Integrand size = 222, antiderivative size = 29 \[ \int \frac {\left (4 e^2 x+8 x^2+e^x \left (e^2 x+2 x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )+e^{4-x} \log \left (x \log \left (4 x+e^x x\right )\right ) \left (4 e^2+4 x+e^x \left (x+x^2+e^2 (1+x)\right )+\left (4 e^2+4 x+e^x \left (e^2+x\right )\right ) \log \left (4 x+e^x x\right )+\left (4 x-4 e^2 x-4 x^2+e^x \left (x-e^2 x-x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )\right )}{\left (4 x+e^x x\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )} \, dx=4+\left (e^2+x\right ) \left (x+e^{4-x} \log \left (x \log \left (\left (4+e^x\right ) x\right )\right )\right ) \]
Time = 0.18 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {\left (4 e^2 x+8 x^2+e^x \left (e^2 x+2 x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )+e^{4-x} \log \left (x \log \left (4 x+e^x x\right )\right ) \left (4 e^2+4 x+e^x \left (x+x^2+e^2 (1+x)\right )+\left (4 e^2+4 x+e^x \left (e^2+x\right )\right ) \log \left (4 x+e^x x\right )+\left (4 x-4 e^2 x-4 x^2+e^x \left (x-e^2 x-x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )\right )}{\left (4 x+e^x x\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )} \, dx=e^2 x+x^2+e^{4-x} \left (e^2+x\right ) \log \left (x \log \left (\left (4+e^x\right ) x\right )\right ) \]
Integrate[((4*E^2*x + 8*x^2 + E^x*(E^2*x + 2*x^2))*Log[4*x + E^x*x]*Log[x* Log[4*x + E^x*x]] + E^(4 - x)*Log[x*Log[4*x + E^x*x]]*(4*E^2 + 4*x + E^x*( x + x^2 + E^2*(1 + x)) + (4*E^2 + 4*x + E^x*(E^2 + x))*Log[4*x + E^x*x] + (4*x - 4*E^2*x - 4*x^2 + E^x*(x - E^2*x - x^2))*Log[4*x + E^x*x]*Log[x*Log [4*x + E^x*x]]))/((4*x + E^x*x)*Log[4*x + E^x*x]*Log[x*Log[4*x + E^x*x]]), x]
Leaf count is larger than twice the leaf count of optimal. \(62\) vs. \(2(29)=58\).
Time = 14.45 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.14, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {7239, 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.
| \(\displaystyle \int \frac {\left (8 x^2+e^x \left (2 x^2+e^2 x\right )+4 e^2 x\right ) \log \left (e^x x+4 x\right ) \log \left (x \log \left (e^x x+4 x\right )\right )+e^{4-x} \left (e^x \left (x^2+x+e^2 (x+1)\right )+\left (-4 x^2+e^x \left (-x^2-e^2 x+x\right )-4 e^2 x+4 x\right ) \log \left (e^x x+4 x\right ) \log \left (x \log \left (e^x x+4 x\right )\right )+4 x+\left (4 x+e^x \left (x+e^2\right )+4 e^2\right ) \log \left (e^x x+4 x\right )+4 e^2\right ) \log \left (x \log \left (e^x x+4 x\right )\right )}{\left (e^x x+4 x\right ) \log \left (e^x x+4 x\right ) \log \left (x \log \left (e^x x+4 x\right )\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {e^{-x} \left (2 e^x x^2+e^{x+2} x+e^4 x-e^4 \left (x+e^2-1\right ) x \log \left (x \log \left (\left (e^x+4\right ) x\right )\right )+\frac {e^4 \left (x+e^2\right ) \left (e^x (x+1)+4\right )}{\left (e^x+4\right ) \log \left (\left (e^x+4\right ) x\right )}+e^6\right )}{x}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {e^{4-x} \left (x^2+x^2 \left (-\log \left (\left (e^x+4\right ) x\right )\right ) \log \left (x \log \left (\left (e^x+4\right ) x\right )\right )+\left (1+e^2\right ) x+x \log \left (\left (e^x+4\right ) x\right )+\left (1-e^2\right ) x \log \left (\left (e^x+4\right ) x\right ) \log \left (x \log \left (\left (e^x+4\right ) x\right )\right )+e^2 \log \left (\left (e^x+4\right ) x\right )+e^2\right )}{x \log \left (e^x x+4 x\right )}+2 x+\frac {4 e^{4-x} \left (-x-e^2\right )}{\left (e^x+4\right ) \log \left (e^x x+4 x\right )}+e^2\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle x^2+e^2 x+e^{4-x} \log \left (x \log \left (e^x x+4 x\right )\right )-e^{4-x} \left (-x-e^2+1\right ) \log \left (x \log \left (e^x x+4 x\right )\right )\) |
Int[((4*E^2*x + 8*x^2 + E^x*(E^2*x + 2*x^2))*Log[4*x + E^x*x]*Log[x*Log[4* x + E^x*x]] + E^(4 - x)*Log[x*Log[4*x + E^x*x]]*(4*E^2 + 4*x + E^x*(x + x^ 2 + E^2*(1 + x)) + (4*E^2 + 4*x + E^x*(E^2 + x))*Log[4*x + E^x*x] + (4*x - 4*E^2*x - 4*x^2 + E^x*(x - E^2*x - x^2))*Log[4*x + E^x*x]*Log[x*Log[4*x + E^x*x]]))/((4*x + E^x*x)*Log[4*x + E^x*x]*Log[x*Log[4*x + E^x*x]]),x]
E^2*x + x^2 + E^(4 - x)*Log[x*Log[4*x + E^x*x]] - E^(4 - x)*(1 - E^2 - x)* Log[x*Log[4*x + E^x*x]]
3.27.57.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.10 (sec) , antiderivative size = 2626, normalized size of antiderivative = 90.55
\[\text {Expression too large to display}\]
int(((((-exp(2)*x-x^2+x)*exp(x)-4*exp(2)*x-4*x^2+4*x)*ln(exp(x)*x+4*x)*ln( x*ln(exp(x)*x+4*x))+((x+exp(2))*exp(x)+4*exp(2)+4*x)*ln(exp(x)*x+4*x)+((1+ x)*exp(2)+x^2+x)*exp(x)+4*exp(2)+4*x)*exp(ln(ln(x*ln(exp(x)*x+4*x)))-x+4)+ ((exp(2)*x+2*x^2)*exp(x)+4*exp(2)*x+8*x^2)*ln(exp(x)*x+4*x)*ln(x*ln(exp(x) *x+4*x)))/(exp(x)*x+4*x)/ln(exp(x)*x+4*x)/ln(x*ln(exp(x)*x+4*x)),x)
ln(-Pi*csgn(I*x)*csgn(I*(exp(x)+4)*x)^2-Pi*csgn(I*(exp(x)+4))*csgn(I*(exp( x)+4)*x)^2+Pi*csgn(I*x)*csgn(I*(exp(x)+4))*csgn(I*(exp(x)+4)*x)+2*I*ln(x)+ 2*I*ln(exp(x)+4)+Pi*csgn(I*(exp(x)+4)*x)^3)*(x+exp(2))*exp(-x+4)+1/2*(I*ex p(6)*Pi*csgn(I*x*(-Pi*csgn(I*x)*csgn(I*(exp(x)+4)*x)^2-Pi*csgn(I*(exp(x)+4 ))*csgn(I*(exp(x)+4)*x)^2+Pi*csgn(I*x)*csgn(I*(exp(x)+4))*csgn(I*(exp(x)+4 )*x)+2*I*ln(x)+2*I*ln(exp(x)+4)+Pi*csgn(I*(exp(x)+4)*x)^3))*csgn(x*(-Pi*cs gn(I*x)*csgn(I*(exp(x)+4)*x)^2-Pi*csgn(I*(exp(x)+4))*csgn(I*(exp(x)+4)*x)^ 2+Pi*csgn(I*x)*csgn(I*(exp(x)+4))*csgn(I*(exp(x)+4)*x)+2*I*ln(x)+2*I*ln(ex