Optimal. Leaf size=76 \[ -\frac {2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {4 x^2}{b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {16 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac {32 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^4} \]
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Rubi [A]
time = 0.03, 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 = {2199, 2188, 30}
\begin {gather*} -\frac {32 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^4}+\frac {16 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac {4 x^2}{b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 30
Rule 2188
Rule 2199
Rubi steps
\begin {align*} \int \frac {x^3}{\tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx &=-\frac {2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2 \int \frac {x^2}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{b}\\ &=-\frac {2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {4 x^2}{b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {8 \int \frac {x}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{b^2}\\ &=-\frac {2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {4 x^2}{b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {16 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac {16 \int \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^3}\\ &=-\frac {2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {4 x^2}{b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {16 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac {16 \text {Subst}\left (\int \sqrt {x} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}\\ &=-\frac {2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {4 x^2}{b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {16 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac {32 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^4}\\ \end {align*}
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Mathematica [A]
time = 0.03, size = 65, normalized size = 0.86 \begin {gather*} -\frac {2 \left (b^3 x^3+6 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-24 b x \tanh ^{-1}(\tanh (a+b x))^2+16 \tanh ^{-1}(\tanh (a+b x))^3\right )}{3 b^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(185\) vs.
\(2(64)=128\).
time = 0.07, size = 186, normalized size = 2.45
method | result | size |
default | \(\frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}-6 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\, a -6 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {2 \left (3 a^{2}+6 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right )}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {2 \left (-a^{3}-3 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )-3 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}-\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}\right )}{3 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}}{b^{4}}\) | \(186\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.54, size = 52, normalized size = 0.68 \begin {gather*} \frac {2 \, {\left (b^{4} x^{4} - 5 \, a b^{3} x^{3} - 30 \, a^{2} b^{2} x^{2} - 40 \, a^{3} b x - 16 \, a^{4}\right )}}{3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.34, size = 62, normalized size = 0.82 \begin {gather*} \frac {2 \, {\left (b^{3} x^{3} - 6 \, a b^{2} x^{2} - 24 \, a^{2} b x - 16 \, a^{3}\right )} \sqrt {b x + a}}{3 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 72.69, size = 90, normalized size = 1.18 \begin {gather*} \begin {cases} - \frac {2 x^{3}}{3 b \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}} - \frac {4 x^{2}}{b^{2} \sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}} + \frac {16 x \sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}}{b^{3}} - \frac {32 \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{3 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 \operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a \right )} \right )}} & \text {otherwise} \end {cases} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.40, size = 59, normalized size = 0.78 \begin {gather*} -\frac {2 \, {\left (9 \, {\left (b x + a\right )} a^{2} - a^{3}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{4}} + \frac {2 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} b^{8} - 9 \, \sqrt {b x + a} a b^{8}\right )}}{3 \, b^{12}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.30, size = 533, normalized size = 7.01 \begin {gather*} \frac {2\,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}}}{3\,b^3}+\frac {\left (\frac {2\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\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 )}{b^3}+\frac {4\,\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\,b^3}\right )\,\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}}}{b}-\frac {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 (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\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}{b^4\,\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )}-\frac {\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 (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\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}{3\,b^4\,{\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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