Integrand size = 101, antiderivative size = 22 \[ \int \frac {e^{4+4 e^{-6+x} x+e^{-12+2 x} x^2} \left (-1+e^{-6+x} \left (16 x+16 x^2\right )+e^{-12+2 x} \left (8 x^2+8 x^3\right )+\left (e^{-6+x} \left (-4 x-4 x^2\right )+e^{-12+2 x} \left (-2 x^2-2 x^3\right )\right ) \log (x)\right )}{x} \, dx=-4+e^{\left (2+e^{-6+x} x\right )^2} (4-\log (x)) \] Output:
exp((x*exp(-6+x)+2)^2)*(-ln(x)+4)-4
Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {e^{4+4 e^{-6+x} x+e^{-12+2 x} x^2} \left (-1+e^{-6+x} \left (16 x+16 x^2\right )+e^{-12+2 x} \left (8 x^2+8 x^3\right )+\left (e^{-6+x} \left (-4 x-4 x^2\right )+e^{-12+2 x} \left (-2 x^2-2 x^3\right )\right ) \log (x)\right )}{x} \, dx=-e^{\frac {\left (2 e^6+e^x x\right )^2}{e^{12}}} (-4+\log (x)) \] Input:
Integrate[(E^(4 + 4*E^(-6 + x)*x + E^(-12 + 2*x)*x^2)*(-1 + E^(-6 + x)*(16 *x + 16*x^2) + E^(-12 + 2*x)*(8*x^2 + 8*x^3) + (E^(-6 + x)*(-4*x - 4*x^2) + E^(-12 + 2*x)*(-2*x^2 - 2*x^3))*Log[x]))/x,x]
Output:
-(E^((2*E^6 + E^x*x)^2/E^12)*(-4 + Log[x]))
Leaf count is larger than twice the leaf count of optimal. \(126\) vs. \(2(22)=44\).
Time = 0.71 (sec) , antiderivative size = 126, normalized size of antiderivative = 5.73, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, 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.
\(\displaystyle \int \frac {e^{e^{2 x-12} x^2+4 e^{x-6} x+4} \left (e^{x-6} \left (16 x^2+16 x\right )+e^{2 x-12} \left (8 x^3+8 x^2\right )+\left (e^{x-6} \left (-4 x^2-4 x\right )+e^{2 x-12} \left (-2 x^3-2 x^2\right )\right ) \log (x)-1\right )}{x} \, dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle \frac {e^{e^{2 x-12} x^2+4 e^{x-6} x+4} \left (8 e^{x-6} \left (x^2+x\right )+4 e^{2 x-12} \left (x^3+x^2\right )-\left (2 e^{x-6} \left (x^2+x\right )+e^{2 x-12} \left (x^3+x^2\right )\right ) \log (x)\right )}{x \left (e^{2 x-12} x^2+2 e^{x-6} x+e^{2 x-12} x+2 e^{x-6}\right )}\) |
Input:
Int[(E^(4 + 4*E^(-6 + x)*x + E^(-12 + 2*x)*x^2)*(-1 + E^(-6 + x)*(16*x + 1 6*x^2) + E^(-12 + 2*x)*(8*x^2 + 8*x^3) + (E^(-6 + x)*(-4*x - 4*x^2) + E^(- 12 + 2*x)*(-2*x^2 - 2*x^3))*Log[x]))/x,x]
Output:
(E^(4 + 4*E^(-6 + x)*x + E^(-12 + 2*x)*x^2)*(8*E^(-6 + x)*(x + x^2) + 4*E^ (-12 + 2*x)*(x^2 + x^3) - (2*E^(-6 + x)*(x + x^2) + E^(-12 + 2*x)*(x^2 + x ^3))*Log[x]))/(x*(2*E^(-6 + x) + 2*E^(-6 + x)*x + E^(-12 + 2*x)*x + E^(-12 + 2*x)*x^2))
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]
Time = 0.45 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27
method | result | size |
risch | \(\left (-\ln \left (x \right )+4\right ) {\mathrm e}^{{\mathrm e}^{2 x -12} x^{2}+4 x \,{\mathrm e}^{-6+x}+4}\) | \(28\) |
