Optimal. Leaf size=1089 \[ \text{result too large to display} \]
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Rubi [A] time = 2.82718, antiderivative size = 1094, normalized size of antiderivative = 1., number of steps used = 30, number of rules used = 18, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.9, Rules used = {6109, 6741, 6121, 6688, 12, 6725, 72, 6742, 5918, 2402, 2315, 5920, 2447, 5948, 6058, 6610, 6056, 5922} \[ \frac{3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{-c-d x+1}\right ) b^3}{2 f (d e-c f+f)}-\frac{3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{c+d x+1}\right ) b^3}{2 f (d e-c f-f)}+\frac{3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{c+d x+1}\right ) b^3}{(d e-c f+f) (d e-(c+1) f)}-\frac{3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2 d (e+f x)}{(d e-c f+f) (c+d x+1)}\right ) b^3}{(d e-c f+f) (d e-(c+1) f)}+\frac{3 d \tanh ^{-1}(c+d x) \text{PolyLog}\left (2,1-\frac{2}{-c-d x+1}\right ) b^3}{2 f (d e-c f+f)}+\frac{3 d \tanh ^{-1}(c+d x) \text{PolyLog}\left (2,1-\frac{2}{c+d x+1}\right ) b^3}{2 f (d e-c f-f)}-\frac{3 d \tanh ^{-1}(c+d x) \text{PolyLog}\left (2,1-\frac{2}{c+d x+1}\right ) b^3}{(d e-c f+f) (d e-(c+1) f)}+\frac{3 d \tanh ^{-1}(c+d x) \text{PolyLog}\left (2,1-\frac{2 d (e+f x)}{(d e-c f+f) (c+d x+1)}\right ) b^3}{(d e-c f+f) (d e-(c+1) f)}-\frac{3 d \text{PolyLog}\left (3,1-\frac{2}{-c-d x+1}\right ) b^3}{4 f (d e-c f+f)}+\frac{3 d \text{PolyLog}\left (3,1-\frac{2}{c+d x+1}\right ) b^3}{4 f (d e-c f-f)}-\frac{3 d \text{PolyLog}\left (3,1-\frac{2}{c+d x+1}\right ) b^3}{2 (d e-c f+f) (d e-(c+1) f)}+\frac{3 d \text{PolyLog}\left (3,1-\frac{2 d (e+f x)}{(d e-c f+f) (c+d x+1)}\right ) b^3}{2 (d e-c f+f) (d e-(c+1) f)}+\frac{3 a d \tanh ^{-1}(c+d x) \log \left (\frac{2}{-c-d x+1}\right ) b^2}{f (d e-c f+f)}-\frac{3 a d \tanh ^{-1}(c+d x) \log \left (\frac{2}{c+d x+1}\right ) b^2}{f (d e-c f-f)}+\frac{6 a d \tanh ^{-1}(c+d x) \log \left (\frac{2}{c+d x+1}\right ) b^2}{(d e-c f+f) (d e-(c+1) f)}-\frac{6 a d \tanh ^{-1}(c+d x) \log \left (\frac{2 d (e+f x)}{(d e-c f+f) (c+d x+1)}\right ) b^2}{(d e-c f+f) (d e-(c+1) f)}+\frac{3 a d \text{PolyLog}\left (2,-\frac{c+d x+1}{-c-d x+1}\right ) b^2}{2 f (d e-c f+f)}+\frac{3 a d \text{PolyLog}\left (2,1-\frac{2}{c+d x+1}\right ) b^2}{2 f (d e-c f-f)}-\frac{3 a d \text{PolyLog}\left (2,1-\frac{2}{c+d x+1}\right ) b^2}{(d e-c f+f) (d e-(c+1) f)}+\frac{3 a d \text{PolyLog}\left (2,1-\frac{2 d (e+f x)}{(d e-c f+f) (c+d x+1)}\right ) b^2}{(d e-c f+f) (d e-(c+1) f)}-\frac{3 a^2 d \log (-c-d x+1) b}{2 f (d e-c f+f)}+\frac{3 a^2 d \log (c+d x+1) b}{2 f (d e-c f-f)}-\frac{3 a^2 d \log (e+f x) b}{(d e-c f+f) (d e-(c+1) f)}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)} \]
Antiderivative was successfully verified.
