Integrand size = 119, antiderivative size = 27 \begin {dmath*} \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=\left (3-\frac {5}{3} e^{\frac {e^x}{x \left (x+x \log \left (\frac {1}{x}\right )\right )}}\right )^2 \end {dmath*}
Time = 5.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.70 \begin {dmath*} \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {10}{9} \left (-9 e^{\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}+\frac {5}{2} e^{\frac {2 e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}\right ) \end {dmath*}
Integrate[(E^(E^x/(x^2 + x^2*Log[x^(-1)]))*(E^x*(90 - 90*x) + E^x*(180 - 9 0*x)*Log[x^(-1)]) + E^((2*E^x)/(x^2 + x^2*Log[x^(-1)]))*(E^x*(-50 + 50*x) + E^x*(-100 + 50*x)*Log[x^(-1)]))/(9*x^3 + 18*x^3*Log[x^(-1)] + 9*x^3*Log[ x^(-1)]^2),x]
Time = 3.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {7239, 27, 7259, 17}
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.
\(\displaystyle \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (50 x-50)+e^x (50 x-100) \log \left (\frac {1}{x}\right )\right )}{9 x^3+9 x^3 \log ^2\left (\frac {1}{x}\right )+18 x^3 \log \left (\frac {1}{x}\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {10 e^{\frac {e^x}{x^2 \left (\log \left (\frac {1}{x}\right )+1\right )}+x} \left (9-5 e^{\frac {e^x}{x^2 \left (\log \left (\frac {1}{x}\right )+1\right )}}\right ) \left (-x-(x-2) \log \left (\frac {1}{x}\right )+1\right )}{9 x^3 \left (\log \left (\frac {1}{x}\right )+1\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {10}{9} \int \frac {e^{x+\frac {e^x}{\left (\log \left (\frac {1}{x}\right )+1\right ) x^2}} \left (9-5 e^{\frac {e^x}{x^2 \left (\log \left (\frac {1}{x}\right )+1\right )}}\right ) \left (-x+(2-x) \log \left (\frac {1}{x}\right )+1\right )}{x^3 \left (\log \left (\frac {1}{x}\right )+1\right )^2}dx\) |
\(\Big \downarrow \) 7259 |
\(\displaystyle \frac {2}{9} \int \left (9-5 e^{\frac {e^x}{x^2 \left (\log \left (\frac {1}{x}\right )+1\right )}}\right )d\left (-5 e^{\frac {e^x}{x^2 \left (\log \left (\frac {1}{x}\right )+1\right )}}\right )\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {1}{9} \left (9-5 e^{\frac {e^x}{x^2 \left (\log \left (\frac {1}{x}\right )+1\right )}}\right )^2\) |
Int[(E^(E^x/(x^2 + x^2*Log[x^(-1)]))*(E^x*(90 - 90*x) + E^x*(180 - 90*x)*L og[x^(-1)]) + E^((2*E^x)/(x^2 + x^2*Log[x^(-1)]))*(E^x*(-50 + 50*x) + E^x* (-100 + 50*x)*Log[x^(-1)]))/(9*x^3 + 18*x^3*Log[x^(-1)] + 9*x^3*Log[x^(-1) ]^2),x]
3.8.76.3.1 Defintions of rubi rules used
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Int[(u_)*((a_) + (b_.)*(v_)^(p_.)*(w_)^(p_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(w*D[v, x] + v*D[w, x])]}, Simp[c Subst[Int[(a + b*x^p)^m, x] , x, v*w], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p}, x] && IntegerQ[p]
Time = 7.40 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26
method | result | size |
