Optimal. Leaf size=110 \[ \frac {6 b^2 x^{m+3} \tanh ^{-1}(\tanh (a+b x))}{m^3+6 m^2+11 m+6}-\frac {3 b x^{m+2} \tanh ^{-1}(\tanh (a+b x))^2}{m^2+3 m+2}+\frac {x^{m+1} \tanh ^{-1}(\tanh (a+b x))^3}{m+1}-\frac {6 b^3 x^{m+4}}{(m+1) \left (m^3+9 m^2+26 m+24\right )} \]
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Rubi [A] time = 0.06, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2168, 30} \[ \frac {6 b^2 x^{m+3} \tanh ^{-1}(\tanh (a+b x))}{m^3+6 m^2+11 m+6}-\frac {3 b x^{m+2} \tanh ^{-1}(\tanh (a+b x))^2}{m^2+3 m+2}+\frac {x^{m+1} \tanh ^{-1}(\tanh (a+b x))^3}{m+1}-\frac {6 b^3 x^{m+4}}{(m+1) \left (m^3+9 m^2+26 m+24\right )} \]
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))^3 \, dx &=\frac {x^{1+m} \tanh ^{-1}(\tanh (a+b x))^3}{1+m}-\frac {(3 b) \int x^{1+m} \tanh ^{-1}(\tanh (a+b x))^2 \, dx}{1+m}\\ &=-\frac {3 b x^{2+m} \tanh ^{-1}(\tanh (a+b x))^2}{2+3 m+m^2}+\frac {x^{1+m} \tanh ^{-1}(\tanh (a+b x))^3}{1+m}+\frac {\left (6 b^2\right ) \int x^{2+m} \tanh ^{-1}(\tanh (a+b x)) \, dx}{2+3 m+m^2}\\ &=\frac {6 b^2 x^{3+m} \tanh ^{-1}(\tanh (a+b x))}{6+11 m+6 m^2+m^3}-\frac {3 b x^{2+m} \tanh ^{-1}(\tanh (a+b x))^2}{2+3 m+m^2}+\frac {x^{1+m} \tanh ^{-1}(\tanh (a+b x))^3}{1+m}-\frac {\left (6 b^3\right ) \int x^{3+m} \, dx}{6+11 m+6 m^2+m^3}\\ &=-\frac {6 b^3 x^{4+m}}{(4+m) \left (6+11 m+6 m^2+m^3\right )}+\frac {6 b^2 x^{3+m} \tanh ^{-1}(\tanh (a+b x))}{6+11 m+6 m^2+m^3}-\frac {3 b x^{2+m} \tanh ^{-1}(\tanh (a+b x))^2}{2+3 m+m^2}+\frac {x^{1+m} \tanh ^{-1}(\tanh (a+b x))^3}{1+m}\\ \end {align*}
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Mathematica [A] time = 0.24, size = 97, normalized size = 0.88 \[ \frac {x^{m+1} \left (6 b^2 (m+4) x^2 \tanh ^{-1}(\tanh (a+b x))-3 b \left (m^2+7 m+12\right ) x \tanh ^{-1}(\tanh (a+b x))^2+\left (m^3+9 m^2+26 m+24\right ) \tanh ^{-1}(\tanh (a+b x))^3-6 b^3 x^3\right )}{(m+1) (m+2) (m+3) (m+4)} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.40, size = 300, normalized size = 2.73 \[ \frac {{\left ({\left (b^{3} m^{3} + 6 \, b^{3} m^{2} + 11 \, b^{3} m + 6 \, b^{3}\right )} x^{4} + 3 \, {\left (a b^{2} m^{3} + 7 \, a b^{2} m^{2} + 14 \, a b^{2} m + 8 \, a b^{2}\right )} x^{3} + 3 \, {\left (a^{2} b m^{3} + 8 \, a^{2} b m^{2} + 19 \, a^{2} b m + 12 \, a^{2} b\right )} x^{2} + {\left (a^{3} m^{3} + 9 \, a^{3} m^{2} + 26 \, a^{3} m + 24 \, a^{3}\right )} x\right )} \cosh \left (m \log \relax (x)\right ) + {\left ({\left (b^{3} m^{3} + 6 \, b^{3} m^{2} + 11 \, b^{3} m + 6 \, b^{3}\right )} x^{4} + 3 \, {\left (a b^{2} m^{3} + 7 \, a b^{2} m^{2} + 14 \, a b^{2} m + 8 \, a b^{2}\right )} x^{3} + 3 \, {\left (a^{2} b m^{3} + 8 \, a^{2} b m^{2} + 19 \, a^{2} b m + 12 \, a^{2} b\right )} x^{2} + {\left (a^{3} m^{3} + 9 \, a^{3} m^{2} + 26 \, a^{3} m + 24 \, a^{3}\right )} x\right )} \sinh \left (m \log \relax (x)\right )}{m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24} \]
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 \[ \int x^{m} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.14, size = 177, normalized size = 1.61 \[ \frac {b^{3} x^{4} {\mathrm e}^{m \ln \relax (x )}}{4+m}+\frac {\left (a^{3}+3 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+3 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}+\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}\right ) x \,{\mathrm e}^{m \ln \relax (x )}}{1+m}+\frac {3 b \left (a^{2}+2 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right ) x^{2} {\mathrm e}^{m \ln \relax (x )}}{2+m}+\frac {3 b^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{3} {\mathrm e}^{m \ln \relax (x )}}{3+m} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.23, size = 332, normalized size = 3.02 \[ \frac {8\,b^3\,x^m\,x^4\,\left (m^3+6\,m^2+11\,m+6\right )}{8\,m^4+80\,m^3+280\,m^2+400\,m+192}-\frac {x\,x^m\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3\,\left (m^3+9\,m^2+26\,m+24\right )}{8\,m^4+80\,m^3+280\,m^2+400\,m+192}-\frac {12\,b^2\,x^m\,x^3\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )\,\left (m^3+7\,m^2+14\,m+8\right )}{8\,m^4+80\,m^3+280\,m^2+400\,m+192}+\frac {6\,b\,x^m\,x^2\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2\,\left (m^3+8\,m^2+19\,m+12\right )}{8\,m^4+80\,m^3+280\,m^2+400\,m+192} \]
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^{3} \log {\relax (x )} - \frac {b^{2} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{x} - \frac {b \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{2 x^{2}} - \frac {\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{3 x^{3}} & \text {for}\: m = -4 \\\int \frac {\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{3}}\, dx & \text {for}\: m = -3 \\\int \frac {\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{2}}\, dx & \text {for}\: m = -2 \\\int \frac {\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{x}\, dx & \text {for}\: m = -1 \\- \frac {6 b^{3} x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {6 b^{2} m x^{3} x^{m} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {24 b^{2} x^{3} x^{m} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac {3 b m^{2} x^{2} x^{m} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac {21 b m x^{2} x^{m} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac {36 b x^{2} x^{m} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {m^{3} x x^{m} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {9 m^{2} x x^{m} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {26 m x x^{m} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {24 x x^{m} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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