Integrand size = 68, antiderivative size = 22 \[ \int \frac {50-50 x+e^{e^x+3 x} \left (-12 x^2+12 x^3-3 x^4+e^x \left (-4 x^2+4 x^3-x^4\right )\right )}{4 x^2-4 x^3+x^4} \, dx=-e^{e^x+3 x}+\frac {25}{(-2+x) x} \]
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Time = 0.56 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09, number of steps used = 15, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1608, 27, 6820, 6874, 2320, 2207, 2225, 75} \[ \int \frac {50-50 x+e^{e^x+3 x} \left (-12 x^2+12 x^3-3 x^4+e^x \left (-4 x^2+4 x^3-x^4\right )\right )}{4 x^2-4 x^3+x^4} \, dx=-e^{3 x+e^x}-\frac {25}{(2-x) x} \]
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Rule 27
Rule 75
Rule 1608
Rule 2207
Rule 2225
Rule 2320
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {50-50 x+e^{e^x+3 x} \left (-12 x^2+12 x^3-3 x^4+e^x \left (-4 x^2+4 x^3-x^4\right )\right )}{x^2 \left (4-4 x+x^2\right )} \, dx \\ & = \int \frac {50-50 x+e^{e^x+3 x} \left (-12 x^2+12 x^3-3 x^4+e^x \left (-4 x^2+4 x^3-x^4\right )\right )}{(-2+x)^2 x^2} \, dx \\ & = \int \frac {50-50 x-e^{e^x+3 x} \left (3+e^x\right ) (-2+x)^2 x^2}{(2-x)^2 x^2} \, dx \\ & = \int \left (-3 e^{e^x+3 x}-e^{e^x+4 x}-\frac {50 (-1+x)}{(-2+x)^2 x^2}\right ) \, dx \\ & = -\left (3 \int e^{e^x+3 x} \, dx\right )-50 \int \frac {-1+x}{(-2+x)^2 x^2} \, dx-\int e^{e^x+4 x} \, dx \\ & = -\frac {25}{(2-x) x}-3 \text {Subst}\left (\int e^x x^2 \, dx,x,e^x\right )-\text {Subst}\left (\int e^x x^3 \, dx,x,e^x\right ) \\ & = -3 e^{e^x+2 x}-e^{e^x+3 x}-\frac {25}{(2-x) x}+3 \text {Subst}\left (\int e^x x^2 \, dx,x,e^x\right )+6 \text {Subst}\left (\int e^x x \, dx,x,e^x\right ) \\ & = 6 e^{e^x+x}-e^{e^x+3 x}-\frac {25}{(2-x) x}-6 \text {Subst}\left (\int e^x \, dx,x,e^x\right )-6 \text {Subst}\left (\int e^x x \, dx,x,e^x\right ) \\ & = -6 e^{e^x}-e^{e^x+3 x}-\frac {25}{(2-x) x}+6 \text {Subst}\left (\int e^x \, dx,x,e^x\right ) \\ & = -e^{e^x+3 x}-\frac {25}{(2-x) x} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {50-50 x+e^{e^x+3 x} \left (-12 x^2+12 x^3-3 x^4+e^x \left (-4 x^2+4 x^3-x^4\right )\right )}{4 x^2-4 x^3+x^4} \, dx=-e^{e^x+3 x}+\frac {25}{(-2+x) x} \]
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Time = 0.17 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95
method | result | size |
risch | \(\frac {25}{\left (-2+x \right ) x}-{\mathrm e}^{3 x +{\mathrm e}^{x}}\) | \(21\) |
parts | \(-{\mathrm e}^{3 x} {\mathrm e}^{{\mathrm e}^{x}}-\frac {25}{2 x}+\frac {25}{2 \left (-2+x \right )}\) | \(23\) |
norman | \(\frac {25+2 x \,{\mathrm e}^{3 x +{\mathrm e}^{x}}-{\mathrm e}^{3 x +{\mathrm e}^{x}} x^{2}}{\left (-2+x \right ) x}\) | \(34\) |
parallelrisch | \(-\frac {{\mathrm e}^{3 x +{\mathrm e}^{x}} x^{2}-2 x \,{\mathrm e}^{3 x +{\mathrm e}^{x}}-25}{\left (-2+x \right ) x}\) | \(34\) |
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Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {50-50 x+e^{e^x+3 x} \left (-12 x^2+12 x^3-3 x^4+e^x \left (-4 x^2+4 x^3-x^4\right )\right )}{4 x^2-4 x^3+x^4} \, dx=-\frac {{\left (x^{2} - 2 \, x\right )} e^{\left (3 \, x + e^{x}\right )} - 25}{x^{2} - 2 \, x} \]
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Time = 0.12 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {50-50 x+e^{e^x+3 x} \left (-12 x^2+12 x^3-3 x^4+e^x \left (-4 x^2+4 x^3-x^4\right )\right )}{4 x^2-4 x^3+x^4} \, dx=- e^{3 x + e^{x}} + \frac {25}{x^{2} - 2 x} \]
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Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {50-50 x+e^{e^x+3 x} \left (-12 x^2+12 x^3-3 x^4+e^x \left (-4 x^2+4 x^3-x^4\right )\right )}{4 x^2-4 x^3+x^4} \, dx=-\frac {25 \, {\left (x - 1\right )}}{x^{2} - 2 \, x} + \frac {25}{x - 2} - e^{\left (3 \, x + e^{x}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (20) = 40\).
Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.18 \[ \int \frac {50-50 x+e^{e^x+3 x} \left (-12 x^2+12 x^3-3 x^4+e^x \left (-4 x^2+4 x^3-x^4\right )\right )}{4 x^2-4 x^3+x^4} \, dx=-\frac {x^{2} e^{\left (7 \, x + e^{x}\right )} - 2 \, x e^{\left (7 \, x + e^{x}\right )} - 25 \, e^{\left (4 \, x\right )}}{x^{2} e^{\left (4 \, x\right )} - 2 \, x e^{\left (4 \, x\right )}} \]
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Time = 0.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {50-50 x+e^{e^x+3 x} \left (-12 x^2+12 x^3-3 x^4+e^x \left (-4 x^2+4 x^3-x^4\right )\right )}{4 x^2-4 x^3+x^4} \, dx=-{\mathrm {e}}^{3\,x+{\mathrm {e}}^x}-\frac {25}{2\,x-x^2} \]
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