parallelrisch | \(-{\mathrm e}^{{\mathrm e}^{2 x -12} x^{2}+4 x \,{\mathrm e}^{-6+x}+4} \ln \left (x \right )+4 \,{\mathrm e}^{{\mathrm e}^{2 x -12} x^{2}+4 x \,{\mathrm e}^{-6+x}+4}\) | \(48\) |
Input:
int((((-2*x^3-2*x^2)*exp(-6+x)^2+(-4*x^2-4*x)*exp(-6+x))*ln(x)+(8*x^3+8*x^ 2)*exp(-6+x)^2+(16*x^2+16*x)*exp(-6+x)-1)*exp(x^2*exp(-6+x)^2+4*x*exp(-6+x )+4)/x,x,method=_RETURNVERBOSE)
Output:
(-ln(x)+4)*exp(exp(2*x-12)*x^2+4*x*exp(-6+x)+4)
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^{4+4 e^{-6+x} x+e^{-12+2 x} x^2} \left (-1+e^{-6+x} \left (16 x+16 x^2\right )+e^{-12+2 x} \left (8 x^2+8 x^3\right )+\left (e^{-6+x} \left (-4 x-4 x^2\right )+e^{-12+2 x} \left (-2 x^2-2 x^3\right )\right ) \log (x)\right )}{x} \, dx=-{\left (\log \left (x\right ) - 4\right )} e^{\left (x^{2} e^{\left (2 \, x - 12\right )} + 4 \, x e^{\left (x - 6\right )} + 4\right )} \] Input:
integrate((((-2*x^3-2*x^2)*exp(-6+x)^2+(-4*x^2-4*x)*exp(-6+x))*log(x)+(8*x ^3+8*x^2)*exp(-6+x)^2+(16*x^2+16*x)*exp(-6+x)-1)*exp(x^2*exp(-6+x)^2+4*x*e xp(-6+x)+4)/x,x, algorithm="fricas")
Output:
-(log(x) - 4)*e^(x^2*e^(2*x - 12) + 4*x*e^(x - 6) + 4)
Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^{4+4 e^{-6+x} x+e^{-12+2 x} x^2} \left (-1+e^{-6+x} \left (16 x+16 x^2\right )+e^{-12+2 x} \left (8 x^2+8 x^3\right )+\left (e^{-6+x} \left (-4 x-4 x^2\right )+e^{-12+2 x} \left (-2 x^2-2 x^3\right )\right ) \log (x)\right )}{x} \, dx=\left (4 - \log {\left (x \right )}\right ) e^{x^{2} e^{2 x - 12} + 4 x e^{x - 6} + 4} \] Input:
integrate((((-2*x**3-2*x**2)*exp(-6+x)**2+(-4*x**2-4*x)*exp(-6+x))*ln(x)+( 8*x**3+8*x**2)*exp(-6+x)**2+(16*x**2+16*x)*exp(-6+x)-1)*exp(x**2*exp(-6+x) **2+4*x*exp(-6+x)+4)/x,x)
Output:
(4 - log(x))*exp(x**2*exp(2*x - 12) + 4*x*exp(x - 6) + 4)
Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {e^{4+4 e^{-6+x} x+e^{-12+2 x} x^2} \left (-1+e^{-6+x} \left (16 x+16 x^2\right )+e^{-12+2 x} \left (8 x^2+8 x^3\right )+\left (e^{-6+x} \left (-4 x-4 x^2\right )+e^{-12+2 x} \left (-2 x^2-2 x^3\right )\right ) \log (x)\right )}{x} \, dx=-{\left (e^{4} \log \left (x\right ) - 4 \, e^{4}\right )} e^{\left (x^{2} e^{\left (2 \, x - 12\right )} + 4 \, x e^{\left (x - 6\right )}\right )} \] Input:
integrate((((-2*x^3-2*x^2)*exp(-6+x)^2+(-4*x^2-4*x)*exp(-6+x))*log(x)+(8*x ^3+8*x^2)*exp(-6+x)^2+(16*x^2+16*x)*exp(-6+x)-1)*exp(x^2*exp(-6+x)^2+4*x*e xp(-6+x)+4)/x,x, algorithm="maxima")
Output:
-(e^4*log(x) - 4*e^4)*e^(x^2*e^(2*x - 12) + 4*x*e^(x - 6))
