Integrand size = 63, antiderivative size = 35 \[ \int \frac {-252+180 x+e^{2 x} \left (-50+150 x-150 x^2+50 x^3\right )+e^x \left (-270+630 x-510 x^2+150 x^3\right )}{-25+75 x-75 x^2+25 x^3} \, dx=\left (e^x+x-\frac {(-1+x)^2-x-\frac {5+x}{5 (1-x)}}{x}\right )^2 \]
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Time = 0.12 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.23, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {6820, 2225, 37, 2230, 2208, 2209} \[ \int \frac {-252+180 x+e^{2 x} \left (-50+150 x-150 x^2+50 x^3\right )+e^x \left (-270+630 x-510 x^2+150 x^3\right )}{-25+75 x-75 x^2+25 x^3} \, dx=\frac {9 (7-5 x)^2}{25 (1-x)^2}+6 e^x+e^{2 x}+\frac {12 e^x}{5 (1-x)} \]
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Rule 37
Rule 2208
Rule 2209
Rule 2225
Rule 2230
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \left (2 e^{2 x}+\frac {36 (-7+5 x)}{25 (-1+x)^3}+\frac {6 e^x \left (9-12 x+5 x^2\right )}{5 (-1+x)^2}\right ) \, dx \\ & = \frac {6}{5} \int \frac {e^x \left (9-12 x+5 x^2\right )}{(-1+x)^2} \, dx+\frac {36}{25} \int \frac {-7+5 x}{(-1+x)^3} \, dx+2 \int e^{2 x} \, dx \\ & = e^{2 x}+\frac {9 (7-5 x)^2}{25 (1-x)^2}+\frac {6}{5} \int \left (5 e^x+\frac {2 e^x}{(-1+x)^2}-\frac {2 e^x}{-1+x}\right ) \, dx \\ & = e^{2 x}+\frac {9 (7-5 x)^2}{25 (1-x)^2}+\frac {12}{5} \int \frac {e^x}{(-1+x)^2} \, dx-\frac {12}{5} \int \frac {e^x}{-1+x} \, dx+6 \int e^x \, dx \\ & = 6 e^x+e^{2 x}+\frac {9 (7-5 x)^2}{25 (1-x)^2}+\frac {12 e^x}{5 (1-x)}-\frac {12}{5} e \operatorname {ExpIntegralEi}(-1+x)+\frac {12}{5} \int \frac {e^x}{-1+x} \, dx \\ & = 6 e^x+e^{2 x}+\frac {9 (7-5 x)^2}{25 (1-x)^2}+\frac {12 e^x}{5 (1-x)} \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.14 \[ \int \frac {-252+180 x+e^{2 x} \left (-50+150 x-150 x^2+50 x^3\right )+e^x \left (-270+630 x-510 x^2+150 x^3\right )}{-25+75 x-75 x^2+25 x^3} \, dx=e^{2 x}+\frac {6}{5} e^x \left (5-\frac {2}{-1+x}\right )+\frac {36}{25 (-1+x)^2}-\frac {36}{5 (-1+x)} \]
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Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94
method | result | size |
default | \(\frac {36}{25 \left (-1+x \right )^{2}}-\frac {36}{5 \left (-1+x \right )}-\frac {12 \,{\mathrm e}^{x}}{5 \left (-1+x \right )}+6 \,{\mathrm e}^{x}+{\mathrm e}^{2 x}\) | \(33\) |
parts | \(\frac {36}{25 \left (-1+x \right )^{2}}-\frac {36}{5 \left (-1+x \right )}-\frac {12 \,{\mathrm e}^{x}}{5 \left (-1+x \right )}+6 \,{\mathrm e}^{x}+{\mathrm e}^{2 x}\) | \(33\) |
risch | \(\frac {-\frac {36 x}{5}+\frac {216}{25}}{x^{2}-2 x +1}+{\mathrm e}^{2 x}+\frac {6 \left (5 x -7\right ) {\mathrm e}^{x}}{5 \left (-1+x \right )}\) | \(36\) |
