Optimal. Leaf size=591 \[ -\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 d}-\frac {i b e^2 \text {Li}_2\left (1-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}-\frac {i b e^2 \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}+\frac {i b e^2 \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}+\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d^3}+\frac {e^2 \log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^3}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{d^3}+\frac {2 e^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^3}+\frac {i c e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}+\frac {e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}-\frac {2 b c e \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac {b^2 c^2 \log \left (c^2 x^2+1\right )}{2 d}+\frac {b^2 c^2 \log (x)}{d}+\frac {b^2 e^2 \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 d^3}-\frac {b^2 e^2 \text {Li}_3\left (1-\frac {2}{i c x+1}\right )}{2 d^3}+\frac {b^2 e^2 \text {Li}_3\left (\frac {2}{i c x+1}-1\right )}{2 d^3}-\frac {b^2 e^2 \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 d^3}+\frac {i b^2 c e \text {Li}_2\left (\frac {2}{1-i c x}-1\right )}{d^2} \]
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Rubi [A] time = 0.84, antiderivative size = 591, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 16, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.762, Rules used = {4876, 4852, 4918, 266, 36, 29, 31, 4884, 4924, 4868, 2447, 4850, 4988, 4994, 6610, 4858} \[ -\frac {i b e^2 \text {PolyLog}\left (2,1-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}-\frac {i b e^2 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}+\frac {i b e^2 \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}+\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{d^3}+\frac {b^2 e^2 \text {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 d^3}-\frac {b^2 e^2 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 d^3}+\frac {b^2 e^2 \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d^3}-\frac {b^2 e^2 \text {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 d^3}+\frac {i b^2 c e \text {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )}{d^2}-\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 d}+\frac {e^2 \log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^3}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{d^3}+\frac {2 e^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^3}+\frac {i c e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}+\frac {e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}-\frac {2 b c e \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac {b^2 c^2 \log \left (c^2 x^2+1\right )}{2 d}+\frac {b^2 c^2 \log (x)}{d} \]
Antiderivative was successfully verified.
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Rule 29
Rule 31
Rule 36
Rule 266
Rule 2447
Rule 4850
Rule 4852
Rule 4858
Rule 4868
Rule 4876
Rule 4884
Rule 4918
Rule 4924
Rule 4988
Rule 4994
Rule 6610
Rubi steps
\begin {align*} \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^3 (d+e x)} \, dx &=\int \left (\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{d x^3}-\frac {e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x^2}+\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 x}-\frac {e^3 \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (d+e x)}\right ) \, dx\\ &=\frac {\int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^3} \, dx}{d}-\frac {e \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^2} \, dx}{d^2}+\frac {e^2 \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx}{d^3}-\frac {e^3 \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{d+e x} \, dx}{d^3}\\ &=-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}+\frac {e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}+\frac {2 e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^3}+\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{d^3}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d^3}-\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{d^3}+\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d^3}+\frac {b^2 e^2 \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 d^3}-\frac {b^2 e^2 \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 d^3}+\frac {(b c) \int \frac {a+b \tan ^{-1}(c x)}{x^2 \left (1+c^2 x^2\right )} \, dx}{d}-\frac {(2 b c e) \int \frac {a+b \tan ^{-1}(c x)}{x \left (1+c^2 x^2\right )} \, dx}{d^2}-\frac {\left (4 b c e^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}{d^3}\\ &=\frac {i c e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}+\frac {e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}+\frac {2 e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^3}+\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{d^3}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d^3}-\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{d^3}+\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d^3}+\frac {b^2 e^2 \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 d^3}-\frac {b^2 e^2 \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 d^3}+\frac {(b c) \int \frac {a+b \tan ^{-1}(c x)}{x^2} \, dx}{d}-\frac {\left (b c^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{d}-\frac {(2 i b c e) \int \frac {a+b \tan ^{-1}(c x)}{x (i+c x)} \, dx}{d^2}+\frac {\left (2 b c e^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}{d^3}-\frac {\left (2 b c e^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}{d^3}\\ &=-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 d}+\frac {i c e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}+\frac {e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}+\frac {2 e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^3}+\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{d^3}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d^3}-\frac {2 b c e \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )}{d^2}-\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{d^3}-\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{d^3}+\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d^3}+\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d^3}+\frac {b^2 e^2 \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 d^3}-\frac {b^2 e^2 \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 d^3}+\frac {\left (b^2 c^2\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx}{d}+\frac {\left (2 b^2 c^2 e\right ) \int \frac {\log \left (2-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{d^2}+\frac {\left (i b^2 c e^2\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d^3}-\frac {\left (i b^2 c e^2\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d^3}\\ &=-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 d}+\frac {i c e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}+\frac {e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}+\frac {2 e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^3}+\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{d^3}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d^3}-\frac {2 b c e \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )}{d^2}-\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{d^3}+\frac {i b^2 c e \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )}{d^2}-\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{d^3}+\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d^3}+\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d^3}+\frac {b^2 e^2 \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 d^3}-\frac {b^2 e^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 d^3}+\frac {b^2 e^2 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d^3}-\frac {b^2 e^2 \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 d^3}+\frac {\left (b^2 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )}{2 d}\\ &=-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 d}+\frac {i c e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}+\frac {e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}+\frac {2 e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^3}+\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{d^3}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d^3}-\frac {2 b c e \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )}{d^2}-\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{d^3}+\frac {i b^2 c e \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )}{d^2}-\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{d^3}+\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d^3}+\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d^3}+\frac {b^2 e^2 \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 d^3}-\frac {b^2 e^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 d^3}+\frac {b^2 e^2 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d^3}-\frac {b^2 e^2 \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 d^3}+\frac {\left (b^2 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 d}-\frac {\left (b^2 c^4\right ) \operatorname {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )}{2 d}\\ &=-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 d}+\frac {i c e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}+\frac {e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}+\frac {2 e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^3}+\frac {b^2 c^2 \log (x)}{d}+\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{d^3}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d^3}-\frac {b^2 c^2 \log \left (1+c^2 x^2\right )}{2 d}-\frac {2 b c e \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )}{d^2}-\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{d^3}+\frac {i b^2 c e \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )}{d^2}-\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{d^3}+\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d^3}+\frac {i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d^3}+\frac {b^2 e^2 \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 d^3}-\frac {b^2 e^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 d^3}+\frac {b^2 e^2 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d^3}-\frac {b^2 e^2 \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 d^3}\\ \end {align*}
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Mathematica [A] time = 21.88, size = 1173, normalized size = 1.98 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \arctan \left (c x\right )^{2} + 2 \, a b \arctan \left (c x\right ) + a^{2}}{e x^{4} + d x^{3}}, x\right ) \]
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 = 65.50, size = 2861, normalized size = 4.84 \[ \text {output too large to display} \]
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 \[ -\frac {1}{2} \, a^{2} {\left (\frac {2 \, e^{2} \log \left (e x + d\right )}{d^{3}} - \frac {2 \, e^{2} \log \relax (x)}{d^{3}} - \frac {2 \, e x - d}{d^{2} x^{2}}\right )} + \frac {2 \, d^{2} x^{2} \int \frac {12 \, {\left (b^{2} c^{2} d^{2} x^{2} + b^{2} d^{2}\right )} \arctan \left (c x\right )^{2} + {\left (b^{2} c^{2} d^{2} x^{2} + b^{2} d^{2}\right )} \log \left (c^{2} x^{2} + 1\right )^{2} - 4 \, {\left (2 \, b^{2} c e^{2} x^{3} - b^{2} c d^{2} x - 8 \, a b d^{2} - {\left (8 \, a b c^{2} d^{2} - b^{2} c d e\right )} x^{2}\right )} \arctan \left (c x\right ) + 2 \, {\left (2 \, b^{2} c^{2} e^{2} x^{4} + b^{2} c^{2} d e x^{3} - b^{2} c^{2} d^{2} x^{2}\right )} \log \left (c^{2} x^{2} + 1\right )}{c^{2} d^{2} e x^{6} + c^{2} d^{3} x^{5} + d^{2} e x^{4} + d^{3} x^{3}}\,{d x} + 4 \, {\left (2 \, b^{2} e x - b^{2} d\right )} \arctan \left (c x\right )^{2} - {\left (2 \, b^{2} e x - b^{2} d\right )} \log \left (c^{2} x^{2} + 1\right )^{2}}{32 \, d^{2} x^{2}} \]
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 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{x^3\,\left (d+e\,x\right )} \,d x \]
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 \[ \int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right )^{2}}{x^{3} \left (d + e x\right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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