Optimal. Leaf size=57 \[ \frac {2 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {8 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac {16 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^3} \]
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Rubi [A]
time = 0.03, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps
used = 4, 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 {16 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^3}-\frac {8 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac {2 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b} \end {gather*}
Antiderivative was successfully verified.
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Rule 30
Rule 2188
Rule 2199
Rubi steps
\begin {align*} \int \frac {x^2}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx &=\frac {2 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {4 \int x \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx}{b}\\ &=\frac {2 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {8 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac {8 \int \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx}{3 b^2}\\ &=\frac {2 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {8 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac {8 \text {Subst}\left (\int x^{3/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^3}\\ &=\frac {2 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {8 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac {16 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^3}\\ \end {align*}
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Mathematica [A]
time = 0.03, size = 49, normalized size = 0.86 \begin {gather*} \frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (15 b^2 x^2-20 b x \tanh ^{-1}(\tanh (a+b x))+8 \tanh ^{-1}(\tanh (a+b x))^2\right )}{15 b^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.06, size = 68, normalized size = 1.19
method | result | size |
default | \(\frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (-2 \arctanh \left (\tanh \left (b x +a \right )\right )+2 b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}+2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{b^{3}}\) | \(68\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.53, size = 42, normalized size = 0.74 \begin {gather*} \frac {2 \, {\left (3 \, b^{3} x^{3} - a b^{2} x^{2} + 4 \, a^{2} b x + 8 \, a^{3}\right )}}{15 \, \sqrt {b x + a} b^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.34, size = 31, normalized size = 0.54 \begin {gather*} \frac {2 \, {\left (3 \, b^{2} x^{2} - 4 \, a b x + 8 \, a^{2}\right )} \sqrt {b x + a}}{15 \, b^{3}} \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 \frac {x^{2}}{\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 = 0.38, size = 37, normalized size = 0.65 \begin {gather*} \frac {2 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )}}{15 \, b^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.14, size = 211, normalized size = 3.70 \begin {gather*} \frac {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 (15\,b^2\,x^2-10\,b\,x\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+10\,b\,x\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,{\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2-4\,\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 )+2\,{\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\right )}{15\,b^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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