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Rule 6109
Rule 6741
Rule 6121
Rule 6688
Rule 12
Rule 6725
Rule 72
Rule 6742
Rule 5918
Rule 2402
Rule 2315
Rule 5920
Rule 2447
Rule 5948
Rule 6058
Rule 6610
Rule 6056
Rule 5922
Rubi steps
\begin{align*} \int \frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{(e+f x)^2} \, dx &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{(3 b d) \int \frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{(e+f x) \left (1-(c+d x)^2\right )} \, dx}{f}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{(3 b d) \int \frac{\left (a+b \tanh ^{-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 \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-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 \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{d \left (a+b \tanh ^{-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 \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{(3 b d) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-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 \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{(3 b d) \operatorname{Subst}\left (\int \left (-\frac{a^2}{(-1+x) (1+x) (d e-c f+f x)}-\frac{2 a b \tanh ^{-1}(x)}{(-1+x) (1+x) (d e-c f+f x)}-\frac{b^2 \tanh ^{-1}(x)^2}{(-1+x) (1+x) (d e-c f+f x)}\right ) \, dx,x,c+d x\right )}{f}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}-\frac{\left (3 a^2 b d\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) (1+x) (d e-c f+f x)} \, dx,x,c+d x\right )}{f}-\frac{\left (6 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x)}{(-1+x) (1+x) (d e-c f+f x)} \, dx,x,c+d x\right )}{f}-\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x)^2}{(-1+x) (1+x) (d e-c f+f x)} \, dx,x,c+d x\right )}{f}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}-\frac{\left (3 a^2 b d\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2 (d e+f-c f) (-1+x)}+\frac{1}{2 (-d e+(1+c) f) (1+x)}+\frac{f^2}{(d e+(1-c) f) (d e-f-c f) (d e-c f+f x)}\right ) \, dx,x,c+d x\right )}{f}-\frac{\left (6 a b^2 d\right ) \operatorname{Subst}\left (\int \left (\frac{\tanh ^{-1}(x)}{2 (d e+f-c f) (-1+x)}+\frac{\tanh ^{-1}(x)}{2 (-d e+(1+c) f) (1+x)}+\frac{f^2 \tanh ^{-1}(x)}{(d e+(1-c) f) (d e-f-c f) (d e-c f+f x)}\right ) \, dx,x,c+d x\right )}{f}-\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \left (\frac{\tanh ^{-1}(x)^2}{2 (d e+f-c f) (-1+x)}+\frac{\tanh ^{-1}(x)^2}{2 (-d e+(1+c) f) (1+x)}+\frac{f^2 \tanh ^{-1}(x)^2}{(d e+(1-c) f) (d e-f-c f) (d e-c f+f x)}\right ) \, dx,x,c+d x\right )}{f}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}-\frac{3 a^2 b d \log (1-c-d x)}{2 f (d e+f-c f)}+\frac{3 a^2 b d \log (1+c+d x)}{2 f (d e-f-c f)}-\frac{3 a^2 b d \log (e+f x)}{(d e+f-c f) (d e-(1+c) f)}+\frac{\left (3 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x)}{1+x} \, dx,x,c+d x\right )}{f (d e-f-c f)}+\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x)^2}{1+x} \, dx,x,c+d x\right )}{2 f (d e-f-c f)}-\frac{\left (3 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x)}{-1+x} \, dx,x,c+d x\right )}{f (d e+f-c f)}-\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x)^2}{-1+x} \, dx,x,c+d x\right )}{2 f (d e+f-c f)}-\frac{\left (6 a b^2 d f\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x)}{d e-c f+f x} \, dx,x,c+d x\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{\left (3 b^3 d f\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x)^2}{d e-c f+f x} \, dx,x,c+d x\right )}{(d e+f-c f) (d e-(1+c) f)}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{3 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2}{1-c-d x}\right )}{f (d e+f-c f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{1-c-d x}\right )}{2 f (d e+f-c f)}-\frac{3 a^2 b d \log (1-c-d x)}{2 f (d e+f-c f)}-\frac{3 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2}{1+c+d x}\right )}{f (d e-f-c f)}+\frac{6 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{1+c+d x}\right )}{2 f (d e-f-c f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 a^2 b d \log (1+c+d x)}{2 f (d e-f-c f)}-\frac{3 a^2 b d \log (e+f x)}{(d e+f-c f) (d e-(1+c) f)}-\frac{6 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{3 b^3 d \tanh ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{3 b^3 d \text{Li}_3\left (1-\frac{2}{1+c+d x}\right )}{2 (d e+f-c f) (d e-(1+c) f)}+\frac{3 b^3 d \text{Li}_3\left (1-\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{2 (d e+f-c f) (d e-(1+c) f)}+\frac{\left (3 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{f (d e-f-c f)}+\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x) \log \left (\frac{2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{f (d e-f-c