p(x)+4)+Pi*csgn(I*(exp(x)+4)*x)^3))^2+I*exp(4)*x*Pi*csgn(I*x*(-Pi*csgn(I*x )*csgn(I*(exp(x)+4)*x)^2-Pi*csgn(I*(exp(x)+4))*csgn(I*(exp(x)+4)*x)^2+Pi*c sgn(I*x)*csgn(I*(exp(x)+4))*csgn(I*(exp(x)+4)*x)+2*I*ln(x)+2*I*ln(exp(x)+4 )+Pi*csgn(I*(exp(x)+4)*x)^3))*csgn(x*(-Pi*csgn(I*x)*csgn(I*(exp(x)+4)*x)^2 -Pi*csgn(I*(exp(x)+4))*csgn(I*(exp(x)+4)*x)^2+Pi*csgn(I*x)*csgn(I*(exp(x)+ 4))*csgn(I*(exp(x)+4)*x)+2*I*ln(x)+2*I*ln(exp(x)+4)+Pi*csgn(I*(exp(x)+4)*x )^3))^2-I*exp(6)*Pi+I*exp(6)*Pi*csgn(x*(-Pi*csgn(I*x)*csgn(I*(exp(x)+4)*x) ^2-Pi*csgn(I*(exp(x)+4))*csgn(I*(exp(x)+4)*x)^2+Pi*csgn(I*x)*csgn(I*(exp(x )+4))*csgn(I*(exp(x)+4)*x)+2*I*ln(x)+2*I*ln(exp(x)+4)+Pi*csgn(I*(exp(x)+4) *x)^3))^2-I*exp(6)*Pi*csgn(x*(-Pi*csgn(I*x)*csgn(I*(exp(x)+4)*x)^2-Pi*csgn (I*(exp(x)+4))*csgn(I*(exp(x)+4)*x)^2+Pi*csgn(I*x)*csgn(I*(exp(x)+4))*csgn (I*(exp(x)+4)*x)+2*I*ln(x)+2*I*ln(exp(x)+4)+Pi*csgn(I*(exp(x)+4)*x)^3))...
Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {\left (4 e^2 x+8 x^2+e^x \left (e^2 x+2 x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )+e^{4-x} \log \left (x \log \left (4 x+e^x x\right )\right ) \left (4 e^2+4 x+e^x \left (x+x^2+e^2 (1+x)\right )+\left (4 e^2+4 x+e^x \left (e^2+x\right )\right ) \log \left (4 x+e^x x\right )+\left (4 x-4 e^2 x-4 x^2+e^x \left (x-e^2 x-x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )\right )}{\left (4 x+e^x x\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )} \, dx={\left ({\left (x^{2} + x e^{2}\right )} e^{x} + {\left (x e^{4} + e^{6}\right )} \log \left (x \log \left (x e^{x} + 4 \, x\right )\right )\right )} e^{\left (-x\right )} \]
integrate(((((-exp(2)*x-x^2+x)*exp(x)-4*exp(2)*x-4*x^2+4*x)*log(exp(x)*x+4 *x)*log(x*log(exp(x)*x+4*x))+((x+exp(2))*exp(x)+4*exp(2)+4*x)*log(exp(x)*x +4*x)+((1+x)*exp(2)+x^2+x)*exp(x)+4*exp(2)+4*x)*exp(log(log(x*log(exp(x)*x +4*x)))-x+4)+((exp(2)*x+2*x^2)*exp(x)+4*exp(2)*x+8*x^2)*log(exp(x)*x+4*x)* log(x*log(exp(x)*x+4*x)))/(exp(x)*x+4*x)/log(exp(x)*x+4*x)/log(x*log(exp(x )*x+4*x)),x, algorithm=\
Time = 0.68 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {\left (4 e^2 x+8 x^2+e^x \left (e^2 x+2 x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )+e^{4-x} \log \left (x \log \left (4 x+e^x x\right )\right ) \left (4 e^2+4 x+e^x \left (x+x^2+e^2 (1+x)\right )+\left (4 e^2+4 x+e^x \left (e^2+x\right )\right ) \log \left (4 x+e^x x\right )+\left (4 x-4 e^2 x-4 x^2+e^x \left (x-e^2 x-x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )\right )}{\left (4 x+e^x x\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )} \, dx=x^{2} + x e^{2} + \left (x e^{4} + e^{6}\right ) e^{- x} \log {\left (x \log {\left (x e^{x} + 4 x \right )} \right )} \]
