Integrand size = 80, antiderivative size = 26 \[ \int \frac {e^{\sqrt {\frac {200-25 \log \left (e^x x\right )}{x}}} \left (16+\sqrt {\frac {200-25 \log \left (e^x x\right )}{x}} \left (9+x-\log \left (e^x x\right )\right )-2 \log \left (e^x x\right )\right )}{-16 x^2+2 x^2 \log \left (e^x x\right )} \, dx=\frac {e^{5 \sqrt {\frac {8-\log \left (e^x x\right )}{x}}}}{x} \]
Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\sqrt {\frac {200-25 \log \left (e^x x\right )}{x}}} \left (16+\sqrt {\frac {200-25 \log \left (e^x x\right )}{x}} \left (9+x-\log \left (e^x x\right )\right )-2 \log \left (e^x x\right )\right )}{-16 x^2+2 x^2 \log \left (e^x x\right )} \, dx=\frac {e^{5 \sqrt {\frac {8-\log \left (e^x x\right )}{x}}}}{x} \]
Integrate[(E^Sqrt[(200 - 25*Log[E^x*x])/x]*(16 + Sqrt[(200 - 25*Log[E^x*x] )/x]*(9 + x - Log[E^x*x]) - 2*Log[E^x*x]))/(-16*x^2 + 2*x^2*Log[E^x*x]),x]
Leaf count is larger than twice the leaf count of optimal. \(101\) vs. \(2(26)=52\).
Time = 0.42 (sec) , antiderivative size = 101, normalized size of antiderivative = 3.88, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {2726}
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.
| \(\displaystyle \int \frac {e^{\sqrt {\frac {200-25 \log \left (e^x x\right )}{x}}} \left (\sqrt {\frac {200-25 \log \left (e^x x\right )}{x}} \left (x-\log \left (e^x x\right )+9\right )-2 \log \left (e^x x\right )+16\right )}{2 x^2 \log \left (e^x x\right )-16 x^2} \, dx\) |
| \(\Big \downarrow \) 2726 |
| \(\displaystyle \frac {e^{5 \sqrt {\frac {8-\log \left (e^x x\right )}{x}}} \left (8-\log \left (e^x x\right )\right ) \left (x-\log \left (e^x x\right )+9\right )}{x \left (\frac {e^{-x} \left (e^x x+e^x\right )}{x^2}+\frac {8-\log \left (e^x x\right )}{x^2}\right ) \left (8 x^2-x^2 \log \left (e^x x\right )\right )}\) |
Int[(E^Sqrt[(200 - 25*Log[E^x*x])/x]*(16 + Sqrt[(200 - 25*Log[E^x*x])/x]*( 9 + x - Log[E^x*x]) - 2*Log[E^x*x]))/(-16*x^2 + 2*x^2*Log[E^x*x]),x]
(E^(5*Sqrt[(8 - Log[E^x*x])/x])*(8 - Log[E^x*x])*(9 + x - Log[E^x*x]))/(x* ((E^x + E^x*x)/(E^x*x^2) + (8 - Log[E^x*x])/x^2)*(8*x^2 - x^2*Log[E^x*x]))
3.27.42.3.1 Defintions of rubi rules used
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
\[\int \frac {\left (\left (-\ln \left ({\mathrm e}^{x} x \right )+x +9\right ) \sqrt {\frac {-25 \ln \left ({\mathrm e}^{x} x \right )+200}{x}}-2 \ln \left ({\mathrm e}^{x} x \right )+16\right ) {\mathrm e}^{\sqrt {\frac {-25 \ln \left ({\mathrm e}^{x} x \right )+200}{x}}}}{2 x^{2} \ln \left ({\mathrm e}^{x} x \right )-16 x^{2}}d x\]
int(((-ln(exp(x)*x)+x+9)*((-25*ln(exp(x)*x)+200)/x)^(1/2)-2*ln(exp(x)*x)+1 6)*exp(((-25*ln(exp(x)*x)+200)/x)^(1/2))/(2*x^2*ln(exp(x)*x)-16*x^2),x)
int(((-ln(exp(x)*x)+x+9)*((-25*ln(exp(x)*x)+200)/x)^(1/2)-2*ln(exp(x)*x)+1 6)*exp(((-25*ln(exp(x)*x)+200)/x)^(1/2))/(2*x^2*ln(exp(x)*x)-16*x^2),x)
Exception generated. \[ \int \frac {e^{\sqrt {\frac {200-25 \log \left (e^x x\right )}{x}}} \left (16+\sqrt {\frac {200-25 \log \left (e^x x\right )}{x}} \left (9+x-\log \left (e^x x\right )\right )-2 \log \left (e^x x\right )\right )}{-16 x^2+2 x^2 \log \left (e^x x\right )} \, dx=\text {Exception raised: TypeError} \]
integrate(((-log(exp(x)*x)+x+9)*((-25*log(exp(x)*x)+200)/x)^(1/2)-2*log(ex p(x)*x)+16)*exp(((-25*log(exp(x)*x)+200)/x)^(1/2))/(2*x^2*log(exp(x)*x)-16 *x^2),x, algorithm=\
Exception raised: TypeError >> Error detected within library code: do_a lg_rde: unimplemented kernel
