Optimal. Leaf size=200 \[ -\frac {b c d^3}{2 x}-\frac {3 b d e^2 x}{2 c}+\frac {b e^3 x}{4 c^3}-\frac {b e^3 x^3}{12 c}-\frac {1}{2} b c^2 d^3 \text {ArcTan}(c x)+\frac {3 b d e^2 \text {ArcTan}(c x)}{2 c^2}-\frac {b e^3 \text {ArcTan}(c x)}{4 c^4}-\frac {d^3 (a+b \text {ArcTan}(c x))}{2 x^2}+\frac {3}{2} d e^2 x^2 (a+b \text {ArcTan}(c x))+\frac {1}{4} e^3 x^4 (a+b \text {ArcTan}(c x))+3 a d^2 e \log (x)+\frac {3}{2} i b d^2 e \text {PolyLog}(2,-i c x)-\frac {3}{2} i b d^2 e \text {PolyLog}(2,i c x) \]
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Rubi [A]
time = 0.14, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps
used = 15, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {5100, 4946,
331, 209, 4940, 2438, 327, 308} \begin {gather*} -\frac {d^3 (a+b \text {ArcTan}(c x))}{2 x^2}+\frac {3}{2} d e^2 x^2 (a+b \text {ArcTan}(c x))+\frac {1}{4} e^3 x^4 (a+b \text {ArcTan}(c x))+3 a d^2 e \log (x)-\frac {b e^3 \text {ArcTan}(c x)}{4 c^4}-\frac {1}{2} b c^2 d^3 \text {ArcTan}(c x)+\frac {3 b d e^2 \text {ArcTan}(c x)}{2 c^2}+\frac {b e^3 x}{4 c^3}-\frac {b c d^3}{2 x}+\frac {3}{2} i b d^2 e \text {Li}_2(-i c x)-\frac {3}{2} i b d^2 e \text {Li}_2(i c x)-\frac {3 b d e^2 x}{2 c}-\frac {b e^3 x^3}{12 c} \end {gather*}
Antiderivative was successfully verified.
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Rule 209
Rule 308
Rule 327
Rule 331
Rule 2438
Rule 4940
Rule 4946
Rule 5100
Rubi steps
\begin {align*} \int \frac {\left (d+e x^2\right )^3 \left (a+b \tan ^{-1}(c x)\right )}{x^3} \, dx &=\int \left (\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^3}+\frac {3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \tan ^{-1}(c x)\right )+e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )\right ) \, dx\\ &=d^3 \int \frac {a+b \tan ^{-1}(c x)}{x^3} \, dx+\left (3 d^2 e\right ) \int \frac {a+b \tan ^{-1}(c x)}{x} \, dx+\left (3 d e^2\right ) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx+e^3 \int x^3 \left (a+b \tan ^{-1}(c x)\right ) \, dx\\ &=-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{4} e^3 x^4 \left (a+b \tan ^{-1}(c x)\right )+3 a d^2 e \log (x)+\frac {1}{2} \left (b c d^3\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx+\frac {1}{2} \left (3 i b d^2 e\right ) \int \frac {\log (1-i c x)}{x} \, dx-\frac {1}{2} \left (3 i b d^2 e\right ) \int \frac {\log (1+i c x)}{x} \, dx-\frac {1}{2} \left (3 b c d e^2\right ) \int \frac {x^2}{1+c^2 x^2} \, dx-\frac {1}{4} \left (b c e^3\right ) \int \frac {x^4}{1+c^2 x^2} \, dx\\ &=-\frac {b c d^3}{2 x}-\frac {3 b d e^2 x}{2 c}-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{4} e^3 x^4 \left (a+b \tan ^{-1}(c x)\right )+3 a d^2 e \log (x)+\frac {3}{2} i b d^2 e \text {Li}_2(-i c x)-\frac {3}{2} i b d^2 e \text {Li}_2(i c x)-\frac {1}{2} \left (b c^3 d^3\right ) \int \frac {1}{1+c^2 x^2} \, dx+\frac {\left (3 b d e^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 c}-\frac {1}{4} \left (b c e^3\right ) \int \left (-\frac {1}{c^4}+\frac {x^2}{c^2}+\frac {1}{c^4 \left (1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac {b c d^3}{2 x}-\frac {3 b d e^2 x}{2 c}+\frac {b e^3 x}{4 c^3}-\frac {b e^3 x^3}{12 c}-\frac {1}{2} b c^2 d^3 \tan ^{-1}(c x)+\frac {3 b d e^2 \tan ^{-1}(c x)}{2 c^2}-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{4} e^3 x^4 \left (a+b \tan ^{-1}(c x)\right )+3 a d^2 e \log (x)+\frac {3}{2} i b d^2 e \text {Li}_2(-i c x)-\frac {3}{2} i b d^2 e \text {Li}_2(i c x)-\frac {\left (b e^3\right ) \int \frac {1}{1+c^2 x^2} \, dx}{4 c^3}\\ &=-\frac {b c d^3}{2 x}-\frac {3 b d e^2 x}{2 c}+\frac {b e^3 x}{4 c^3}-\frac {b e^3 x^3}{12 c}-\frac {1}{2} b c^2 d^3 \tan ^{-1}(c x)+\frac {3 b d e^2 \tan ^{-1}(c x)}{2 c^2}-\frac {b e^3 \tan ^{-1}(c x)}{4 c^4}-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{4} e^3 x^4 \left (a+b \tan ^{-1}(c x)\right )+3 a d^2 e \log (x)+\frac {3}{2} i b d^2 e \text {Li}_2(-i c x)-\frac {3}{2} i b d^2 e \text {Li}_2(i c x)\\ \end {align*}
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Mathematica [A]
time = 0.09, size = 198, normalized size = 0.99 \begin {gather*} \frac {1}{4} \left (-\frac {2 a d^3}{x^2}-\frac {2 b c d^3}{x}-\frac {6 b d e^2 x}{c}+\frac {b e^3 x}{c^3}+6 a d e^2 x^2-\frac {b e^3 x^3}{3 c}+a e^3 x^4-2 b c^2 d^3 \text {ArcTan}(c x)+\frac {6 b d e^2 \text {ArcTan}(c x)}{c^2}-\frac {b e^3 \text {ArcTan}(c x)}{c^4}-\frac {2 b d^3 \text {ArcTan}(c x)}{x^2}+6 b d e^2 x^2 \text {ArcTan}(c x)+b e^3 x^4 \text {ArcTan}(c x)+12 a d^2 e \log (x)+6 i b d^2 e \text {PolyLog}(2,-i c x)-6 i b d^2 e \text {PolyLog}(2,i c x)\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.21, size = 290, normalized size = 1.45
method | result | size |
derivativedivides | \(c^{2} \left (\frac {3 a d \,e^{2} x^{2}}{2 c^{2}}+\frac {a \,e^{3} x^{4}}{4 c^{2}}-\frac {a \,d^{3}}{2 c^{2} x^{2}}+\frac {3 a \,d^{2} e \ln \left (c x \right )}{c^{2}}+\frac {3 b \arctan \left (c x \right ) d \,e^{2} x^{2}}{2 c^{2}}+\frac {b \arctan \left (c x \right ) e^{3} x^{4}}{4 c^{2}}-\frac {b \arctan \left (c x \right ) d^{3}}{2 c^{2} x^{2}}+\frac {3 b \arctan \left (c x \right ) d^{2} e \ln \left (c x \right )}{c^{2}}-\frac {3 b d \,e^{2} x}{2 c^{3}}-\frac {b \,e^{3} x^{3}}{12 c^{3}}+\frac {b \,e^{3} x}{4 c^{5}}-\frac {b \,d^{3} \arctan \left (c x \right )}{2}+\frac {3 b d \,e^{2} \arctan \left (c x \right )}{2 c^{4}}-\frac {b \,e^{3} \arctan \left (c x \right )}{4 c^{6}}-\frac {b \,d^{3}}{2 c x}-\frac {3 i b \,d^{2} e \dilog \left (-i c x +1\right )}{2 c^{2}}-\frac {3 i b \,d^{2} e \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2 c^{2}}+\frac {3 i b \,d^{2} e \dilog \left (i c x +1\right )}{2 c^{2}}+\frac {3 i b \,d^{2} e \ln \left (c x \right ) \ln \left (i c x +1\right )}{2 c^{2}}\right )\) | \(290\) |
default | \(c^{2} \left (\frac {3 a d \,e^{2} x^{2}}{2 c^{2}}+\frac {a \,e^{3} x^{4}}{4 c^{2}}-\frac {a \,d^{3}}{2 c^{2} x^{2}}+\frac {3 a \,d^{2} e \ln \left (c x \right )}{c^{2}}+\frac {3 b \arctan \left (c x \right ) d \,e^{2} x^{2}}{2 c^{2}}+\frac {b \arctan \left (c x \right ) e^{3} x^{4}}{4 c^{2}}-\frac {b \arctan \left (c x \right ) d^{3}}{2 c^{2} x^{2}}+\frac {3 b \arctan \left (c x \right ) d^{2} e \ln \left (c x \right )}{c^{2}}-\frac {3 b d \,e^{2} x}{2 c^{3}}-\frac {b \,e^{3} x^{3}}{12 c^{3}}+\frac {b \,e^{3} x}{4 c^{5}}-\frac {b \,d^{3} \arctan \left (c x \right )}{2}+\frac {3 b d \,e^{2} \arctan \left (c x \right )}{2 c^{4}}-\frac {b \,e^{3} \arctan \left (c x \right )}{4 c^{6}}-\frac {b \,d^{3}}{2 c x}-\frac {3 i b \,d^{2} e \dilog \left (-i c x +1\right )}{2 c^{2}}-\frac {3 i b \,d^{2} e \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2 c^{2}}+\frac {3 i b \,d^{2} e \dilog \left (i c x +1\right )}{2 c^{2}}+\frac {3 i b \,d^{2} e \ln \left (c x \right ) \ln \left (i c x +1\right )}{2 c^{2}}\right )\) | \(290\) |
