Integrand size = 117, antiderivative size = 34 \[ \int \frac {270-150 x+399 x^2-180 x^3+25 x^4+e^{2 x} \left (324-180 x+25 x^2\right )+e^x \left (465+303 x-285 x^2+50 x^3\right )}{e^{5+2 x} \left (324-180 x+25 x^2\right )+e^{5+x} \left (648 x-360 x^2+50 x^3\right )+e^5 \left (324 x^2-180 x^3+25 x^4\right )} \, dx=10-\frac {-x+\frac {3 (1-x)}{\left (\frac {18}{5}-x\right ) \left (e^x+x\right )}}{e^5} \] Output:
10-(3*(1-x)/(18/5-x)/(exp(x)+x)-x)/exp(5)
Time = 2.50 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74 \[ \int \frac {270-150 x+399 x^2-180 x^3+25 x^4+e^{2 x} \left (324-180 x+25 x^2\right )+e^x \left (465+303 x-285 x^2+50 x^3\right )}{e^{5+2 x} \left (324-180 x+25 x^2\right )+e^{5+x} \left (648 x-360 x^2+50 x^3\right )+e^5 \left (324 x^2-180 x^3+25 x^4\right )} \, dx=\frac {x-\frac {15 (-1+x)}{\left (e^x+x\right ) (-18+5 x)}}{e^5} \] Input:
Integrate[(270 - 150*x + 399*x^2 - 180*x^3 + 25*x^4 + E^(2*x)*(324 - 180*x + 25*x^2) + E^x*(465 + 303*x - 285*x^2 + 50*x^3))/(E^(5 + 2*x)*(324 - 180 *x + 25*x^2) + E^(5 + x)*(648*x - 360*x^2 + 50*x^3) + E^5*(324*x^2 - 180*x ^3 + 25*x^4)),x]
Output:
(x - (15*(-1 + x))/((E^x + x)*(-18 + 5*x)))/E^5
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {25 x^4-180 x^3+399 x^2+e^{2 x} \left (25 x^2-180 x+324\right )+e^x \left (50 x^3-285 x^2+303 x+465\right )-150 x+270}{e^{2 x+5} \left (25 x^2-180 x+324\right )+e^{x+5} \left (50 x^3-360 x^2+648 x\right )+e^5 \left (25 x^4-180 x^3+324 x^2\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {25 x^4-180 x^3+399 x^2+e^x \left (50 x^3-285 x^2+303 x+465\right )-150 x+e^{2 x} (18-5 x)^2+270}{e^5 (18-5 x)^2 \left (x+e^x\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {25 x^4-180 x^3+399 x^2-150 x+e^{2 x} (18-5 x)^2+e^x \left (50 x^3-285 x^2+303 x+465\right )+270}{(18-5 x)^2 \left (x+e^x\right )^2}dx}{e^5}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (-\frac {15 (x-1)^2}{\left (x+e^x\right )^2 (5 x-18)}+\frac {15 \left (5 x^2-23 x+31\right )}{\left (x+e^x\right ) (5 x-18)^2}+1\right )dx}{e^5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {39}{5} \int \frac {1}{\left (x+e^x\right )^2}dx+195 \int \frac {1}{\left (x+e^x\right ) (5 x-18)^2}dx-\frac {507}{5} \int \frac {1}{\left (x+e^x\right )^2 (5 x-18)}dx+39 \int \frac {1}{\left (x+e^x\right ) (5 x-18)}dx+x-\frac {3}{x+e^x}}{e^5}\) |
Input:
Int[(270 - 150*x + 399*x^2 - 180*x^3 + 25*x^4 + E^(2*x)*(324 - 180*x + 25* x^2) + E^x*(465 + 303*x - 285*x^2 + 50*x^3))/(E^(5 + 2*x)*(324 - 180*x + 2 5*x^2) + E^(5 + x)*(648*x - 360*x^2 + 50*x^3) + E^5*(324*x^2 - 180*x^3 + 2 5*x^4)),x]
Output:
$Aborted
Time = 0.45 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76
method | result | size |
risch | \(x \,{\mathrm e}^{-5}-\frac {15 \left (-1+x \right ) {\mathrm e}^{-5}}{\left (5 x -18\right ) \left ({\mathrm e}^{x}+x \right )}\) | \(26\) |
parallelrisch | \(\frac {\left (25 x^{3}+25 \,{\mathrm e}^{x} x^{2}+75-399 x -324 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-5}}{25 \,{\mathrm e}^{x} x +25 x^{2}-90 \,{\mathrm e}^{x}-90 x}\) | \(48\) |
norman | \(\frac {-\frac {399 x \,{\mathrm e}^{-5}}{5}-\frac {324 \,{\mathrm e}^{-5} {\mathrm e}^{x}}{5}+15 \,{\mathrm e}^{-5}+5 \,{\mathrm e}^{-5} x^{3}+5 \,{\mathrm e}^{x} {\mathrm e}^{-5} x^{2}}{5 \,{\mathrm e}^{x} x +5 x^{2}-18 \,{\mathrm e}^{x}-18 x}\) | \(64\) |
Input:
int(((25*x^2-180*x+324)*exp(x)^2+(50*x^3-285*x^2+303*x+465)*exp(x)+25*x^4- 180*x^3+399*x^2-150*x+270)/((25*x^2-180*x+324)*exp(5)*exp(x)^2+(50*x^3-360 *x^2+648*x)*exp(5)*exp(x)+(25*x^4-180*x^3+324*x^2)*exp(5)),x,method=_RETUR NVERBOSE)
Output:
x*exp(-5)-15*(-1+x)*exp(-5)/(5*x-18)/(exp(x)+x)
Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (25) = 50\).
