Integrand size = 219, antiderivative size = 30 \[ \int \frac {5 x^3+135 e^{12-3 e^x} x^3+135 e^{8-2 e^x} x^3+45 e^{4-e^x} x^3+e^{\frac {2}{5 x^2+45 e^{8-2 e^x} x^2+30 e^{4-e^x} x^2}} \left (-4+e^{4-e^x} \left (-12+12 e^x x\right )\right )+e^{\frac {1}{5 x^2+45 e^{8-2 e^x} x^2+30 e^{4-e^x} x^2}} \left (-8+e^{4-e^x} \left (-24+24 e^x x\right )\right )}{5 x^3+135 e^{12-3 e^x} x^3+135 e^{8-2 e^x} x^3+45 e^{4-e^x} x^3} \, dx=\left (2+e^{\frac {1}{5 \left (1+3 e^{4-e^x}\right )^2 x^2}}\right )^2+x \]
Leaf count is larger than twice the leaf count of optimal. \(124\) vs. \(2(30)=60\).
Time = 1.93 (sec) , antiderivative size = 124, normalized size of antiderivative = 4.13 \[ \int \frac {5 x^3+135 e^{12-3 e^x} x^3+135 e^{8-2 e^x} x^3+45 e^{4-e^x} x^3+e^{\frac {2}{5 x^2+45 e^{8-2 e^x} x^2+30 e^{4-e^x} x^2}} \left (-4+e^{4-e^x} \left (-12+12 e^x x\right )\right )+e^{\frac {1}{5 x^2+45 e^{8-2 e^x} x^2+30 e^{4-e^x} x^2}} \left (-8+e^{4-e^x} \left (-24+24 e^x x\right )\right )}{5 x^3+135 e^{12-3 e^x} x^3+135 e^{8-2 e^x} x^3+45 e^{4-e^x} x^3} \, dx=\frac {1}{5} \left (5 e^{\frac {2}{5 x^2}+\frac {18 e^8}{5 \left (3 e^4+e^{e^x}\right )^2 x^2}-\frac {12 e^4}{5 \left (3 e^4+e^{e^x}\right ) x^2}}+20 e^{\frac {1}{5 x^2}+\frac {9 e^8}{5 \left (3 e^4+e^{e^x}\right )^2 x^2}-\frac {6 e^4}{5 \left (3 e^4+e^{e^x}\right ) x^2}}+5 x\right ) \]
Integrate[(5*x^3 + 135*E^(12 - 3*E^x)*x^3 + 135*E^(8 - 2*E^x)*x^3 + 45*E^( 4 - E^x)*x^3 + E^(2/(5*x^2 + 45*E^(8 - 2*E^x)*x^2 + 30*E^(4 - E^x)*x^2))*( -4 + E^(4 - E^x)*(-12 + 12*E^x*x)) + E^(5*x^2 + 45*E^(8 - 2*E^x)*x^2 + 30* E^(4 - E^x)*x^2)^(-1)*(-8 + E^(4 - E^x)*(-24 + 24*E^x*x)))/(5*x^3 + 135*E^ (12 - 3*E^x)*x^3 + 135*E^(8 - 2*E^x)*x^3 + 45*E^(4 - E^x)*x^3),x]
(5*E^(2/(5*x^2) + (18*E^8)/(5*(3*E^4 + E^E^x)^2*x^2) - (12*E^4)/(5*(3*E^4 + E^E^x)*x^2)) + 20*E^(1/(5*x^2) + (9*E^8)/(5*(3*E^4 + E^E^x)^2*x^2) - (6* E^4)/(5*(3*E^4 + E^E^x)*x^2)) + 5*x)/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.