Timed out. \[ \int \frac {e^{\sqrt {\frac {200-25 \log \left (e^x x\right )}{x}}} \left (16+\sqrt {\frac {200-25 \log \left (e^x x\right )}{x}} \left (9+x-\log \left (e^x x\right )\right )-2 \log \left (e^x x\right )\right )}{-16 x^2+2 x^2 \log \left (e^x x\right )} \, dx=\text {Timed out} \]
integrate(((-ln(exp(x)*x)+x+9)*((-25*ln(exp(x)*x)+200)/x)**(1/2)-2*ln(exp( x)*x)+16)*exp(((-25*ln(exp(x)*x)+200)/x)**(1/2))/(2*x**2*ln(exp(x)*x)-16*x **2),x)
\[ \int \frac {e^{\sqrt {\frac {200-25 \log \left (e^x x\right )}{x}}} \left (16+\sqrt {\frac {200-25 \log \left (e^x x\right )}{x}} \left (9+x-\log \left (e^x x\right )\right )-2 \log \left (e^x x\right )\right )}{-16 x^2+2 x^2 \log \left (e^x x\right )} \, dx=\int { \frac {{\left (5 \, {\left (x - \log \left (x e^{x}\right ) + 9\right )} \sqrt {-\frac {\log \left (x e^{x}\right ) - 8}{x}} - 2 \, \log \left (x e^{x}\right ) + 16\right )} e^{\left (5 \, \sqrt {-\frac {\log \left (x e^{x}\right ) - 8}{x}}\right )}}{2 \, {\left (x^{2} \log \left (x e^{x}\right ) - 8 \, x^{2}\right )}} \,d x } \]
integrate(((-log(exp(x)*x)+x+9)*((-25*log(exp(x)*x)+200)/x)^(1/2)-2*log(ex p(x)*x)+16)*exp(((-25*log(exp(x)*x)+200)/x)^(1/2))/(2*x^2*log(exp(x)*x)-16 *x^2),x, algorithm=\
((x - 1)*x - x^2 + (x - 17)*log(x) + log(x)^2 - 8*x + 72)*e^(5*sqrt(-x - l og(x) + 8)/sqrt(x))/(((x - 17)*log(x) + log(x)^2 - 9*x + 72)*x) + 1/2*inte grate(-2*(x*log(x)^4 - (log(x)^2 - x - 19*log(x) + 89)*x^3 + 2*(x^2 - 17*x )*log(x)^3 + (x*log(x)^2 - x^2 - 2*(9*x - 1)*log(x) + 81*x - 17)*x^2 + 81* x^3 + (x^3 - 52*x^2 + 433*x)*log(x)^2 - (2*(x - 17)*log(x)^3 + log(x)^4 + (x^2 - 52*x + 433)*log(x)^2 + 73*x^2 - (17*x^2 - 452*x + 2448)*log(x) - 13 13*x + 5184)*x - 1296*x^2 - 18*(x^3 - 25*x^2 + 136*x)*log(x) + 5184*x)*e^( 5*sqrt(-x - log(x) + 8)/sqrt(x))/((x*log(x)^4 + 2*(x^2 - 17*x)*log(x)^3 + 81*x^3 + (x^3 - 52*x^2 + 433*x)*log(x)^2 - 1296*x^2 - 18*(x^3 - 25*x^2 + 1 36*x)*log(x) + 5184*x)*x^2), x)
\[ \int \frac {e^{\sqrt {\frac {200-25 \log \left (e^x x\right )}{x}}} \left (16+\sqrt {\frac {200-25 \log \left (e^x x\right )}{x}} \left (9+x-\log \left (e^x x\right )\right )-2 \log \left (e^x x\right )\right )}{-16 x^2+2 x^2 \log \left (e^x x\right )} \, dx=\int { \frac {{\left (5 \, {\left (x - \log \left (x e^{x}\right ) + 9\right )} \sqrt {-\frac {\log \left (x e^{x}\right ) - 8}{x}} - 2 \, \log \left (x e^{x}\right ) + 16\right )} e^{\left (5 \, \sqrt {-\frac {\log \left (x e^{x}\right ) - 8}{x}}\right )}}{2 \, {\left (x^{2} \log \left (x e^{x}\right ) - 8 \, x^{2}\right )}} \,d x } \]
integrate(((-log(exp(x)*x)+x+9)*((-25*log(exp(x)*x)+200)/x)^(1/2)-2*log(ex p(x)*x)+16)*exp(((-25*log(exp(x)*x)+200)/x)^(1/2))/(2*x^2*log(exp(x)*x)-16 *x^2),x, algorithm=\
integrate(1/2*(5*(x - log(x*e^x) + 9)*sqrt(-(log(x*e^x) - 8)/x) - 2*log(x* e^x) + 16)*e^(5*sqrt(-(log(x*e^x) - 8)/x))/(x^2*log(x*e^x) - 8*x^2), x)
Timed out. \[ \int \frac {e^{\sqrt {\frac {200-25 \log \left (e^x x\right )}{x}}} \left (16+\sqrt {\frac {200-25 \log \left (e^x x\right )}{x}} \left (9+x-\log \left (e^x x\right )\right )-2 \log \left (e^x x\right )\right )}{-16 x^2+2 x^2 \log \left (e^x x\right )} \, dx=\int \frac {{\mathrm {e}}^{\sqrt {-\frac {25\,\ln \left (x\,{\mathrm {e}}^x\right )-200}{x}}}\,\left (\sqrt {-\frac {25\,\ln \left (x\,{\mathrm {e}}^x\right )-200}{x}}\,\left (x-\ln \left (x\,{\mathrm {e}}^x\right )+9\right )-2\,\ln \left (x\,{\mathrm {e}}^x\right )+16\right )}{2\,x^2\,\ln \left (x\,{\mathrm {e}}^x\right )-16\,x^2} \,d x \]
int((exp((-(25*log(x*exp(x)) - 200)/x)^(1/2))*((-(25*log(x*exp(x)) - 200)/ x)^(1/2)*(x - log(x*exp(x)) + 9) - 2*log(x*exp(x)) + 16))/(2*x^2*log(x*exp (x)) - 16*x^2),x)