Optimal. Leaf size=80 \[ \frac {2 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^2}+\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^3}-\frac {32 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^4} \]
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Rubi [A]
time = 0.03, antiderivative size = 80, 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 = {2199, 2188, 30}
\begin {gather*} -\frac {32 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^4}+\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^3}-\frac {4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^2}+\frac {2 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b} \end {gather*}
Antiderivative was successfully verified.
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Rule 30
Rule 2188
Rule 2199
Rubi steps
\begin {align*} \int x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac {2 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {2 \int x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx}{b}\\ &=\frac {2 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^2}+\frac {8 \int x \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx}{5 b^2}\\ &=\frac {2 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^2}+\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^3}-\frac {16 \int \tanh ^{-1}(\tanh (a+b x))^{7/2} \, dx}{35 b^3}\\ &=\frac {2 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^2}+\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^3}-\frac {16 \text {Subst}\left (\int x^{7/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{35 b^4}\\ &=\frac {2 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^2}+\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^3}-\frac {32 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^4}\\ \end {align*}
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Mathematica [A]
time = 0.02, size = 66, normalized size = 0.82 \begin {gather*} \frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (105 b^3 x^3-126 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+72 b x \tanh ^{-1}(\tanh (a+b x))^2-16 \tanh ^{-1}(\tanh (a+b x))^3\right )}{315 b^4} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.08, size = 124, normalized size = 1.55
method | result | size |
default | \(\frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (-3 \arctanh \left (\tanh \left (b x +a \right )\right )+3 b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right ) \left (-2 \arctanh \left (\tanh \left (b x +a \right )\right )+2 b x \right )+\left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{2}\right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{3} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}}{b^{4}}\) | \(124\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.52, size = 53, normalized size = 0.66 \begin {gather*} \frac {2 \, {\left (35 \, b^{4} x^{4} + 5 \, a b^{3} x^{3} - 6 \, a^{2} b^{2} x^{2} + 8 \, a^{3} b x - 16 \, a^{4}\right )} \sqrt {b x + a}}{315 \, b^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.34, size = 53, normalized size = 0.66 \begin {gather*} \frac {2 \, {\left (35 \, b^{4} x^{4} + 5 \, a b^{3} x^{3} - 6 \, a^{2} b^{2} x^{2} + 8 \, a^{3} b x - 16 \, a^{4}\right )} \sqrt {b x + a}}{315 \, b^{4}} \end {gather*}
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 \begin {gather*} \int x^{3} \sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.39, size = 125, normalized size = 1.56 \begin {gather*} \frac {\sqrt {2} {\left (\frac {9 \, \sqrt {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 )} a}{b^{3}} + \frac {\sqrt {2} {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )}}{b^{3}}\right )}}{315 \, b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.99, size = 648, normalized size = 8.10 \begin {gather*} \frac {2\,x^4\,\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}}}{9}-\frac {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}}\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{9}-\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 )}{9}+\frac {2\,b\,x}{9}\right )}{7\,b}-\frac {16\,\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\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{9}-\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 )}{9}+\frac {2\,b\,x}{9}\right )}{35\,b^4}-\frac {6\,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 )\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{9}-\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 )}{9}+\frac {2\,b\,x}{9}\right )}{35\,b^2}-\frac {8\,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\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{9}-\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 )}{9}+\frac {2\,b\,x}{9}\right )}{35\,b^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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