Integrand size = 88, antiderivative size = 27 \[ \int \frac {e^{\frac {8-3 x+e^2 x}{x}} \left (e^2 x-x^2\right )+e^{\frac {8-3 x+e^2 x}{x}} \left (e^2 (-8-3 x)+8 x+4 x^2\right ) \log (x)}{e^4 x^5-2 e^2 x^6+x^7} \, dx=\frac {e^{-3+e^2+\frac {8}{x}} \log (x)}{\left (e^2-x\right ) x^3} \] Output:
exp(exp(2)+8/x-3)/x^3*ln(x)/(exp(2)-x)
Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {8-3 x+e^2 x}{x}} \left (e^2 x-x^2\right )+e^{\frac {8-3 x+e^2 x}{x}} \left (e^2 (-8-3 x)+8 x+4 x^2\right ) \log (x)}{e^4 x^5-2 e^2 x^6+x^7} \, dx=\frac {e^{-3+e^2+\frac {8}{x}} \log (x)}{\left (e^2-x\right ) x^3} \] Input:
Integrate[(E^((8 - 3*x + E^2*x)/x)*(E^2*x - x^2) + E^((8 - 3*x + E^2*x)/x) *(E^2*(-8 - 3*x) + 8*x + 4*x^2)*Log[x])/(E^4*x^5 - 2*E^2*x^6 + x^7),x]
Output:
(E^(-3 + E^2 + 8/x)*Log[x])/((E^2 - x)*x^3)
Leaf count is larger than twice the leaf count of optimal. \(272\) vs. \(2(27)=54\).
Time = 3.82 (sec) , antiderivative size = 272, normalized size of antiderivative = 10.07, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {2026, 7277, 27, 7292, 7293, 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.
\(\displaystyle \int \frac {e^{\frac {e^2 x-3 x+8}{x}} \left (e^2 x-x^2\right )+e^{\frac {e^2 x-3 x+8}{x}} \left (4 x^2+8 x+e^2 (-3 x-8)\right ) \log (x)}{x^7-2 e^2 x^6+e^4 x^5} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{\frac {e^2 x-3 x+8}{x}} \left (e^2 x-x^2\right )+e^{\frac {e^2 x-3 x+8}{x}} \left (4 x^2+8 x+e^2 (-3 x-8)\right ) \log (x)}{x^5 \left (x^2-2 e^2 x+e^4\right )}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 \int \frac {e^{\frac {e^2 x-3 x+8}{x}} \left (e^2 x-x^2\right )+e^{\frac {e^2 x-3 x+8}{x}} \left (4 x^2+8 x-e^2 (3 x+8)\right ) \log (x)}{4 \left (e^2-x\right )^2 x^5}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {e^{\frac {e^2 x-3 x+8}{x}} \left (e^2 x-x^2\right )+e^{\frac {e^2 x-3 x+8}{x}} \left (4 x^2+8 x-e^2 (3 x+8)\right ) \log (x)}{\left (e^2-x\right )^2 x^5}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\frac {8}{x}+e^2-3} \left (-x^2+4 x^2 \log (x)+e^2 x+8 \left (1-\frac {3 e^2}{8}\right ) x \log (x)-8 e^2 \log (x)\right )}{\left (e^2-x\right )^2 x^5}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{\frac {8}{x}+e^2-3}}{\left (e^2-x\right ) x^4}+\frac {e^{\frac {8}{x}+e^2-3} \left (4 x^2+\left (8-3 e^2\right ) x-8 e^2\right ) \log (x)}{\left (e^2-x\right )^2 x^5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^{\frac {8}{x}+e^2-5} \log (x)}{x^3}-\frac {3 e^{\frac {8}{x}+e^2-5} \log (x)}{8 x^2}+\frac {\left (8+3 e^2\right ) e^{\frac {8}{x}+e^2-7} \log (x)}{8 x^2}-e^{\frac {8}{x}+e^2-11} \log (x)-\frac {3}{256} e^{\frac {8}{x}+e^2-5} \log (x)+\frac {e^{\frac {8}{x}+e^2-9} \log (x)}{e^2-x}+\frac {3 e^{\frac {8}{x}+e^2-5} \log (x)}{32 x}-\frac {\left (8+3 e^2\right ) e^{\frac {8}{x}+e^2-7} \log (x)}{32 x}+\frac {\left (4+e^2\right ) e^{\frac {8}{x}+e^2-9} \log (x)}{4 x}+\frac {1}{256} \left (8+3 e^2\right ) e^{\frac {8}{x}+e^2-7} \log (x)+\frac {1}{8} \left (8+e^2\right ) e^{\frac {8}{x}+e^2-11} \log (x)-\frac {1}{32} \left (4+e^2\right ) e^{\frac {8}{x}+e^2-9} \log (x)\) |
Input:
Int[(E^((8 - 3*x + E^2*x)/x)*(E^2*x - x^2) + E^((8 - 3*x + E^2*x)/x)*(E^2* (-8 - 3*x) + 8*x + 4*x^2)*Log[x])/(E^4*x^5 - 2*E^2*x^6 + x^7),x]
Output:
-(E^(-11 + E^2 + 8/x)*Log[x]) - (3*E^(-5 + E^2 + 8/x)*Log[x])/256 - (E^(-9 + E^2 + 8/x)*(4 + E^2)*Log[x])/32 + (E^(-11 + E^2 + 8/x)*(8 + E^2)*Log[x] )/8 + (E^(-7 + E^2 + 8/x)*(8 + 3*E^2)*Log[x])/256 + (E^(-9 + E^2 + 8/x)*Lo g[x])/(E^2 - x) + (E^(-5 + E^2 + 8/x)*Log[x])/x^3 - (3*E^(-5 + E^2 + 8/x)* Log[x])/(8*x^2) + (E^(-7 + E^2 + 8/x)*(8 + 3*E^2)*Log[x])/(8*x^2) + (3*E^( -5 + E^2 + 8/x)*Log[x])/(32*x) + (E^(-9 + E^2 + 8/x)*(4 + E^2)*Log[x])/(4* x) - (E^(-7 + E^2 + 8/x)*(8 + 3*E^2)*Log[x])/(32*x)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 2.69 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07
method | result | size |
parallelrisch | \(\frac {\ln \left (x \right ) {\mathrm e}^{\frac {{\mathrm e}^{2} x -3 x +8}{x}}}{x^{3} \left ({\mathrm e}^{2}-x \right )}\) | \(29\) |
Input:
int((((-3*x-8)*exp(2)+4*x^2+8*x)*exp((exp(2)*x-3*x+8)/x)*ln(x)+(exp(2)*x-x ^2)*exp((exp(2)*x-3*x+8)/x))/(x^5*exp(2)^2-2*x^6*exp(2)+x^7),x,method=_RET URNVERBOSE)
Output:
1/x^3*ln(x)*exp((exp(2)*x-3*x+8)/x)/(exp(2)-x)
Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\frac {8-3 x+e^2 x}{x}} \left (e^2 x-x^2\right )+e^{\frac {8-3 x+e^2 x}{x}} \left (e^2 (-8-3 x)+8 x+4 x^2\right ) \log (x)}{e^4 x^5-2 e^2 x^6+x^7} \, dx=-\frac {e^{\left (\frac {x e^{2} - 3 \, x + 8}{x}\right )} \log \left (x\right )}{x^{4} - x^{3} e^{2}} \] Input:
integrate((((-3*x-8)*exp(2)+4*x^2+8*x)*exp((exp(2)*x-3*x+8)/x)*log(x)+(exp (2)*x-x^2)*exp((exp(2)*x-3*x+8)/x))/(x^5*exp(2)^2-2*x^6*exp(2)+x^7),x, alg orithm="fricas")
Output:
-e^((x*e^2 - 3*x + 8)/x)*log(x)/(x^4 - x^3*e^2)
Time = 0.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {8-3 x+e^2 x}{x}} \left (e^2 x-x^2\right )+e^{\frac {8-3 x+e^2 x}{x}} \left (e^2 (-8-3 x)+8 x+4 x^2\right ) \log (x)}{e^4 x^5-2 e^2 x^6+x^7} \, dx=- \frac {e^{\frac {- 3 x + x e^{2} + 8}{x}} \log {\left (x \right )}}{x^{4} - x^{3} e^{2}} \] Input:
