Optimal. Leaf size=263 \[ \frac{a \left (a^2+1\right ) \text{PolyLog}\left (2,-\frac{a+b x+1}{-a-b x+1}\right )}{b^4}+\frac{\left (6 a^2+1\right ) \log \left (1-(a+b x)^2\right )}{4 b^4}-\frac{a \left (a^2+1\right ) \coth ^{-1}(a+b x)^2}{b^4}-\frac{\left (a^4+6 a^2+1\right ) \coth ^{-1}(a+b x)^2}{4 b^4}+\frac{\left (6 a^2+1\right ) (a+b x) \coth ^{-1}(a+b x)}{2 b^4}+\frac{2 a \left (a^2+1\right ) \log \left (\frac{2}{-a-b x+1}\right ) \coth ^{-1}(a+b x)}{b^4}-\frac{a x}{b^3}+\frac{(a+b x)^2}{12 b^4}+\frac{\log \left (1-(a+b x)^2\right )}{12 b^4}+\frac{a \tanh ^{-1}(a+b x)}{b^4}+\frac{(a+b x)^3 \coth ^{-1}(a+b x)}{6 b^4}-\frac{a (a+b x)^2 \coth ^{-1}(a+b x)}{b^4}+\frac{1}{4} x^4 \coth ^{-1}(a+b x)^2 \]
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Rubi [A] time = 0.348779, antiderivative size = 263, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 15, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.25, Rules used = {6112, 5929, 5911, 260, 5917, 321, 206, 266, 43, 6049, 5949, 5985, 5919, 2402, 2315} \[ \frac{a \left (a^2+1\right ) \text{PolyLog}\left (2,-\frac{a+b x+1}{-a-b x+1}\right )}{b^4}+\frac{\left (6 a^2+1\right ) \log \left (1-(a+b x)^2\right )}{4 b^4}-\frac{a \left (a^2+1\right ) \coth ^{-1}(a+b x)^2}{b^4}-\frac{\left (a^4+6 a^2+1\right ) \coth ^{-1}(a+b x)^2}{4 b^4}+\frac{\left (6 a^2+1\right ) (a+b x) \coth ^{-1}(a+b x)}{2 b^4}+\frac{2 a \left (a^2+1\right ) \log \left (\frac{2}{-a-b x+1}\right ) \coth ^{-1}(a+b x)}{b^4}-\frac{a x}{b^3}+\frac{(a+b x)^2}{12 b^4}+\frac{\log \left (1-(a+b x)^2\right )}{12 b^4}+\frac{a \tanh ^{-1}(a+b x)}{b^4}+\frac{(a+b x)^3 \coth ^{-1}(a+b x)}{6 b^4}-\frac{a (a+b x)^2 \coth ^{-1}(a+b x)}{b^4}+\frac{1}{4} x^4 \coth ^{-1}(a+b x)^2 \]
Antiderivative was successfully verified.
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Rule 6112
Rule 5929
Rule 5911
Rule 260
Rule 5917
Rule 321
Rule 206
Rule 266
Rule 43
Rule 6049
Rule 5949
Rule 5985
Rule 5919
Rule 2402
Rule 2315
Rubi steps
\begin{align*} \int x^3 \coth ^{-1}(a+b x)^2 \, dx &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a}{b}+\frac{x}{b}\right )^3 \coth ^{-1}(x)^2 \, dx,x,a+b x\right )}{b}\\ &=\frac{1}{4} x^4 \coth ^{-1}(a+b x)^2-\frac{1}{2} \operatorname{Subst}\left (\int \left (-\frac{\left (1+6 a^2\right ) \coth ^{-1}(x)}{b^4}+\frac{4 a x \coth ^{-1}(x)}{b^4}-\frac{x^2 \coth ^{-1}(x)}{b^4}+\frac{\left (1+6 a^2+a^4-4 a \left (1+a^2\right ) x\right ) \coth ^{-1}(x)}{b^4 \left (1-x^2\right )}\right ) \, dx,x,a+b x\right )\\ &=\frac{1}{4} x^4 \coth ^{-1}(a+b x)^2+\frac{\operatorname{Subst}\left (\int x^2 \coth ^{-1}(x) \, dx,x,a+b x\right )}{2 b^4}-\frac{\operatorname{Subst}\left (\int \frac{\left (1+6 a^2+a^4-4 a \left (1+a^2\right ) x\right ) \coth ^{-1}(x)}{1-x^2} \, dx,x,a+b x\right )}{2 b^4}-\frac{(2 a) \operatorname{Subst}\left (\int x \coth ^{-1}(x) \, dx,x,a+b x\right )}{b^4}+\frac{\left (1+6 a^2\right ) \operatorname{Subst}\left (\int \coth ^{-1}(x) \, dx,x,a+b x\right )}{2 b^4}\\ &=\frac{\left (1+6 a^2\right ) (a+b x) \coth ^{-1}(a+b x)}{2 b^4}-\frac{a (a+b x)^2 \coth ^{-1}(a+b x)}{b^4}+\frac{(a+b x)^3 \coth ^{-1}(a+b x)}{6 b^4}+\frac{1}{4} x^4 \coth ^{-1}(a+b x)^2-\frac{\operatorname{Subst}\left (\int \frac{x^3}{1-x^2} \, dx,x,a+b x\right )}{6 