Integrand size = 19, antiderivative size = 39 \[ \int \frac {(1-x)^2}{\left (1-x^3\right )^{4/3}} \, dx=\frac {1+(1-2 x) x}{\sqrt [3]{1-x^3}}+x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},x^3\right ) \]
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Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1868, 12, 371} \[ \int \frac {(1-x)^2}{\left (1-x^3\right )^{4/3}} \, dx=x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},x^3\right )+\frac {(1-2 x) x+1}{\sqrt [3]{1-x^3}} \]
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Rule 12
Rule 371
Rule 1868
Rubi steps \begin{align*} \text {integral}& = \frac {1+(1-2 x) x}{\sqrt [3]{1-x^3}}-\int -\frac {2 x}{\sqrt [3]{1-x^3}} \, dx \\ & = \frac {1+(1-2 x) x}{\sqrt [3]{1-x^3}}+2 \int \frac {x}{\sqrt [3]{1-x^3}} \, dx \\ & = \frac {1+(1-2 x) x}{\sqrt [3]{1-x^3}}+x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},x^3\right ) \\ \end{align*}
Time = 10.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.10 \[ \int \frac {(1-x)^2}{\left (1-x^3\right )^{4/3}} \, dx=\frac {1}{\sqrt [3]{1-x^3}}+\frac {x}{\sqrt [3]{1-x^3}}-x^2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {4}{3},\frac {5}{3},x^3\right ) \]
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Time = 0.53 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.87
method | result | size |
risch | \(-\frac {\left (-1+x \right ) \left (1+2 x \right )}{\left (-x^{3}+1\right )^{\frac {1}{3}}}+x^{2} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};x^{3}\right )\) | \(34\) |
meijerg | \(\frac {x}{\left (-x^{3}+1\right )^{\frac {1}{3}}}-x^{2} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {2}{3},\frac {4}{3};\frac {5}{3};x^{3}\right )+\frac {x^{3} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (1,\frac {4}{3};2;x^{3}\right )}{3}\) | \(41\) |
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\[ \int \frac {(1-x)^2}{\left (1-x^3\right )^{4/3}} \, dx=\int { \frac {{\left (x - 1\right )}^{2}}{{\left (-x^{3} + 1\right )}^{\frac {4}{3}}} \,d x } \]
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\[ \int \frac {(1-x)^2}{\left (1-x^3\right )^{4/3}} \, dx=\int \frac {\left (x - 1\right )^{2}}{\left (- \left (x - 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac {4}{3}}}\, dx \]
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\[ \int \frac {(1-x)^2}{\left (1-x^3\right )^{4/3}} \, dx=\int { \frac {{\left (x - 1\right )}^{2}}{{\left (-x^{3} + 1\right )}^{\frac {4}{3}}} \,d x } \]
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\[ \int \frac {(1-x)^2}{\left (1-x^3\right )^{4/3}} \, dx=\int { \frac {{\left (x - 1\right )}^{2}}{{\left (-x^{3} + 1\right )}^{\frac {4}{3}}} \,d x } \]
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Timed out. \[ \int \frac {(1-x)^2}{\left (1-x^3\right )^{4/3}} \, dx=\int \frac {{\left (x-1\right )}^2}{{\left (1-x^3\right )}^{4/3}} \,d x \]
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