Integrand size = 75, antiderivative size = 38 \[ \int \frac {e^{8-x} \left (100+100 x+100 x^2\right )+e^5 \left (2 x^2+4 x^3\right )+e^{3 x} \left (100 e^{8-x} x^2+e^5 \left (-2 x^3-3 x^4\right )\right )}{25 x^2} \, dx=e^5 \left (4+\left (2-e^{3 x}+\frac {2}{x}\right ) \left (-2 e^{3-x}+\frac {x^2}{25}\right )\right ) \] Output:
((1/25*x^2-2*exp(3-x))*(2/x-exp(3*x)+2)+4)*exp(5)
Time = 0.11 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \frac {e^{8-x} \left (100+100 x+100 x^2\right )+e^5 \left (2 x^2+4 x^3\right )+e^{3 x} \left (100 e^{8-x} x^2+e^5 \left (-2 x^3-3 x^4\right )\right )}{25 x^2} \, dx=-\frac {e^{5-x} \left (-2+\left (-2+e^{3 x}\right ) x\right ) \left (-50 e^3+e^x x^2\right )}{25 x} \] Input:
Integrate[(E^(8 - x)*(100 + 100*x + 100*x^2) + E^5*(2*x^2 + 4*x^3) + E^(3* x)*(100*E^(8 - x)*x^2 + E^5*(-2*x^3 - 3*x^4)))/(25*x^2),x]
Output:
-1/25*(E^(5 - x)*(-2 + (-2 + E^(3*x))*x)*(-50*E^3 + E^x*x^2))/x
Time = 0.37 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.61, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {27, 2010, 2009}
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.
\(\displaystyle \int \frac {e^{8-x} \left (100 x^2+100 x+100\right )+e^5 \left (4 x^3+2 x^2\right )+e^{3 x} \left (100 e^{8-x} x^2+e^5 \left (-3 x^4-2 x^3\right )\right )}{25 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{25} \int \frac {100 e^{8-x} \left (x^2+x+1\right )+2 e^5 \left (2 x^3+x^2\right )+e^{3 x} \left (100 e^{8-x} x^2-e^5 \left (3 x^4+2 x^3\right )\right )}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {1}{25} \int \left (2 e^5 (2 x+1)+100 e^{2 x+8}-e^{3 x+5} x (3 x+2)+\frac {100 e^{8-x} \left (x^2+x+1\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{25} \left (-e^{3 x+5} x^2-100 e^{8-x}+50 e^{2 x+8}+\frac {1}{2} e^5 (2 x+1)^2-\frac {100 e^{8-x}}{x}\right )\) |
Input:
Int[(E^(8 - x)*(100 + 100*x + 100*x^2) + E^5*(2*x^2 + 4*x^3) + E^(3*x)*(10 0*E^(8 - x)*x^2 + E^5*(-2*x^3 - 3*x^4)))/(25*x^2),x]
Output:
(-100*E^(8 - x) + 50*E^(8 + 2*x) - (100*E^(8 - x))/x - E^(5 + 3*x)*x^2 + ( E^5*(1 + 2*x)^2)/2)/25
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 0.53 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24
method | result | size |
risch | \(\frac {2 x^{2} {\mathrm e}^{5}}{25}+\frac {2 x \,{\mathrm e}^{5}}{25}-\frac {x^{2} {\mathrm e}^{3 x +5}}{25}+2 \,{\mathrm e}^{2 x +8}-\frac {4 \left (1+x \right ) {\mathrm e}^{8-x}}{x}\) | \(47\) |
norman | \(\frac {\left (-4 \,{\mathrm e}^{5} {\mathrm e}^{3}+\frac {2 x^{2} {\mathrm e}^{5} {\mathrm e}^{x}}{25}+\frac {2 x^{3} {\mathrm e}^{5} {\mathrm e}^{x}}{25}-\frac {x^{3} {\mathrm e}^{5} {\mathrm e}^{4 x}}{25}-4 \,{\mathrm e}^{5} {\mathrm e}^{3} x +2 \,{\mathrm e}^{5} {\mathrm e}^{3} x \,{\mathrm e}^{3 x}\right ) {\mathrm e}^{-x}}{x}\) | \(63\) |