integrate(((((-exp(2)*x-x**2+x)*exp(x)-4*exp(2)*x-4*x**2+4*x)*ln(exp(x)*x+ 4*x)*ln(x*ln(exp(x)*x+4*x))+((x+exp(2))*exp(x)+4*exp(2)+4*x)*ln(exp(x)*x+4 *x)+((1+x)*exp(2)+x**2+x)*exp(x)+4*exp(2)+4*x)*exp(ln(ln(x*ln(exp(x)*x+4*x )))-x+4)+((exp(2)*x+2*x**2)*exp(x)+4*exp(2)*x+8*x**2)*ln(exp(x)*x+4*x)*ln( x*ln(exp(x)*x+4*x)))/(exp(x)*x+4*x)/ln(exp(x)*x+4*x)/ln(x*ln(exp(x)*x+4*x) ),x)
Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48 \[ \int \frac {\left (4 e^2 x+8 x^2+e^x \left (e^2 x+2 x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )+e^{4-x} \log \left (x \log \left (4 x+e^x x\right )\right ) \left (4 e^2+4 x+e^x \left (x+x^2+e^2 (1+x)\right )+\left (4 e^2+4 x+e^x \left (e^2+x\right )\right ) \log \left (4 x+e^x x\right )+\left (4 x-4 e^2 x-4 x^2+e^x \left (x-e^2 x-x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )\right )}{\left (4 x+e^x x\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )} \, dx={\left (x e^{4} + e^{6}\right )} e^{\left (-x\right )} \log \left (x\right ) + {\left (x e^{4} + e^{6}\right )} e^{\left (-x\right )} \log \left (\log \left (x\right ) + \log \left (e^{x} + 4\right )\right ) + x^{2} + x e^{2} \]
integrate(((((-exp(2)*x-x^2+x)*exp(x)-4*exp(2)*x-4*x^2+4*x)*log(exp(x)*x+4 *x)*log(x*log(exp(x)*x+4*x))+((x+exp(2))*exp(x)+4*exp(2)+4*x)*log(exp(x)*x +4*x)+((1+x)*exp(2)+x^2+x)*exp(x)+4*exp(2)+4*x)*exp(log(log(x*log(exp(x)*x +4*x)))-x+4)+((exp(2)*x+2*x^2)*exp(x)+4*exp(2)*x+8*x^2)*log(exp(x)*x+4*x)* log(x*log(exp(x)*x+4*x)))/(exp(x)*x+4*x)/log(exp(x)*x+4*x)/log(x*log(exp(x )*x+4*x)),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (26) = 52\).
Time = 0.44 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.86 \[ \int \frac {\left (4 e^2 x+8 x^2+e^x \left (e^2 x+2 x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )+e^{4-x} \log \left (x \log \left (4 x+e^x x\right )\right ) \left (4 e^2+4 x+e^x \left (x+x^2+e^2 (1+x)\right )+\left (4 e^2+4 x+e^x \left (e^2+x\right )\right ) \log \left (4 x+e^x x\right )+\left (4 x-4 e^2 x-4 x^2+e^x \left (x-e^2 x-x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )\right )}{\left (4 x+e^x x\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )} \, dx={\left (x^{2} e^{x} + x e^{4} \log \left (x\right ) + x e^{4} \log \left (\log \left (x\right ) + \log \left (e^{x} + 4\right )\right ) + x e^{\left (x + 2\right )} + e^{6} \log \left (x\right ) + e^{6} \log \left (\log \left (x\right ) + \log \left (e^{x} + 4\right )\right )\right )} e^{\left (-x\right )} \]
integrate(((((-exp(2)*x-x^2+x)*exp(x)-4*exp(2)*x-4*x^2+4*x)*log(exp(x)*x+4 *x)*log(x*log(exp(x)*x+4*x))+((x+exp(2))*exp(x)+4*exp(2)+4*x)*log(exp(x)*x +4*x)+((1+x)*exp(2)+x^2+x)*exp(x)+4*exp(2)+4*x)*exp(log(log(x*log(exp(x)*x +4*x)))-x+4)+((exp(2)*x+2*x^2)*exp(x)+4*exp(2)*x+8*x^2)*log(exp(x)*x+4*x)* log(x*log(exp(x)*x+4*x)))/(exp(x)*x+4*x)/log(exp(x)*x+4*x)/log(x*log(exp(x )*x+4*x)),x, algorithm=\
