Integrand size = 20, antiderivative size = 91 \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^3} \, dx=-\frac {(1-x)^{3/4} \sqrt [4]{1+x}}{4 x}-\frac {(1-x)^{3/4} (1+x)^{5/4}}{2 x^2}-\frac {1}{4} \arctan \left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\frac {1}{4} \text {arctanh}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right ) \]
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Time = 0.01 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {98, 96, 95, 218, 212, 209} \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^3} \, dx=-\frac {1}{4} \arctan \left (\frac {\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right )-\frac {1}{4} \text {arctanh}\left (\frac {\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right )-\frac {(1-x)^{3/4} (x+1)^{5/4}}{2 x^2}-\frac {(1-x)^{3/4} \sqrt [4]{x+1}}{4 x} \]
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Rule 95
Rule 96
Rule 98
Rule 209
Rule 212
Rule 218
Rubi steps \begin{align*} \text {integral}& = -\frac {(1-x)^{3/4} (1+x)^{5/4}}{2 x^2}+\frac {1}{4} \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^2} \, dx \\ & = -\frac {(1-x)^{3/4} \sqrt [4]{1+x}}{4 x}-\frac {(1-x)^{3/4} (1+x)^{5/4}}{2 x^2}+\frac {1}{8} \int \frac {1}{\sqrt [4]{1-x} x (1+x)^{3/4}} \, dx \\ & = -\frac {(1-x)^{3/4} \sqrt [4]{1+x}}{4 x}-\frac {(1-x)^{3/4} (1+x)^{5/4}}{2 x^2}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{-1+x^4} \, dx,x,\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right ) \\ & = -\frac {(1-x)^{3/4} \sqrt [4]{1+x}}{4 x}-\frac {(1-x)^{3/4} (1+x)^{5/4}}{2 x^2}-\frac {1}{4} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right ) \\ & = -\frac {(1-x)^{3/4} \sqrt [4]{1+x}}{4 x}-\frac {(1-x)^{3/4} (1+x)^{5/4}}{2 x^2}-\frac {1}{4} \tan ^{-1}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\frac {1}{4} \tanh ^{-1}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right ) \\ \end{align*}
Time = 0.32 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^3} \, dx=\frac {1}{4} \left (-\frac {(1-x)^{3/4} \sqrt [4]{1+x} (2+3 x)}{x^2}-\arctan \left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\text {arctanh}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.06 (sec) , antiderivative size = 384, normalized size of antiderivative = 4.22
method | result | size |
risch | \(\frac {\left (-1+x \right ) \left (1+x \right )^{\frac {1}{4}} \left (2+3 x \right ) \left (\left (1-x \right ) \left (1+x \right )^{3}\right )^{\frac {1}{4}}}{4 x^{2} \left (-\left (-1+x \right ) \left (1+x \right )^{3}\right )^{\frac {1}{4}} \left (1-x \right )^{\frac {1}{4}}}+\frac {\left (-\frac {\ln \left (\frac {\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {3}{4}}+\sqrt {-x^{4}-2 x^{3}+2 x +1}\, x +\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x^{2}+\sqrt {-x^{4}-2 x^{3}+2 x +1}+2 \left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x +x^{2}+\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}}+2 x +1}{\left (1+x \right )^{2} x}\right )}{8}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{4}-2 x^{3}+2 x +1}\, x +\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{4}-2 x^{3}+2 x +1}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}+\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {3}{4}}-\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )-\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}}}{x \left (1+x \right )^{2}}\right )}{8}\right ) \left (\left (1-x \right ) \left (1+x \right )^{3}\right )^{\frac {1}{4}}}{\left (1+x \right )^{\frac {3}{4}} \left (1-x \right )^{\frac {1}{4}}}\) | \(384\) |
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Time = 0.23 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.16 \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^3} \, dx=\frac {2 \, x^{2} \arctan \left (\frac {{\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}}}{x - 1}\right ) + x^{2} \log \left (\frac {x + {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - 1}{x - 1}\right ) - x^{2} \log \left (-\frac {x - {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - 1}{x - 1}\right ) - 2 \, {\left (3 \, x + 2\right )} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}}}{8 \, x^{2}} \]
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\[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^3} \, dx=\int \frac {\sqrt [4]{x + 1}}{x^{3} \sqrt [4]{1 - x}}\, dx \]
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\[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^3} \, dx=\int { \frac {{\left (x + 1\right )}^{\frac {1}{4}}}{x^{3} {\left (-x + 1\right )}^{\frac {1}{4}}} \,d x } \]
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\[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^3} \, dx=\int { \frac {{\left (x + 1\right )}^{\frac {1}{4}}}{x^{3} {\left (-x + 1\right )}^{\frac {1}{4}}} \,d x } \]
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Timed out. \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^3} \, dx=\int \frac {{\left (x+1\right )}^{1/4}}{x^3\,{\left (1-x\right )}^{1/4}} \,d x \]
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