Optimal. Leaf size=71 \[ -\frac {2 b x^{m+2} \tanh ^{-1}(\tanh (a+b x))}{m^2+3 m+2}+\frac {x^{m+1} \tanh ^{-1}(\tanh (a+b x))^2}{m+1}+\frac {2 b^2 x^{m+3}}{m^3+6 m^2+11 m+6} \]
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Rubi [A] time = 0.03, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2168, 30} \[ -\frac {2 b x^{m+2} \tanh ^{-1}(\tanh (a+b x))}{m^2+3 m+2}+\frac {x^{m+1} \tanh ^{-1}(\tanh (a+b x))^2}{m+1}+\frac {2 b^2 x^{m+3}}{m^3+6 m^2+11 m+6} \]
Antiderivative was successfully verified.
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Rule 30
Rule 2168
Rubi steps
\begin {align*} \int x^m \tanh ^{-1}(\tanh (a+b x))^2 \, dx &=\frac {x^{1+m} \tanh ^{-1}(\tanh (a+b x))^2}{1+m}-\frac {(2 b) \int x^{1+m} \tanh ^{-1}(\tanh (a+b x)) \, dx}{1+m}\\ &=-\frac {2 b x^{2+m} \tanh ^{-1}(\tanh (a+b x))}{2+3 m+m^2}+\frac {x^{1+m} \tanh ^{-1}(\tanh (a+b x))^2}{1+m}+\frac {\left (2 b^2\right ) \int x^{2+m} \, dx}{2+3 m+m^2}\\ &=\frac {2 b^2 x^{3+m}}{6+11 m+6 m^2+m^3}-\frac {2 b x^{2+m} \tanh ^{-1}(\tanh (a+b x))}{2+3 m+m^2}+\frac {x^{1+m} \tanh ^{-1}(\tanh (a+b x))^2}{1+m}\\ \end {align*}
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Mathematica [A] time = 0.16, size = 62, normalized size = 0.87 \[ \frac {x^{m+1} \left (\left (m^2+5 m+6\right ) \tanh ^{-1}(\tanh (a+b x))^2-2 b (m+3) x \tanh ^{-1}(\tanh (a+b x))+2 b^2 x^2\right )}{(m+1) (m+2) (m+3)} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.68, size = 161, normalized size = 2.27 \[ \frac {{\left ({\left (b^{2} m^{2} + 3 \, b^{2} m + 2 \, b^{2}\right )} x^{3} + 2 \, {\left (a b m^{2} + 4 \, a b m + 3 \, a b\right )} x^{2} + {\left (a^{2} m^{2} + 5 \, a^{2} m + 6 \, a^{2}\right )} x\right )} \cosh \left (m \log \relax (x)\right ) + {\left ({\left (b^{2} m^{2} + 3 \, b^{2} m + 2 \, b^{2}\right )} x^{3} + 2 \, {\left (a b m^{2} + 4 \, a b m + 3 \, a b\right )} x^{2} + {\left (a^{2} m^{2} + 5 \, a^{2} m + 6 \, a^{2}\right )} x\right )} \sinh \left (m \log \relax (x)\right )}{m^{3} + 6 \, m^{2} + 11 \, m + 6} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.15, size = 98, normalized size = 1.38 \[ \frac {b^{2} x^{3} {\mathrm e}^{m \ln \relax (x )}}{3+m}+\frac {\left (a^{2}+2 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right ) x \,{\mathrm e}^{m \ln \relax (x )}}{1+m}+\frac {2 b \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{2} {\mathrm e}^{m \ln \relax (x )}}{2+m} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.40, size = 73, normalized size = 1.03 \[ \frac {2 \, b^{2} x^{3} x^{m}}{{\left (m + 3\right )} {\left (m + 2\right )} {\left (m + 1\right )}} - \frac {2 \, b x^{2} x^{m} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}{{\left (m + 2\right )} {\left (m + 1\right )}} + \frac {x^{m + 1} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{m + 1} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.13, size = 203, normalized size = 2.86 \[ \frac {4\,b^2\,x^m\,x^3\,\left (m^2+3\,m+2\right )}{4\,m^3+24\,m^2+44\,m+24}+\frac {x\,x^m\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2\,\left (m^2+5\,m+6\right )}{4\,m^3+24\,m^2+44\,m+24}-\frac {4\,b\,x^m\,x^2\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )\,\left (m^2+4\,m+3\right )}{4\,m^3+24\,m^2+44\,m+24} \]
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 \[ \begin {cases} b^{2} \log {\relax (x )} - \frac {b \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{x} - \frac {\operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{2 x^{2}} & \text {for}\: m = -3 \\\int \frac {\operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{2}}\, dx & \text {for}\: m = -2 \\\int \frac {\operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{x}\, dx & \text {for}\: m = -1 \\\frac {2 b^{2} x^{3} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} - \frac {2 b m x^{2} x^{m} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{3} + 6 m^{2} + 11 m + 6} - \frac {6 b x^{2} x^{m} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {m^{2} x x^{m} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {5 m x x^{m} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac {6 x x^{m} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{3} + 6 m^{2} + 11 m + 6} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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