Optimal. Leaf size=1233 \[ \text{result too large to display} \]
[Out]
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Rubi [A] time = 2.31078, antiderivative size = 1233, normalized size of antiderivative = 1., number of steps used = 35, number of rules used = 22, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.1, Rules used = {5045, 6741, 5057, 6688, 12, 6725, 706, 31, 635, 203, 260, 4856, 2402, 2315, 2447, 4984, 4884, 4920, 4854, 4858, 4994, 6610} \[ \frac{i d \tan ^{-1}(c+d x)^3 b^3}{d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2}+\frac{d (d e-c f) \tan ^{-1}(c+d x)^3 b^3}{f \left (d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2\right )}-\frac{3 d \tan ^{-1}(c+d x)^2 \log \left (\frac{2}{1-i (c+d x)}\right ) b^3}{d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2}+\frac{3 d \tan ^{-1}(c+d x)^2 \log \left (\frac{2 d (e+f x)}{(d e+(i-c) f) (1-i (c+d x))}\right ) b^3}{d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2}+\frac{3 d \tan ^{-1}(c+d x)^2 \log \left (\frac{2}{i (c+d x)+1}\right ) b^3}{d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2}+\frac{3 i d \tan ^{-1}(c+d x) \text{PolyLog}\left (2,1-\frac{2}{1-i (c+d x)}\right ) b^3}{d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2}-\frac{3 i d \tan ^{-1}(c+d x) \text{PolyLog}\left (2,1-\frac{2 d (e+f x)}{(d e+(i-c) f) (1-i (c+d x))}\right ) b^3}{d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2}+\frac{3 i d \tan ^{-1}(c+d x) \text{PolyLog}\left (2,1-\frac{2}{i (c+d x)+1}\right ) b^3}{d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2}-\frac{3 d \text{PolyLog}\left (3,1-\frac{2}{1-i (c+d x)}\right ) b^3}{2 \left (d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2\right )}+\frac{3 d \text{PolyLog}\left (3,1-\frac{2 d (e+f x)}{(d e+(i-c) f) (1-i (c+d x))}\right ) b^3}{2 \left (d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2\right )}+\frac{3 d \text{PolyLog}\left (3,1-\frac{2}{i (c+d x)+1}\right ) b^3}{2 \left (d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2\right )}+\frac{3 i a d \tan ^{-1}(c+d x)^2 b^2}{d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2}+\frac{3 a d (d e-c f) \tan ^{-1}(c+d x)^2 b^2}{f \left (d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2\right )}-\frac{6 a d \tan ^{-1}(c+d x) \log \left (\frac{2}{1-i (c+d x)}\right ) b^2}{d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2}+\frac{6 a d \tan ^{-1}(c+d x) \log \left (\frac{2 d (e+f x)}{(d e+(i-c) f) (1-i (c+d x))}\right ) b^2}{d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2}+\frac{6 a d \tan ^{-1}(c+d x) \log \left (\frac{2}{i (c+d x)+1}\right ) b^2}{d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2}+\frac{3 i a d \text{PolyLog}\left (2,1-\frac{2}{1-i (c+d x)}\right ) b^2}{d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2}-\frac{3 i a d \text{PolyLog}\left (2,1-\frac{2 d (e+f x)}{(d e+(i-c) f) (1-i (c+d x))}\right ) b^2}{d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2}+\frac{3 i a d \text{PolyLog}\left (2,1-\frac{2}{i (c+d x)+1}\right ) b^2}{d^2 e^2-2 c d f e+\left (c^2+1\right ) f^2}+\frac{3 a^2 d (d e-c f) \tan ^{-1}(c+d x) b}{f \left (f^2+(d e-c f)^2\right )}+\frac{3 a^2 d \log (e+f x) b}{f^2+(d e-c f)^2}-\frac{3 a^2 d \log \left ((c+d x)^2+1\right ) b}{2 \left (f^2+(d e-c f)^2\right )}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{f (e+f x)} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 5045
Rule 6741
Rule 5057
Rule 6688
Rule 12
Rule 6725
Rule 706
Rule 31
Rule 635
Rule 203
Rule 260
Rule 4856
Rule 2402
Rule 2315
Rule 2447
Rule 4984
Rule 4884
Rule 4920
Rule 4854
Rule 4858
Rule 4994
Rule 6610
Rubi steps
\begin{align*} \int \frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{(e+f x)^2} \, dx &=-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{(3 b d) \int \frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{(e+f x) \left (1+(c+d x)^2\right )} \, dx}{f}\\ &=-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{(3 b d) \int \frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{(e+f x) \left (1+c^2+2 c d x+d^2 x^2\right )} \, dx}{f}\\ &=-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right )^2}{\left (\frac{d e-c f}{d}+\frac{f x}{d}\right ) \left (1+x^2\right )} \, dx,x,c+d x\right )}{f}\\ &=-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{d \left (a+b \tan ^{-1}(x)\right )^2}{(d e-c f+f x) \left (1+x^2\right )} \, dx,x,c+d x\right )}{f}\\ &=-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{(3 b d) \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right )^2}{(d e-c f+f x) \left (1+x^2\right )} \, dx,x,c+d x\right )}{f}\\ &=-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{(3 b d) \operatorname{Subst}\left (\int \left (\frac{a^2}{(d e-c f+f x) \left (1+x^2\right )}+\frac{2 a b \tan ^{-1}(x)}{(d e-c f+f x) \left (1+x^2\right )}+\frac{b^2 \tan ^{-1}(x)^2}{(d e-c f+f x) \left (1+x^2\right )}\right ) \, dx,x,c+d x\right )}{f}\\ &=-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{\left (3 a^2 b d\right ) \operatorname{Subst}\left (\int \frac{1}{(d e-c f+f x) \left (1+x^2\right )} \, dx,x,c+d x\right )}{f}+\frac{\left (6 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\tan ^{-1}(x)}{(d e-c f+f x) \left (1+x^2\right )} \, dx,x,c+d x\right )}{f}+\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \frac{\tan ^{-1}(x)^2}{(d e-c f+f x) \left (1+x^2\right )} \, dx,x,c+d x\right )}{f}\\ &=-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{\left (6 a b^2 d\right ) \operatorname{Subst}\left (\int \left (\frac{f^2 \tan ^{-1}(x)}{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) (d e-c f+f x)}+\frac{(d e-c f-f x) \tan ^{-1}(x)}{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) \left (1+x^2\right )}\right ) \, dx,x,c+d x\right )}{f}+\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \left (\frac{f^2 \tan ^{-1}(x)^2}{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) (d e-c f+f x)}+\frac{(d e-c f-f x) \tan ^{-1}(x)^2}{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) \left (1+x^2\right )}\right ) \, dx,x,c+d x\right )}{f}+\frac{\left (3 a^2 b d\right ) \operatorname{Subst}\left (\int \frac{d e-c f-f x}{1+x^2} \, dx,x,c+d x\right )}{f \left (f^2+(d e-c f)^2\right )}+\frac{\left (3 a^2 b d f\right ) \operatorname{Subst}\left (\int \frac{1}{d e-c f+f x} \, dx,x,c+d x\right )}{f^2+(d e-c f)^2}\\ &=-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{3 a^2 b d \log (e+f x)}{f^2+(d e-c f)^2}+\frac{\left (6 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{(d e-c f-f x) \tan ^{-1}(x)}{1+x^2} \, dx,x,c+d x\right )}{f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}+\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \frac{(d e-c f-f x) \tan ^{-1}(x)^2}{1+x^2} \, dx,x,c+d x\right )}{f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}+\frac{\left (6 a b^2 d f\right ) \operatorname{Subst}\left (\int \frac{\tan ^{-1}(x)}{d e-c f+f x} \, dx,x,c+d x\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{\left (3 b^3 d f\right ) \operatorname{Subst}\left (\int \frac{\tan ^{-1}(x)^2}{d e-c f+f x} \, dx,x,c+d x\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{\left (3 a^2 b d\right ) \operatorname{Subst}\left (\int \frac{x}{1+x^2} \, dx,x,c+d x\right )}{f^2+(d e-c f)^2}+\frac{\left (3 a^2 b d (d e-c f)\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,c+d x\right )}{f \left (f^2+(d e-c f)^2\right )}\\ &=\frac{3 a^2 b d (d e-c f) \tan ^{-1}(c+d x)}{f \left (f^2+(d e-c f)^2\right )}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{3 a^2 b d \log (e+f x)}{f^2+(d e-c f)^2}-\frac{6 a b^2 d \tan ^{-1}(c+d x) \log \left (\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 b^3 d \tan ^{-1}(c+d x)^2 \log \left (\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{6 a b^2 d \tan ^{-1}(c+d x) \log \left (\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{3 b^3 d \tan ^{-1}(c+d x)^2 \log \left (\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 a^2 b d \log \left (1+(c+d x)^2\right )}{2 \left (f^2+(d e-c f)^2\right )}+\frac{3 i b^3 d \tan ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 i b^3 d \tan ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 b^3 d \text{Li}_3\left (1-\frac{2}{1-i (c+d x)}\right )}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}+\frac{3 b^3 d \text{Li}_3\left (1-\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}+\frac{\left (6 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{2}{1-i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{\left (6 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{2 (d e-c f+f x)}{(d e+i f-c f) (1-i x)}\right )}{1+x^2} \, dx,x,c+d x\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{\left (6 a b^2 d\right ) \operatorname{Subst}\left (\int \left (\frac{d e \left (1-\frac{c f}{d e}\right ) \tan ^{-1}(x)}{1+x^2}-\frac{f x \tan ^{-1}(x)}{1+x^2}\right ) \, dx,x,c+d x\right )}{f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}+\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \left (\frac{d e \left (1-\frac{c f}{d e}\right ) \tan ^{-1}(x)^2}{1+x^2}-\frac{f x \tan ^{-1}(x)^2}{1+x^2}\right ) \, dx,x,c+d x\right )}{f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}\\ &=\frac{3 a^2 b d (d e-c f) \tan ^{-1}(c+d x)}{f \left (f^2+(d e-c f)^2\right )}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{3 a^2 b d \log (e+f x)}{f^2+(d e-c f)^2}-\frac{6 a b^2 d \tan ^{-1}(c+d x) \log \left (\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 b^3 d \tan ^{-1}(c+d x)^2 \log \left (\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{6 a b^2 d \tan ^{-1}(c+d x) \log \left (\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{3 b^3 d \tan ^{-1}(c+d x)^2 \log \left (\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 a^2 b d \log \left (1+(c+d x)^2\right )}{2 \left (f^2+(d e-c f)^2\right )}+\frac{3 i b^3 d \tan ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 i a b^2 d \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 i b^3 d \tan ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 b^3 d \text{Li}_3\left (1-\frac{2}{1-i (c+d x)}\right )}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}+\frac{3 b^3 d \text{Li}_3\left (1-\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}+\frac{\left (6 i a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{\left (6 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{x \tan ^{-1}(x)}{1+x^2} \, dx,x,c+d x\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \frac{x \tan ^{-1}(x)^2}{1+x^2} \, dx,x,c+d x\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{\left (6 a b^2 d (d e-c f)\right ) \operatorname{Subst}\left (\int \frac{\tan ^{-1}(x)}{1+x^2} \, dx,x,c+d x\right )}{f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}+\frac{\left (3 b^3 d (d e-c f)\right ) \operatorname{Subst}\left (\int \frac{\tan ^{-1}(x)^2}{1+x^2} \, dx,x,c+d x\right )}{f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}\\ &=\frac{3 a^2 b d (d e-c f) \tan ^{-1}(c+d x)}{f \left (f^2+(d e-c f)^2\right )}+\frac{3 i a b^2 d \tan ^{-1}(c+d x)^2}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{3 a b^2 d (d e-c f) \tan ^{-1}(c+d x)^2}{f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}+\frac{i b^3 d \tan ^{-1}(c+d x)^3}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{b^3 d (d e-c f) \tan ^{-1}(c+d x)^3}{f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{3 a^2 b d \log (e+f x)}{f^2+(d e-c f)^2}-\frac{6 a b^2 d \tan ^{-1}(c+d x) \log \left (\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 b^3 d \tan ^{-1}(c+d x)^2 \log \left (\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{6 a b^2 d \tan ^{-1}(c+d x) \log \left (\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{3 b^3 d \tan ^{-1}(c+d x)^2 \log \left (\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 a^2 b d \log \left (1+(c+d x)^2\right )}{2 \left (f^2+(d e-c f)^2\right )}+\frac{3 i a b^2 d \text{Li}_2\left (1-\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{3 i b^3 d \tan ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 i a b^2 d \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 i b^3 d \tan ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 b^3 d \text{Li}_3\left (1-\frac{2}{1-i (c+d x)}\right )}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}+\frac{3 b^3 d \text{Li}_3\left (1-\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}+\frac{\left (6 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\tan ^{-1}(x)}{i-x} \, dx,x,c+d x\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \frac{\tan ^{-1}(x)^2}{i-x} \, dx,x,c+d x\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}\\ &=\frac{3 a^2 b d (d e-c f) \tan ^{-1}(c+d x)}{f \left (f^2+(d e-c f)^2\right )}+\frac{3 i a b^2 d \tan ^{-1}(c+d x)^2}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{3 a b^2 d (d e-c f) \tan ^{-1}(c+d x)^2}{f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}+\frac{i b^3 d \tan ^{-1}(c+d x)^3}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{b^3 d (d e-c f) \tan ^{-1}(c+d x)^3}{f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{3 a^2 b d \log (e+f x)}{f^2+(d e-c f)^2}-\frac{6 a b^2 d \tan ^{-1}(c+d x) \log \left (\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 b^3 d \tan ^{-1}(c+d x)^2 \log \left (\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{6 a b^2 d \tan ^{-1}(c+d x) \log \left (\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{3 b^3 d \tan ^{-1}(c+d x)^2 \log \left (\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{6 a b^2 d \tan ^{-1}(c+d x) \log \left (\frac{2}{1+i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{3 b^3 d \tan ^{-1}(c+d x)^2 \log \left (\frac{2}{1+i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 a^2 b d \log \left (1+(c+d x)^2\right )}{2 \left (f^2+(d e-c f)^2\right )}+\frac{3 i a b^2 d \text{Li}_2\left (1-\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{3 i b^3 d \tan ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 i a b^2 d \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 i b^3 d \tan ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 b^3 d \text{Li}_3\left (1-\frac{2}{1-i (c+d x)}\right )}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}+\frac{3 b^3 d \text{Li}_3\left (1-\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}-\frac{\left (6 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{\left (6 b^3 d\right ) \operatorname{Subst}\left (\int \frac{\tan ^{-1}(x) \log \left (\frac{2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}\\ &=\frac{3 a^2 b d (d e-c f) \tan ^{-1}(c+d x)}{f \left (f^2+(d e-c f)^2\right )}+\frac{3 i a b^2 d \tan ^{-1}(c+d x)^2}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{3 a b^2 d (d e-c f) \tan ^{-1}(c+d x)^2}{f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}+\frac{i b^3 d \tan ^{-1}(c+d x)^3}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{b^3 d (d e-c f) \tan ^{-1}(c+d x)^3}{f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{3 a^2 b d \log (e+f x)}{f^2+(d e-c f)^2}-\frac{6 a b^2 d \tan ^{-1}(c+d x) \log \left (\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 b^3 d \tan ^{-1}(c+d x)^2 \log \left (\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{6 a b^2 d \tan ^{-1}(c+d x) \log \left (\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{3 b^3 d \tan ^{-1}(c+d x)^2 \log \left (\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{6 a b^2 d \tan ^{-1}(c+d x) \log \left (\frac{2}{1+i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{3 b^3 d \tan ^{-1}(c+d x)^2 \log \left (\frac{2}{1+i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 a^2 b d \log \left (1+(c+d x)^2\right )}{2 \left (f^2+(d e-c f)^2\right )}+\frac{3 i a b^2 d \text{Li}_2\left (1-\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{3 i b^3 d \tan ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 i a b^2 d \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 i b^3 d \tan ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{3 i b^3 d \tan ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1+i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 b^3 d \text{Li}_3\left (1-\frac{2}{1-i (c+d x)}\right )}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}+\frac{3 b^3 d \text{Li}_3\left (1-\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}+\frac{\left (6 i a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{\left (3 i b^3 d\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (1-\frac{2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}\\ &=\frac{3 a^2 b d (d e-c f) \tan ^{-1}(c+d x)}{f \left (f^2+(d e-c f)^2\right )}+\frac{3 i a b^2 d \tan ^{-1}(c+d x)^2}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{3 a b^2 d (d e-c f) \tan ^{-1}(c+d x)^2}{f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}+\frac{i b^3 d \tan ^{-1}(c+d x)^3}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{b^3 d (d e-c f) \tan ^{-1}(c+d x)^3}{f \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{3 a^2 b d \log (e+f x)}{f^2+(d e-c f)^2}-\frac{6 a b^2 d \tan ^{-1}(c+d x) \log \left (\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 b^3 d \tan ^{-1}(c+d x)^2 \log \left (\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{6 a b^2 d \tan ^{-1}(c+d x) \log \left (\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{3 b^3 d \tan ^{-1}(c+d x)^2 \log \left (\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{6 a b^2 d \tan ^{-1}(c+d x) \log \left (\frac{2}{1+i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{3 b^3 d \tan ^{-1}(c+d x)^2 \log \left (\frac{2}{1+i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 a^2 b d \log \left (1+(c+d x)^2\right )}{2 \left (f^2+(d e-c f)^2\right )}+\frac{3 i a b^2 d \text{Li}_2\left (1-\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{3 i b^3 d \tan ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1-i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 i a b^2 d \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 i b^3 d \tan ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{3 i a b^2 d \text{Li}_2\left (1-\frac{2}{1+i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}+\frac{3 i b^3 d \tan ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1+i (c+d x)}\right )}{d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2}-\frac{3 b^3 d \text{Li}_3\left (1-\frac{2}{1-i (c+d x)}\right )}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}+\frac{3 b^3 d \text{Li}_3\left (1-\frac{2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}+\frac{3 b^3 d \text{Li}_3\left (1-\frac{2}{1+i (c+d x)}\right )}{2 \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )}\\ \end{align*}
Mathematica [F] time = 70.2675, size = 0, normalized size = 0. \[ \int \frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{(e+f x)^2} \, dx \]
Verification is Not applicable to the result.