Time = 0.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.74 \[ \int \frac {270-150 x+399 x^2-180 x^3+25 x^4+e^{2 x} \left (324-180 x+25 x^2\right )+e^x \left (465+303 x-285 x^2+50 x^3\right )}{e^{5+2 x} \left (324-180 x+25 x^2\right )+e^{5+x} \left (648 x-360 x^2+50 x^3\right )+e^5 \left (324 x^2-180 x^3+25 x^4\right )} \, dx=\frac {{\left (5 \, x^{3} - 18 \, x^{2} - 15 \, x + 15\right )} e^{5} + {\left (5 \, x^{2} - 18 \, x\right )} e^{\left (x + 5\right )}}{{\left (5 \, x^{2} - 18 \, x\right )} e^{10} + {\left (5 \, x - 18\right )} e^{\left (x + 10\right )}} \] Input:
integrate(((25*x^2-180*x+324)*exp(x)^2+(50*x^3-285*x^2+303*x+465)*exp(x)+2 5*x^4-180*x^3+399*x^2-150*x+270)/((25*x^2-180*x+324)*exp(5)*exp(x)^2+(50*x ^3-360*x^2+648*x)*exp(5)*exp(x)+(25*x^4-180*x^3+324*x^2)*exp(5)),x, algori thm="fricas")
Output:
((5*x^3 - 18*x^2 - 15*x + 15)*e^5 + (5*x^2 - 18*x)*e^(x + 5))/((5*x^2 - 18 *x)*e^10 + (5*x - 18)*e^(x + 10))
Time = 0.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {270-150 x+399 x^2-180 x^3+25 x^4+e^{2 x} \left (324-180 x+25 x^2\right )+e^x \left (465+303 x-285 x^2+50 x^3\right )}{e^{5+2 x} \left (324-180 x+25 x^2\right )+e^{5+x} \left (648 x-360 x^2+50 x^3\right )+e^5 \left (324 x^2-180 x^3+25 x^4\right )} \, dx=\frac {x}{e^{5}} + \frac {15 - 15 x}{5 x^{2} e^{5} - 18 x e^{5} + \left (5 x e^{5} - 18 e^{5}\right ) e^{x}} \] Input:
integrate(((25*x**2-180*x+324)*exp(x)**2+(50*x**3-285*x**2+303*x+465)*exp( x)+25*x**4-180*x**3+399*x**2-150*x+270)/((25*x**2-180*x+324)*exp(5)*exp(x) **2+(50*x**3-360*x**2+648*x)*exp(5)*exp(x)+(25*x**4-180*x**3+324*x**2)*exp (5)),x)
Output:
x*exp(-5) + (15 - 15*x)/(5*x**2*exp(5) - 18*x*exp(5) + (5*x*exp(5) - 18*ex p(5))*exp(x))
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (25) = 50\).
Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.65 \[ \int \frac {270-150 x+399 x^2-180 x^3+25 x^4+e^{2 x} \left (324-180 x+25 x^2\right )+e^x \left (465+303 x-285 x^2+50 x^3\right )}{e^{5+2 x} \left (324-180 x+25 x^2\right )+e^{5+x} \left (648 x-360 x^2+50 x^3\right )+e^5 \left (324 x^2-180 x^3+25 x^4\right )} \, dx=\frac {5 \, x^{3} - 18 \, x^{2} + {\left (5 \, x^{2} - 18 \, x\right )} e^{x} - 15 \, x + 15}{5 \, x^{2} e^{5} - 18 \, x e^{5} + {\left (5 \, x e^{5} - 18 \, e^{5}\right )} e^{x}} \] Input:
integrate(((25*x^2-180*x+324)*exp(x)^2+(50*x^3-285*x^2+303*x+465)*exp(x)+2 5*x^4-180*x^3+399*x^2-150*x+270)/((25*x^2-180*x+324)*exp(5)*exp(x)^2+(50*x ^3-360*x^2+648*x)*exp(5)*exp(x)+(25*x^4-180*x^3+324*x^2)*exp(5)),x, algori thm="maxima")
Output:
(5*x^3 - 18*x^2 + (5*x^2 - 18*x)*e^x - 15*x + 15)/(5*x^2*e^5 - 18*x*e^5 + (5*x*e^5 - 18*e^5)*e^x)
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (25) = 50\).
