Integrand size = 18, antiderivative size = 29 \[ \int \frac {1}{(-1+x) \sqrt [4]{2-2 x+x^2}} \, dx=\arctan \left (\sqrt [4]{2-2 x+x^2}\right )-\text {arctanh}\left (\sqrt [4]{2-2 x+x^2}\right ) \]
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Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {708, 272, 65, 304, 209, 212} \[ \int \frac {1}{(-1+x) \sqrt [4]{2-2 x+x^2}} \, dx=\arctan \left (\sqrt [4]{(x-1)^2+1}\right )-\text {arctanh}\left (\sqrt [4]{(x-1)^2+1}\right ) \]
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Rule 65
Rule 209
Rule 212
Rule 272
Rule 304
Rule 708
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{x \sqrt [4]{1+x^2}} \, dx,x,-1+x\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x \sqrt [4]{1+x}} \, dx,x,(-1+x)^2\right ) \\ & = 2 \text {Subst}\left (\int \frac {x^2}{-1+x^4} \, dx,x,\sqrt [4]{1+(-1+x)^2}\right ) \\ & = -\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt [4]{1+(-1+x)^2}\right )+\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [4]{1+(-1+x)^2}\right ) \\ & = \arctan \left (\sqrt [4]{1+(-1+x)^2}\right )-\text {arctanh}\left (\sqrt [4]{1+(-1+x)^2}\right ) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(-1+x) \sqrt [4]{2-2 x+x^2}} \, dx=\arctan \left (\sqrt [4]{2-2 x+x^2}\right )-\text {arctanh}\left (\sqrt [4]{2-2 x+x^2}\right ) \]
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Time = 2.80 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48
method | result | size |
pseudoelliptic | \(\frac {\ln \left (\left (x^{2}-2 x +2\right )^{\frac {1}{4}}-1\right )}{2}-\frac {\ln \left (\left (x^{2}-2 x +2\right )^{\frac {1}{4}}+1\right )}{2}+\arctan \left (\left (x^{2}-2 x +2\right )^{\frac {1}{4}}\right )\) | \(43\) |
trager | \(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \left (x^{2}-2 x +2\right )^{\frac {3}{4}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{2}-2 x +2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \left (x^{2}-2 x +2\right )^{\frac {1}{4}}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{\left (x -1\right )^{2}}\right )}{2}-\frac {\ln \left (-\frac {2 \left (x^{2}-2 x +2\right )^{\frac {3}{4}}+2 \sqrt {x^{2}-2 x +2}+x^{2}+2 \left (x^{2}-2 x +2\right )^{\frac {1}{4}}-2 x +3}{\left (x -1\right )^{2}}\right )}{2}\) | \(143\) |
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Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {1}{(-1+x) \sqrt [4]{2-2 x+x^2}} \, dx=\arctan \left ({\left (x^{2} - 2 \, x + 2\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \log \left ({\left (x^{2} - 2 \, x + 2\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} - 2 \, x + 2\right )}^{\frac {1}{4}} - 1\right ) \]
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\[ \int \frac {1}{(-1+x) \sqrt [4]{2-2 x+x^2}} \, dx=\int \frac {1}{\left (x - 1\right ) \sqrt [4]{x^{2} - 2 x + 2}}\, dx \]
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\[ \int \frac {1}{(-1+x) \sqrt [4]{2-2 x+x^2}} \, dx=\int { \frac {1}{{\left (x^{2} - 2 \, x + 2\right )}^{\frac {1}{4}} {\left (x - 1\right )}} \,d x } \]
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\[ \int \frac {1}{(-1+x) \sqrt [4]{2-2 x+x^2}} \, dx=\int { \frac {1}{{\left (x^{2} - 2 \, x + 2\right )}^{\frac {1}{4}} {\left (x - 1\right )}} \,d x } \]
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Timed out. \[ \int \frac {1}{(-1+x) \sqrt [4]{2-2 x+x^2}} \, dx=\int \frac {1}{\left (x-1\right )\,{\left (x^2-2\,x+2\right )}^{1/4}} \,d x \]
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