(x^2*e^x + x*e^4*log(x) + x*e^4*log(log(x) + log(e^x + 4)) + x*e^(x + 2) + e^6*log(x) + e^6*log(log(x) + log(e^x + 4)))*e^(-x)
Timed out. \[ \int \frac {\left (4 e^2 x+8 x^2+e^x \left (e^2 x+2 x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )+e^{4-x} \log \left (x \log \left (4 x+e^x x\right )\right ) \left (4 e^2+4 x+e^x \left (x+x^2+e^2 (1+x)\right )+\left (4 e^2+4 x+e^x \left (e^2+x\right )\right ) \log \left (4 x+e^x x\right )+\left (4 x-4 e^2 x-4 x^2+e^x \left (x-e^2 x-x^2\right )\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )\right )}{\left (4 x+e^x x\right ) \log \left (4 x+e^x x\right ) \log \left (x \log \left (4 x+e^x x\right )\right )} \, dx=\int \frac {{\mathrm {e}}^{\ln \left (\ln \left (x\,\ln \left (4\,x+x\,{\mathrm {e}}^x\right )\right )\right )-x+4}\,\left (4\,x+4\,{\mathrm {e}}^2+\ln \left (4\,x+x\,{\mathrm {e}}^x\right )\,\left (4\,x+4\,{\mathrm {e}}^2+{\mathrm {e}}^x\,\left (x+{\mathrm {e}}^2\right )\right )+{\mathrm {e}}^x\,\left (x+{\mathrm {e}}^2\,\left (x+1\right )+x^2\right )-\ln \left (4\,x+x\,{\mathrm {e}}^x\right )\,\ln \left (x\,\ln \left (4\,x+x\,{\mathrm {e}}^x\right )\right )\,\left (4\,x\,{\mathrm {e}}^2-4\,x+{\mathrm {e}}^x\,\left (x\,{\mathrm {e}}^2-x+x^2\right )+4\,x^2\right )\right )+\ln \left (4\,x+x\,{\mathrm {e}}^x\right )\,\ln \left (x\,\ln \left (4\,x+x\,{\mathrm {e}}^x\right )\right )\,\left ({\mathrm {e}}^x\,\left (2\,x^2+{\mathrm {e}}^2\,x\right )+4\,x\,{\mathrm {e}}^2+8\,x^2\right )}{\ln \left (4\,x+x\,{\mathrm {e}}^x\right )\,\ln \left (x\,\ln \left (4\,x+x\,{\mathrm {e}}^x\right )\right )\,\left (4\,x+x\,{\mathrm {e}}^x\right )} \,d x \]
int((exp(log(log(x*log(4*x + x*exp(x)))) - x + 4)*(4*x + 4*exp(2) + log(4* x + x*exp(x))*(4*x + 4*exp(2) + exp(x)*(x + exp(2))) + exp(x)*(x + exp(2)* (x + 1) + x^2) - log(4*x + x*exp(x))*log(x*log(4*x + x*exp(x)))*(4*x*exp(2 ) - 4*x + exp(x)*(x*exp(2) - x + x^2) + 4*x^2)) + log(4*x + x*exp(x))*log( x*log(4*x + x*exp(x)))*(exp(x)*(x*exp(2) + 2*x^2) + 4*x*exp(2) + 8*x^2))/( log(4*x + x*exp(x))*log(x*log(4*x + x*exp(x)))*(4*x + x*exp(x))),x)
int((exp(log(log(x*log(4*x + x*exp(x)))) - x + 4)*(4*x + 4*exp(2) + log(4* x + x*exp(x))*(4*x + 4*exp(2) + exp(x)*(x + exp(2))) + exp(x)*(x + exp(2)* (x + 1) + x^2) - log(4*x + x*exp(x))*log(x*log(4*x + x*exp(x)))*(4*x*exp(2 ) - 4*x + exp(x)*(x*exp(2) - x + x^2) + 4*x^2)) + log(4*x + x*exp(x))*log( x*log(4*x + x*exp(x)))*(exp(x)*(x*exp(2) + 2*x^2) + 4*x*exp(2) + 8*x^2))/( log(4*x + x*exp(x))*log(x*log(4*x + x*exp(x)))*(4*x + x*exp(x))), x)