Optimal. Leaf size=27 \[ -\frac {4}{15} b x^{5/2}+\frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x)) \]
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Rubi [A]
time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps
used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2199, 30}
\begin {gather*} \frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))-\frac {4}{15} b x^{5/2} \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)) \, dx &=\frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))-\frac {1}{3} (2 b) \int x^{3/2} \, dx\\ &=-\frac {4}{15} b x^{5/2}+\frac {2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))\\ \end {align*}
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Mathematica [A]
time = 0.02, size = 23, normalized size = 0.85 \begin {gather*} \frac {2}{15} x^{3/2} \left (-2 b x+5 \tanh ^{-1}(\tanh (a+b x))\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.10, size = 20, normalized size = 0.74
method | result | size |
derivativedivides | \(-\frac {4 b \,x^{\frac {5}{2}}}{15}+\frac {2 x^{\frac {3}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )}{3}\) | \(20\) |
default | \(-\frac {4 b \,x^{\frac {5}{2}}}{15}+\frac {2 x^{\frac {3}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )}{3}\) | \(20\) |
risch | \(\frac {2 x^{\frac {3}{2}} \ln \left ({\mathrm e}^{b x +a}\right )}{3}-\frac {\left (-10 i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+5 i \pi x \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )+5 i \pi x \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-5 i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+5 i \pi x \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-5 i \pi x \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+5 i \pi x \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+8 b \,x^{2}\right ) \sqrt {x}}{30}\) | \(296\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.27, size = 19, normalized size = 0.70 \begin {gather*} -\frac {4}{15} \, b x^{\frac {5}{2}} + \frac {2}{3} \, x^{\frac {3}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.32, size = 16, normalized size = 0.59 \begin {gather*} \frac {2}{15} \, {\left (3 \, b x^{2} + 5 \, a 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}{\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 = 13, normalized size = 0.48 \begin {gather*} \frac {2}{5} \, b x^{\frac {5}{2}} + \frac {2}{3} \, a x^{\frac {3}{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.09, size = 57, normalized size = 2.11 \begin {gather*} \frac {x^{3/2}\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{3}-\frac {4\,b\,x^{5/2}}{15}-\frac {x^{3/2}\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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