\[ \int \frac {e^{4+4 e^{-6+x} x+e^{-12+2 x} x^2} \left (-1+e^{-6+x} \left (16 x+16 x^2\right )+e^{-12+2 x} \left (8 x^2+8 x^3\right )+\left (e^{-6+x} \left (-4 x-4 x^2\right )+e^{-12+2 x} \left (-2 x^2-2 x^3\right )\right ) \log (x)\right )}{x} \, dx=\int { \frac {{\left (8 \, {\left (x^{3} + x^{2}\right )} e^{\left (2 \, x - 12\right )} + 16 \, {\left (x^{2} + x\right )} e^{\left (x - 6\right )} - 2 \, {\left ({\left (x^{3} + x^{2}\right )} e^{\left (2 \, x - 12\right )} + 2 \, {\left (x^{2} + x\right )} e^{\left (x - 6\right )}\right )} \log \left (x\right ) - 1\right )} e^{\left (x^{2} e^{\left (2 \, x - 12\right )} + 4 \, x e^{\left (x - 6\right )} + 4\right )}}{x} \,d x } \] Input:
integrate((((-2*x^3-2*x^2)*exp(-6+x)^2+(-4*x^2-4*x)*exp(-6+x))*log(x)+(8*x ^3+8*x^2)*exp(-6+x)^2+(16*x^2+16*x)*exp(-6+x)-1)*exp(x^2*exp(-6+x)^2+4*x*e xp(-6+x)+4)/x,x, algorithm="giac")
Output:
integrate((8*(x^3 + x^2)*e^(2*x - 12) + 16*(x^2 + x)*e^(x - 6) - 2*((x^3 + x^2)*e^(2*x - 12) + 2*(x^2 + x)*e^(x - 6))*log(x) - 1)*e^(x^2*e^(2*x - 12 ) + 4*x*e^(x - 6) + 4)/x, x)
Timed out. \[ \int \frac {e^{4+4 e^{-6+x} x+e^{-12+2 x} x^2} \left (-1+e^{-6+x} \left (16 x+16 x^2\right )+e^{-12+2 x} \left (8 x^2+8 x^3\right )+\left (e^{-6+x} \left (-4 x-4 x^2\right )+e^{-12+2 x} \left (-2 x^2-2 x^3\right )\right ) \log (x)\right )}{x} \, dx=\int \frac {{\mathrm {e}}^{4\,x\,{\mathrm {e}}^{x-6}+x^2\,{\mathrm {e}}^{2\,x-12}+4}\,\left ({\mathrm {e}}^{x-6}\,\left (16\,x^2+16\,x\right )-\ln \left (x\right )\,\left ({\mathrm {e}}^{x-6}\,\left (4\,x^2+4\,x\right )+{\mathrm {e}}^{2\,x-12}\,\left (2\,x^3+2\,x^2\right )\right )+{\mathrm {e}}^{2\,x-12}\,\left (8\,x^3+8\,x^2\right )-1\right )}{x} \,d x \] Input:
int((exp(4*x*exp(x - 6) + x^2*exp(2*x - 12) + 4)*(exp(x - 6)*(16*x + 16*x^ 2) - log(x)*(exp(x - 6)*(4*x + 4*x^2) + exp(2*x - 12)*(2*x^2 + 2*x^3)) + e xp(2*x - 12)*(8*x^2 + 8*x^3) - 1))/x,x)
Output:
int((exp(4*x*exp(x - 6) + x^2*exp(2*x - 12) + 4)*(exp(x - 6)*(16*x + 16*x^ 2) - log(x)*(exp(x - 6)*(4*x + 4*x^2) + exp(2*x - 12)*(2*x^2 + 2*x^3)) + e xp(2*x - 12)*(8*x^2 + 8*x^3) - 1))/x, x)
Time = 0.16 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59 \[ \int \frac {e^{4+4 e^{-6+x} x+e^{-12+2 x} x^2} \left (-1+e^{-6+x} \left (16 x+16 x^2\right )+e^{-12+2 x} \left (8 x^2+8 x^3\right )+\left (e^{-6+x} \left (-4 x-4 x^2\right )+e^{-12+2 x} \left (-2 x^2-2 x^3\right )\right ) \log (x)\right )}{x} \, dx=e^{\frac {e^{2 x} x^{2}+4 e^{x} e^{6} x}{e^{12}}} e^{4} \left (-\mathrm {log}\left (x \right )+4\right ) \] Input:
int((((-2*x^3-2*x^2)*exp(-6+x)^2+(-4*x^2-4*x)*exp(-6+x))*log(x)+(8*x^3+8*x ^2)*exp(-6+x)^2+(16*x^2+16*x)*exp(-6+x)-1)*exp(x^2*exp(-6+x)^2+4*x*exp(-6+ x)+4)/x,x)
Output:
e**((e**(2*x)*x**2 + 4*e**x*e**6*x)/e**12)*e**4*( - log(x) + 4)