Integrand size = 26, antiderivative size = 29 \[ \int \frac {-1+x}{(-3+x) (1+x) \sqrt [4]{-2-2 x+x^2}} \, dx=\arctan \left (\sqrt [4]{-2-2 x+x^2}\right )-\text {arctanh}\left (\sqrt [4]{-2-2 x+x^2}\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(260\) vs. \(2(29)=58\).
Time = 0.75 (sec) , antiderivative size = 260, normalized size of antiderivative = 8.97, number of steps used = 42, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {6874, 763, 762, 760, 408, 504, 1227, 551, 455, 65, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {-1+x}{(-3+x) (1+x) \sqrt [4]{-2-2 x+x^2}} \, dx=-\frac {\sqrt [4]{-x^2+2 x+2} \arctan \left (1-\sqrt {2} \sqrt [4]{3-(1-x)^2}\right )}{\sqrt {2} \sqrt [4]{x^2-2 x-2}}+\frac {\sqrt [4]{-x^2+2 x+2} \arctan \left (\sqrt {2} \sqrt [4]{3-(1-x)^2}+1\right )}{\sqrt {2} \sqrt [4]{x^2-2 x-2}}+\frac {\sqrt [4]{-x^2+2 x+2} \log \left (\sqrt {-x^2+2 x+2}-\sqrt {2} \sqrt [4]{-x^2+2 x+2}+1\right )}{2 \sqrt {2} \sqrt [4]{x^2-2 x-2}}-\frac {\sqrt [4]{-x^2+2 x+2} \log \left (\sqrt {-x^2+2 x+2}+\sqrt {2} \sqrt [4]{-x^2+2 x+2}+1\right )}{2 \sqrt {2} \sqrt [4]{x^2-2 x-2}} \]
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Rule 65
Rule 210
Rule 303
Rule 408
Rule 455
Rule 504
Rule 551
Rule 631
Rule 642
Rule 760
Rule 762
Rule 763
Rule 1176
Rule 1179
Rule 1227
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{2 (-3+x) \sqrt [4]{-2-2 x+x^2}}+\frac {1}{2 (1+x) \sqrt [4]{-2-2 x+x^2}}\right ) \, dx \\ & = \frac {1}{2} \int \frac {1}{(-3+x) \sqrt [4]{-2-2 x+x^2}} \, dx+\frac {1}{2} \int \frac {1}{(1+x) \sqrt [4]{-2-2 x+x^2}} \, dx \\ & = \frac {\sqrt [4]{2+2 x-x^2} \int \frac {1}{(-3+x) \sqrt [4]{\frac {1}{6}+\frac {x}{6}-\frac {x^2}{12}}} \, dx}{2 \sqrt {2} \sqrt [4]{3} \sqrt [4]{-2-2 x+x^2}}+\frac {\sqrt [4]{2+2 x-x^2} \int \frac {1}{(1+x) \sqrt [4]{\frac {1}{6}+\frac {x}{6}-\frac {x^2}{12}}} \, dx}{2 \sqrt {2} \sqrt [4]{3} \sqrt [4]{-2-2 x+x^2}} \\ & = \frac {\sqrt [4]{2+2 x-x^2} \text {Subst}\left (\int \frac {1}{\left (-\frac {1}{3}+x\right ) \sqrt [4]{1-12 x^2}} \, dx,x,\frac {1}{6}-\frac {x}{6}\right )}{2 \sqrt [4]{3} \sqrt [4]{-2-2 x+x^2}}+\frac {\sqrt [4]{2+2 x-x^2} \text {Subst}\left (\int \frac {1}{\left (\frac {1}{3}+x\right ) \sqrt [4]{1-12 x^2}} \, dx,x,\frac {1}{6}-\frac {x}{6}\right )}{2 \sqrt [4]{3} \sqrt [4]{-2-2 x+x^2}} \\ & = -2 \frac {\sqrt [4]{2+2 x-x^2} \text {Subst}\left (\int \frac {x}{\sqrt [4]{1-12 x^2} \left (\frac {1}{9}-x^2\right )} \, dx,x,\frac {1}{6}-\frac {x}{6}\right )}{2 \sqrt [4]{3} \sqrt [4]{-2-2 x+x^2}} \\ & = -2 \frac {\sqrt [4]{2+2 x-x^2} \text {Subst}\left (\int \frac {1}{\sqrt [4]{1-12 x} \left (\frac {1}{9}-x\right )} \, dx,x,\left (\frac {1}{6}-\frac {x}{6}\right )^2\right )}{4 \sqrt [4]{3} \sqrt [4]{-2-2 x+x^2}} \\ & = 2 \frac {\sqrt [4]{2+2 x-x^2} \text {Subst}\left (\int \frac {x^2}{\frac {1}{36}+\frac {x^4}{12}} \, dx,x,\frac {\sqrt [4]{3-(-1+x)^2}}{\sqrt [4]{3}}\right )}{12 \sqrt [4]{3} \sqrt [4]{-2-2 x+x^2}} \\ & = 2 \left (-\frac {\sqrt [4]{2+2 x-x^2} \text {Subst}\left (\int \frac {1-\sqrt {3} x^2}{\frac {1}{36}+\frac {x^4}{12}} \, dx,x,\frac {\sqrt [4]{3-(-1+x)^2}}{\sqrt [4]{3}}\right )}{24\ 3^{3/4} \sqrt [4]{-2-2 x+x^2}}+\frac {\sqrt [4]{2+2 x-x^2} \text {Subst}\left (\int \frac {1+\sqrt {3} x^2}{\frac {1}{36}+\frac {x^4}{12}} \, dx,x,\frac {\sqrt [4]{3-(-1+x)^2}}{\sqrt [4]{3}}\right )}{24\ 3^{3/4} \sqrt [4]{-2-2 x+x^2}}\right ) \\ & = 2 \left (\frac {\sqrt [4]{2+2 x-x^2} \text {Subst}\left (\int \frac {\frac {\sqrt {2}}{\sqrt [4]{3}}+2 x}{-\frac {1}{\sqrt {3}}-\frac {\sqrt {2} x}{\sqrt [4]{3}}-x^2} \, dx,x,\frac {\sqrt [4]{3-(-1+x)^2}}{\sqrt [4]{3}}\right )}{4 \sqrt {2} \sqrt [4]{-2-2 x+x^2}}+\frac {\sqrt [4]{2+2 x-x^2} \text {Subst}\left (\int \frac {\frac {\sqrt {2}}{\sqrt [4]{3}}-2 x}{-\frac {1}{\sqrt {3}}+\frac {\sqrt {2} x}{\sqrt [4]{3}}-x^2} \, dx,x,\frac {\sqrt [4]{3-(-1+x)^2}}{\sqrt [4]{3}}\right )}{4 \sqrt {2} \sqrt [4]{-2-2 x+x^2}}+\frac {\sqrt [4]{2+2 x-x^2} \text {Subst}\left (\int \frac {1}{\frac {1}{\sqrt {3}}-\frac {\sqrt {2} x}{\sqrt [4]{3}}+x^2} \, dx,x,\frac {\sqrt [4]{3-(-1+x)^2}}{\sqrt [4]{3}}\right )}{4 \sqrt [4]{3} \sqrt [4]{-2-2 x+x^2}}+\frac {\sqrt [4]{2+2 x-x^2} \text {Subst}\left (\int \frac {1}{\frac {1}{\sqrt {3}}+\frac {\sqrt {2} x}{\sqrt [4]{3}}+x^2} \, dx,x,\frac {\sqrt [4]{3-(-1+x)^2}}{\sqrt [4]{3}}\right )}{4 \sqrt [4]{3} \sqrt [4]{-2-2 x+x^2}}\right ) \\ & = 2 \left (\frac {\sqrt [4]{2+2 x-x^2} \log \left (\sqrt {3}-\sqrt {6} \sqrt [4]{3-(1-x)^2}+\sqrt {3} \sqrt {3-(1-x)^2}\right )}{4 \sqrt {2} \sqrt [4]{-2-2 x+x^2}}-\frac {\sqrt [4]{2+2 x-x^2} \log \left (\sqrt {3}+\sqrt {6} \sqrt [4]{3-(1-x)^2}+\sqrt {3} \sqrt {3-(1-x)^2}\right )}{4 \sqrt {2} \sqrt [4]{-2-2 x+x^2}}+\frac {\sqrt [4]{2+2 x-x^2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt [4]{3-(-1+x)^2}\right )}{2 \sqrt {2} \sqrt [4]{-2-2 x+x^2}}-\frac {\sqrt [4]{2+2 x-x^2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt [4]{3-(-1+x)^2}\right )}{2 \sqrt {2} \sqrt [4]{-2-2 x+x^2}}\right ) \\ & = 2 \left (-\frac {\sqrt [4]{2+2 x-x^2} \arctan \left (1-\sqrt {2} \sqrt [4]{3-(1-x)^2}\right )}{2 \sqrt {2} \sqrt [4]{-2-2 x+x^2}}+\frac {\sqrt [4]{2+2 x-x^2} \arctan \left (1+\sqrt {2} \sqrt [4]{3-(1-x)^2}\right )}{2 \sqrt {2} \sqrt [4]{-2-2 x+x^2}}+\frac {\sqrt [4]{2+2 x-x^2} \log \left (\sqrt {3}-\sqrt {6} \sqrt [4]{3-(1-x)^2}+\sqrt {3} \sqrt {3-(1-x)^2}\right )}{4 \sqrt {2} \sqrt [4]{-2-2 x+x^2}}-\frac {\sqrt [4]{2+2 x-x^2} \log \left (\sqrt {3}+\sqrt {6} \sqrt [4]{3-(1-x)^2}+\sqrt {3} \sqrt {3-(1-x)^2}\right )}{4 \sqrt {2} \sqrt [4]{-2-2 x+x^2}}\right ) \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {-1+x}{(-3+x) (1+x) \sqrt [4]{-2-2 x+x^2}} \, dx=\arctan \left (\sqrt [4]{-2-2 x+x^2}\right )-\text {arctanh}\left (\sqrt [4]{-2-2 x+x^2}\right ) \]
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Time = 2.54 (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 | \(\frac {\ln \left (\frac {2 \left (x^{2}-2 x -2\right )^{\frac {3}{4}}-2 \sqrt {x^{2}-2 x -2}-x^{2}+2 \left (x^{2}-2 x -2\right )^{\frac {1}{4}}+2 x +1}{\left (-3+x \right ) \left (1+x \right )}\right )}{2}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{2}-2 x -2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}-2 \left (x^{2}-2 x -2\right )^{\frac {3}{4}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+2 \left (x^{2}-2 x -2\right )^{\frac {1}{4}}}{\left (-3+x \right ) \left (1+x \right )}\right )}{2}\) | \(152\) |
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Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {-1+x}{(-3+x) (1+x) \sqrt [4]{-2-2 x+x^2}} \, dx=\arctan \left ({\left (x^{2} - 2 \, x - 2\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \log \left ({\left (x^{2} - 2 \, x - 2\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} - 2 \, x - 2\right )}^{\frac {1}{4}} - 1\right ) \]
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\[ \int \frac {-1+x}{(-3+x) (1+x) \sqrt [4]{-2-2 x+x^2}} \, dx=\int \frac {x - 1}{\left (x - 3\right ) \left (x + 1\right ) \sqrt [4]{x^{2} - 2 x - 2}}\, dx \]
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\[ \int \frac {-1+x}{(-3+x) (1+x) \sqrt [4]{-2-2 x+x^2}} \, dx=\int { \frac {x - 1}{{\left (x^{2} - 2 \, x - 2\right )}^{\frac {1}{4}} {\left (x + 1\right )} {\left (x - 3\right )}} \,d x } \]
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\[ \int \frac {-1+x}{(-3+x) (1+x) \sqrt [4]{-2-2 x+x^2}} \, dx=\int { \frac {x - 1}{{\left (x^{2} - 2 \, x - 2\right )}^{\frac {1}{4}} {\left (x + 1\right )} {\left (x - 3\right )}} \,d x } \]
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Time = 5.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {-1+x}{(-3+x) (1+x) \sqrt [4]{-2-2 x+x^2}} \, dx=\mathrm {atan}\left ({\left (x^2-2\,x-2\right )}^{1/4}\right )-\mathrm {atanh}\left ({\left (x^2-2\,x-2\right )}^{1/4}\right ) \]
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