f)}-\frac{\left (3 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{f (d e+f-c f)}-\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}(x) \log \left (\frac{2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{f (d e+f-c f)}-\frac{\left (6 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{(d e+f-c f) (d e-(1+c) f)}+\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+f-c f) (1+x)}\right )}{1-x^2} \, dx,x,c+d x\right )}{(d e+f-c f) (d e-(1+c) f)}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{3 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2}{1-c-d x}\right )}{f (d e+f-c f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{1-c-d x}\right )}{2 f (d e+f-c f)}-\frac{3 a^2 b d \log (1-c-d x)}{2 f (d e+f-c f)}-\frac{3 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2}{1+c+d x}\right )}{f (d e-f-c f)}+\frac{6 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{1+c+d x}\right )}{2 f (d e-f-c f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 a^2 b d \log (1+c+d x)}{2 f (d e-f-c f)}-\frac{3 a^2 b d \log (e+f x)}{(d e+f-c f) (d e-(1+c) f)}-\frac{6 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1-c-d x}\right )}{2 f (d e+f-c f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1+c+d x}\right )}{2 f (d e-f-c f)}-\frac{3 b^3 d \tanh ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 a b^2 d \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{3 b^3 d \text{Li}_3\left (1-\frac{2}{1+c+d x}\right )}{2 (d e+f-c f) (d e-(1+c) f)}+\frac{3 b^3 d \text{Li}_3\left (1-\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{2 (d e+f-c f) (d e-(1+c) f)}+\frac{\left (3 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+c+d x}\right )}{f (d e-f-c f)}-\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (1-\frac{2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{2 f (d e-f-c f)}+\frac{\left (3 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-c-d x}\right )}{f (d e+f-c f)}-\frac{\left (3 b^3 d\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (1-\frac{2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{2 f (d e+f-c f)}-\frac{\left (6 a b^2 d\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{f (e+f x)}+\frac{3 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2}{1-c-d x}\right )}{f (d e+f-c f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{1-c-d x}\right )}{2 f (d e+f-c f)}-\frac{3 a^2 b d \log (1-c-d x)}{2 f (d e+f-c f)}-\frac{3 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2}{1+c+d x}\right )}{f (d e-f-c f)}+\frac{6 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{1+c+d x}\right )}{2 f (d e-f-c f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 a^2 b d \log (1+c+d x)}{2 f (d e-f-c f)}-\frac{3 a^2 b d \log (e+f x)}{(d e+f-c f) (d e-(1+c) f)}-\frac{6 a b^2 d \tanh ^{-1}(c+d x) \log \left (\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{3 b^3 d \tanh ^{-1}(c+d x)^2 \log \left (\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 a b^2 d \text{Li}_2\left (1-\frac{2}{1-c-d x}\right )}{2 f (d e+f-c f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1-c-d x}\right )}{2 f (d e+f-c f)}+\frac{3 a b^2 d \text{Li}_2\left (1-\frac{2}{1+c+d x}\right )}{2 f (d e-f-c f)}-\frac{3 a b^2 d \text{Li}_2\left (1-\frac{2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1+c+d x}\right )}{2 f (d e-f-c f)}-\frac{3 b^3 d \tanh ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2}{1+c+d x}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 a b^2 d \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}+\frac{3 b^3 d \tanh ^{-1}(c+d x) \text{Li}_2\left (1-\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{(d e+f-c f) (d e-(1+c) f)}-\frac{3 b^3 d \text{Li}_3\left (1-\frac{2}{1-c-d x}\right )}{4 f (d e+f-c f)}+\frac{3 b^3 d \text{Li}_3\left (1-\frac{2}{1+c+d x}\right )}{4 f (d e-f-c f)}-\frac{3 b^3 d \text{Li}_3\left (1-\frac{2}{1+c+d x}\right )}{2 (d e+f-c f) (d e-(1+c) f)}+\frac{3 b^3 d \text{Li}_3\left (1-\frac{2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{2 (d e+f-c f) (d e-(1+c) f)}\\ \end{align*}
Mathematica [C] time = 23.1682, size = 2683, normalized size = 2.46 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 0.84, size = 5728, normalized size = 5.3 \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*} \text{result too large to display} \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} \operatorname{artanh}\left (d x + c\right )^{3} + 3 \, a b^{2} \operatorname{artanh}\left (d x + c\right )^{2} + 3 \, a^{2} b \operatorname{artanh}\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 \operatorname{artanh}\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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