\(\displaystyle \int \frac {\left (e^{4-e^x} \left (12 e^x x-12\right )-4\right ) \exp \left (\frac {2}{45 e^{8-2 e^x} x^2+30 e^{4-e^x} x^2+5 x^2}\right )+\left (e^{4-e^x} \left (24 e^x x-24\right )-8\right ) \exp \left (\frac {1}{45 e^{8-2 e^x} x^2+30 e^{4-e^x} x^2+5 x^2}\right )+135 e^{12-3 e^x} x^3+135 e^{8-2 e^x} x^3+45 e^{4-e^x} x^3+5 x^3}{135 e^{12-3 e^x} x^3+135 e^{8-2 e^x} x^3+45 e^{4-e^x} x^3+5 x^3} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{3 e^x} \left (\left (e^{4-e^x} \left (12 e^x x-12\right )-4\right ) \exp \left (\frac {2}{45 e^{8-2 e^x} x^2+30 e^{4-e^x} x^2+5 x^2}\right )+\left (e^{4-e^x} \left (24 e^x x-24\right )-8\right ) \exp \left (\frac {1}{45 e^{8-2 e^x} x^2+30 e^{4-e^x} x^2+5 x^2}\right )+135 e^{12-3 e^x} x^3+135 e^{8-2 e^x} x^3+45 e^{4-e^x} x^3+5 x^3\right )}{5 \left (e^{e^x}+3 e^4\right )^3 x^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {e^{3 e^x} \left (135 e^{12-3 e^x} x^3+135 e^{8-2 e^x} x^3+45 e^{4-e^x} x^3+5 x^3-4 \exp \left (\frac {2}{5 \left (9 e^{8-2 e^x} x^2+6 e^{4-e^x} x^2+x^2\right )}\right ) \left (3 e^{4-e^x} \left (1-e^x x\right )+1\right )-8 \exp \left (\frac {1}{45 e^{8-2 e^x} x^2+30 e^{4-e^x} x^2+5 x^2}\right ) \left (3 e^{4-e^x} \left (1-e^x x\right )+1\right )\right )}{\left (3 e^4+e^{e^x}\right )^3 x^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{5} \int \left (-\frac {8 e^{2 e^x+\frac {e^{2 e^x}}{5 \left (3 e^4+e^{e^x}\right )^2 x^2}} \left (-3 e^{x+4} x+e^{e^x}+3 e^4\right )}{\left (3 e^4+e^{e^x}\right )^3 x^3}-\frac {4 \exp \left (2 e^x+\frac {2 e^{2 e^x}}{5 \left (3 e^4+e^{e^x}\right )^2 x^2}\right ) \left (-3 e^{x+4} x+e^{e^x}+3 e^4\right )}{\left (3 e^4+e^{e^x}\right )^3 x^3}+\frac {5 e^{3 e^x}}{\left (3 e^4+e^{e^x}\right )^3}+\frac {45 e^{4+2 e^x}}{\left (3 e^4+e^{e^x}\right )^3}+\frac {135 e^{3 e^x-3 \left (-4+e^x\right )}}{\left (3 e^4+e^{e^x}\right )^3}+\frac {135 e^{3 e^x-2 \left (-4+e^x\right )}}{\left (3 e^4+e^{e^x}\right )^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (24 \int \frac {\exp \left (x+2 e^x+4+\frac {e^{2 e^x}}{5 \left (3 e^4+e^{e^x}\right )^2 x^2}\right )}{\left (3 e^4+e^{e^x}\right )^3 x^2}dx+12 \int \frac {\exp \left (x+2 e^x+4+\frac {2 e^{2 e^x}}{5 \left (3 e^4+e^{e^x}\right )^2 x^2}\right )}{\left (3 e^4+e^{e^x}\right )^3 x^2}dx-4 \int \frac {\exp \left (2 e^x+\frac {2 e^{2 e^x}}{5 \left (3 e^4+e^{e^x}\right )^2 x^2}\right )}{\left (3 e^4+e^{e^x}\right )^2 x^3}dx-\frac {135}{2} e^8 \text {Subst}\left (\int \frac {1}{\left (3 e^4+e^x\right )^2 x^2}dx,x,e^x\right )+45 \text {Subst}\left (\int \frac {e^{2 x+4}}{\left (3 e^4+e^x\right )^3 x}dx,x,e^x\right )+135 e^8 \text {Subst}\left (\int \frac {1}{\left (3 e^4+e^x\right )^2 x}dx,x,e^x\right )-45 e^4 \text {Subst}\left (\int \frac {1}{\left (3 e^4+e^x\right ) x}dx,x,e^x\right )-8 \int \frac {e^{2 e^x+\frac {e^{2 e^x}}{5 \left (3 e^4+e^{e^x}\right )^2 x^2}}}{\left (3 e^4+e^{e^x}\right )^2 x^3}dx+5 x-\frac {135 e^{8-x}}{2 \left (e^{e^x}+3 e^4\right )^2}\right )\) |
Int[(5*x^3 + 135*E^(12 - 3*E^x)*x^3 + 135*E^(8 - 2*E^x)*x^3 + 45*E^(4 - E^ x)*x^3 + E^(2/(5*x^2 + 45*E^(8 - 2*E^x)*x^2 + 30*E^(4 - E^x)*x^2))*(-4 + E ^(4 - E^x)*(-12 + 12*E^x*x)) + E^(5*x^2 + 45*E^(8 - 2*E^x)*x^2 + 30*E^(4 - E^x)*x^2)^(-1)*(-8 + E^(4 - E^x)*(-24 + 24*E^x*x)))/(5*x^3 + 135*E^(12 - 3*E^x)*x^3 + 135*E^(8 - 2*E^x)*x^3 + 45*E^(4 - E^x)*x^3),x]
3.19.67.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Leaf count of result is larger than twice the leaf count of optimal. \(60\) vs. \(2(25)=50\).