risch | \(\frac {25 \,{\mathrm e}^{-\frac {2 \,{\mathrm e}^{x}}{x^{2} \left (\ln \left (x \right )-1\right )}}}{9}-10 \,{\mathrm e}^{-\frac {{\mathrm e}^{x}}{x^{2} \left (\ln \left (x \right )-1\right )}}\) | \(34\) |
int((((50*x-100)*exp(x)*ln(1/x)+(50*x-50)*exp(x))*exp(exp(x)/(x^2*ln(1/x)+ x^2))^2+((-90*x+180)*exp(x)*ln(1/x)+(-90*x+90)*exp(x))*exp(exp(x)/(x^2*ln( 1/x)+x^2)))/(9*x^3*ln(1/x)^2+18*x^3*ln(1/x)+9*x^3),x,method=_RETURNVERBOSE )
Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \begin {dmath*} \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {25}{9} \, e^{\left (\frac {2 \, e^{x}}{x^{2} \log \left (\frac {1}{x}\right ) + x^{2}}\right )} - 10 \, e^{\left (\frac {e^{x}}{x^{2} \log \left (\frac {1}{x}\right ) + x^{2}}\right )} \end {dmath*}
integrate((((50*x-100)*exp(x)*log(1/x)+(50*x-50)*exp(x))*exp(exp(x)/(x^2*l og(1/x)+x^2))^2+((-90*x+180)*exp(x)*log(1/x)+(-90*x+90)*exp(x))*exp(exp(x) /(x^2*log(1/x)+x^2)))/(9*x^3*log(1/x)^2+18*x^3*log(1/x)+9*x^3),x, algorith m=\
Time = 0.50 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \begin {dmath*} \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {25 e^{\frac {2 e^{x}}{x^{2} \log {\left (\frac {1}{x} \right )} + x^{2}}}}{9} - 10 e^{\frac {e^{x}}{x^{2} \log {\left (\frac {1}{x} \right )} + x^{2}}} \end {dmath*}
integrate((((50*x-100)*exp(x)*ln(1/x)+(50*x-50)*exp(x))*exp(exp(x)/(x**2*l n(1/x)+x**2))**2+((-90*x+180)*exp(x)*ln(1/x)+(-90*x+90)*exp(x))*exp(exp(x) /(x**2*ln(1/x)+x**2)))/(9*x**3*ln(1/x)**2+18*x**3*ln(1/x)+9*x**3),x)
Time = 0.32 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \begin {dmath*} \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=-\frac {5}{9} \, {\left (18 \, e^{\left (\frac {e^{x}}{x^{2} \log \left (x\right ) - x^{2}}\right )} - 5\right )} e^{\left (-\frac {2 \, e^{x}}{x^{2} \log \left (x\right ) - x^{2}}\right )} \end {dmath*}
integrate((((50*x-100)*exp(x)*log(1/x)+(50*x-50)*exp(x))*exp(exp(x)/(x^2*l og(1/x)+x^2))^2+((-90*x+180)*exp(x)*log(1/x)+(-90*x+90)*exp(x))*exp(exp(x) /(x^2*log(1/x)+x^2)))/(9*x^3*log(1/x)^2+18*x^3*log(1/x)+9*x^3),x, algorith m=\
Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \begin {dmath*} \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=-10 \, e^{\left (-\frac {e^{x}}{x^{2} \log \left (x\right ) - x^{2}}\right )} + \frac {25}{9} \, e^{\left (-\frac {2 \, e^{x}}{x^{2} \log \left (x\right ) - x^{2}}\right )} \end {dmath*}
integrate((((50*x-100)*exp(x)*log(1/x)+(50*x-50)*exp(x))*exp(exp(x)/(x^2*l og(1/x)+x^2))^2+((-90*x+180)*exp(x)*log(1/x)+(-90*x+90)*exp(x))*exp(exp(x) /(x^2*log(1/x)+x^2)))/(9*x^3*log(1/x)^2+18*x^3*log(1/x)+9*x^3),x, algorith m=\
Time = 16.58 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \begin {dmath*} \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {5\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{x^2\,\ln \left (\frac {1}{x}\right )+x^2}}\,\left (5\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{x^2\,\ln \left (\frac {1}{x}\right )+x^2}}-18\right )}{9} \end {dmath*}