Time = 0.13 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.65 \[ \int \frac {270-150 x+399 x^2-180 x^3+25 x^4+e^{2 x} \left (324-180 x+25 x^2\right )+e^x \left (465+303 x-285 x^2+50 x^3\right )}{e^{5+2 x} \left (324-180 x+25 x^2\right )+e^{5+x} \left (648 x-360 x^2+50 x^3\right )+e^5 \left (324 x^2-180 x^3+25 x^4\right )} \, dx=\frac {5 \, x^{3} + 5 \, x^{2} e^{x} - 18 \, x^{2} - 18 \, x e^{x} - 15 \, x + 15}{5 \, x^{2} e^{5} - 18 \, x e^{5} + 5 \, x e^{\left (x + 5\right )} - 18 \, e^{\left (x + 5\right )}} \] Input:
integrate(((25*x^2-180*x+324)*exp(x)^2+(50*x^3-285*x^2+303*x+465)*exp(x)+2 5*x^4-180*x^3+399*x^2-150*x+270)/((25*x^2-180*x+324)*exp(5)*exp(x)^2+(50*x ^3-360*x^2+648*x)*exp(5)*exp(x)+(25*x^4-180*x^3+324*x^2)*exp(5)),x, algori thm="giac")
Output:
(5*x^3 + 5*x^2*e^x - 18*x^2 - 18*x*e^x - 15*x + 15)/(5*x^2*e^5 - 18*x*e^5 + 5*x*e^(x + 5) - 18*e^(x + 5))
Time = 0.15 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.24 \[ \int \frac {270-150 x+399 x^2-180 x^3+25 x^4+e^{2 x} \left (324-180 x+25 x^2\right )+e^x \left (465+303 x-285 x^2+50 x^3\right )}{e^{5+2 x} \left (324-180 x+25 x^2\right )+e^{5+x} \left (648 x-360 x^2+50 x^3\right )+e^5 \left (324 x^2-180 x^3+25 x^4\right )} \, dx=x\,{\mathrm {e}}^{-5}-\frac {15\,{\mathrm {e}}^{-5}\,\left (5\,x^3-28\,x^2+41\,x-18\right )}{{\left (5\,x-18\right )}^2\,\left (x+{\mathrm {e}}^x\right )\,\left (x-1\right )} \] Input:
int((exp(2*x)*(25*x^2 - 180*x + 324) - 150*x + 399*x^2 - 180*x^3 + 25*x^4 + exp(x)*(303*x - 285*x^2 + 50*x^3 + 465) + 270)/(exp(5)*(324*x^2 - 180*x^ 3 + 25*x^4) + exp(5)*exp(x)*(648*x - 360*x^2 + 50*x^3) + exp(2*x)*exp(5)*( 25*x^2 - 180*x + 324)),x)
Output:
x*exp(-5) - (15*exp(-5)*(41*x - 28*x^2 + 5*x^3 - 18))/((5*x - 18)^2*(x + e xp(x))*(x - 1))
Time = 0.15 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.62 \[ \int \frac {270-150 x+399 x^2-180 x^3+25 x^4+e^{2 x} \left (324-180 x+25 x^2\right )+e^x \left (465+303 x-285 x^2+50 x^3\right )}{e^{5+2 x} \left (324-180 x+25 x^2\right )+e^{5+x} \left (648 x-360 x^2+50 x^3\right )+e^5 \left (324 x^2-180 x^3+25 x^4\right )} \, dx=\frac {5 e^{x} x^{2}-18 e^{x} x +5 x^{3}-18 x^{2}-15 x +15}{e^{5} \left (5 e^{x} x -18 e^{x}+5 x^{2}-18 x \right )} \] Input:
int(((25*x^2-180*x+324)*exp(x)^2+(50*x^3-285*x^2+303*x+465)*exp(x)+25*x^4- 180*x^3+399*x^2-150*x+270)/((25*x^2-180*x+324)*exp(5)*exp(x)^2+(50*x^3-360 *x^2+648*x)*exp(5)*exp(x)+(25*x^4-180*x^3+324*x^2)*exp(5)),x)
Output:
(5*e**x*x**2 - 18*e**x*x + 5*x**3 - 18*x**2 - 15*x + 15)/(e**5*(5*e**x*x - 18*e**x + 5*x**2 - 18*x))