Optimal. Leaf size=144 \[ -\frac {b e \left (6 c^2 d^2-e^2\right ) x}{4 c^3}-\frac {b d e^2 x^2}{2 c}-\frac {b e^3 x^3}{12 c}-\frac {b \left (c^4 d^4-6 c^2 d^2 e^2+e^4\right ) \text {ArcTan}(c x)}{4 c^4 e}+\frac {(d+e x)^4 (a+b \text {ArcTan}(c x))}{4 e}-\frac {b d (c d-e) (c d+e) \log \left (1+c^2 x^2\right )}{2 c^3} \]
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Rubi [A]
time = 0.09, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {4972, 716, 649,
209, 266} \begin {gather*} \frac {(d+e x)^4 (a+b \text {ArcTan}(c x))}{4 e}-\frac {b \text {ArcTan}(c x) \left (c^4 d^4-6 c^2 d^2 e^2+e^4\right )}{4 c^4 e}-\frac {b e x \left (6 c^2 d^2-e^2\right )}{4 c^3}-\frac {b d (c d-e) (c d+e) \log \left (c^2 x^2+1\right )}{2 c^3}-\frac {b d e^2 x^2}{2 c}-\frac {b e^3 x^3}{12 c} \end {gather*}
Antiderivative was successfully verified.
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Rule 209
Rule 266
Rule 649
Rule 716
Rule 4972
Rubi steps
\begin {align*} \int (d+e x)^3 \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac {(d+e x)^4 \left (a+b \tan ^{-1}(c x)\right )}{4 e}-\frac {(b c) \int \frac {(d+e x)^4}{1+c^2 x^2} \, dx}{4 e}\\ &=\frac {(d+e x)^4 \left (a+b \tan ^{-1}(c x)\right )}{4 e}-\frac {(b c) \int \left (\frac {e^2 \left (6 c^2 d^2-e^2\right )}{c^4}+\frac {4 d e^3 x}{c^2}+\frac {e^4 x^2}{c^2}+\frac {c^4 d^4-6 c^2 d^2 e^2+e^4+4 c^2 d (c d-e) e (c d+e) x}{c^4 \left (1+c^2 x^2\right )}\right ) \, dx}{4 e}\\ &=-\frac {b e \left (6 c^2 d^2-e^2\right ) x}{4 c^3}-\frac {b d e^2 x^2}{2 c}-\frac {b e^3 x^3}{12 c}+\frac {(d+e x)^4 \left (a+b \tan ^{-1}(c x)\right )}{4 e}-\frac {b \int \frac {c^4 d^4-6 c^2 d^2 e^2+e^4+4 c^2 d (c d-e) e (c d+e) x}{1+c^2 x^2} \, dx}{4 c^3 e}\\ &=-\frac {b e \left (6 c^2 d^2-e^2\right ) x}{4 c^3}-\frac {b d e^2 x^2}{2 c}-\frac {b e^3 x^3}{12 c}+\frac {(d+e x)^4 \left (a+b \tan ^{-1}(c x)\right )}{4 e}-\frac {(b d (c d-e) (c d+e)) \int \frac {x}{1+c^2 x^2} \, dx}{c}-\frac {\left (b \left (c^4 d^4-6 c^2 d^2 e^2+e^4\right )\right ) \int \frac {1}{1+c^2 x^2} \, dx}{4 c^3 e}\\ &=-\frac {b e \left (6 c^2 d^2-e^2\right ) x}{4 c^3}-\frac {b d e^2 x^2}{2 c}-\frac {b e^3 x^3}{12 c}-\frac {b \left (c^4 d^4-6 c^2 d^2 e^2+e^4\right ) \tan ^{-1}(c x)}{4 c^4 e}+\frac {(d+e x)^4 \left (a+b \tan ^{-1}(c x)\right )}{4 e}-\frac {b d (c d-e) (c d+e) \log \left (1+c^2 x^2\right )}{2 c^3}\\ \end {align*}
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Mathematica [A]
time = 0.30, size = 218, normalized size = 1.51 \begin {gather*} \frac {(d+e x)^4 (a+b \text {ArcTan}(c x))-\frac {b c \left (2 \sqrt {-c^2} e^2 x \left (-3 e^2+c^2 \left (18 d^2+6 d e x+e^2 x^2\right )\right )-3 \left (c^4 d^4+e^3 \left (4 \sqrt {-c^2} d+e\right )-2 c^2 d^2 e \left (2 \sqrt {-c^2} d+3 e\right )\right ) \log \left (1-\sqrt {-c^2} x\right )+3 \left (c^4 d^4+2 c^2 d^2 \left (2 \sqrt {-c^2} d-3 e\right ) e+e^3 \left (-4 \sqrt {-c^2} d+e\right )\right ) \log \left (1+\sqrt {-c^2} x\right )\right )}{6 \left (-c^2\right )^{5/2}}}{4 e} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.12, size = 179, normalized size = 1.24