norman | \(\frac {{\mathrm e}^{2 x}-\frac {36 x}{5}+{\mathrm e}^{2 x} x^{2}-2 x \,{\mathrm e}^{2 x}-\frac {72 \,{\mathrm e}^{x} x}{5}+6 \,{\mathrm e}^{x} x^{2}+\frac {42 \,{\mathrm e}^{x}}{5}+\frac {216}{25}}{\left (-1+x \right )^{2}}\) | \(47\) |
parallelrisch | \(\frac {150 \,{\mathrm e}^{x} x^{2}+25 \,{\mathrm e}^{2 x} x^{2}-360 \,{\mathrm e}^{x} x -50 x \,{\mathrm e}^{2 x}+210 \,{\mathrm e}^{x}+25 \,{\mathrm e}^{2 x}-180 x +216}{25 x^{2}-50 x +25}\) | \(56\) |
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Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.29 \[ \int \frac {-252+180 x+e^{2 x} \left (-50+150 x-150 x^2+50 x^3\right )+e^x \left (-270+630 x-510 x^2+150 x^3\right )}{-25+75 x-75 x^2+25 x^3} \, dx=\frac {25 \, {\left (x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} + 30 \, {\left (5 \, x^{2} - 12 \, x + 7\right )} e^{x} - 180 \, x + 216}{25 \, {\left (x^{2} - 2 \, x + 1\right )}} \]
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Time = 0.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int \frac {-252+180 x+e^{2 x} \left (-50+150 x-150 x^2+50 x^3\right )+e^x \left (-270+630 x-510 x^2+150 x^3\right )}{-25+75 x-75 x^2+25 x^3} \, dx=\frac {216 - 180 x}{25 x^{2} - 50 x + 25} + \frac {\left (5 x - 5\right ) e^{2 x} + \left (30 x - 42\right ) e^{x}}{5 x - 5} \]
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\[ \int \frac {-252+180 x+e^{2 x} \left (-50+150 x-150 x^2+50 x^3\right )+e^x \left (-270+630 x-510 x^2+150 x^3\right )}{-25+75 x-75 x^2+25 x^3} \, dx=\int { \frac {2 \, {\left (25 \, {\left (x^{3} - 3 \, x^{2} + 3 \, x - 1\right )} e^{\left (2 \, x\right )} + 15 \, {\left (5 \, x^{3} - 17 \, x^{2} + 21 \, x - 9\right )} e^{x} + 90 \, x - 126\right )}}{25 \, {\left (x^{3} - 3 \, x^{2} + 3 \, x - 1\right )}} \,d x } \]
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Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.57 \[ \int \frac {-252+180 x+e^{2 x} \left (-50+150 x-150 x^2+50 x^3\right )+e^x \left (-270+630 x-510 x^2+150 x^3\right )}{-25+75 x-75 x^2+25 x^3} \, dx=\frac {25 \, x^{2} e^{\left (2 \, x\right )} + 150 \, x^{2} e^{x} - 50 \, x e^{\left (2 \, x\right )} - 360 \, x e^{x} - 180 \, x + 25 \, e^{\left (2 \, x\right )} + 210 \, e^{x} + 216}{25 \, {\left (x^{2} - 2 \, x + 1\right )}} \]
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Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86 \[ \int \frac {-252+180 x+e^{2 x} \left (-50+150 x-150 x^2+50 x^3\right )+e^x \left (-270+630 x-510 x^2+150 x^3\right )}{-25+75 x-75 x^2+25 x^3} \, dx={\mathrm {e}}^{2\,x}+6\,{\mathrm {e}}^x+\frac {\frac {12\,{\mathrm {e}}^x}{5}-x\,\left (\frac {12\,{\mathrm {e}}^x}{5}+\frac {36}{5}\right )+\frac {216}{25}}{{\left (x-1\right )}^2} \]
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