Integrand size = 43, antiderivative size = 25 \[ \int \frac {500 x^2+e^{5/x} \left (150000+33000 x+1815 x^2\right )}{30000 x^2+6600 x^3+363 x^4} \, dx=8-e^{5/x}+\frac {x}{3 \left (20+\frac {11 x}{5}\right )} \]
[Out]
Time = 0.17 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {1608, 27, 12, 6874, 2240} \[ \int \frac {500 x^2+e^{5/x} \left (150000+33000 x+1815 x^2\right )}{30000 x^2+6600 x^3+363 x^4} \, dx=-e^{5/x}-\frac {500}{33 (11 x+100)} \]
[In]
[Out]
Rule 12
Rule 27
Rule 1608
Rule 2240
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {500 x^2+e^{5/x} \left (150000+33000 x+1815 x^2\right )}{x^2 \left (30000+6600 x+363 x^2\right )} \, dx \\ & = \int \frac {500 x^2+e^{5/x} \left (150000+33000 x+1815 x^2\right )}{3 x^2 (100+11 x)^2} \, dx \\ & = \frac {1}{3} \int \frac {500 x^2+e^{5/x} \left (150000+33000 x+1815 x^2\right )}{x^2 (100+11 x)^2} \, dx \\ & = \frac {1}{3} \int \left (\frac {15 e^{5/x}}{x^2}+\frac {500}{(100+11 x)^2}\right ) \, dx \\ & = -\frac {500}{33 (100+11 x)}+5 \int \frac {e^{5/x}}{x^2} \, dx \\ & = -e^{5/x}-\frac {500}{33 (100+11 x)} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {500 x^2+e^{5/x} \left (150000+33000 x+1815 x^2\right )}{30000 x^2+6600 x^3+363 x^4} \, dx=\frac {5}{3} \left (-\frac {3 e^{5/x}}{5}-\frac {100}{11 (100+11 x)}\right ) \]
[In]
[Out]
Time = 0.64 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68
method | result | size |
risch | \(-\frac {500}{363 \left (x +\frac {100}{11}\right )}-{\mathrm e}^{\frac {5}{x}}\) | \(17\) |
parts | \(-\frac {500}{33 \left (11 x +100\right )}-{\mathrm e}^{\frac {5}{x}}\) | \(19\) |
derivativedivides | \(\frac {5}{3 \left (\frac {100}{x}+11\right )}-{\mathrm e}^{\frac {5}{x}}\) | \(21\) |
default | \(\frac {5}{3 \left (\frac {100}{x}+11\right )}-{\mathrm e}^{\frac {5}{x}}\) | \(21\) |
parallelrisch | \(-\frac {363 x \,{\mathrm e}^{\frac {5}{x}}+3300 \,{\mathrm e}^{\frac {5}{x}}+500}{33 \left (11 x +100\right )}\) | \(29\) |
norman | \(\frac {-\frac {500 x}{33}-100 x \,{\mathrm e}^{\frac {5}{x}}-11 x^{2} {\mathrm e}^{\frac {5}{x}}}{x \left (11 x +100\right )}\) | \(36\) |
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {500 x^2+e^{5/x} \left (150000+33000 x+1815 x^2\right )}{30000 x^2+6600 x^3+363 x^4} \, dx=-\frac {33 \, {\left (11 \, x + 100\right )} e^{\frac {5}{x}} + 500}{33 \, {\left (11 \, x + 100\right )}} \]
[In]
[Out]
Time = 0.09 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.48 \[ \int \frac {500 x^2+e^{5/x} \left (150000+33000 x+1815 x^2\right )}{30000 x^2+6600 x^3+363 x^4} \, dx=- e^{\frac {5}{x}} - \frac {500}{363 x + 3300} \]
[In]
[Out]
none
Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {500 x^2+e^{5/x} \left (150000+33000 x+1815 x^2\right )}{30000 x^2+6600 x^3+363 x^4} \, dx=-\frac {500}{33 \, {\left (11 \, x + 100\right )}} - e^{\frac {5}{x}} \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {500 x^2+e^{5/x} \left (150000+33000 x+1815 x^2\right )}{30000 x^2+6600 x^3+363 x^4} \, dx=-\frac {\frac {300 \, e^{\frac {5}{x}}}{x} + 33 \, e^{\frac {5}{x}} - 5}{3 \, {\left (\frac {100}{x} + 11\right )}} \]
[In]
[Out]
Time = 13.66 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {500 x^2+e^{5/x} \left (150000+33000 x+1815 x^2\right )}{30000 x^2+6600 x^3+363 x^4} \, dx=-{\mathrm {e}}^{5/x}-\frac {500}{33\,\left (11\,x+100\right )} \]
[In]
[Out]