\,B\,a\,c\,e^4\,f^2-2\,A\,a\,c\,e^3\,f^3+4\,B\,b^2\,d^2\,f^4-5\,B\,b^2\,d\,e^2\,f^3+4\,A\,b^2\,d\,e\,f^4+B\,b^2\,e^4\,f^2-A\,b^2\,e^3\,f^3-16\,B\,b\,c\,d^2\,e\,f^3+8\,A\,b\,c\,d^2\,f^4+12\,B\,b\,c\,d\,e^3\,f^2-10\,A\,b\,c\,d\,e^2\,f^3-2\,B\,b\,c\,e^5\,f+2\,A\,b\,c\,e^4\,f^2-4\,B\,c^2\,d^3\,f^3+13\,B\,c^2\,d^2\,e^2\,f^2-8\,A\,c^2\,d^2\,e\,f^3-7\,B\,c^2\,d\,e^4\,f+6\,A\,c^2\,d\,e^3\,f^2+B\,c^2\,e^6-A\,c^2\,e^5\,f\right )}{2\,\left (4\,d\,f^6-e^2\,f^5\right )}+\frac {B\,c^2\,x^4}{4\,f}+\frac {\mathrm {atan}\left (\frac {e}{\sqrt {4\,d\,f-e^2}}+\frac {2\,f\,x}{\sqrt {4\,d\,f-e^2}}\right )\,\left (-B\,a^2\,e\,f^4+2\,A\,a^2\,f^5-4\,B\,a\,b\,d\,f^4+2\,B\,a\,b\,e^2\,f^3-2\,A\,a\,b\,e\,f^4+6\,B\,a\,c\,d\,e\,f^3-4\,A\,a\,c\,d\,f^4-2\,B\,a\,c\,e^3\,f^2+2\,A\,a\,c\,e^2\,f^3+3\,B\,b^2\,d\,e\,f^3-2\,A\,b^2\,d\,f^4-B\,b^2\,e^3\,f^2+A\,b^2\,e^2\,f^3+4\,B\,b\,c\,d^2\,f^3-8\,B\,b\,c\,d\,e^2\,f^2+6\,A\,b\,c\,d\,e\,f^3+2\,B\,b\,c\,e^4\,f-2\,A\,b\,c\,e^3\,f^2-5\,B\,c^2\,d^2\,e\,f^2+2\,A\,c^2\,d^2\,f^3+5\,B\,c^2\,d\,e^3\,f-4\,A\,c^2\,d\,e^2\,f^2-B\,c^2\,e^5+A\,c^2\,e^4\,f\right )}{f^5\,\sqrt {4\,d\,f-e^2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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