Optimal. Leaf size=355 \[ -\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4}+\frac {a b e^2 x}{2 c^3}+\frac {d e \left (a+b \tan ^{-1}(c x)\right )^2}{c^2}-i b d^2 \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )+i b d^2 \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) \left (a+b \tan ^{-1}(c x)\right )+2 d^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2+d e x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {2 a b d e x}{c}+\frac {1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {b e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac {b^2 e^2 x \tan ^{-1}(c x)}{2 c^3}+\frac {b^2 d e \log \left (c^2 x^2+1\right )}{c^2}+\frac {b^2 e^2 x^2}{12 c^2}-\frac {b^2 e^2 \log \left (c^2 x^2+1\right )}{3 c^4}-\frac {1}{2} b^2 d^2 \text {Li}_3\left (1-\frac {2}{i c x+1}\right )+\frac {1}{2} b^2 d^2 \text {Li}_3\left (\frac {2}{i c x+1}-1\right )-\frac {2 b^2 d e x \tan ^{-1}(c x)}{c} \]
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Rubi [A] time = 0.69, antiderivative size = 355, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 12, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {4980, 4850, 4988, 4884, 4994, 6610, 4852, 4916, 4846, 260, 266, 43} \[ -i b d^2 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+i b d^2 \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{2} b^2 d^2 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d^2 \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )+\frac {d e \left (a+b \tan ^{-1}(c x)\right )^2}{c^2}+\frac {a b e^2 x}{2 c^3}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4}+2 d^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2+d e x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {2 a b d e x}{c}+\frac {1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {b e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac {b^2 d e \log \left (c^2 x^2+1\right )}{c^2}+\frac {b^2 e^2 x^2}{12 c^2}-\frac {b^2 e^2 \log \left (c^2 x^2+1\right )}{3 c^4}+\frac {b^2 e^2 x \tan ^{-1}(c x)}{2 c^3}-\frac {2 b^2 d e x \tan ^{-1}(c x)}{c} \]
Antiderivative was successfully verified.
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Rule 43
Rule 260
Rule 266
Rule 4846
Rule 4850
Rule 4852
Rule 4884
Rule 4916
Rule 4980
Rule 4988
Rule 4994
Rule 6610
Rubi steps
\begin {align*} \int \frac {\left (d+e x^2\right )^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx &=\int \left (\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x}+2 d e x \left (a+b \tan ^{-1}(c x)\right )^2+e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )^2\right ) \, dx\\ &=d^2 \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx+(2 d e) \int x \left (a+b \tan ^{-1}(c x)\right )^2 \, dx+e^2 \int x^3 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx\\ &=d e x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+2 d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-\left (4 b c d^2\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-(2 b c d e) \int \frac {x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx-\frac {1}{2} \left (b c e^2\right ) \int \frac {x^4 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=d e x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+2 d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+\left (2 b c d^2\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (2 b c d^2\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\frac {(2 b d e) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c}+\frac {(2 b d e) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{c}-\frac {\left (b e^2\right ) \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx}{2 c}+\frac {\left (b e^2\right ) \int \frac {x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{2 c}\\ &=-\frac {2 a b d e x}{c}-\frac {b e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac {d e \left (a+b \tan ^{-1}(c x)\right )^2}{c^2}+d e x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+2 d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )+\left (i b^2 c d^2\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (i b^2 c d^2\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\frac {\left (2 b^2 d e\right ) \int \tan ^{-1}(c x) \, dx}{c}+\frac {1}{6} \left (b^2 e^2\right ) \int \frac {x^3}{1+c^2 x^2} \, dx+\frac {\left (b e^2\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{2 c^3}-\frac {\left (b e^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{2 c^3}\\ &=-\frac {2 a b d e x}{c}+\frac {a b e^2 x}{2 c^3}-\frac {2 b^2 d e x \tan ^{-1}(c x)}{c}-\frac {b e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac {d e \left (a+b \tan ^{-1}(c x)\right )^2}{c^2}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4}+d e x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+2 d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 d^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d^2 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )+\left (2 b^2 d e\right ) \int \frac {x}{1+c^2 x^2} \, dx+\frac {1}{12} \left (b^2 e^2\right ) \operatorname {Subst}\left (\int \frac {x}{1+c^2 x} \, dx,x,x^2\right )+\frac {\left (b^2 e^2\right ) \int \tan ^{-1}(c x) \, dx}{2 c^3}\\ &=-\frac {2 a b d e x}{c}+\frac {a b e^2 x}{2 c^3}-\frac {2 b^2 d e x \tan ^{-1}(c x)}{c}+\frac {b^2 e^2 x \tan ^{-1}(c x)}{2 c^3}-\frac {b e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac {d e \left (a+b \tan ^{-1}(c x)\right )^2}{c^2}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4}+d e x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+2 d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+\frac {b^2 d e \log \left (1+c^2 x^2\right )}{c^2}-i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 d^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d^2 