Integrand size = 11, antiderivative size = 23 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^4} \, dx=-\frac {b}{6 x^2}-\frac {\coth ^{-1}(\tanh (a+b x))}{3 x^3} \]
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Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2199, 30} \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^4} \, dx=-\frac {\coth ^{-1}(\tanh (a+b x))}{3 x^3}-\frac {b}{6 x^2} \]
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Rule 30
Rule 2199
Rubi steps \begin{align*} \text {integral}& = -\frac {\coth ^{-1}(\tanh (a+b x))}{3 x^3}+\frac {1}{3} b \int \frac {1}{x^3} \, dx \\ & = -\frac {b}{6 x^2}-\frac {\coth ^{-1}(\tanh (a+b x))}{3 x^3} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^4} \, dx=-\frac {b x+2 \coth ^{-1}(\tanh (a+b x))}{6 x^3} \]
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Time = 0.32 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83
method | result | size |
parallelrisch | \(-\frac {b x +2 \,\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{6 x^{3}}\) | \(19\) |
default | \(-\frac {b}{6 x^{2}}-\frac {\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{3 x^{3}}\) | \(20\) |
parts | \(-\frac {b}{6 x^{2}}-\frac {\operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}{3 x^{3}}\) | \(20\) |
risch | \(-\frac {\ln \left ({\mathrm e}^{b x +a}\right )}{3 x^{3}}-\frac {2 b x +2 i \pi \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+i \pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-i \pi \operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )+2 i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}-2 i \pi \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}-i \pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )-i \pi \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-2 i \pi }{12 x^{3}}\) | \(337\) |
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Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^4} \, dx=\frac {-i \, \pi - 3 \, b x - 2 \, a}{6 \, x^{3}} \]
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Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^4} \, dx=- \frac {b}{6 x^{2}} - \frac {\operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}{3 x^{3}} \]
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none
Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^4} \, dx=-\frac {b}{6 \, x^{2}} - \frac {\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )}{3 \, x^{3}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (19) = 38\).
Time = 0.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 3.09 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^4} \, dx=-\frac {b}{6 \, x^{2}} - \frac {\log \left (-\frac {\frac {e^{\left (2 \, b x + 2 \, a\right )} + 1}{e^{\left (2 \, b x + 2 \, a\right )} - 1} + 1}{\frac {e^{\left (2 \, b x + 2 \, a\right )} + 1}{e^{\left (2 \, b x + 2 \, a\right )} - 1} - 1}\right )}{6 \, x^{3}} \]
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Time = 3.77 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {\coth ^{-1}(\tanh (a+b x))}{x^4} \, dx=-\frac {\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{3\,x^3}-\frac {b}{6\,x^2} \]
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