b^4}-\frac{\operatorname{Subst}\left (\int \left (\frac{\left (1+a^2 \left (6+a^2\right )\right ) \coth ^{-1}(x)}{1-x^2}-\frac{4 a \left (1+a^2\right ) x \coth ^{-1}(x)}{1-x^2}\right ) \, dx,x,a+b x\right )}{2 b^4}+\frac{a \operatorname{Subst}\left (\int \frac{x^2}{1-x^2} \, dx,x,a+b x\right )}{b^4}-\frac{\left (1+6 a^2\right ) \operatorname{Subst}\left (\int \frac{x}{1-x^2} \, dx,x,a+b x\right )}{2 b^4}\\ &=-\frac{a x}{b^3}+\frac{\left (1+6 a^2\right ) (a+b x) \coth ^{-1}(a+b x)}{2 b^4}-\frac{a (a+b x)^2 \coth ^{-1}(a+b x)}{b^4}+\frac{(a+b x)^3 \coth ^{-1}(a+b x)}{6 b^4}+\frac{1}{4} x^4 \coth ^{-1}(a+b x)^2+\frac{\left (1+6 a^2\right ) \log \left (1-(a+b x)^2\right )}{4 b^4}-\frac{\operatorname{Subst}\left (\int \frac{x}{1-x} \, dx,x,(a+b x)^2\right )}{12 b^4}+\frac{a \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,a+b x\right )}{b^4}+\frac{\left (2 a \left (1+a^2\right )\right ) \operatorname{Subst}\left (\int \frac{x \coth ^{-1}(x)}{1-x^2} \, dx,x,a+b x\right )}{b^4}-\frac{\left (1+6 a^2+a^4\right ) \operatorname{Subst}\left (\int \frac{\coth ^{-1}(x)}{1-x^2} \, dx,x,a+b x\right )}{2 b^4}\\ &=-\frac{a x}{b^3}+\frac{\left (1+6 a^2\right ) (a+b x) \coth ^{-1}(a+b x)}{2 b^4}-\frac{a (a+b x)^2 \coth ^{-1}(a+b x)}{b^4}+\frac{(a+b x)^3 \coth ^{-1}(a+b x)}{6 b^4}-\frac{a \left (1+a^2\right ) \coth ^{-1}(a+b x)^2}{b^4}-\frac{\left (1+6 a^2+a^4\right ) \coth ^{-1}(a+b x)^2}{4 b^4}+\frac{1}{4} x^4 \coth ^{-1}(a+b x)^2+\frac{a \tanh ^{-1}(a+b x)}{b^4}+\frac{\left (1+6 a^2\right ) \log \left (1-(a+b x)^2\right )}{4 b^4}-\frac{\operatorname{Subst}\left (\int \left (-1+\frac{1}{1-x}\right ) \, dx,x,(a+b x)^2\right )}{12 b^4}+\frac{\left (2 a \left (1+a^2\right )\right ) \operatorname{Subst}\left (\int \frac{\coth ^{-1}(x)}{1-x} \, dx,x,a+b x\right )}{b^4}\\ &=-\frac{a x}{b^3}+\frac{(a+b x)^2}{12 b^4}+\frac{\left (1+6 a^2\right ) (a+b x) \coth ^{-1}(a+b x)}{2 b^4}-\frac{a (a+b x)^2 \coth ^{-1}(a+b x)}{b^4}+\frac{(a+b x)^3 \coth ^{-1}(a+b x)}{6 b^4}-\frac{a \left (1+a^2\right ) \coth ^{-1}(a+b x)^2}{b^4}-\frac{\left (1+6 a^2+a^4\right ) \coth ^{-1}(a+b x)^2}{4 b^4}+\frac{1}{4} x^4 \coth ^{-1}(a+b x)^2+\frac{a \tanh ^{-1}(a+b x)}{b^4}+\frac{2 a \left (1+a^2\right ) \coth ^{-1}(a+b x) \log \left (\frac{2}{1-a-b x}\right )}{b^4}+\frac{\log \left (1-(a+b x)^2\right )}{12 b^4}+\frac{\left (1+6 a^2\right ) \log \left (1-(a+b x)^2\right )}{4 b^4}-\frac{\left (2 a \left (1+a^2\right )\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{2}{1-x}\right )}{1-x^2} \, dx,x,a+b x\right )}{b^4}\\ &=-\frac{a x}{b^3}+\frac{(a+b x)^2}{12 b^4}+\frac{\left (1+6 a^2\right ) (a+b x) \coth ^{-1}(a+b x)}{2 b^4}-\frac{a (a+b x)^2 \coth ^{-1}(a+b x)}{b^4}+\frac{(a+b x)^3 \coth ^{-1}(a+b x)}{6 b^4}-\frac{a \left (1+a^2\right ) \coth ^{-1}(a+b x)^2}{b^4}-\frac{\left (1+6 a^2+a^4\right ) \coth ^{-1}(a+b x)^2}{4 b^4}+\frac{1}{4} x^4 \coth ^{-1}(a+b x)^2+\frac{a \tanh ^{-1}(a+b x)}{b^4}+\frac{2 a \left (1+a^2\right ) \coth ^{-1}(a+b x) \log \left (\frac{2}{1-a-b x}\right )}{b^4}+\frac{\log \left (1-(a+b x)^2\right )}{12 b^4}+\frac{\left (1+6 a^2\right ) \log \left (1-(a+b x)^2\right )}{4 b^4}+\frac{\left (2 a \left (1+a^2\right )\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-a-b x}\right )}{b^4}\\ &=-\frac{a x}{b^3}+\frac{(a+b