Optimal. Leaf size=157 \[ \frac {1}{2} a b e^3 x+\frac {b^2 e^3 (c+d x)^2}{12 d}+\frac {b^2 e^3 (c+d x) \text {ArcTan}(c+d x)}{2 d}-\frac {b e^3 (c+d x)^3 (a+b \text {ArcTan}(c+d x))}{6 d}-\frac {e^3 (a+b \text {ArcTan}(c+d x))^2}{4 d}+\frac {e^3 (c+d x)^4 (a+b \text {ArcTan}(c+d x))^2}{4 d}-\frac {b^2 e^3 \log \left (1+(c+d x)^2\right )}{3 d} \]
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Rubi [A]
time = 0.16, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps
used = 13, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {5151, 12,
4946, 5036, 272, 45, 4930, 266, 5004} \begin {gather*} \frac {e^3 (c+d x)^4 (a+b \text {ArcTan}(c+d x))^2}{4 d}-\frac {b e^3 (c+d x)^3 (a+b \text {ArcTan}(c+d x))}{6 d}-\frac {e^3 (a+b \text {ArcTan}(c+d x))^2}{4 d}+\frac {1}{2} a b e^3 x+\frac {b^2 e^3 (c+d x) \text {ArcTan}(c+d x)}{2 d}+\frac {b^2 e^3 (c+d x)^2}{12 d}-\frac {b^2 e^3 \log \left ((c+d x)^2+1\right )}{3 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 45
Rule 266
Rule 272
Rule 4930
Rule 4946
Rule 5004
Rule 5036
Rule 5151
Rubi steps
\begin {align*} \int (c e+d e x)^3 \left (a+b \tan ^{-1}(c+d x)\right )^2 \, dx &=\frac {\text {Subst}\left (\int e^3 x^3 \left (a+b \tan ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 \text {Subst}\left (\int x^3 \left (a+b \tan ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 (c+d x)^4 \left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d}-\frac {\left (b e^3\right ) \text {Subst}\left (\int \frac {x^4 \left (a+b \tan ^{-1}(x)\right )}{1+x^2} \, dx,x,c+d x\right )}{2 d}\\ &=\frac {e^3 (c+d x)^4 \left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d}-\frac {\left (b e^3\right ) \text {Subst}\left (\int x^2 \left (a+b \tan ^{-1}(x)\right ) \, dx,x,c+d x\right )}{2 d}+\frac {\left (b e^3\right ) \text {Subst}\left (\int \frac {x^2 \left (a+b \tan ^{-1}(x)\right )}{1+x^2} \, dx,x,c+d x\right )}{2 d}\\ &=-\frac {b e^3 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )}{6 d}+\frac {e^3 (c+d x)^4 \left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d}+\frac {\left (b e^3\right ) \text {Subst}\left (\int \left (a+b \tan ^{-1}(x)\right ) \, dx,x,c+d x\right )}{2 d}-\frac {\left (b e^3\right ) \text {Subst}\left (\int \frac {a+b \tan ^{-1}(x)}{1+x^2} \, dx,x,c+d x\right )}{2 d}+\frac {\left (b^2 e^3\right ) \text {Subst}\left (\int \frac {x^3}{1+x^2} \, dx,x,c+d x\right )}{6 d}\\ &=\frac {1}{2} a b e^3 x-\frac {b e^3 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )}{6 d}-\frac {e^3 \left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d}+\frac {e^3 (c+d x)^4 \left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d}+\frac {\left (b^2 e^3\right ) \text {Subst}\left (\int \frac {x}{1+x} \, dx,x,(c+d x)^2\right )}{12 d}+\frac {\left (b^2 e^3\right ) \text {Subst}\left (\int \tan ^{-1}(x) \, dx,x,c+d x\right )}{2 d}\\ &=\frac {1}{2} a b e^3 x+\frac {b^2 e^3 (c+d x) \tan ^{-1}(c+d x)}{2 d}-\frac {b e^3 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )}{6 d}-\frac {e^3 \left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d}+\frac {e^3 (c+d x)^4 \left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d}+\frac {\left (b^2 e^3\right ) \text {Subst}\left (\int \left (1+\frac {1}{-1-x}\right ) \, dx,x,(c+d x)^2\right )}{12 d}-\frac {\left (b^2 e^3\right ) \text {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,c+d x\right )}{2 d}\\ &=\frac {1}{2} a b e^3 x+\frac {b^2 e^3 (c+d x)^2}{12 d}+\frac {b^2 e^3 (c+d x) \tan ^{-1}(c+d x)}{2 d}-\frac {b e^3 (c+d x)^3 \left (a+b \tan ^{-1}(c+d x)\right )}{6 d}-\frac {e^3 \left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d}+\frac {e^3 (c+d x)^4 \left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d}-\frac {b^2 e^3 \log \left (1+(c+d x)^2\right )}{3 d}\\ \end {align*}
