Integrand size = 17, antiderivative size = 29 \[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{2/3}} \, dx=\frac {3 (3-x) (1+x)}{4 \left (9+3 x-5 x^2+x^3\right )^{2/3}} \]
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Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2092, 2089, 37} \[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{2/3}} \, dx=\frac {3 (3-x) (x+1)}{4 \left (x^3-5 x^2+3 x+9\right )^{2/3}} \]
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Rule 37
Rule 2089
Rule 2092
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{\left (\frac {128}{27}-\frac {16 x}{3}+x^3\right )^{2/3}} \, dx,x,-\frac {5}{3}+x\right ) \\ & = \frac {\left (512 \sqrt [3]{2} (3-x)^{4/3} (1+x)^{2/3}\right ) \text {Subst}\left (\int \frac {1}{\left (\frac {128}{9}-\frac {32 x}{3}\right )^{4/3} \left (\frac {128}{9}+\frac {16 x}{3}\right )^{2/3}} \, dx,x,-\frac {5}{3}+x\right )}{9 \left (9+3 x-5 x^2+x^3\right )^{2/3}} \\ & = \frac {3 (3-x) (1+x)}{4 \left (9+3 x-5 x^2+x^3\right )^{2/3}} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{2/3}} \, dx=-\frac {3 (-3+x) (1+x)}{4 \left ((-3+x)^2 (1+x)\right )^{2/3}} \]
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Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69
method | result | size |
risch | \(-\frac {3 \left (-3+x \right ) \left (1+x \right )}{4 \left (\left (1+x \right ) \left (-3+x \right )^{2}\right )^{\frac {2}{3}}}\) | \(20\) |
trager | \(-\frac {3 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {1}{3}}}{4 \left (-3+x \right )}\) | \(23\) |
gosper | \(-\frac {3 \left (1+x \right ) \left (-3+x \right )}{4 \left (x^{3}-5 x^{2}+3 x +9\right )^{\frac {2}{3}}}\) | \(24\) |
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Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{2/3}} \, dx=-\frac {3 \, {\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {1}{3}}}{4 \, {\left (x - 3\right )}} \]
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\[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{2/3}} \, dx=\int \frac {1}{\left (x^{3} - 5 x^{2} + 3 x + 9\right )^{\frac {2}{3}}}\, dx \]
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\[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{2/3}} \, dx=\int { \frac {1}{{\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {2}{3}}} \,d x } \]
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\[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{2/3}} \, dx=\int { \frac {1}{{\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )}^{\frac {2}{3}}} \,d x } \]
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Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\left (9+3 x-5 x^2+x^3\right )^{2/3}} \, dx=-\frac {3\,{\left (x^3-5\,x^2+3\,x+9\right )}^{1/3}}{4\,\left (x-3\right )} \]
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