Optimal. Leaf size=275 \[ -\frac{11 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{48 c (c x+1)}-\frac{5 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{48 c (c x+1)^2}-\frac{b^2 \left (a+b \tanh ^{-1}(c x)\right )}{18 c (c x+1)^3}+\frac{11 b \left (a+b \tanh ^{-1}(c x)\right )^2}{96 c}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (c x+1)}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (c x+1)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c (c x+1)^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{24 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{3 c (c x+1)^3}-\frac{85 b^3}{576 c (c x+1)}-\frac{19 b^3}{576 c (c x+1)^2}-\frac{b^3}{108 c (c x+1)^3}+\frac{85 b^3 \tanh ^{-1}(c x)}{576 c} \]
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Rubi [A] time = 0.613865, antiderivative size = 275, normalized size of antiderivative = 1., number of steps used = 42, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5928, 5926, 627, 44, 207, 5948} \[ -\frac{11 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{48 c (c x+1)}-\frac{5 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{48 c (c x+1)^2}-\frac{b^2 \left (a+b \tanh ^{-1}(c x)\right )}{18 c (c x+1)^3}+\frac{11 b \left (a+b \tanh ^{-1}(c x)\right )^2}{96 c}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (c x+1)}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (c x+1)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c (c x+1)^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{24 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{3 c (c x+1)^3}-\frac{85 b^3}{576 c (c x+1)}-\frac{19 b^3}{576 c (c x+1)^2}-\frac{b^3}{108 c (c x+1)^3}+\frac{85 b^3 \tanh ^{-1}(c x)}{576 c} \]
Antiderivative was successfully verified.
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Rule 5928
Rule 5926
Rule 627
Rule 44
Rule 207
Rule 5948
Rubi steps
\begin{align*} \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{(1+c x)^4} \, dx &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{3 c (1+c x)^3}+b \int \left (\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 (1+c x)^4}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{4 (1+c x)^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{8 (1+c x)^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{8 \left (-1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{3 c (1+c x)^3}+\frac{1}{8} b \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^2} \, dx-\frac{1}{8} b \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{-1+c^2 x^2} \, dx+\frac{1}{4} b \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^3} \, dx+\frac{1}{2} b \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^4} \, dx\\ &=-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c (1+c x)^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{24 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{3 c (1+c x)^3}+\frac{1}{4} b^2 \int \left (\frac{a+b \tanh ^{-1}(c x)}{2 (1+c x)^2}-\frac{a+b \tanh ^{-1}(c x)}{2 \left (-1+c^2 x^2\right )}\right ) \, dx+\frac{1}{4} b^2 \int \left (\frac{a+b \tanh ^{-1}(c x)}{2 (1+c x)^3}+\frac{a+b \tanh ^{-1}(c x)}{4 (1+c x)^2}-\frac{a+b \tanh ^{-1}(c x)}{4 \left (-1+c^2 x^2\right )}\right ) \, dx+\frac{1}{3} b^2 \int \left (\frac{a+b \tanh ^{-1}(c x)}{2 (1+c x)^4}+\frac{a+b \tanh ^{-1}(c x)}{4 (1+c x)^3}+\frac{a+b \tanh ^{-1}(c x)}{8 (1+c x)^2}-\frac{a+b \tanh ^{-1}(c x)}{8 \left (-1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c (1+c x)^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{24 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{3 c (1+c x)^3}+\frac{1}{24} b^2 \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx-\frac{1}{24} b^2 \int \frac{a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx+\frac{1}{16} b^2 \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx-\frac{1}{16} b^2 \int \frac{a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx+\frac{1}{12} b^2 \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx+\frac{1}{8} b^2 \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx+\frac{1}{8} b^2 \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx-\frac{1}{8} b^2 \int \frac{a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx+\frac{1}{6} b^2 \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^4} \, dx\\ &=-\frac{b^2 \left (a+b \tanh ^{-1}(c x)\right )}{18 c (1+c x)^3}-\frac{5 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{48 c (1+c x)^2}-\frac{11 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{48 c (1+c x)}+\frac{11 b \left (a+b \tanh ^{-1}(c x)\right )^2}{96 c}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c (1+c x)^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{24 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{3 c (1+c x)^3}+\frac{1}{24} b^3 \int \frac{1}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx+\frac{1}{24} b^3 \int \frac{1}{(1+c x) \left (1-c^2 x^2\right )} \, dx+\frac{1}{18} b^3 \int \frac{1}{(1+c x)^3 \left (1-c^2 x^2\right )} \, dx+\frac{1}{16} b^3 \int \frac{1}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx+\frac{1}{16} b^3 \int \frac{1}{(1+c x) \left (1-c^2 x^2\right )} \, dx+\frac{1}{8} b^3 \int \frac{1}{(1+c x) \left (1-c^2 x^2\right )} \, dx\\ &=-\frac{b^2 \left (a+b \tanh ^{-1}(c x)\right )}{18 c (1+c x)^3}-\frac{5 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{48 c (1+c x)^2}-\frac{11 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{48 c (1+c x)}+\frac{11 b \left (a+b \tanh ^{-1}(c x)\right )^2}{96 c}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c (1+c x)^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{24 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{3 c (1+c x)^3}+\frac{1}{24} b^3 \int \frac{1}{(1-c x) (1+c x)^3} \, dx+\frac{1}{24} b^3 \int \frac{1}{(1-c x) (1+c x)^2} \, dx+\frac{1}{18} b^3 \int \frac{1}{(1-c x) (1+c x)^4} \, dx+\frac{1}{16} b^3 \int \frac{1}{(1-c x) (1+c x)^3} \, dx+\frac{1}{16} b^3 \int \frac{1}{(1-c x) (1+c x)^2} \, dx+\frac{1}{8} b^3 \int \frac{1}{(1-c x) (1+c x)^2} \, dx\\ &=-\frac{b^2 \left (a+b \tanh ^{-1}(c x)\right )}{18 c (1+c x)^3}-\frac{5 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{48 c (1+c x)^2}-\frac{11 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{48 c (1+c x)}+\frac{11 b \left (a+b \tanh ^{-1}(c x)\right )^2}{96 c}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c (1+c x)^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{24 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{3 c (1+c x)^3}+\frac{1}{24} b^3 \int \left (\frac{1}{2 (1+c x)^2}-\frac{1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx+\frac{1}{24} b^3 \int \left (\frac{1}{2 (1+c x)^3}+\frac{1}{4 (1+c x)^2}-\frac{1}{4 \left (-1+c^2 x^2\right )}\right ) \, dx+\frac{1}{18} b^3 \int \left (\frac{1}{2 (1+c x)^4}+\frac{1}{4 (1+c x)^3}+\frac{1}{8 (1+c x)^2}-\frac{1}{8 \left (-1+c^2 x^2\right )}\right ) \, dx+\frac{1}{16} b^3 \int \left (\frac{1}{2 (1+c x)^2}-\frac{1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx+\frac{1}{16} b^3 \int \left (\frac{1}{2 (1+c x)^3}+\frac{1}{4 (1+c x)^2}-\frac{1}{4 \left (-1+c^2 x^2\right )}\right ) \, dx+\frac{1}{8} b^3 \int \left (\frac{1}{2 (1+c x)^2}-\frac{1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac{b^3}{108 c (1+c x)^3}-\frac{19 b^3}{576 c (1+c x)^2}-\frac{85 b^3}{576 c (1+c x)}-\frac{b^2 \left (a+b \tanh ^{-1}(c x)\right )}{18 c (1+c x)^3}-\frac{5 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{48 c (1+c x)^2}-\frac{11 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{48 c (1+c x)}+\frac{11 b \left (a+b \tanh ^{-1}(c x)\right )^2}{96 c}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c (1+c x)^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{24 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{3 c (1+c x)^3}-\frac{1}{144} b^3 \int \frac{1}{-1+c^2 x^2} \, dx-\frac{1}{96} b^3 \int \frac{1}{-1+c^2 x^2} \, dx-\frac{1}{64} b^3 \int \frac{1}{-1+c^2 x^2} \, dx-\frac{1}{48} b^3 \int \frac{1}{-1+c^2 x^2} \, dx-\frac{1}{32} b^3 \int \frac{1}{-1+c^2 x^2} \, dx-\frac{1}{16} b^3 \int \frac{1}{-1+c^2 x^2} \, dx\\ &=-\frac{b^3}{108 c (1+c x)^3}-\frac{19 b^3}{576 c (1+c x)^2}-\frac{85 b^3}{576 c (1+c x)}+\frac{85 b^3 \tanh ^{-1}(c x)}{576 c}-\frac{b^2 \left (a+b \tanh ^{-1}(c x)\right )}{18 c (1+c x)^3}-\frac{5 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{48 c (1+c x)^2}-\frac{11 b^2 \left (a+b \tanh ^{-1}(c x)\right )}{48 c (1+c x)}+\frac{11 b \left (a+b \tanh ^{-1}(c x)\right )^2}{96 c}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c (1+c x)^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{24 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{3 c (1+c x)^3}\\ \end{align*}
Mathematica [A] time = 0.217635, size = 279, normalized size = 1.01 \[ -\frac{24 b \tanh ^{-1}(c x) \left (144 a^2+12 a b \left (3 c^2 x^2+9 c x+10\right )+b^2 \left (33 c^2 x^2+81 c x+56\right )\right )+6 b \left (72 a^2+132 a b+85 b^2\right ) (c x+1)^2+6 b \left (72 a^2+60 a b+19 b^2\right ) (c x+1)+3 b \left (72 a^2+132 a b+85 b^2\right ) (c x+1)^3 \log (1-c x)-3 b \left (72 a^2+132 a b+85 b^2\right ) (c x+1)^3 \log (c x+1)+32 \left (18 a^2 b+36 a^3+6 a b^2+b^3\right )-36 b^2 (c x-1) \tanh ^{-1}(c x)^2 \left (12 a \left (c^2 x^2+4 c x+7\right )+b \left (11 c^2 x^2+32 c x+29\right )\right )-144 b^3 \left (c^3 x^3+3 c^2 x^2+3 c x-7\right ) \tanh ^{-1}(c x)^3}{3456 c (c x+1)^3} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.454, size = 3637, normalized size = 13.2 \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 [B] time = 1.18677, size = 1465, normalized size = 5.33 \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 [A] time = 1.99753, size = 788, normalized size = 2.87 \begin{align*} -\frac{6 \,{\left (72 \, a^{2} b + 132 \, a b^{2} + 85 \, b^{3}\right )} c^{2} x^{2} - 18 \,{\left (b^{3} c^{3} x^{3} + 3 \, b^{3} c^{2} x^{2} + 3 \, b^{3} c x - 7 \, b^{3}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )^{3} + 1152 \, a^{3} + 1440 \, a^{2} b + 1344 \, a b^{2} + 656 \, b^{3} + 162 \,{\left (8 \, a^{2} b + 12 \, a b^{2} + 7 \, b^{3}\right )} c x - 9 \,{\left ({\left (12 \, a b^{2} + 11 \, b^{3}\right )} c^{3} x^{3} + 3 \,{\left (12 \, a b^{2} + 7 \, b^{3}\right )} c^{2} x^{2} - 84 \, a b^{2} - 29 \, b^{3} + 3 \,{\left (12 \, a b^{2} - b^{3}\right )} c x\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )^{2} - 3 \,{\left ({\left (72 \, a^{2} b + 132 \, a b^{2} + 85 \, b^{3}\right )} c^{3} x^{3} + 3 \,{\left (72 \, a^{2} b + 84 \, a b^{2} + 41 \, b^{3}\right )} c^{2} x^{2} - 504 \, a^{2} b - 348 \, a b^{2} - 139 \, b^{3} + 3 \,{\left (72 \, a^{2} b - 12 \, a b^{2} - 23 \, b^{3}\right )} c x\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{3456 \,{\left (c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atanh}{\left (c x \right )}\right )^{3}}{\left (c x + 1\right )^{4}}\, dx \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 (c x\right ) + a\right )}^{3}}{{\left (c x + 1\right )}^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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