method | result | size |
derivativedivides | \(\frac {\frac {\left (c e x +c d \right )^{4} a}{4 c^{3} e}+b \arctan \left (c x \right ) d^{3} c x +\frac {3 b c e \arctan \left (c x \right ) d^{2} x^{2}}{2}+b c \,e^{2} \arctan \left (c x \right ) d \,x^{3}+\frac {b c \,e^{3} \arctan \left (c x \right ) x^{4}}{4}-\frac {3 b e \,d^{2} x}{2}-\frac {b \,e^{2} d \,x^{2}}{2}-\frac {b \,e^{3} x^{3}}{12}+\frac {b \,e^{3} x}{4 c^{2}}-\frac {b \ln \left (c^{2} x^{2}+1\right ) d^{3}}{2}+\frac {b \,e^{2} \ln \left (c^{2} x^{2}+1\right ) d}{2 c^{2}}+\frac {3 b e \arctan \left (c x \right ) d^{2}}{2 c}-\frac {b \,e^{3} \arctan \left (c x \right )}{4 c^{3}}}{c}\) | \(179\) |
default | \(\frac {\frac {\left (c e x +c d \right )^{4} a}{4 c^{3} e}+b \arctan \left (c x \right ) d^{3} c x +\frac {3 b c e \arctan \left (c x \right ) d^{2} x^{2}}{2}+b c \,e^{2} \arctan \left (c x \right ) d \,x^{3}+\frac {b c \,e^{3} \arctan \left (c x \right ) x^{4}}{4}-\frac {3 b e \,d^{2} x}{2}-\frac {b \,e^{2} d \,x^{2}}{2}-\frac {b \,e^{3} x^{3}}{12}+\frac {b \,e^{3} x}{4 c^{2}}-\frac {b \ln \left (c^{2} x^{2}+1\right ) d^{3}}{2}+\frac {b \,e^{2} \ln \left (c^{2} x^{2}+1\right ) d}{2 c^{2}}+\frac {3 b e \arctan \left (c x \right ) d^{2}}{2 c}-\frac {b \,e^{3} \arctan \left (c x \right )}{4 c^{3}}}{c}\) | \(179\) |
risch | \(\frac {i b \,d^{4} \ln \left (c^{2} x^{2}+1\right )}{16 e}+\frac {3 e b \,d^{2} \arctan \left (c x \right )}{2 c^{2}}+\frac {i e^{2} b d \,x^{3} \ln \left (-i c x +1\right )}{2}-\frac {i \left (e x +d \right )^{4} b \ln \left (i c x +1\right )}{8 e}+\frac {i b \,d^{3} x \ln \left (-i c x +1\right )}{2}+\frac {i e^{3} b \,x^{4} \ln \left (-i c x +1\right )}{8}-\frac {b \,d^{4} \arctan \left (c x \right )}{8 e}+\frac {x^{4} e^{3} a}{4}-\frac {b \,e^{3} x^{3}}{12 c}+\frac {e^{3} b x}{4 c^{3}}+x^{3} e^{2} d a +\frac {3 x^{2} e \,d^{2} a}{2}+x \,d^{3} a -\frac {b d \,e^{2} x^{2}}{2 c}-\frac {3 e b \,d^{2} x}{2 c}+\frac {e^{2} b d \ln \left (c^{2} x^{2}+1\right )}{2 c^{3}}-\frac {b \,d^{3} \ln \left (c^{2} x^{2}+1\right )}{2 c}-\frac {e^{3} b \arctan \left (c x \right )}{4 c^{4}}+\frac {3 i e b \,d^{2} x^{2} \ln \left (-i c x +1\right )}{4}\) | \(275\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.49, size = 184, normalized size = 1.28 \begin {gather*} \frac {1}{4} \, a x^{4} e^{3} + a d x^{3} e^{2} + \frac {3}{2} \, a d^{2} x^{2} e + a d^{3} x + \frac {3}{2} \, {\left (x^{2} \arctan \left (c x\right ) - c {\left (\frac {x}{c^{2}} - \frac {\arctan \left (c x\right )}{c^{3}}\right )}\right )} b d^{2} e + \frac {{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d^{3}}{2 \, c} + \frac {1}{2} \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b d e^{2} + \frac {1}{12} \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b e^{3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 1.49, size = 190, normalized size = 1.32 \begin {gather*} \frac {12 \, a c^{4} d^{3} x + 3 \, {\left (4 \, b c^{4} d x^{3} e^{2} + 4 \, b c^{4} d^{3} x + {\left (b c^{4} x^{4} - b\right )} e^{3} + 6 \, {\left (b c^{4} d^{2} x^{2} + b c^{2} d^{2}\right )} e\right )} \arctan \left (c x\right ) + {\left (3 \, a c^{4} x^{4} - b c^{3} x^{3} + 3 \, b c x\right )} e^{3} + 6 \, {\left (2 \, a c^{4} d x^{3} - b c^{3} d x^{2}\right )} e^{2} + 18 \, {\left (a c^{4} d^{2} x^{2} - b c^{3} d^{2} x\right )} e - 6 \, {\left (b c^{3} d^{3} - b c d e^{2}\right )} \log \left (c^{2} x^{2} + 1\right )}{12 \, c^{4}} \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. 262 vs.
\(2 (129) = 258\).
time = 0.35, size = 262, normalized size = 1.82 \begin {gather*} \begin {cases} a d^{3} x + \frac {3 a d^{2} e x^{2}}{2} + a d e^{2} x^{3} + \frac {a e^{3} x^{4}}{4} + b d^{3} x \operatorname {atan}{\left (c x \right )} + \frac {3 b d^{2} e x^{2} \operatorname {atan}{\left (c x \right )}}{2} + b d e^{2} x^{3} \operatorname {atan}{\left (c x \right )} + \frac {b e^{3} x^{4} \operatorname {atan}{\left (c x \right )}}{4} - \frac {b d^{3} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c} - \frac {3 b d^{2} e x}{2 c} - \frac {b d e^{2} x^{2}}{2 c} - \frac {b e^{3} x^{3}}{12 c} + \frac {3 b d^{2} e \operatorname {atan}{\left (c x \right )}}{2 c^{2}} + \frac {b d e^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c^{3}} + \frac {b e^{3} x}{4 c^{3}} - \frac {b e^{3} \operatorname {atan}{\left (c x \right )}}{4 c^{4}} & \text {for}\: c \neq 0 \\a \left (d^{3} x + \frac {3 d^{2} e x^{2}}{2} + d e^{2} x^{3} + \frac {e^{3} x^{4}}{4}\right ) & \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 = 0.80, size = 197, normalized size = 1.37 \begin {gather*} \frac {a\,e^3\,x^4}{4}+a\,d^3\,x-\frac {b\,d^3\,\ln \left (c^2\,x^2+1\right )}{2\,c}-\frac {b\,e^3\,x^3}{12\,c}+b\,d^3\,x\,\mathrm {atan}\left (c\,x\right )+\frac {3\,a\,d^2\,e\,x^2}{2}+a\,d\,e^2\,x^3+\frac {b\,e^3\,x}{4\,c^3}-\frac {b\,e^3\,\mathrm {atan}\left (c\,x\right )}{4\,c^4}+\frac {b\,e^3\,x^4\,\mathrm {atan}\left (c\,x\right )}{4}-\frac {3\,b\,d^2\,e\,x}{2\,c}+\frac {3\,b\,d^2\,e\,\mathrm {atan}\left (c\,x\right )}{2\,c^2}+\frac {3\,b\,d^2\,e\,x^2\,\mathrm {atan}\left (c\,x\right )}{2}+b\,d\,e^2\,x^3\,\mathrm {atan}\left (c\,x\right )+\frac {b\,d\,e^2\,\ln \left (c^2\,x^2+1\right )}{2\,c^3}-\frac {b\,d\,e^2\,x^2}{2\,c} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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