Integrand size = 54, antiderivative size = 26 \[ \int \frac {497-139 x^2+10 x^4+e^x \left (245+245 x-70 x^2-70 x^3+5 x^4+5 x^5\right )}{49-14 x^2+x^4} \, dx=\frac {5}{4} \left (4+4 \left (2+e^x\right ) x\right )+\frac {x}{7-x^2} \]
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Time = 0.16 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {28, 6874, 2207, 2225, 1171, 21, 8} \[ \int \frac {497-139 x^2+10 x^4+e^x \left (245+245 x-70 x^2-70 x^3+5 x^4+5 x^5\right )}{49-14 x^2+x^4} \, dx=\frac {x}{7-x^2}+10 x-5 e^x+5 e^x (x+1) \]
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Rule 8
Rule 21
Rule 28
Rule 1171
Rule 2207
Rule 2225
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {497-139 x^2+10 x^4+e^x \left (245+245 x-70 x^2-70 x^3+5 x^4+5 x^5\right )}{\left (-7+x^2\right )^2} \, dx \\ & = \int \left (5 e^x (1+x)+\frac {497-139 x^2+10 x^4}{\left (-7+x^2\right )^2}\right ) \, dx \\ & = 5 \int e^x (1+x) \, dx+\int \frac {497-139 x^2+10 x^4}{\left (-7+x^2\right )^2} \, dx \\ & = 5 e^x (1+x)+\frac {x}{7-x^2}+\frac {1}{14} \int \frac {-980+140 x^2}{-7+x^2} \, dx-5 \int e^x \, dx \\ & = -5 e^x+5 e^x (1+x)+\frac {x}{7-x^2}+10 \int 1 \, dx \\ & = -5 e^x+10 x+5 e^x (1+x)+\frac {x}{7-x^2} \\ \end{align*}
Time = 0.87 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int \frac {497-139 x^2+10 x^4+e^x \left (245+245 x-70 x^2-70 x^3+5 x^4+5 x^5\right )}{49-14 x^2+x^4} \, dx=x \left (10+5 e^x+\frac {1}{7-x^2}\right ) \]
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Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77
method | result | size |
risch | \(-\frac {x}{x^{2}-7}+10 x +5 \,{\mathrm e}^{x} x\) | \(20\) |
parts | \(-\frac {x}{x^{2}-7}+10 x +5 \,{\mathrm e}^{x} x\) | \(20\) |
default | \(-\frac {x}{x^{2}-7}+10 x +5 \left (-1+x \right ) {\mathrm e}^{x}+5 \,{\mathrm e}^{x}\) | \(26\) |
norman | \(\frac {-71 x +10 x^{3}-35 \,{\mathrm e}^{x} x +5 \,{\mathrm e}^{x} x^{3}}{x^{2}-7}\) | \(30\) |
parallelrisch | \(\frac {-71 x +10 x^{3}-35 \,{\mathrm e}^{x} x +5 \,{\mathrm e}^{x} x^{3}}{x^{2}-7}\) | \(30\) |
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Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {497-139 x^2+10 x^4+e^x \left (245+245 x-70 x^2-70 x^3+5 x^4+5 x^5\right )}{49-14 x^2+x^4} \, dx=\frac {10 \, x^{3} + 5 \, {\left (x^{3} - 7 \, x\right )} e^{x} - 71 \, x}{x^{2} - 7} \]
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Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.58 \[ \int \frac {497-139 x^2+10 x^4+e^x \left (245+245 x-70 x^2-70 x^3+5 x^4+5 x^5\right )}{49-14 x^2+x^4} \, dx=5 x e^{x} + 10 x - \frac {x}{x^{2} - 7} \]
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Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {497-139 x^2+10 x^4+e^x \left (245+245 x-70 x^2-70 x^3+5 x^4+5 x^5\right )}{49-14 x^2+x^4} \, dx=5 \, x e^{x} + 10 \, x - \frac {x}{x^{2} - 7} \]
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Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {497-139 x^2+10 x^4+e^x \left (245+245 x-70 x^2-70 x^3+5 x^4+5 x^5\right )}{49-14 x^2+x^4} \, dx=\frac {5 \, x^{3} e^{x} + 10 \, x^{3} - 35 \, x e^{x} - 71 \, x}{x^{2} - 7} \]
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Time = 8.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {497-139 x^2+10 x^4+e^x \left (245+245 x-70 x^2-70 x^3+5 x^4+5 x^5\right )}{49-14 x^2+x^4} \, dx=10\,x-\frac {x}{x^2-7}+5\,x\,{\mathrm {e}}^x \]
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