Optimal. Leaf size=76 \[ -\frac {32 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^4}+\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^3}-\frac {4 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^2}+\frac {2 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b} \]
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Rubi [A] time = 0.05, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2157, 30} \[ -\frac {4 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^2}-\frac {32 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^4}+\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^3}+\frac {2 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b} \]
Antiderivative was successfully verified.
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Rule 30
Rule 2157
Rule 2168
Rubi steps
\begin {align*} \int \frac {x^3}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx &=\frac {2 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {6 \int x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx}{b}\\ &=\frac {2 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {4 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^2}+\frac {8 \int x \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx}{b^2}\\ &=\frac {2 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {4 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^2}+\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^3}-\frac {16 \int \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx}{5 b^3}\\ &=\frac {2 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {4 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^2}+\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^3}-\frac {16 \operatorname {Subst}\left (\int x^{5/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{5 b^4}\\ &=\frac {2 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {4 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^2}+\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^3}-\frac {32 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^4}\\ \end {align*}
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Mathematica [A] time = 0.04, size = 66, normalized size = 0.87 \[ \frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (-70 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+56 b x \tanh ^{-1}(\tanh (a+b x))^2-16 \tanh ^{-1}(\tanh (a+b x))^3+35 b^3 x^3\right )}{35 b^4} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.47, size = 42, normalized size = 0.55 \[ \frac {2 \, {\left (5 \, b^{3} x^{3} - 6 \, a b^{2} x^{2} + 8 \, a^{2} b x - 16 \, a^{3}\right )} \sqrt {b x + a}}{35 \, b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.13, size = 49, normalized size = 0.64 \[ \frac {2 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )}}{35 \, b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.18, size = 123, normalized size = 1.62 \[ \frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}+\frac {2 \left (-3 \arctanh \left (\tanh \left (b x +a \right )\right )+3 b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (\left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right ) \left (-2 \arctanh \left (\tanh \left (b x +a \right )\right )+2 b x \right )+\left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{2}\right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}+2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{3} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.52, size = 53, normalized size = 0.70 \[ \frac {2 \, {\left (5 \, b^{4} x^{4} - a b^{3} x^{3} + 2 \, a^{2} b^{2} x^{2} - 8 \, a^{3} b x - 16 \, a^{4}\right )}}{35 \, \sqrt {b x + a} b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.07, size = 385, normalized size = 5.07 \[ \frac {2\,x^3\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{7\,b}+\frac {32\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,{\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}+b\,x\right )}^3}{35\,b^4}+\frac {12\,x^2\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}+b\,x\right )}{35\,b^2}+\frac {16\,x\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,{\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}+b\,x\right )}^2}{35\,b^3} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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