risch | \(-\frac {a \,e^{3}}{4 c^{4}}-\frac {a \,d^{3}}{2 x^{2}}+\frac {a \,x^{4} e^{3}}{4}-\frac {3 b d \,e^{2} x}{2 c}+\frac {3 b d \,e^{2} \arctan \left (c x \right )}{2 c^{2}}+\frac {i b \,d^{3} \ln \left (i c x +1\right )}{4 x^{2}}-\frac {i b \,c^{2} d^{3} \ln \left (i c x \right )}{4}+\frac {b \,e^{3} x}{4 c^{3}}-\frac {b \,e^{3} x^{3}}{12 c}-\frac {b \,e^{3} \arctan \left (c x \right )}{4 c^{4}}+\frac {3 a d \,e^{2}}{2 c^{2}}+3 a \,d^{2} e \ln \left (-i c x \right )+\frac {3 a d \,e^{2} x^{2}}{2}-\frac {b c \,d^{3}}{2 x}-\frac {b \,c^{2} d^{3} \arctan \left (c x \right )}{2}+\frac {3 i b \,d^{2} e \dilog \left (i c x +1\right )}{2}+\frac {i c^{2} b \,d^{3} \ln \left (-i c x \right )}{4}-\frac {i b \,d^{3} \ln \left (-i c x +1\right )}{4 x^{2}}+\frac {i b \ln \left (-i c x +1\right ) e^{3} x^{4}}{8}+\frac {3 i b d \,e^{2} \ln \left (-i c x +1\right ) x^{2}}{4}-\frac {3 i b \,d^{2} e \dilog \left (-i c x +1\right )}{2}-\frac {i b \,e^{3} \ln \left (i c x +1\right ) x^{4}}{8}-\frac {3 i b \,e^{2} d \ln \left (i c x +1\right ) x^{2}}{4}\) | \(319\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.61, size = 219, normalized size = 1.10 \begin {gather*} \frac {1}{4} \, a x^{4} e^{3} - \frac {1}{2} \, {\left ({\left (c \arctan \left (c x\right ) + \frac {1}{x}\right )} c + \frac {\arctan \left (c x\right )}{x^{2}}\right )} b d^{3} + \frac {3}{2} \, a d x^{2} e^{2} + 3 \, a d^{2} e \log \left (x\right ) - \frac {a d^{3}}{2 \, x^{2}} - \frac {9 \, \pi b c^{4} d^{2} e \log \left (c^{2} x^{2} + 1\right ) - 36 \, b c^{4} d^{2} \arctan \left (c x\right ) e \log \left (c x\right ) + 18 i \, b c^{4} d^{2} {\rm Li}_2\left (i \, c x + 1\right ) e - 18 i \, b c^{4} d^{2} {\rm Li}_2\left (-i \, c x + 1\right ) e + b c^{3} x^{3} e^{3} + 3 \, {\left (6 \, b c^{3} d e^{2} - b c e^{3}\right )} x - 3 \, {\left (b c^{4} x^{4} e^{3} + 6 \, b c^{4} d x^{2} e^{2} + 6 \, b c^{2} d e^{2} - b e^{3}\right )} \arctan \left (c x\right )}{12 \, c^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{3}}{x^{3}}\, dx \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 = 0.73, size = 224, normalized size = 1.12 \begin {gather*} \left \{\begin {array}{cl} \frac {a\,e^3\,x^4}{4}-\frac {a\,d^3}{2\,x^2}+\frac {3\,a\,d\,e^2\,x^2}{2}+3\,a\,d^2\,e\,\ln \left (x\right ) & \text {\ if\ \ }c=0\\ \frac {a\,e^3\,x^4}{4}-\frac {a\,d^3}{2\,x^2}-\frac {b\,d^3\,\left (c^3\,\mathrm {atan}\left (c\,x\right )+\frac {c^2}{x}\right )}{2\,c}-3\,b\,d\,e^2\,\left (\frac {x}{2\,c}-\mathrm {atan}\left (c\,x\right )\,\left (\frac {1}{2\,c^2}+\frac {x^2}{2}\right )\right )+\frac {3\,a\,d\,e^2\,x^2}{2}+3\,a\,d^2\,e\,\ln \left (x\right )-\frac {b\,e^3\,\left (3\,\mathrm {atan}\left (c\,x\right )-3\,c\,x+c^3\,x^3\right )}{12\,c^4}-\frac {b\,d^3\,\mathrm {atan}\left (c\,x\right )}{2\,x^2}+\frac {b\,e^3\,x^4\,\mathrm {atan}\left (c\,x\right )}{4}-\frac {b\,d^2\,e\,{\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2}+\frac {b\,d^2\,e\,{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2} & \text {\ if\ \ }c\neq 0 \end {array}\right . \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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