Optimal. Leaf size=69 \[ -\frac {32}{315} b^3 x^{9/2}+\frac {16}{35} b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))-\frac {4}{5} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^3 \]
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Rubi [A]
time = 0.03, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2199, 30}
\begin {gather*} \frac {16}{35} b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))-\frac {4}{5} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^3-\frac {32}{315} b^3 x^{9/2} \end {gather*}
Antiderivative was successfully verified.
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Rule 30
Rule 2199
Rubi steps
\begin {align*} \int \sqrt {x} \tanh ^{-1}(\tanh (a+b x))^3 \, dx &=\frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^3-(2 b) \int x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2 \, dx\\ &=-\frac {4}{5} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{5} \left (8 b^2\right ) \int x^{5/2} \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=\frac {16}{35} b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))-\frac {4}{5} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^3-\frac {1}{35} \left (16 b^3\right ) \int x^{7/2} \, dx\\ &=-\frac {32}{315} b^3 x^{9/2}+\frac {16}{35} b^2 x^{7/2} \tanh ^{-1}(\tanh (a+b x))-\frac {4}{5} b x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))^3\\ \end {align*}
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Mathematica [A]
time = 0.02, size = 57, normalized size = 0.83 \begin {gather*} -\frac {2}{315} x^{3/2} \left (16 b^3 x^3-72 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+126 b x \tanh ^{-1}(\tanh (a+b x))^2-105 \tanh ^{-1}(\tanh (a+b x))^3\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 1.19, size = 56, normalized size = 0.81
method | result | size |
derivativedivides | \(\frac {2 x^{\frac {3}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{3}-4 b \left (\frac {x^{\frac {5}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{5}-\frac {4 b \left (\frac {x^{\frac {7}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )}{7}-\frac {2 b \,x^{\frac {9}{2}}}{63}\right )}{5}\right )\) | \(56\) |
default | \(\frac {2 x^{\frac {3}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{3}-4 b \left (\frac {x^{\frac {5}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{5}-\frac {4 b \left (\frac {x^{\frac {7}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )}{7}-\frac {2 b \,x^{\frac {9}{2}}}{63}\right )}{5}\right )\) | \(56\) |
risch | \(\text {Expression too large to display}\) | \(7981\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.28, size = 55, normalized size = 0.80 \begin {gather*} -\frac {4}{5} \, b x^{\frac {5}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + \frac {2}{3} \, x^{\frac {3}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} - \frac {16}{315} \, {\left (2 \, b^{2} x^{\frac {9}{2}} - 9 \, b x^{\frac {7}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.37, size = 38, normalized size = 0.55 \begin {gather*} \frac {2}{315} \, {\left (35 \, b^{3} x^{4} + 135 \, a b^{2} x^{3} + 189 \, a^{2} b x^{2} + 105 \, a^{3} x\right )} \sqrt {x} \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 \sqrt {x} \operatorname {atanh}^{3}{\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 = 0.38, size = 35, normalized size = 0.51 \begin {gather*} \frac {2}{9} \, b^{3} x^{\frac {9}{2}} + \frac {6}{7} \, a b^{2} x^{\frac {7}{2}} + \frac {6}{5} \, a^{2} b x^{\frac {5}{2}} + \frac {2}{3} \, a^{3} x^{\frac {3}{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.16, size = 182, normalized size = 2.64 \begin {gather*} \frac {2\,b^3\,x^{9/2}}{9}-\frac {x^{3/2}\,{\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}{12}+\frac {3\,b\,x^{5/2}\,{\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}{10}-\frac {3\,b^2\,x^{7/2}\,\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 )}{7} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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