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Mathematica [A]
time = 0.08, size = 216, normalized size = 1.38 \begin {gather*} \frac {e^3 \left ((c+d x) \left (b^2 (c+d x)+3 a^2 (c+d x)^3-2 a b \left (-3+c^2+2 c d x+d^2 x^2\right )\right )+2 b \left (-b \left (-3 c+c^3-3 d x+3 c^2 d x+3 c d^2 x^2+d^3 x^3\right )+3 a \left (-1+c^4+4 c^3 d x+6 c^2 d^2 x^2+4 c d^3 x^3+d^4 x^4\right )\right ) \text {ArcTan}(c+d x)+3 b^2 \left (-1+c^4+4 c^3 d x+6 c^2 d^2 x^2+4 c d^3 x^3+d^4 x^4\right ) \text {ArcTan}(c+d x)^2-4 b^2 \log \left (1+(c+d x)^2\right )\right )}{12 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.41, size = 192, normalized size = 1.22
method | result | size |
derivativedivides | \(\frac {\frac {e^{3} \left (d x +c \right )^{4} a^{2}}{4}+\frac {e^{3} b^{2} \left (d x +c \right )^{4} \arctan \left (d x +c \right )^{2}}{4}-\frac {e^{3} b^{2} \arctan \left (d x +c \right ) \left (d x +c \right )^{3}}{6}+\frac {e^{3} b^{2} \arctan \left (d x +c \right ) \left (d x +c \right )}{2}-\frac {e^{3} b^{2} \arctan \left (d x +c \right )^{2}}{4}+\frac {e^{3} b^{2} \left (d x +c \right )^{2}}{12}-\frac {e^{3} b^{2} \ln \left (1+\left (d x +c \right )^{2}\right )}{3}+\frac {e^{3} a b \left (d x +c \right )^{4} \arctan \left (d x +c \right )}{2}-\frac {e^{3} \left (d x +c \right )^{3} a b}{6}+\frac {e^{3} a b \left (d x +c \right )}{2}-\frac {e^{3} a b \arctan \left (d x +c \right )}{2}}{d}\) | \(192\) |
default | \(\frac {\frac {e^{3} \left (d x +c \right )^{4} a^{2}}{4}+\frac {e^{3} b^{2} \left (d x +c \right )^{4} \arctan \left (d x +c \right )^{2}}{4}-\frac {e^{3} b^{2} \arctan \left (d x +c \right ) \left (d x +c \right )^{3}}{6}+\frac {e^{3} b^{2} \arctan \left (d x +c \right ) \left (d x +c \right )}{2}-\frac {e^{3} b^{2} \arctan \left (d x +c \right )^{2}}{4}+\frac {e^{3} b^{2} \left (d x +c \right )^{2}}{12}-\frac {e^{3} b^{2} \ln \left (1+\left (d x +c \right )^{2}\right )}{3}+\frac {e^{3} a b \left (d x +c \right )^{4} \arctan \left (d x +c \right )}{2}-\frac {e^{3} \left (d x +c \right )^{3} a b}{6}+\frac {e^{3} a b \left (d x +c \right )}{2}-\frac {e^{3} a b \arctan \left (d x +c \right )}{2}}{d}\) | \(192\) |
risch | \(\frac {3 i e^{3} d a b \,c^{2} x^{2} \ln \left (1-i \left (d x +c \right )\right )}{2}+i e^{3} d^{2} a b c \,x^{3} \ln \left (1-i \left (d x +c \right )\right )+i e^{3} a b \,c^{3} x \ln \left (1-i \left (d x +c \right )\right )+\frac {i e^{3} d^{3} a b \,x^{4} \ln \left (1-i \left (d x +c \right )\right )}{4}-\frac {i e^{3} d \,b^{2} c \,x^{2} \ln \left (1-i \left (d x +c \right )\right )}{4}+\frac {i e^{3} b^{2} x \ln \left (1-i \left (d x +c \right )\right )}{4}-\frac {e^{3} b^{2} \left (d^{4} x^{4}+4 c \,d^{3} x^{3}+6 c^{2} d^{2} x^{2}+4 c^{3} d x +c^{4}-1\right ) \ln \left (1+i \left (d x +c \right )\right )^{2}}{16 d}+\frac {e^{3} b \left (-6 i a \,d^{4} x^{4}+3 b \,d^{4} x^{4} \ln \left (1-i \left (d x +c \right )\right )-24 i a c \,d^{3} x^{3}+12 b c \,d^{3} x^{3} \ln \left (1-i \left (d x +c \right )\right )-36 i a \,c^{2} d^{2} x^{2}+2 i b \,d^{3} x^{3}+18 b \,c^{2} d^{2} x^{2} \ln \left (1-i \left (d x +c \right )\right )-24 i a \,c^{3} d x +6 i b c \,d^{2} x^{2}+12 b \,c^{3} d x \ln \left (1-i \left (d x +c \right )\right )+6 i b \,c^{2} d x +3 b \,c^{4} \ln \left (1-i \left (d x +c \right )\right )-6 i b d x -3 b \ln \left (1-i \left (d x +c \right )\right )\right ) \ln \left (1+i \left (d x +c \right )\right )}{24 d}-\frac {e^{3} b^{2} c^{3} x \ln \left (1-i \left (d x +c \right )\right )^{2}}{4}-\frac {e^{3} d^{3} b^{2} x^{4} \ln \left (1-i \left (d x +c \right )\right )^{2}}{16}-\frac {e^{3} b^{2} c^{4} \ln \left (1-i \left (d x +c \right )\right )^{2}}{16 d}+\frac {e^{3} b^{2} \ln \left (1-i \left (d x +c \right )\right )^{2}}{16 d}-\frac {e^{3} b^{2} \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right )}{3 d}-\frac {3 e^{3} d \,b^{2} c^{2} x^{2} \ln \left (1-i \left (d x +c \right )\right )^{2}}{8}-\frac {e^{3} d^{2} b^{2} c \,x^{3} \ln \left (1-i \left (d x +c \right )\right )^{2}}{4}-\frac {i e^{3} b^{2} c^{2} x \ln \left (1-i \left (d x +c \right )\right )}{4}-\frac {i e^{3} d^{2} b^{2} x^{3} \ln \left (1-i \left (d x +c \right )\right )}{12}+\frac {e^{3} a b \,c^{4} \arctan \left (d x +c \right )}{2 d}-\frac {e^{3} b^{2} c^{3} \arctan \left (d x +c \right )}{6 d}+\frac {e^{3} b^{2} c \arctan \left (d x +c \right )}{2 d}-\frac {e^{3} a b \arctan \left (d x +c \right )}{2 d}+\frac {a b \,e^{3} x}{2}+\frac {e^{3} b^{2} c x}{6}+e^{3} a^{2} c \,d^{2} x^{3}+\frac {3 e^{3} a^{2} c^{2} d \,x^{2}}{2}-\frac {e^{3} a b \,d^{2} x^{3}}{6}-\frac {e^{3} a b \,c^{2} x}{2}+\frac {e^{3} a^{2} d^{3} x^{4}}{4}+e^{3} c^{3} a^{2} x +\frac {e^{3} b^{2} d \,x^{2}}{12}-\frac {e^{3} a b c d \,x^{2}}{2}\) | \(856\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 579 vs.
\(2 (136) = 272\).
time = 1.24, size = 579, normalized size = 3.69 \begin {gather*} \frac {1}{4} \, a^{2} d^{3} x^{4} e^{3} + a^{2} c d^{2} x^{3} e^{3} + \frac {3}{2} \, a^{2} c^{2} d x^{2} e^{3} + 3 \, {\left (x^{2} \arctan \left (d x + c\right ) - d {\left (\frac {x}{d^{2}} + \frac {{\left (c^{2} - 1\right )} \arctan \left (\frac {d^{2} x + c d}{d}\right )}{d^{3}} - \frac {c \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{3}}\right )}\right )} a b c^{2} d e^{3} + {\left (2 \, x^{3} \arctan \left (d x + c\right ) - d {\left (\frac {d x^{2} - 4 \, c x}{d^{3}} - \frac {2 \, {\left (c^{3} - 3 \, c\right )} \arctan \left (\frac {d^{2} x + c d}{d}\right )}{d^{4}} + \frac {{\left (3 \, c^{2} - 1\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{4}}\right )}\right )} a b c d^{2} e^{3} + \frac {1}{6} \, {\left (3 \, x^{4} \arctan \left (d x + c\right ) - d {\left (\frac {d^{2} x^{3} - 3 \, c d x^{2} + 3 \, {\left (3 \, c^{2} - 1\right )} x}{d^{4}} + \frac {3 \, {\left (c^{4} - 6 \, c^{2} + 1\right )} \arctan \left (\frac {d^{2} x + c d}{d}\right )}{d^{5}} - \frac {6 \, {\left (c^{3} - c\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{5}}\right )}\right )} a b d^{3} e^{3} + a^{2} c^{3} x e^{3} + \frac {{\left (2 \, {\left (d x + c\right )} \arctan \left (d x + c\right ) - \log \left ({\left (d x + c\right )}^{2} + 1\right )\right )} a b c^{3} e^{3}}{d} + \frac {b^{2} d^{2} x^{2} e^{3} + 2 \, b^{2} c d x e^{3} - 4 \, b^{2} e^{3} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) + 3 \, {\left (b^{2} d^{4} x^{4} e^{3} + 4 \, b^{2} c d^{3} x^{3} e^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} e^{3} + 4 \, b^{2} c^{3} d x e^{3} + b^{2} c^{4} e^{3} - b^{2} e^{3}\right )} \arctan \left (d x + c\right )^{2} - 2 \, {\left (b^{2} d^{3} x^{3} e^{3} + 3 \, b^{2} c d^{2} x^{2} e^{3} + b^{2} c^{3} e^{3} - 3 \, b^{2} c e^{3} + 3 \, {\left (b^{2} c^{2} e^{3} - b^{2} e^{3}\right )} d x\right )} \arctan \left (d x + c\right )}{12 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 298 vs.
