Integrand size = 13, antiderivative size = 110 \[ \int x^m \coth ^{-1}(\tanh (a+b x))^3 \, dx=-\frac {6 b^3 x^{4+m}}{(1+m) \left (24+26 m+9 m^2+m^3\right )}+\frac {6 b^2 x^{3+m} \coth ^{-1}(\tanh (a+b x))}{6+11 m+6 m^2+m^3}-\frac {3 b x^{2+m} \coth ^{-1}(\tanh (a+b x))^2}{2+3 m+m^2}+\frac {x^{1+m} \coth ^{-1}(\tanh (a+b x))^3}{1+m} \]
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Time = 0.04 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2199, 30} \[ \int x^m \coth ^{-1}(\tanh (a+b x))^3 \, dx=\frac {6 b^2 x^{m+3} \coth ^{-1}(\tanh (a+b x))}{m^3+6 m^2+11 m+6}-\frac {3 b x^{m+2} \coth ^{-1}(\tanh (a+b x))^2}{m^2+3 m+2}+\frac {x^{m+1} \coth ^{-1}(\tanh (a+b x))^3}{m+1}-\frac {6 b^3 x^{m+4}}{(m+1) \left (m^3+9 m^2+26 m+24\right )} \]
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Rule 30
Rule 2199
Rubi steps \begin{align*} \text {integral}& = \frac {x^{1+m} \coth ^{-1}(\tanh (a+b x))^3}{1+m}-\frac {(3 b) \int x^{1+m} \coth ^{-1}(\tanh (a+b x))^2 \, dx}{1+m} \\ & = -\frac {3 b x^{2+m} \coth ^{-1}(\tanh (a+b x))^2}{2+3 m+m^2}+\frac {x^{1+m} \coth ^{-1}(\tanh (a+b x))^3}{1+m}+\frac {\left (6 b^2\right ) \int x^{2+m} \coth ^{-1}(\tanh (a+b x)) \, dx}{2+3 m+m^2} \\ & = \frac {6 b^2 x^{3+m} \coth ^{-1}(\tanh (a+b x))}{6+11 m+6 m^2+m^3}-\frac {3 b x^{2+m} \coth ^{-1}(\tanh (a+b x))^2}{2+3 m+m^2}+\frac {x^{1+m} \coth ^{-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} \coth ^{-1}(\tanh (a+b x))}{6+11 m+6 m^2+m^3}-\frac {3 b x^{2+m} \coth ^{-1}(\tanh (a+b x))^2}{2+3 m+m^2}+\frac {x^{1+m} \coth ^{-1}(\tanh (a+b x))^3}{1+m} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.88 \[ \int x^m \coth ^{-1}(\tanh (a+b x))^3 \, dx=\frac {x^{1+m} \left (-6 b^3 x^3+6 b^2 (4+m) x^2 \coth ^{-1}(\tanh (a+b x))-3 b \left (12+7 m+m^2\right ) x \coth ^{-1}(\tanh (a+b x))^2+\left (24+26 m+9 m^2+m^3\right ) \coth ^{-1}(\tanh (a+b x))^3\right )}{(1+m) (2+m) (3+m) (4+m)} \]
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Time = 6.03 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.79
method | result | size |
parallelrisch | \(-\frac {36 b \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2} x^{m} x^{2}-24 b^{2} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right ) x^{m} x^{3}-x \,x^{m} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{3} m^{3}-9 x \,x^{m} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{3} m^{2}-26 x \,x^{m} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{3} m +6 b^{3} x^{m} x^{4}-24 \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{3} x \,x^{m}-6 x^{3} x^{m} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right ) b^{2} m +3 x^{2} x^{m} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2} b \,m^{2}+21 x^{2} x^{m} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2} b m}{\left (1+m \right ) \left (m^{3}+9 m^{2}+26 m +24\right )}\) | \(197\) |
risch | \(\text {Expression too large to display}\) | \(63382\) |
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Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 627, normalized size of antiderivative = 5.70 \[ \int x^m \coth ^{-1}(\tanh (a+b x))^3 \, dx=\frac {{\left (-i \, \pi ^{3} {\left (m^{3} + 9 \, m^{2} + 26 \, m + 24\right )} x + 8 \, {\left (b^{3} m^{3} + 6 \, b^{3} m^{2} + 11 \, b^{3} m + 6 \, b^{3}\right )} x^{4} + 24 \, {\left (a b^{2} m^{3} + 7 \, a b^{2} m^{2} + 14 \, a b^{2} m + 8 \, a b^{2}\right )} x^{3} - 6 \, \pi ^{2} {\left ({\left (b m^{3} + 8 \, b m^{2} + 19 \, b m + 12 \, b\right )} x^{2} + {\left (a m^{3} + 9 \, a m^{2} + 26 \, a m + 24 \, a\right )} x\right )} + 24 \, {\left (a^{2} b m^{3} + 8 \, a^{2} b m^{2} + 19 \, a^{2} b m + 12 \, a^{2} b\right )} x^{2} + 12 i \, \pi {\left ({\left (b^{2} m^{3} + 7 \, b^{2} m^{2} + 14 \, b^{2} m + 8 \, b^{2}\right )} x^{3} + 2 \, {\left (a b m^{3} + 8 \, a b m^{2} + 19 \, a b m + 12 \, a b\right )} x^{2} + {\left (a^{2} m^{3} + 9 \, a^{2} m^{2} + 26 \, a^{2} m + 24 \, a^{2}\right )} x\right )} + 8 \, {\left (a^{3} m^{3} + 9 \, a^{3} m^{2} + 26 \, a^{3} m + 24 \, a^{3}\right )} x\right )} \cosh \left (m \log \left (x\right )\right ) + {\left (-i \, \pi ^{3} {\left (m^{3} + 9 \, m^{2} + 26 \, m + 24\right )} x + 8 \, {\left (b^{3} m^{3} + 6 \, b^{3} m^{2} + 11 \, b^{3} m + 6 \, b^{3}\right )} x^{4} + 24 \, {\left (a b^{2} m^{3} + 7 \, a b^{2} m^{2} + 14 \, a b^{2} m + 8 \, a b^{2}\right )} x^{3} - 6 \, \pi ^{2} {\left ({\left (b m^{3} + 8 \, b m^{2} + 19 \, b m + 12 \, b\right )} x^{2} + {\left (a m^{3} + 9 \, a m^{2} + 26 \, a m + 24 \, a\right )} x\right )} + 24 \, {\left (a^{2} b m^{3} + 8 \, a^{2} b m^{2} + 19 \, a^{2} b m + 12 \, a^{2} b\right )} x^{2} + 12 i \, \pi {\left ({\left (b^{2} m^{3} + 7 \, b^{2} m^{2} + 14 \, b^{2} m + 8 \, b^{2}\right )} x^{3} + 2 \, {\left (a b m^{3} + 8 \, a b m^{2} + 19 \, a b m + 12 \, a b\right )} x^{2} + {\left (a^{2} m^{3} + 9 \, a^{2} m^{2} + 26 \, a^{2} m + 24 \, a^{2}\right )} x\right )} + 8 \, {\left (a^{3} m^{3} + 9 \, a^{3} m^{2} + 26 \, a^{3} m + 24 \, a^{3}\right )} x\right )} \sinh \left (m \log \left (x\right )\right )}{8 \, {\left (m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24\right )}} \]
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\[ \int x^m \coth ^{-1}(\tanh (a+b x))^3 \, dx=\begin {cases} b^{3} \log {\left (x \right )} - \frac {b^{2} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{x} - \frac {b \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{2 x^{2}} - \frac {\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{3 x^{3}} & \text {for}\: m = -4 \\\int \frac {\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{3}}\, dx & \text {for}\: m = -3 \\\int \frac {\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{2}}\, dx & \text {for}\: m = -2 \\\int \frac {\operatorname {acoth}^{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 {acoth}{\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 {acoth}{\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 {acoth}^{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 {acoth}^{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 {acoth}^{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 {acoth}^{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 {acoth}^{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 {acoth}^{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 {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} & \text {otherwise} \end {cases} \]
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Time = 0.31 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.99 \[ \int x^m \coth ^{-1}(\tanh (a+b x))^3 \, dx=-\frac {3 \, b x^{2} x^{m} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}{{\left (m + 2\right )} {\left (m + 1\right )}} + \frac {x^{m + 1} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}{m + 1} - \frac {6 \, {\left (\frac {b^{2} x^{4} x^{m}}{{\left (m + 4\right )} {\left (m + 3\right )} {\left (m + 2\right )}} - \frac {b x^{3} x^{m} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )}{{\left (m + 3\right )} {\left (m + 2\right )}}\right )} b}{m + 1} \]
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\[ \int x^m \coth ^{-1}(\tanh (a+b x))^3 \, dx=\int { x^{m} \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3} \,d x } \]
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Time = 4.04 (sec) , antiderivative size = 332, normalized size of antiderivative = 3.02 \[ \int x^m \coth ^{-1}(\tanh (a+b x))^3 \, dx=\frac {8\,b^3\,x^m\,x^4\,\left (m^3+6\,m^2+11\,m+6\right )}{8\,m^4+80\,m^3+280\,m^2+400\,m+192}-\frac {x\,x^m\,{\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^3\,\left (m^3+9\,m^2+26\,m+24\right )}{8\,m^4+80\,m^3+280\,m^2+400\,m+192}-\frac {12\,b^2\,x^m\,x^3\,\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )\,\left (m^3+7\,m^2+14\,m+8\right )}{8\,m^4+80\,m^3+280\,m^2+400\,m+192}+\frac {6\,b\,x^m\,x^2\,{\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2\,\left (m^3+8\,m^2+19\,m+12\right )}{8\,m^4+80\,m^3+280\,m^2+400\,m+192} \]
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