Time = 27.90 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.03
method | result | size |
risch | \({\mathrm e}^{\frac {2}{5 x^{2} \left (6 \,{\mathrm e}^{-{\mathrm e}^{x}+4}+9 \,{\mathrm e}^{-2 \,{\mathrm e}^{x}+8}+1\right )}}+x +4 \,{\mathrm e}^{\frac {1}{5 x^{2} \left (6 \,{\mathrm e}^{-{\mathrm e}^{x}+4}+9 \,{\mathrm e}^{-2 \,{\mathrm e}^{x}+8}+1\right )}}\) | \(61\) |
int((((12*exp(x)*x-12)*exp(-exp(x)+4)-4)*exp(1/(45*x^2*exp(-exp(x)+4)^2+30 *x^2*exp(-exp(x)+4)+5*x^2))^2+((24*exp(x)*x-24)*exp(-exp(x)+4)-8)*exp(1/(4 5*x^2*exp(-exp(x)+4)^2+30*x^2*exp(-exp(x)+4)+5*x^2))+135*x^3*exp(-exp(x)+4 )^3+135*x^3*exp(-exp(x)+4)^2+45*x^3*exp(-exp(x)+4)+5*x^3)/(135*x^3*exp(-ex p(x)+4)^3+135*x^3*exp(-exp(x)+4)^2+45*x^3*exp(-exp(x)+4)+5*x^3),x,method=_ RETURNVERBOSE)
exp(2/5/x^2/(6*exp(-exp(x)+4)+9*exp(-2*exp(x)+8)+1))+x+4*exp(1/5/x^2/(6*ex p(-exp(x)+4)+9*exp(-2*exp(x)+8)+1))
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (25) = 50\).
Time = 0.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.33 \[ \int \frac {5 x^3+135 e^{12-3 e^x} x^3+135 e^{8-2 e^x} x^3+45 e^{4-e^x} x^3+e^{\frac {2}{5 x^2+45 e^{8-2 e^x} x^2+30 e^{4-e^x} x^2}} \left (-4+e^{4-e^x} \left (-12+12 e^x x\right )\right )+e^{\frac {1}{5 x^2+45 e^{8-2 e^x} x^2+30 e^{4-e^x} x^2}} \left (-8+e^{4-e^x} \left (-24+24 e^x x\right )\right )}{5 x^3+135 e^{12-3 e^x} x^3+135 e^{8-2 e^x} x^3+45 e^{4-e^x} x^3} \, dx=x + e^{\left (\frac {2}{5 \, {\left (6 \, x^{2} e^{\left (-e^{x} + 4\right )} + 9 \, x^{2} e^{\left (-2 \, e^{x} + 8\right )} + x^{2}\right )}}\right )} + 4 \, e^{\left (\frac {1}{5 \, {\left (6 \, x^{2} e^{\left (-e^{x} + 4\right )} + 9 \, x^{2} e^{\left (-2 \, e^{x} + 8\right )} + x^{2}\right )}}\right )} \]
integrate((((12*exp(x)*x-12)*exp(-exp(x)+4)-4)*exp(1/(45*x^2*exp(-exp(x)+4 )^2+30*x^2*exp(-exp(x)+4)+5*x^2))^2+((24*exp(x)*x-24)*exp(-exp(x)+4)-8)*ex p(1/(45*x^2*exp(-exp(x)+4)^2+30*x^2*exp(-exp(x)+4)+5*x^2))+135*x^3*exp(-ex p(x)+4)^3+135*x^3*exp(-exp(x)+4)^2+45*x^3*exp(-exp(x)+4)+5*x^3)/(135*x^3*e xp(-exp(x)+4)^3+135*x^3*exp(-exp(x)+4)^2+45*x^3*exp(-exp(x)+4)+5*x^3),x, a lgorithm=\
x + e^(2/5/(6*x^2*e^(-e^x + 4) + 9*x^2*e^(-2*e^x + 8) + x^2)) + 4*e^(1/5/( 6*x^2*e^(-e^x + 4) + 9*x^2*e^(-2*e^x + 8) + x^2))
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (24) = 48\).