\(2 (136) = 272\).
time = 2.21, size = 298, normalized size = 1.90 \begin {gather*} \frac {3 \, {\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4} - b^{2}\right )} \arctan \left (d x + c\right )^{2} e^{3} - 4 \, b^{2} e^{3} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) + 2 \, {\left (3 \, a b d^{4} x^{4} + {\left (12 \, a b c - b^{2}\right )} d^{3} x^{3} + 3 \, a b c^{4} - b^{2} c^{3} + 3 \, {\left (6 \, a b c^{2} - b^{2} c\right )} d^{2} x^{2} + 3 \, b^{2} c + 3 \, {\left (4 \, a b c^{3} - b^{2} c^{2} + b^{2}\right )} d x - 3 \, a b\right )} \arctan \left (d x + c\right ) e^{3} + {\left (3 \, a^{2} d^{4} x^{4} + 2 \, {\left (6 \, a^{2} c - a b\right )} d^{3} x^{3} + {\left (18 \, a^{2} c^{2} - 6 \, a b c + b^{2}\right )} d^{2} x^{2} + 2 \, {\left (6 \, a^{2} c^{3} - 3 \, a b c^{2} + b^{2} c + 3 \, a b\right )} d x\right )} e^{3}}{12 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [C] Result contains complex when optimal does not.
time = 3.46, size = 583, normalized size = 3.71 \begin {gather*} \begin {cases} a^{2} c^{3} e^{3} x + \frac {3 a^{2} c^{2} d e^{3} x^{2}}{2} + a^{2} c d^{2} e^{3} x^{3} + \frac {a^{2} d^{3} e^{3} x^{4}}{4} + \frac {a b c^{4} e^{3} \operatorname {atan}{\left (c + d x \right )}}{2 d} + 2 a b c^{3} e^{3} x \operatorname {atan}{\left (c + d x \right )} + 3 a b c^{2} d e^{3} x^{2} \operatorname {atan}{\left (c + d x \right )} - \frac {a b c^{2} e^{3} x}{2} + 2 a b c d^{2} e^{3} x^{3} \operatorname {atan}{\left (c + d x \right )} - \frac {a b c d e^{3} x^{2}}{2} + \frac {a b d^{3} e^{3} x^{4} \operatorname {atan}{\left (c + d x \right )}}{2} - \frac {a b d^{2} e^{3} x^{3}}{6} + \frac {a b e^{3} x}{2} - \frac {a b e^{3} \operatorname {atan}{\left (c + d x \right )}}{2 d} + \frac {b^{2} c^{4} e^{3} \operatorname {atan}^{2}{\left (c + d x \right )}}{4 d} + b^{2} c^{3} e^{3} x \operatorname {atan}^{2}{\left (c + d x \right )} - \frac {b^{2} c^{3} e^{3} \operatorname {atan}{\left (c + d x \right )}}{6 d} + \frac {3 b^{2} c^{2} d e^{3} x^{2} \operatorname {atan}^{2}{\left (c + d x \right )}}{2} - \frac {b^{2} c^{2} e^{3} x \operatorname {atan}{\left (c + d x \right )}}{2} + b^{2} c d^{2} e^{3} x^{3} \operatorname {atan}^{2}{\left (c + d x \right )} - \frac {b^{2} c d e^{3} x^{2} \operatorname {atan}{\left (c + d x \right )}}{2} + \frac {b^{2} c e^{3} x}{6} + \frac {b^{2} c e^{3} \operatorname {atan}{\left (c + d x \right )}}{2 d} + \frac {b^{2} d^{3} e^{3} x^{4} \operatorname {atan}^{2}{\left (c + d x \right )}}{4} - \frac {b^{2} d^{2} e^{3} x^{3} \operatorname {atan}{\left (c + d x \right )}}{6} + \frac {b^{2} d e^{3} x^{2}}{12} + \frac {b^{2} e^{3} x \operatorname {atan}{\left (c + d x \right )}}{2} - \frac {2 b^{2} e^{3} \log {\left (\frac {c}{d} + x - \frac {i}{d} \right )}}{3 d} - \frac {b^{2} e^{3} \operatorname {atan}^{2}{\left (c + d x \right )}}{4 d} + \frac {2 i b^{2} e^{3} \operatorname {atan}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\c^{3} e^{3} x \left (a + b \operatorname {atan}{\left (c \right )}\right )^{2} & \text {otherwise} \end {cases} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 3.26, size = 633, normalized size = 4.03 \begin {gather*} x\,\left (\frac {c\,e^3\,\left (20\,a^2\,c^2+6\,a^2-6\,a\,b\,c+b^2\right )}{2}+\frac {\left (6\,c^2+6\right )\,\left (2\,a^2\,c\,d^2\,e^3+\frac {a\,d^2\,e^3\,\left (b-10\,a\,c\right )}{2}\right )}{6\,d^2}-\frac {2\,c\,\left (\frac {2\,c\,\left (2\,a^2\,c\,d^2\,e^3+\frac {a\,d^2\,e^3\,\left (b-10\,a\,c\right )}{2}\right )}{d}+\frac {d\,e^3\,\left (60\,a^2\,c^2+6\,a^2-12\,a\,b\,c+b^2\right )}{6}-\frac {a^2\,d\,e^3\,\left (6\,c^2+6\right )}{6}\right )}{d}\right )+x^2\,\left (\frac {c\,\left (2\,a^2\,c\,d^2\,e^3+\frac {a\,d^2\,e^3\,\left (b-10\,a\,c\right )}{2}\right )}{d}+\frac {d\,e^3\,\left (60\,a^2\,c^2+6\,a^2-12\,a\,b\,c+b^2\right )}{12}-\frac {a^2\,d\,e^3\,\left (6\,c^2+6\right )}{12}\right )-x^3\,\left (\frac {2\,a^2\,c\,d^2\,e^3}{3}+\frac {a\,d^2\,e^3\,\left (b-10\,a\,c\right )}{6}\right )+{\mathrm {atan}\left (c+d\,x\right )}^2\,\left (b^2\,c^3\,e^3\,x-\frac {b^2\,e^3-b^2\,c^4\,e^3}{4\,d}+\frac {b^2\,d^3\,e^3\,x^4}{4}+\frac {3\,b^2\,c^2\,d\,e^3\,x^2}{2}+b^2\,c\,d^2\,e^3\,x^3\right )-d^2\,\mathrm {atan}\left (c+d\,x\right )\,\left (x^3\,\left (\frac {b^2\,e^3}{6}-2\,a\,b\,c\,e^3\right )-\frac {x\,\left (-b^2\,c^2\,e^3+b^2\,e^3+4\,a\,b\,c^3\,e^3\right )}{2\,d^2}+\frac {x^2\,\left (b^2\,c\,e^3-6\,a\,b\,c^2\,e^3\right )}{2\,d}-\frac {a\,b\,d\,e^3\,x^4}{2}\right )+\frac {a^2\,d^3\,e^3\,x^4}{4}-\frac {b^2\,e^3\,\ln \left (c^2+2\,c\,d\,x+d^2\,x^2+1\right )}{3\,d}+\frac {b\,e^3\,\mathrm {atan}\left (\frac {\frac {b\,c\,e^3\,\left (-3\,a\,c^4+b\,c^3-3\,b\,c+3\,a\right )}{6}+\frac {b\,d\,e^3\,x\,\left (-3\,a\,c^4+b\,c^3-3\,b\,c+3\,a\right )}{6}}{-\frac {b^2\,c^3\,e^3}{6}+\frac {b^2\,c\,e^3}{2}+\frac {a\,b\,c^4\,e^3}{2}-\frac {a\,b\,e^3}{2}}\right )\,\left (-3\,a\,c^4+b\,c^3-3\,b\,c+3\,a\right )}{6\,d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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