Integrand size = 13, antiderivative size = 71 \[ \int e^{\coth ^{-1}(x)} (1-x)^2 x \, dx=\frac {1}{8} \sqrt {1-\frac {1}{x^2}} x^2-\frac {1}{3} \left (1-\frac {1}{x^2}\right )^{3/2} x^3+\frac {1}{4} \left (1-\frac {1}{x^2}\right )^{3/2} x^4-\frac {1}{8} \text {arctanh}\left (\sqrt {1-\frac {1}{x^2}}\right ) \]
[Out]
Time = 0.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {6310, 6313, 849, 821, 272, 43, 65, 212} \[ \int e^{\coth ^{-1}(x)} (1-x)^2 x \, dx=-\frac {1}{8} \text {arctanh}\left (\sqrt {1-\frac {1}{x^2}}\right )+\frac {1}{8} \sqrt {1-\frac {1}{x^2}} x^2+\frac {1}{4} \left (1-\frac {1}{x^2}\right )^{3/2} x^4-\frac {1}{3} \left (1-\frac {1}{x^2}\right )^{3/2} x^3 \]
[In]
[Out]
Rule 43
Rule 65
Rule 212
Rule 272
Rule 821
Rule 849
Rule 6310
Rule 6313
Rubi steps \begin{align*} \text {integral}& = \int e^{\coth ^{-1}(x)} \left (1-\frac {1}{x}\right )^2 x^3 \, dx \\ & = -\text {Subst}\left (\int \frac {(1-x) \sqrt {1-x^2}}{x^5} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{4} \left (1-\frac {1}{x^2}\right )^{3/2} x^4+\frac {1}{4} \text {Subst}\left (\int \frac {(4-x) \sqrt {1-x^2}}{x^4} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {1}{3} \left (1-\frac {1}{x^2}\right )^{3/2} x^3+\frac {1}{4} \left (1-\frac {1}{x^2}\right )^{3/2} x^4-\frac {1}{4} \text {Subst}\left (\int \frac {\sqrt {1-x^2}}{x^3} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {1}{3} \left (1-\frac {1}{x^2}\right )^{3/2} x^3+\frac {1}{4} \left (1-\frac {1}{x^2}\right )^{3/2} x^4-\frac {1}{8} \text {Subst}\left (\int \frac {\sqrt {1-x}}{x^2} \, dx,x,\frac {1}{x^2}\right ) \\ & = \frac {1}{8} \sqrt {1-\frac {1}{x^2}} x^2-\frac {1}{3} \left (1-\frac {1}{x^2}\right )^{3/2} x^3+\frac {1}{4} \left (1-\frac {1}{x^2}\right )^{3/2} x^4+\frac {1}{16} \text {Subst}\left (\int \frac {1}{\sqrt {1-x} x} \, dx,x,\frac {1}{x^2}\right ) \\ & = \frac {1}{8} \sqrt {1-\frac {1}{x^2}} x^2-\frac {1}{3} \left (1-\frac {1}{x^2}\right )^{3/2} x^3+\frac {1}{4} \left (1-\frac {1}{x^2}\right )^{3/2} x^4-\frac {1}{8} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1-\frac {1}{x^2}}\right ) \\ & = \frac {1}{8} \sqrt {1-\frac {1}{x^2}} x^2-\frac {1}{3} \left (1-\frac {1}{x^2}\right )^{3/2} x^3+\frac {1}{4} \left (1-\frac {1}{x^2}\right )^{3/2} x^4-\frac {1}{8} \text {arctanh}\left (\sqrt {1-\frac {1}{x^2}}\right ) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.73 \[ \int e^{\coth ^{-1}(x)} (1-x)^2 x \, dx=\frac {1}{24} \sqrt {1-\frac {1}{x^2}} x \left (8-3 x-8 x^2+6 x^3\right )-\frac {1}{8} \log \left (\left (1+\sqrt {1-\frac {1}{x^2}}\right ) x\right ) \]
[In]
[Out]
Time = 0.46 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.99
method | result | size |
default | \(\frac {\left (x -1\right ) \left (6 x \left (x^{2}-1\right )^{\frac {3}{2}}-8 \left (\left (x -1\right ) \left (1+x \right )\right )^{\frac {3}{2}}+3 x \sqrt {x^{2}-1}-3 \ln \left (x +\sqrt {x^{2}-1}\right )\right )}{24 \sqrt {\frac {x -1}{1+x}}\, \sqrt {\left (x -1\right ) \left (1+x \right )}}\) | \(70\) |