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Maple [C] time = 1.013, size = 4764, normalized size = 3.9 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{3}{2} \,{\left (d{\left (\frac{2 \,{\left (d^{2} e - c d f\right )} \arctan \left (\frac{d^{2} x + c d}{d}\right )}{{\left (d^{2} e^{2} f - 2 \, c d e f^{2} +{\left (c^{2} + 1\right )} f^{3}\right )} d} - \frac{\log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{2} e^{2} - 2 \, c d e f +{\left (c^{2} + 1\right )} f^{2}} + \frac{2 \, \log \left (f x + e\right )}{d^{2} e^{2} - 2 \, c d e f +{\left (c^{2} + 1\right )} f^{2}}\right )} - \frac{2 \, \arctan \left (d x + c\right )}{f^{2} x + e f}\right )} a^{2} b - \frac{a^{3}}{f^{2} x + e f} - \frac{\frac{15}{2} \, b^{3} \arctan \left (d x + c\right )^{3} - \frac{21}{8} \, b^{3} \arctan \left (d x + c\right ) \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )^{2} -{\left (f^{2} x + e f\right )} \int \frac{196 \,{\left (b^{3} d^{2} f x^{2} + 2 \, b^{3} c d f x +{\left (b^{3} c^{2} + b^{3}\right )} f\right )} \arctan \left (d x + c\right )^{3} + 12 \,{\left (64 \, a b^{2} d^{2} f x^{2} + 15 \, b^{3} d e +{\left (128 \, a b^{2} c + 15 \, b^{3}\right )} d f x + 64 \,{\left (a b^{2} c^{2} + a b^{2}\right )} f\right )} \arctan \left (d x + c\right )^{2} - 84 \,{\left (b^{3} d^{2} f x^{2} + b^{3} c d e +{\left (b^{3} d^{2} e + b^{3} c d f\right )} x\right )} \arctan \left (d x + c\right ) \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) - 21 \,{\left (b^{3} d f x + b^{3} d e -{\left (b^{3} d^{2} f x^{2} + 2 \, b^{3} c d f x +{\left (b^{3} c^{2} + b^{3}\right )} f\right )} \arctan \left (d x + c\right )\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )^{2}}{8 \,{\left (d^{2} f^{3} x^{4} +{\left (c^{2} + 1\right )} e^{2} f + 2 \,{\left (d^{2} e f^{2} + c d f^{3}\right )} x^{3} +{\left (d^{2} e^{2} f + 4 \, c d e f^{2} +{\left (c^{2} + 1\right )} f^{3}\right )} x^{2} + 2 \,{\left (c d e^{2} f +{\left (c^{2} + 1\right )} e f^{2}\right )} x\right )}}\,{d x}}{32 \,{\left (f^{2} x + e f\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{3} \arctan \left (d x + c\right )^{3} + 3 \, a b^{2} \arctan \left (d x + c\right )^{2} + 3 \, a^{2} b \arctan \left (d x + c\right ) + a^{3}}{f^{2} x^{2} + 2 \, e f x + e^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (d x + c\right ) + a\right )}^{3}}{{\left (f x + e\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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