parallelrisch | \(-\frac {{\mathrm e}^{5} x^{3} {\mathrm e}^{3 x}-2 x^{3} {\mathrm e}^{5}-50 \,{\mathrm e}^{5} {\mathrm e}^{-x +3} {\mathrm e}^{3 x} x -2 x^{2} {\mathrm e}^{5}+100 x \,{\mathrm e}^{5} {\mathrm e}^{-x +3}+100 \,{\mathrm e}^{5} {\mathrm e}^{-x +3}}{25 x}\) | \(67\) |
parts | \(-\frac {2 \,{\mathrm e}^{5} \left (\frac {{\mathrm e}^{3 x} x}{3}-\frac {{\mathrm e}^{3 x}}{9}\right )}{25}-\frac {3 \,{\mathrm e}^{5} \left (\frac {{\mathrm e}^{3 x} x^{2}}{3}-\frac {2 \,{\mathrm e}^{3 x} x}{9}+\frac {2 \,{\mathrm e}^{3 x}}{27}\right )}{25}+2 \,{\mathrm e}^{2 x} {\mathrm e}^{5} {\mathrm e}^{3}+\frac {2 \,{\mathrm e}^{5} \left (x^{2}+x \right )}{25}+4 \,{\mathrm e}^{5} \left (-{\mathrm e}^{-x +3}-\frac {{\mathrm e}^{-x +3}}{x}\right )\) | \(90\) |
default | \(\frac {2 x^{2} {\mathrm e}^{5}}{25}-\frac {2 \,{\mathrm e}^{5} \left (\frac {{\mathrm e}^{3 x} x}{3}-\frac {{\mathrm e}^{3 x}}{9}\right )}{25}-\frac {3 \,{\mathrm e}^{5} \left (\frac {{\mathrm e}^{3 x} x^{2}}{3}-\frac {2 \,{\mathrm e}^{3 x} x}{9}+\frac {2 \,{\mathrm e}^{3 x}}{27}\right )}{25}-4 \,{\mathrm e}^{5} {\mathrm e}^{-x} {\mathrm e}^{3}+2 \,{\mathrm e}^{2 x} {\mathrm e}^{5} {\mathrm e}^{3}+4 \,{\mathrm e}^{5} {\mathrm e}^{3} \left (-\frac {{\mathrm e}^{-x}}{x}+\operatorname {expIntegral}_{1}\left (x \right )\right )-4 \,{\mathrm e}^{5} {\mathrm e}^{3} \operatorname {expIntegral}_{1}\left (x \right )+\frac {2 x \,{\mathrm e}^{5}}{25}\) | \(107\) |
orering | \(\text {Expression too large to display}\) | \(1641\) |
Input:
int(1/25*((100*x^2*exp(5)*exp(-x+3)+(-3*x^4-2*x^3)*exp(5))*exp(3*x)+(100*x ^2+100*x+100)*exp(5)*exp(-x+3)+(4*x^3+2*x^2)*exp(5))/x^2,x,method=_RETURNV ERBOSE)
Output:
2/25*x^2*exp(5)+2/25*x*exp(5)-1/25*x^2*exp(3*x+5)+2*exp(2*x+8)-4*(1+x)/x*e xp(8-x)
Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.39 \[ \int \frac {e^{8-x} \left (100+100 x+100 x^2\right )+e^5 \left (2 x^2+4 x^3\right )+e^{3 x} \left (100 e^{8-x} x^2+e^5 \left (-2 x^3-3 x^4\right )\right )}{25 x^2} \, dx=-\frac {{\left (x^{3} e^{29} - 50 \, x e^{\left (-x + 32\right )} - 2 \, {\left (x^{3} + x^{2}\right )} e^{\left (-3 \, x + 29\right )} + 100 \, {\left (x + 1\right )} e^{\left (-4 \, x + 32\right )}\right )} e^{\left (3 \, x - 24\right )}}{25 \, x} \] Input:
integrate(1/25*((100*x^2*exp(5)*exp(3-x)+(-3*x^4-2*x^3)*exp(5))*exp(3*x)+( 100*x^2+100*x+100)*exp(5)*exp(3-x)+(4*x^3+2*x^2)*exp(5))/x^2,x, algorithm= "fricas")
Output:
-1/25*(x^3*e^29 - 50*x*e^(-x + 32) - 2*(x^3 + x^2)*e^(-3*x + 29) + 100*(x + 1)*e^(-4*x + 32))*e^(3*x - 24)/x
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (26) = 52\).
Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.84 \[ \int \frac {e^{8-x} \left (100+100 x+100 x^2\right )+e^5 \left (2 x^2+4 x^3\right )+e^{3 x} \left (100 e^{8-x} x^2+e^5 \left (-2 x^3-3 x^4\right )\right )}{25 x^2} \, dx=\frac {2 x^{2} e^{5}}{25} + \frac {2 x e^{5}}{25} + \frac {- x^{3} e^{5} e^{3 x} + 50 x e^{8} \left (e^{3 x}\right )^{\frac {2}{3}} + \frac {- 100 x e^{8} - 100 e^{8}}{\sqrt [3]{e^{3 x}}}}{25 x} \] Input:
integrate(1/25*((100*x**2*exp(5)*exp(3-x)+(-3*x**4-2*x**3)*exp(5))*exp(3*x )+(100*x**2+100*x+100)*exp(5)*exp(3-x)+(4*x**3+2*x**2)*exp(5))/x**2,x)
Output:
2*x**2*exp(5)/25 + 2*x*exp(5)/25 + (-x**3*exp(5)*exp(3*x) + 50*x*exp(8)*ex p(3*x)**(2/3) + (-100*x*exp(8) - 100*exp(8))/exp(3*x)**(1/3))/(25*x)
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 2.18 \[ \int \frac {e^{8-x} \left (100+100 x+100 x^2\right )+e^5 \left (2 x^2+4 x^3\right )+e^{3 x} \left (100 e^{8-x} x^2+e^5 \left (-2 x^3-3 x^4\right )\right )}{25 x^2} \, dx=\frac {2}{25} \, x^{2} e^{5} + 4 \, {\rm Ei}\left (-x\right ) e^{8} + \frac {2}{25} \, x e^{5} - \frac {1}{225} \, {\left (9 \, x^{2} e^{5} - 6 \, x e^{5} + 2 \, e^{5}\right )} e^{\left (3 \, x\right )} - \frac {2}{225} \, {\left (3 \, x e^{5} - e^{5}\right )} e^{\left (3 \, x\right )} - 4 \, e^{8} \Gamma \left (-1, x\right ) + 2 \, e^{\left (2 \, x + 8\right )} - 4 \, e^{\left (-x + 8\right )} \] Input:
integrate(1/25*((100*x^2*exp(5)*exp(3-x)+(-3*x^4-2*x^3)*exp(5))*exp(3*x)+( 100*x^2+100*x+100)*exp(5)*exp(3-x)+(4*x^3+2*x^2)*exp(5))/x^2,x, algorithm= "maxima")
Output:
2/25*x^2*e^5 + 4*Ei(-x)*e^8 + 2/25*x*e^5 - 1/225*(9*x^2*e^5 - 6*x*e^5 + 2* e^5)*e^(3*x) - 2/225*(3*x*e^5 - e^5)*e^(3*x) - 4*e^8*gamma(-1, x) + 2*e^(2 *x + 8) - 4*e^(-x + 8)
Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (32) = 64\).