risch | \(\frac {\left (6 x^{3}-8 x^{2}-3 x +8\right ) \left (x -1\right )}{24 \sqrt {\frac {x -1}{1+x}}}-\frac {\ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {\left (x -1\right ) \left (1+x \right )}}{8 \sqrt {\frac {x -1}{1+x}}\, \left (1+x \right )}\) | \(70\) |
trager | \(\frac {\left (1+x \right ) \left (6 x^{3}-8 x^{2}-3 x +8\right ) \sqrt {-\frac {1-x}{1+x}}}{24}-\frac {\ln \left (\sqrt {-\frac {1-x}{1+x}}\, x +\sqrt {-\frac {1-x}{1+x}}+x \right )}{8}\) | \(71\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.93 \[ \int e^{\coth ^{-1}(x)} (1-x)^2 x \, dx=\frac {1}{24} \, {\left (6 \, x^{4} - 2 \, x^{3} - 11 \, x^{2} + 5 \, x + 8\right )} \sqrt {\frac {x - 1}{x + 1}} - \frac {1}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) + \frac {1}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]
[In]
[Out]
\[ \int e^{\coth ^{-1}(x)} (1-x)^2 x \, dx=\int \frac {x \left (x - 1\right )^{2}}{\sqrt {\frac {x - 1}{x + 1}}}\, dx \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (55) = 110\).
Time = 0.21 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.94 \[ \int e^{\coth ^{-1}(x)} (1-x)^2 x \, dx=-\frac {3 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {7}{2}} + 53 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {5}{2}} - 11 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {3}{2}} + 3 \, \sqrt {\frac {x - 1}{x + 1}}}{12 \, {\left (\frac {4 \, {\left (x - 1\right )}}{x + 1} - \frac {6 \, {\left (x - 1\right )}^{2}}{{\left (x + 1\right )}^{2}} + \frac {4 \, {\left (x - 1\right )}^{3}}{{\left (x + 1\right )}^{3}} - \frac {{\left (x - 1\right )}^{4}}{{\left (x + 1\right )}^{4}} - 1\right )}} - \frac {1}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) + \frac {1}{8} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.01 \[ \int e^{\coth ^{-1}(x)} (1-x)^2 x \, dx=\frac {1}{24} \, {\left ({\left (2 \, x {\left (\frac {3 \, x}{\mathrm {sgn}\left (x + 1\right )} - \frac {4}{\mathrm {sgn}\left (x + 1\right )}\right )} - \frac {3}{\mathrm {sgn}\left (x + 1\right )}\right )} x + \frac {8}{\mathrm {sgn}\left (x + 1\right )}\right )} \sqrt {x^{2} - 1} + \frac {\log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right )}{8 \, \mathrm {sgn}\left (x + 1\right )} \]
[In]
[Out]
Time = 4.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.66 \[ \int e^{\coth ^{-1}(x)} (1-x)^2 x \, dx=\frac {\frac {\sqrt {\frac {x-1}{x+1}}}{4}-\frac {11\,{\left (\frac {x-1}{x+1}\right )}^{3/2}}{12}+\frac {53\,{\left (\frac {x-1}{x+1}\right )}^{5/2}}{12}+\frac {{\left (\frac {x-1}{x+1}\right )}^{7/2}}{4}}{\frac {6\,{\left (x-1\right )}^2}{{\left (x+1\right )}^2}-\frac {4\,\left (x-1\right )}{x+1}-\frac {4\,{\left (x-1\right )}^3}{{\left (x+1\right )}^3}+\frac {{\left (x-1\right )}^4}{{\left (x+1\right )}^4}+1}-\frac {\mathrm {atanh}\left (\sqrt {\frac {x-1}{x+1}}\right )}{4} \]
[In]
[Out]