Time = 0.50 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.33 \[ \int \frac {5 x^3+135 e^{12-3 e^x} x^3+135 e^{8-2 e^x} x^3+45 e^{4-e^x} x^3+e^{\frac {2}{5 x^2+45 e^{8-2 e^x} x^2+30 e^{4-e^x} x^2}} \left (-4+e^{4-e^x} \left (-12+12 e^x x\right )\right )+e^{\frac {1}{5 x^2+45 e^{8-2 e^x} x^2+30 e^{4-e^x} x^2}} \left (-8+e^{4-e^x} \left (-24+24 e^x x\right )\right )}{5 x^3+135 e^{12-3 e^x} x^3+135 e^{8-2 e^x} x^3+45 e^{4-e^x} x^3} \, dx=x + e^{\frac {2}{30 x^{2} e^{4 - e^{x}} + 45 x^{2} e^{8 - 2 e^{x}} + 5 x^{2}}} + 4 e^{\frac {1}{30 x^{2} e^{4 - e^{x}} + 45 x^{2} e^{8 - 2 e^{x}} + 5 x^{2}}} \]
integrate((((12*exp(x)*x-12)*exp(-exp(x)+4)-4)*exp(1/(45*x**2*exp(-exp(x)+ 4)**2+30*x**2*exp(-exp(x)+4)+5*x**2))**2+((24*exp(x)*x-24)*exp(-exp(x)+4)- 8)*exp(1/(45*x**2*exp(-exp(x)+4)**2+30*x**2*exp(-exp(x)+4)+5*x**2))+135*x* *3*exp(-exp(x)+4)**3+135*x**3*exp(-exp(x)+4)**2+45*x**3*exp(-exp(x)+4)+5*x **3)/(135*x**3*exp(-exp(x)+4)**3+135*x**3*exp(-exp(x)+4)**2+45*x**3*exp(-e xp(x)+4)+5*x**3),x)
x + exp(2/(30*x**2*exp(4 - exp(x)) + 45*x**2*exp(8 - 2*exp(x)) + 5*x**2)) + 4*exp(1/(30*x**2*exp(4 - exp(x)) + 45*x**2*exp(8 - 2*exp(x)) + 5*x**2))
Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (25) = 50\).
Time = 0.25 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.60 \[ \int \frac {5 x^3+135 e^{12-3 e^x} x^3+135 e^{8-2 e^x} x^3+45 e^{4-e^x} x^3+e^{\frac {2}{5 x^2+45 e^{8-2 e^x} x^2+30 e^{4-e^x} x^2}} \left (-4+e^{4-e^x} \left (-12+12 e^x x\right )\right )+e^{\frac {1}{5 x^2+45 e^{8-2 e^x} x^2+30 e^{4-e^x} x^2}} \left (-8+e^{4-e^x} \left (-24+24 e^x x\right )\right )}{5 x^3+135 e^{12-3 e^x} x^3+135 e^{8-2 e^x} x^3+45 e^{4-e^x} x^3} \, dx=x + e^{\left (\frac {2 \, e^{\left (2 \, e^{x}\right )}}{5 \, {\left (9 \, x^{2} e^{8} + x^{2} e^{\left (2 \, e^{x}\right )} + 6 \, x^{2} e^{\left (e^{x} + 4\right )}\right )}}\right )} + 4 \, e^{\left (\frac {e^{\left (2 \, e^{x}\right )}}{5 \, {\left (9 \, x^{2} e^{8} + x^{2} e^{\left (2 \, e^{x}\right )} + 6 \, x^{2} e^{\left (e^{x} + 4\right )}\right )}}\right )} \]
integrate((((12*exp(x)*x-12)*exp(-exp(x)+4)-4)*exp(1/(45*x^2*exp(-exp(x)+4 )^2+30*x^2*exp(-exp(x)+4)+5*x^2))^2+((24*exp(x)*x-24)*exp(-exp(x)+4)-8)*ex p(1/(45*x^2*exp(-exp(x)+4)^2+30*x^2*exp(-exp(x)+4)+5*x^2))+135*x^3*exp(-ex p(x)+4)^3+135*x^3*exp(-exp(x)+4)^2+45*x^3*exp(-exp(x)+4)+5*x^3)/(135*x^3*e xp(-exp(x)+4)^3+135*x^3*exp(-exp(x)+4)^2+45*x^3*exp(-exp(x)+4)+5*x^3),x, a lgorithm=\
