Optimal. Leaf size=101 \[ \frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^3}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^4}+\frac {256 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{3465 b^5} \]
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Rubi [A]
time = 0.05, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps
used = 6, 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 {256 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{3465 b^5}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^4}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^3}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac {2 x^4 \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^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {8 \int x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx}{3 b}\\ &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac {16 \int x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx}{5 b^2}\\ &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^3}-\frac {64 \int x \tanh ^{-1}(\tanh (a+b x))^{7/2} \, dx}{35 b^3}\\ &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^3}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^4}+\frac {128 \int \tanh ^{-1}(\tanh (a+b x))^{9/2} \, dx}{315 b^4}\\ &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^3}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^4}+\frac {128 \text {Subst}\left (\int x^{9/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{315 b^5}\\ &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^3}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^4}+\frac {256 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{3465 b^5}\\ \end {align*}
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Mathematica [A]
time = 0.02, size = 83, normalized size = 0.82 \begin {gather*} \frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (1155 b^4 x^4-1848 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+1584 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-704 b x \tanh ^{-1}(\tanh (a+b x))^3+128 \tanh ^{-1}(\tanh (a+b x))^4\right )}{3465 b^5} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.09, size = 154, normalized size = 1.52
method | result | size |
default | \(\frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {11}{2}}}{11}+\frac {2 \left (-4 \arctanh \left (\tanh \left (b x +a \right )\right )+4 b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{2}+\left (-2 \arctanh \left (\tanh \left (b x +a \right )\right )+2 b x \right )^{2}\right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}+\frac {4 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{2} \left (-2 \arctanh \left (\tanh \left (b x +a \right )\right )+2 b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}}{b^{5}}\) | \(154\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.53, size = 64, normalized size = 0.63 \begin {gather*} \frac {2 \, {\left (315 \, b^{5} x^{5} + 35 \, a b^{4} x^{4} - 40 \, a^{2} b^{3} x^{3} + 48 \, a^{3} b^{2} x^{2} - 64 \, a^{4} b x + 128 \, a^{5}\right )} \sqrt {b x + a}}{3465 \, b^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.32, size = 64, normalized size = 0.63 \begin {gather*} \frac {2 \, {\left (315 \, b^{5} x^{5} + 35 \, a b^{4} x^{4} - 40 \, a^{2} b^{3} x^{3} + 48 \, a^{3} b^{2} x^{2} - 64 \, a^{4} b x + 128 \, a^{5}\right )} \sqrt {b x + a}}{3465 \, b^{5}} \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^{4} \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 = 0.40, size = 150, normalized size = 1.49 \begin {gather*} \frac {\sqrt {2} {\left (\frac {11 \, \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 )} a}{b^{4}} + \frac {5 \, \sqrt {2} {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )}}{b^{4}}\right )}}{3465 \, b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.05, size = 811, normalized size = 8.03 \begin {gather*} \frac {2\,x^5\,\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}}}{11}-\frac {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}}\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{11}-\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 )}{11}+\frac {2\,b\,x}{11}\right )}{9\,b}-\frac {128\,\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 )}^4\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{11}-\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 )}{11}+\frac {2\,b\,x}{11}\right )}{315\,b^5}-\frac {8\,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 )}{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 )}{11}-\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 )}{11}+\frac {2\,b\,x}{11}\right )}{63\,b^2}-\frac {64\,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 )}^3\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{11}-\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 )}{11}+\frac {2\,b\,x}{11}\right )}{315\,b^4}-\frac {16\,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 )}^2\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{11}-\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 )}{11}+\frac {2\,b\,x}{11}\right )}{105\,b^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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