x)^2}{12 b^4}+\frac{\left (1+6 a^2\right ) (a+b x) \coth ^{-1}(a+b x)}{2 b^4}-\frac{a (a+b x)^2 \coth ^{-1}(a+b x)}{b^4}+\frac{(a+b x)^3 \coth ^{-1}(a+b x)}{6 b^4}-\frac{a \left (1+a^2\right ) \coth ^{-1}(a+b x)^2}{b^4}-\frac{\left (1+6 a^2+a^4\right ) \coth ^{-1}(a+b x)^2}{4 b^4}+\frac{1}{4} x^4 \coth ^{-1}(a+b x)^2+\frac{a \tanh ^{-1}(a+b x)}{b^4}+\frac{2 a \left (1+a^2\right ) \coth ^{-1}(a+b x) \log \left (\frac{2}{1-a-b x}\right )}{b^4}+\frac{\log \left (1-(a+b x)^2\right )}{12 b^4}+\frac{\left (1+6 a^2\right ) \log \left (1-(a+b x)^2\right )}{4 b^4}+\frac{a \left (1+a^2\right ) \text{Li}_2\left (1-\frac{2}{1-a-b x}\right )}{b^4}\\ \end{align*}
Mathematica [A] time = 1.56441, size = 203, normalized size = 0.77 \[ -\frac{12 \left (a^3+a\right ) \text{PolyLog}\left (2,e^{-2 \coth ^{-1}(a+b x)}\right )+3 \left (a^4-4 a^3+6 a^2-4 a-b^4 x^4+1\right ) \coth ^{-1}(a+b x)^2-2 \coth ^{-1}(a+b x) \left (9 a^2 b x+12 \left (a^3+a\right ) \log \left (1-e^{-2 \coth ^{-1}(a+b x)}\right )+13 a^3-3 a b^2 x^2+9 a+b^3 x^3+3 b x\right )+36 a^2 \log \left (\frac{1}{(a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}\right )+11 a^2+10 a b x+8 \log \left (\frac{1}{(a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}\right )-b^2 x^2+1}{12 b^4} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.056, size = 967, normalized size = 3.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.988167, size = 432, normalized size = 1.64 \begin{align*} \frac{1}{4} \, x^{4} \operatorname{arcoth}\left (b x + a\right )^{2} + \frac{1}{48} \, b^{2}{\left (\frac{48 \,{\left (a^{3} + a\right )}{\left (\log \left (b x + a - 1\right ) \log \left (\frac{1}{2} \, b x + \frac{1}{2} \, a + \frac{1}{2}\right ) +{\rm Li}_2\left (-\frac{1}{2} \, b x - \frac{1}{2} \, a + \frac{1}{2}\right )\right )}}{b^{6}} + \frac{4 \,{\left (13 \, a^{3} + 18 \, a^{2} + 9 \, a + 4\right )} \log \left (b x + a + 1\right )}{b^{6}} + \frac{4 \, b^{2} x^{2} - 40 \, a b x + 3 \,{\left (a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} \log \left (b x + a + 1\right )^{2} - 6 \,{\left (a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} \log \left (b x + a + 1\right ) \log \left (b x + a - 1\right ) + 3 \,{\left (a^{4} - 4 \, a^{3} + 6 \, a^{2} - 4 \, a + 1\right )} \log \left (b x + a - 1\right )^{2} - 4 \,{\left (13 \, a^{3} - 18 \, a^{2} + 9 \, a - 4\right )} \log \left (b x + a - 1\right )}{b^{6}}\right )} + \frac{1}{12} \, b{\left (\frac{2 \,{\left (b^{2} x^{3} - 3 \, a b x^{2} + 3 \,{\left (3 \, a^{2} + 1\right )} x\right )}}{b^{4}} - \frac{3 \,{\left (a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} \log \left (b x + a + 1\right )}{b^{5}} + \frac{3 \,{\left (a^{4} - 4 \, a^{3} + 6 \, a^{2} - 4 \, a + 1\right )} \log \left (b x + a - 1\right )}{b^{5}}\right )} \operatorname{arcoth}\left (b x + a\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 (x^{3} \operatorname{arcoth}\left (b x + a\right )^{2}, x\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 x^{3} \operatorname{acoth}^{2}{\left (a + b x \right )}\, 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 x^{3} \operatorname{arcoth}\left (b x + a\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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