integrate((((-3*x-8)*exp(2)+4*x**2+8*x)*exp((exp(2)*x-3*x+8)/x)*ln(x)+(exp (2)*x-x**2)*exp((exp(2)*x-3*x+8)/x))/(x**5*exp(2)**2-2*x**6*exp(2)+x**7),x )
Output:
-exp((-3*x + x*exp(2) + 8)/x)*log(x)/(x**4 - x**3*exp(2))
Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{\frac {8-3 x+e^2 x}{x}} \left (e^2 x-x^2\right )+e^{\frac {8-3 x+e^2 x}{x}} \left (e^2 (-8-3 x)+8 x+4 x^2\right ) \log (x)}{e^4 x^5-2 e^2 x^6+x^7} \, dx=-\frac {e^{\left (\frac {8}{x} + e^{2}\right )} \log \left (x\right )}{x^{4} e^{3} - x^{3} e^{5}} \] Input:
integrate((((-3*x-8)*exp(2)+4*x^2+8*x)*exp((exp(2)*x-3*x+8)/x)*log(x)+(exp (2)*x-x^2)*exp((exp(2)*x-3*x+8)/x))/(x^5*exp(2)^2-2*x^6*exp(2)+x^7),x, alg orithm="maxima")
Output:
-e^(8/x + e^2)*log(x)/(x^4*e^3 - x^3*e^5)
Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {8-3 x+e^2 x}{x}} \left (e^2 x-x^2\right )+e^{\frac {8-3 x+e^2 x}{x}} \left (e^2 (-8-3 x)+8 x+4 x^2\right ) \log (x)}{e^4 x^5-2 e^2 x^6+x^7} \, dx=-\frac {e^{\left (\frac {x e^{2} - x + 8}{x}\right )} \log \left (x\right )}{x^{4} e^{2} - x^{3} e^{4}} \] Input:
integrate((((-3*x-8)*exp(2)+4*x^2+8*x)*exp((exp(2)*x-3*x+8)/x)*log(x)+(exp (2)*x-x^2)*exp((exp(2)*x-3*x+8)/x))/(x^5*exp(2)^2-2*x^6*exp(2)+x^7),x, alg orithm="giac")
Output:
-e^((x*e^2 - x + 8)/x)*log(x)/(x^4*e^2 - x^3*e^4)
Timed out. \[ \int \frac {e^{\frac {8-3 x+e^2 x}{x}} \left (e^2 x-x^2\right )+e^{\frac {8-3 x+e^2 x}{x}} \left (e^2 (-8-3 x)+8 x+4 x^2\right ) \log (x)}{e^4 x^5-2 e^2 x^6+x^7} \, dx=\int \frac {{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^2-3\,x+8}{x}}\,\left (x\,{\mathrm {e}}^2-x^2\right )+{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^2-3\,x+8}{x}}\,\ln \left (x\right )\,\left (8\,x+4\,x^2-{\mathrm {e}}^2\,\left (3\,x+8\right )\right )}{x^7-2\,{\mathrm {e}}^2\,x^6+{\mathrm {e}}^4\,x^5} \,d x \] Input:
int((exp((x*exp(2) - 3*x + 8)/x)*(x*exp(2) - x^2) + exp((x*exp(2) - 3*x + 8)/x)*log(x)*(8*x + 4*x^2 - exp(2)*(3*x + 8)))/(x^5*exp(4) - 2*x^6*exp(2) + x^7),x)
Output:
int((exp((x*exp(2) - 3*x + 8)/x)*(x*exp(2) - x^2) + exp((x*exp(2) - 3*x + 8)/x)*log(x)*(8*x + 4*x^2 - exp(2)*(3*x + 8)))/(x^5*exp(4) - 2*x^6*exp(2) + x^7), x)
Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {e^{\frac {8-3 x+e^2 x}{x}} \left (e^2 x-x^2\right )+e^{\frac {8-3 x+e^2 x}{x}} \left (e^2 (-8-3 x)+8 x+4 x^2\right ) \log (x)}{e^4 x^5-2 e^2 x^6+x^7} \, dx=\frac {e^{\frac {e^{2} x +8}{x}} \mathrm {log}\left (x \right )}{e^{3} x^{3} \left (e^{2}-x \right )} \] Input:
int((((-3*x-8)*exp(2)+4*x^2+8*x)*exp((exp(2)*x-3*x+8)/x)*log(x)+(exp(2)*x- x^2)*exp((exp(2)*x-3*x+8)/x))/(x^5*exp(2)^2-2*x^6*exp(2)+x^7),x)
Output:
(e**((e**2*x + 8)/x)*log(x))/(e**3*x**3*(e**2 - x))