x + e^(2/5*e^(2*e^x)/(9*x^2*e^8 + x^2*e^(2*e^x) + 6*x^2*e^(e^x + 4))) + 4* e^(1/5*e^(2*e^x)/(9*x^2*e^8 + x^2*e^(2*e^x) + 6*x^2*e^(e^x + 4)))
Timed out. \[ \int \frac {5 x^3+135 e^{12-3 e^x} x^3+135 e^{8-2 e^x} x^3+45 e^{4-e^x} x^3+e^{\frac {2}{5 x^2+45 e^{8-2 e^x} x^2+30 e^{4-e^x} x^2}} \left (-4+e^{4-e^x} \left (-12+12 e^x x\right )\right )+e^{\frac {1}{5 x^2+45 e^{8-2 e^x} x^2+30 e^{4-e^x} x^2}} \left (-8+e^{4-e^x} \left (-24+24 e^x x\right )\right )}{5 x^3+135 e^{12-3 e^x} x^3+135 e^{8-2 e^x} x^3+45 e^{4-e^x} x^3} \, dx=\text {Timed out} \]
integrate((((12*exp(x)*x-12)*exp(-exp(x)+4)-4)*exp(1/(45*x^2*exp(-exp(x)+4 )^2+30*x^2*exp(-exp(x)+4)+5*x^2))^2+((24*exp(x)*x-24)*exp(-exp(x)+4)-8)*ex p(1/(45*x^2*exp(-exp(x)+4)^2+30*x^2*exp(-exp(x)+4)+5*x^2))+135*x^3*exp(-ex p(x)+4)^3+135*x^3*exp(-exp(x)+4)^2+45*x^3*exp(-exp(x)+4)+5*x^3)/(135*x^3*e xp(-exp(x)+4)^3+135*x^3*exp(-exp(x)+4)^2+45*x^3*exp(-exp(x)+4)+5*x^3),x, a lgorithm=\
Time = 12.06 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.40 \[ \int \frac {5 x^3+135 e^{12-3 e^x} x^3+135 e^{8-2 e^x} x^3+45 e^{4-e^x} x^3+e^{\frac {2}{5 x^2+45 e^{8-2 e^x} x^2+30 e^{4-e^x} x^2}} \left (-4+e^{4-e^x} \left (-12+12 e^x x\right )\right )+e^{\frac {1}{5 x^2+45 e^{8-2 e^x} x^2+30 e^{4-e^x} x^2}} \left (-8+e^{4-e^x} \left (-24+24 e^x x\right )\right )}{5 x^3+135 e^{12-3 e^x} x^3+135 e^{8-2 e^x} x^3+45 e^{4-e^x} x^3} \, dx=x+{\mathrm {e}}^{\frac {2}{5\,x^2+30\,x^2\,{\mathrm {e}}^4\,{\mathrm {e}}^{-{\mathrm {e}}^x}+45\,x^2\,{\mathrm {e}}^8\,{\mathrm {e}}^{-2\,{\mathrm {e}}^x}}}+4\,{\mathrm {e}}^{\frac {1}{5\,x^2+30\,x^2\,{\mathrm {e}}^4\,{\mathrm {e}}^{-{\mathrm {e}}^x}+45\,x^2\,{\mathrm {e}}^8\,{\mathrm {e}}^{-2\,{\mathrm {e}}^x}}} \]
int((exp(1/(30*x^2*exp(4 - exp(x)) + 45*x^2*exp(8 - 2*exp(x)) + 5*x^2))*(e xp(4 - exp(x))*(24*x*exp(x) - 24) - 8) + 45*x^3*exp(4 - exp(x)) + 135*x^3* exp(8 - 2*exp(x)) + 135*x^3*exp(12 - 3*exp(x)) + exp(2/(30*x^2*exp(4 - exp (x)) + 45*x^2*exp(8 - 2*exp(x)) + 5*x^2))*(exp(4 - exp(x))*(12*x*exp(x) - 12) - 4) + 5*x^3)/(45*x^3*exp(4 - exp(x)) + 135*x^3*exp(8 - 2*exp(x)) + 13 5*x^3*exp(12 - 3*exp(x)) + 5*x^3),x)