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )+\frac {1}{12} \left (b^2 e^2\right ) \operatorname {Subst}\left (\int \left (\frac {1}{c^2}-\frac {1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )-\frac {\left (b^2 e^2\right ) \int \frac {x}{1+c^2 x^2} \, dx}{2 c^2}\\ &=-\frac {2 a b d e x}{c}+\frac {a b e^2 x}{2 c^3}+\frac {b^2 e^2 x^2}{12 c^2}-\frac {2 b^2 d e x \tan ^{-1}(c x)}{c}+\frac {b^2 e^2 x \tan ^{-1}(c x)}{2 c^3}-\frac {b e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac {d e \left (a+b \tan ^{-1}(c x)\right )^2}{c^2}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4}+d e x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+2 d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+\frac {b^2 d e \log \left (1+c^2 x^2\right )}{c^2}-\frac {b^2 e^2 \log \left (1+c^2 x^2\right )}{3 c^4}-i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 d^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d^2 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )\\ \end {align*}
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Mathematica [A] time = 0.72, size = 389, normalized size = 1.10 \[ a^2 d^2 \log (x)+a^2 d e x^2+\frac {1}{4} a^2 e^2 x^4+\frac {2 a b d e \left (\left (c^2 x^2+1\right ) \tan ^{-1}(c x)-c x\right )}{c^2}+\frac {a b e^2 \left (3 \left (c^4 x^4-1\right ) \tan ^{-1}(c x)-c^3 x^3+3 c x\right )}{6 c^4}+i a b d^2 (\text {Li}_2(-i c x)-\text {Li}_2(i c x))+\frac {b^2 d e \left (\log \left (c^2 x^2+1\right )+\left (c^2 x^2+1\right ) \tan ^{-1}(c x)^2-2 c x \tan ^{-1}(c x)\right )}{c^2}+\frac {b^2 e^2 \left (3 \left (c^4 x^4-1\right ) \tan ^{-1}(c x)^2+\left (6 c x-2 c^3 x^3\right ) \tan ^{-1}(c x)+c^2 x^2-4 \log \left (c^2 x^2+1\right )+1\right )}{12 c^4}+b^2 d^2 \left (i \tan ^{-1}(c x) \text {Li}_2\left (e^{-2 i \tan ^{-1}(c x)}\right )+i \tan ^{-1}(c x) \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )+\frac {1}{2} \text {Li}_3\left (e^{-2 i \tan ^{-1}(c x)}\right )-\frac {1}{2} \text {Li}_3\left (-e^{2 i \tan ^{-1}(c x)}\right )+\frac {2}{3} i \tan ^{-1}(c x)^3+\tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )-\tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-\frac {i \pi ^3}{24}\right ) \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{2} e^{2} x^{4} + 2 \, a^{2} d e x^{2} + a^{2} d^{2} + {\left (b^{2} e^{2} x^{4} + 2 \, b^{2} d e x^{2} + b^{2} d^{2}\right )} \arctan \left (c x\right )^{2} + 2 \, {\left (a b e^{2} x^{4} + 2 \, a b d e x^{2} + a b d^{2}\right )} \arctan \left (c x\right )}{x}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 7.83, size = 1549, normalized size = 4.36 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} \, a^{2} e^{2} x^{4} + 12 \, b^{2} c^{2} e^{2} \int \frac {x^{6} \arctan \left (c x\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + b^{2} c^{2} e^{2} \int \frac {x^{6} \log \left (c^{2} x^{2} + 1\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 32 \, a b c^{2} e^{2} \int \frac {x^{6} \arctan \left (c x\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + b^{2} c^{2} e^{2} \int \frac {x^{6} \log \left (c^{2} x^{2} + 1\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 24 \, b^{2} c^{2} d e \int \frac {x^{4} \arctan \left (c x\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 2 \, b^{2} c^{2} d e \int \frac {x^{4} \log \left (c^{2} x^{2} + 1\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 64 \, a b c^{2} d e \int \frac {x^{4} \arctan \left (c x\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 4 \, b^{2} c^{2} d e \int \frac {x^{4} \log \left (c^{2} x^{2} + 1\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 12 \, b^{2} c^{2} d^{2} \int \frac {x^{2} \arctan \left (c x\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 32 \, a b c^{2} d^{2} \int \frac {x^{2} \arctan \left (c x\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + \frac {1}{96} \, b^{2} d^{2} \log \left (c^{2} x^{2} + 1\right )^{3} + a^{2} d e x^{2} - 2 \, b^{2} c e^{2} \int \frac {x^{5} \arctan \left (c x\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} - 8 \, b^{2} c d e \int \frac {x^{3} \arctan \left (c x\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 12 \, b^{2} e^{2} \int \frac {x^{4} \arctan \left (c x\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + b^{2} e^{2} \int \frac {x^{4} \log \left (c^{2} x^{2} + 1\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 32 \, a b e^{2} \int \frac {x^{4} \arctan \left (c x\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 24 \, b^{2} d e \int \frac {x^{2} \arctan \left (c x\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 64 \, a b d e \int \frac {x^{2} \arctan \left (c x\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 12 \, b^{2} d^{2} \int \frac {\arctan \left (c x\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + b^{2} d^{2} \int \frac {\log \left (c^{2} x^{2} + 1\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 32 \, a b d^{2} \int \frac {\arctan \left (c x\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + \frac {b^{2} d e \log \left (c^{2} x^{2} + 1\right )^{3}}{48 \, c^{2}} + a^{2} d^{2} \log \relax (x) + \frac {1}{16} \, {\left (b^{2} e^{2} x^{4} + 4 \, b^{2} d e x^{2}\right )} \arctan \left (c x\right )^{2} - \frac {1}{64} \, {\left (b^{2} e^{2} x^{4} + 4 \, b^{2} d e x^{2}\right )} \log \left (c^{2} x^{2} + 1\right )^{2} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,{\left (e\,x^2+d\right )}^2}{x} \,d x \]
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 \[ \int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right )^{2} \left (d + e x^{2}\right )^{2}}{x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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