Time = 0.12 (sec) , antiderivative size = 114, normalized size of antiderivative = 3.00 \[ \int \frac {e^{8-x} \left (100+100 x+100 x^2\right )+e^5 \left (2 x^2+4 x^3\right )+e^{3 x} \left (100 e^{8-x} x^2+e^5 \left (-2 x^3-3 x^4\right )\right )}{25 x^2} \, dx=\frac {2 \, {\left (x - 8\right )}^{3} e^{5} - {\left (x - 8\right )}^{3} e^{\left (3 \, x + 5\right )} + 50 \, {\left (x - 8\right )}^{2} e^{5} - 24 \, {\left (x - 8\right )}^{2} e^{\left (3 \, x + 5\right )} + 272 \, {\left (x - 8\right )} e^{5} - 192 \, {\left (x - 8\right )} e^{\left (3 \, x + 5\right )} + 50 \, {\left (x - 8\right )} e^{\left (2 \, x + 8\right )} - 100 \, {\left (x - 8\right )} e^{\left (-x + 8\right )} - 512 \, e^{\left (3 \, x + 5\right )} + 400 \, e^{\left (2 \, x + 8\right )} - 900 \, e^{\left (-x + 8\right )}}{25 \, x} \] Input:
integrate(1/25*((100*x^2*exp(5)*exp(3-x)+(-3*x^4-2*x^3)*exp(5))*exp(3*x)+( 100*x^2+100*x+100)*exp(5)*exp(3-x)+(4*x^3+2*x^2)*exp(5))/x^2,x, algorithm= "giac")
Output:
1/25*(2*(x - 8)^3*e^5 - (x - 8)^3*e^(3*x + 5) + 50*(x - 8)^2*e^5 - 24*(x - 8)^2*e^(3*x + 5) + 272*(x - 8)*e^5 - 192*(x - 8)*e^(3*x + 5) + 50*(x - 8) *e^(2*x + 8) - 100*(x - 8)*e^(-x + 8) - 512*e^(3*x + 5) + 400*e^(2*x + 8) - 900*e^(-x + 8))/x
Time = 4.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.39 \[ \int \frac {e^{8-x} \left (100+100 x+100 x^2\right )+e^5 \left (2 x^2+4 x^3\right )+e^{3 x} \left (100 e^{8-x} x^2+e^5 \left (-2 x^3-3 x^4\right )\right )}{25 x^2} \, dx=2\,{\mathrm {e}}^{2\,x+8}+\frac {2\,x\,{\mathrm {e}}^5}{25}+\frac {2\,x^2\,{\mathrm {e}}^5}{25}-\frac {x^2\,{\mathrm {e}}^{3\,x+5}}{25}-\frac {{\mathrm {e}}^{3-x}\,\left (4\,{\mathrm {e}}^5+4\,x\,{\mathrm {e}}^5\right )}{x} \] Input:
int(((exp(5)*(2*x^2 + 4*x^3))/25 - (exp(3*x)*(exp(5)*(2*x^3 + 3*x^4) - 100 *x^2*exp(5)*exp(3 - x)))/25 + (exp(5)*exp(3 - x)*(100*x + 100*x^2 + 100))/ 25)/x^2,x)
Output:
2*exp(2*x + 8) + (2*x*exp(5))/25 + (2*x^2*exp(5))/25 - (x^2*exp(3*x + 5))/ 25 - (exp(3 - x)*(4*exp(5) + 4*x*exp(5)))/x
Time = 0.16 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.63 \[ \int \frac {e^{8-x} \left (100+100 x+100 x^2\right )+e^5 \left (2 x^2+4 x^3\right )+e^{3 x} \left (100 e^{8-x} x^2+e^5 \left (-2 x^3-3 x^4\right )\right )}{25 x^2} \, dx=\frac {e^{5} \left (-e^{4 x} x^{3}+50 e^{3 x} e^{3} x +2 e^{x} x^{3}+2 e^{x} x^{2}-100 e^{3} x -100 e^{3}\right )}{25 e^{x} x} \] Input:
int(1/25*((100*x^2*exp(5)*exp(3-x)+(-3*x^4-2*x^3)*exp(5))*exp(3*x)+(100*x^ 2+100*x+100)*exp(5)*exp(3-x)+(4*x^3+2*x^2)*exp(5))/x^2,x)
Output:
(e**5*( - e**(4*x)*x**3 + 50*e**(3*x)*e**3*x + 2*e**x*x**3 + 2*e**x*x**2 - 100*e**3*x - 100*e**3))/(25*e**x*x)