Integrand size = 71, antiderivative size = 28 \[ \int \frac {e^{\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}} \left (4 e^{2+4 x} x+\left (1-2 x^2+2 x \log (x)+x \log ^2(x)\right ) \log (\log (5))\right )}{\log (\log (5))} \, dx=e^{x \left (-x+\log ^2(x)\right )+\frac {e^{2+4 x}}{\log (\log (5))}} x \] Output:
exp(x*(ln(x)^2-x)+exp(2)*exp(4*x)/ln(ln(5)))*x
Time = 0.67 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}} \left (4 e^{2+4 x} x+\left (1-2 x^2+2 x \log (x)+x \log ^2(x)\right ) \log (\log (5))\right )}{\log (\log (5))} \, dx=e^{-x^2+x \log ^2(x)+\frac {e^{2+4 x}}{\log (\log (5))}} x \] Input:
Integrate[(E^((E^(2 + 4*x) + (-x^2 + x*Log[x]^2)*Log[Log[5]])/Log[Log[5]]) *(4*E^(2 + 4*x)*x + (1 - 2*x^2 + 2*x*Log[x] + x*Log[x]^2)*Log[Log[5]]))/Lo g[Log[5]],x]
Output:
E^(-x^2 + x*Log[x]^2 + E^(2 + 4*x)/Log[Log[5]])*x
Time = 0.43 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {27, 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.
\(\displaystyle \int \frac {\left (\log (\log (5)) \left (-2 x^2+x \log ^2(x)+2 x \log (x)+1\right )+4 e^{4 x+2} x\right ) \exp \left (\frac {\log (\log (5)) \left (x \log ^2(x)-x^2\right )+e^{4 x+2}}{\log (\log (5))}\right )}{\log (\log (5))} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int e^{\frac {e^{4 x+2}}{\log (\log (5))}} \log ^{-\frac {x^2-x \log ^2(x)}{\log (\log (5))}}(5) \left (4 e^{4 x+2} x+\left (-2 x^2+\log ^2(x) x+2 \log (x) x+1\right ) \log (\log (5))\right )dx}{\log (\log (5))}\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle x e^{\frac {e^{4 x+2}}{\log (\log (5))}} \log ^{-\frac {x^2-x \log ^2(x)}{\log (\log (5))}}(5)\) |
Input:
Int[(E^((E^(2 + 4*x) + (-x^2 + x*Log[x]^2)*Log[Log[5]])/Log[Log[5]])*(4*E^ (2 + 4*x)*x + (1 - 2*x^2 + 2*x*Log[x] + x*Log[x]^2)*Log[Log[5]]))/Log[Log[ 5]],x]
Output:
(E^(E^(2 + 4*x)/Log[Log[5]])*x)/Log[5]^((x^2 - x*Log[x]^2)/Log[Log[5]])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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 = 1.34 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21
method | result | size |
risch | \(x \,{\mathrm e}^{\frac {\ln \left (x \right )^{2} \ln \left (\ln \left (5\right )\right ) x -\ln \left (\ln \left (5\right )\right ) x^{2}+{\mathrm e}^{4 x +2}}{\ln \left (\ln \left (5\right )\right )}}\) | \(34\) |
parallelrisch | \(x \,{\mathrm e}^{\frac {\left (x \ln \left (x \right )^{2}-x^{2}\right ) \ln \left (\ln \left (5\right )\right )+{\mathrm e}^{2} {\mathrm e}^{4 x}}{\ln \left (\ln \left (5\right )\right )}}\) | \(34\) |
Input:
int(((x*ln(x)^2+2*x*ln(x)-2*x^2+1)*ln(ln(5))+4*x*exp(2)*exp(4*x))*exp(((x* ln(x)^2-x^2)*ln(ln(5))+exp(2)*exp(4*x))/ln(ln(5)))/ln(ln(5)),x,method=_RET URNVERBOSE)
Output:
x*exp((ln(x)^2*ln(ln(5))*x-ln(ln(5))*x^2+exp(4*x+2))/ln(ln(5)))
Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}} \left (4 e^{2+4 x} x+\left (1-2 x^2+2 x \log (x)+x \log ^2(x)\right ) \log (\log (5))\right )}{\log (\log (5))} \, dx=x e^{\left (\frac {{\left (x \log \left (x\right )^{2} - x^{2}\right )} \log \left (\log \left (5\right )\right ) + e^{\left (4 \, x + 2\right )}}{\log \left (\log \left (5\right )\right )}\right )} \] Input:
integrate(((x*log(x)^2+2*x*log(x)-2*x^2+1)*log(log(5))+4*x*exp(2)*exp(4*x) )*exp(((x*log(x)^2-x^2)*log(log(5))+exp(2)*exp(4*x))/log(log(5)))/log(log( 5)),x, algorithm="fricas")
Output:
x*e^(((x*log(x)^2 - x^2)*log(log(5)) + e^(4*x + 2))/log(log(5)))
Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}} \left (4 e^{2+4 x} x+\left (1-2 x^2+2 x \log (x)+x \log ^2(x)\right ) \log (\log (5))\right )}{\log (\log (5))} \, dx=x e^{\frac {\left (- x^{2} + x \log {\left (x \right )}^{2}\right ) \log {\left (\log {\left (5 \right )} \right )} + e^{2} e^{4 x}}{\log {\left (\log {\left (5 \right )} \right )}}} \] Input:
integrate(((x*ln(x)**2+2*x*ln(x)-2*x**2+1)*ln(ln(5))+4*x*exp(2)*exp(4*x))* exp(((x*ln(x)**2-x**2)*ln(ln(5))+exp(2)*exp(4*x))/ln(ln(5)))/ln(ln(5)),x)
Output:
x*exp(((-x**2 + x*log(x)**2)*log(log(5)) + exp(2)*exp(4*x))/log(log(5)))
Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}} \left (4 e^{2+4 x} x+\left (1-2 x^2+2 x \log (x)+x \log ^2(x)\right ) \log (\log (5))\right )}{\log (\log (5))} \, dx=x e^{\left (x \log \left (x\right )^{2} - x^{2} + \frac {e^{\left (4 \, x + 2\right )}}{\log \left (\log \left (5\right )\right )}\right )} \] Input:
integrate(((x*log(x)^2+2*x*log(x)-2*x^2+1)*log(log(5))+4*x*exp(2)*exp(4*x) )*exp(((x*log(x)^2-x^2)*log(log(5))+exp(2)*exp(4*x))/log(log(5)))/log(log( 5)),x, algorithm="maxima")
Output:
x*e^(x*log(x)^2 - x^2 + e^(4*x + 2)/log(log(5)))
\[ \int \frac {e^{\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}} \left (4 e^{2+4 x} x+\left (1-2 x^2+2 x \log (x)+x \log ^2(x)\right ) \log (\log (5))\right )}{\log (\log (5))} \, dx=\int { \frac {{\left (4 \, x e^{\left (4 \, x + 2\right )} + {\left (x \log \left (x\right )^{2} - 2 \, x^{2} + 2 \, x \log \left (x\right ) + 1\right )} \log \left (\log \left (5\right )\right )\right )} e^{\left (\frac {{\left (x \log \left (x\right )^{2} - x^{2}\right )} \log \left (\log \left (5\right )\right ) + e^{\left (4 \, x + 2\right )}}{\log \left (\log \left (5\right )\right )}\right )}}{\log \left (\log \left (5\right )\right )} \,d x } \] Input:
integrate(((x*log(x)^2+2*x*log(x)-2*x^2+1)*log(log(5))+4*x*exp(2)*exp(4*x) )*exp(((x*log(x)^2-x^2)*log(log(5))+exp(2)*exp(4*x))/log(log(5)))/log(log( 5)),x, algorithm="giac")
Output:
undef
Time = 2.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}} \left (4 e^{2+4 x} x+\left (1-2 x^2+2 x \log (x)+x \log ^2(x)\right ) \log (\log (5))\right )}{\log (\log (5))} \, dx=x\,{\mathrm {e}}^{x\,{\ln \left (x\right )}^2}\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^2}{\ln \left (\ln \left (5\right )\right )}} \] Input:
int((exp((exp(4*x)*exp(2) + log(log(5))*(x*log(x)^2 - x^2))/log(log(5)))*( log(log(5))*(x*log(x)^2 + 2*x*log(x) - 2*x^2 + 1) + 4*x*exp(4*x)*exp(2)))/ log(log(5)),x)
Output:
x*exp(x*log(x)^2)*exp(-x^2)*exp((exp(4*x)*exp(2))/log(log(5)))
Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {e^{\frac {e^{2+4 x}+\left (-x^2+x \log ^2(x)\right ) \log (\log (5))}{\log (\log (5))}} \left (4 e^{2+4 x} x+\left (1-2 x^2+2 x \log (x)+x \log ^2(x)\right ) \log (\log (5))\right )}{\log (\log (5))} \, dx=\frac {e^{\frac {e^{4 x} e^{2}+\mathrm {log}\left (\mathrm {log}\left (5\right )\right ) \mathrm {log}\left (x \right )^{2} x}{\mathrm {log}\left (\mathrm {log}\left (5\right )\right )}} x}{e^{x^{2}}} \] Input:
int(((x*log(x)^2+2*x*log(x)-2*x^2+1)*log(log(5))+4*x*exp(2)*exp(4*x))*exp( ((x*log(x)^2-x^2)*log(log(5))+exp(2)*exp(4*x))/log(log(5)))/log(log(5)),x)
Output:
(e**((e**(4*x)*e**2 + log(log(5))*log(x)**2*x)/log(log(5)))*x)/e**(x**2)