Optimal. Leaf size=101 \[ \frac {\left (6 a^2+1\right ) x}{4 b^3}+\frac {(a+b x)^3}{12 b^4}-\frac {a (a+b x)^2}{2 b^4}+\frac {(1-a)^4 \log (-a-b x+1)}{8 b^4}-\frac {(a+1)^4 \log (a+b x+1)}{8 b^4}+\frac {1}{4} x^4 \coth ^{-1}(a+b x) \]
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Rubi [A] time = 0.13, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6112, 5927, 702, 633, 31} \[ \frac {\left (6 a^2+1\right ) x}{4 b^3}+\frac {(a+b x)^3}{12 b^4}-\frac {a (a+b x)^2}{2 b^4}+\frac {(1-a)^4 \log (-a-b x+1)}{8 b^4}-\frac {(a+1)^4 \log (a+b x+1)}{8 b^4}+\frac {1}{4} x^4 \coth ^{-1}(a+b x) \]
Antiderivative was successfully verified.
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Rule 31
Rule 633
Rule 702
Rule 5927
Rule 6112
Rubi steps
\begin {align*} \int x^3 \coth ^{-1}(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \left (-\frac {a}{b}+\frac {x}{b}\right )^3 \coth ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac {1}{4} x^4 \coth ^{-1}(a+b x)-\frac {1}{4} \operatorname {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^4}{1-x^2} \, dx,x,a+b x\right )\\ &=\frac {1}{4} x^4 \coth ^{-1}(a+b x)-\frac {1}{4} \operatorname {Subst}\left (\int \left (-\frac {1+6 a^2}{b^4}+\frac {4 a x}{b^4}-\frac {x^2}{b^4}+\frac {1+6 a^2+a^4-4 a \left (1+a^2\right ) x}{b^4 \left (1-x^2\right )}\right ) \, dx,x,a+b x\right )\\ &=\frac {\left (1+6 a^2\right ) x}{4 b^3}-\frac {a (a+b x)^2}{2 b^4}+\frac {(a+b x)^3}{12 b^4}+\frac {1}{4} x^4 \coth ^{-1}(a+b x)-\frac {\operatorname {Subst}\left (\int \frac {1+6 a^2+a^4-4 a \left (1+a^2\right ) x}{1-x^2} \, dx,x,a+b x\right )}{4 b^4}\\ &=\frac {\left (1+6 a^2\right ) x}{4 b^3}-\frac {a (a+b x)^2}{2 b^4}+\frac {(a+b x)^3}{12 b^4}+\frac {1}{4} x^4 \coth ^{-1}(a+b x)-\frac {(1-a)^4 \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,a+b x\right )}{8 b^4}+\frac {(1+a)^4 \operatorname {Subst}\left (\int \frac {1}{-1-x} \, dx,x,a+b x\right )}{8 b^4}\\ &=\frac {\left (1+6 a^2\right ) x}{4 b^3}-\frac {a (a+b x)^2}{2 b^4}+\frac {(a+b x)^3}{12 b^4}+\frac {1}{4} x^4 \coth ^{-1}(a+b x)+\frac {(1-a)^4 \log (1-a-b x)}{8 b^4}-\frac {(1+a)^4 \log (1+a+b x)}{8 b^4}\\ \end {align*}
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Mathematica [A] time = 0.04, size = 81, normalized size = 0.80 \[ \frac {6 \left (3 a^2+1\right ) b x+6 b^4 x^4 \coth ^{-1}(a+b x)-6 a b^2 x^2+3 (a-1)^4 \log (-a-b x+1)-3 (a+1)^4 \log (a+b x+1)+2 b^3 x^3}{24 b^4} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.50, size = 112, normalized size = 1.11 \[ \frac {3 \, b^{4} x^{4} \log \left (\frac {b x + a + 1}{b x + a - 1}\right ) + 2 \, b^{3} x^{3} - 6 \, a b^{2} x^{2} + 6 \, {\left (3 \, a^{2} + 1\right )} b x - 3 \, {\left (a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} \log \left (b x + a + 1\right ) + 3 \, {\left (a^{4} - 4 \, a^{3} + 6 \, a^{2} - 4 \, a + 1\right )} \log \left (b x + a - 1\right )}{24 \, b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \operatorname {arcoth}\left (b x + a\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.04, size = 199, normalized size = 1.97 \[ \frac {a}{4 b^{4}}+\frac {x}{4 b^{3}}+\frac {x^{4} \mathrm {arccoth}\left (b x +a \right )}{4}+\frac {\ln \left (b x +a -1\right )}{8 b^{4}}-\frac {\ln \left (b x +a +1\right )}{8 b^{4}}-\frac {\ln \left (b x +a +1\right ) a^{4}}{8 b^{4}}-\frac {\ln \left (b x +a +1\right ) a^{3}}{2 b^{4}}-\frac {3 \ln \left (b x +a +1\right ) a^{2}}{4 b^{4}}-\frac {\ln \left (b x +a +1\right ) a}{2 b^{4}}+\frac {x^{3}}{12 b}-\frac {x^{2} a}{4 b^{2}}+\frac {3 x \,a^{2}}{4 b^{3}}+\frac {13 a^{3}}{12 b^{4}}+\frac {\ln \left (b x +a -1\right ) a^{4}}{8 b^{4}}-\frac {\ln \left (b x +a -1\right ) a^{3}}{2 b^{4}}+\frac {3 \ln \left (b x +a -1\right ) a^{2}}{4 b^{4}}-\frac {\ln \left (b x +a -1\right ) a}{2 b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.31, size = 106, normalized size = 1.05 \[ \frac {1}{4} \, x^{4} \operatorname {arcoth}\left (b x + a\right ) + \frac {1}{24} \, b {\left (\frac {2 \, {\left (b^{2} x^{3} - 3 \, a b x^{2} + 3 \, {\left (3 \, a^{2} + 1\right )} x\right )}}{b^{4}} - \frac {3 \, {\left (a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} \log \left (b x + a + 1\right )}{b^{5}} + \frac {3 \, {\left (a^{4} - 4 \, a^{3} + 6 \, a^{2} - 4 \, a + 1\right )} \log \left (b x + a - 1\right )}{b^{5}}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.47, size = 134, normalized size = 1.33 \[ \frac {x^4\,\ln \left (\frac {1}{a+b\,x}+1\right )}{8}-x\,\left (\frac {4\,a^2-4}{16\,b^3}-\frac {a^2}{b^3}\right )-\frac {x^4\,\ln \left (1-\frac {1}{a+b\,x}\right )}{8}+\frac {x^3}{12\,b}-\frac {a\,x^2}{4\,b^2}+\frac {\ln \left (a+b\,x-1\right )\,\left (a^4-4\,a^3+6\,a^2-4\,a+1\right )}{8\,b^4}-\frac {\ln \left (a+b\,x+1\right )\,\left (a^4+4\,a^3+6\,a^2+4\,a+1\right )}{8\,b^4} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.84, size = 153, normalized size = 1.51 \[ \begin {cases} - \frac {a^{4} \operatorname {acoth}{\left (a + b x \right )}}{4 b^{4}} - \frac {a^{3} \log {\left (a + b x + 1 \right )}}{b^{4}} + \frac {a^{3} \operatorname {acoth}{\left (a + b x \right )}}{b^{4}} + \frac {3 a^{2} x}{4 b^{3}} - \frac {3 a^{2} \operatorname {acoth}{\left (a + b x \right )}}{2 b^{4}} - \frac {a x^{2}}{4 b^{2}} - \frac {a \log {\left (a + b x + 1 \right )}}{b^{4}} + \frac {a \operatorname {acoth}{\left (a + b x \right )}}{b^{4}} + \frac {x^{4} \operatorname {acoth}{\left (a + b x \right )}}{4} + \frac {x^{3}}{12 b} + \frac {x}{4 b^{3}} - \frac {\operatorname {acoth}{\left (a + b x \right )}}{4 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {acoth}{\relax (a )}}{4} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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