Integrand size = 118, antiderivative size = 26 \[ \int \frac {e^{\frac {x^4 \log ^2\left (-e+x-\log \left (2 x^2\right )\right )}{\log ^2(2)}} \left (\left (4 x^3-2 x^4\right ) \log \left (-e+x-\log \left (2 x^2\right )\right )+\left (4 e x^3-4 x^4+4 x^3 \log \left (2 x^2\right )\right ) \log ^2\left (-e+x-\log \left (2 x^2\right )\right )\right )}{(e-x) \log ^2(2)+\log ^2(2) \log \left (2 x^2\right )} \, dx=e^{\frac {x^4 \log ^2\left (-e+x-\log \left (2 x^2\right )\right )}{\log ^2(2)}} \] Output:
exp(x^4*ln(-ln(2*x^2)+x-exp(1))^2/ln(2)^2)
Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x^4 \log ^2\left (-e+x-\log \left (2 x^2\right )\right )}{\log ^2(2)}} \left (\left (4 x^3-2 x^4\right ) \log \left (-e+x-\log \left (2 x^2\right )\right )+\left (4 e x^3-4 x^4+4 x^3 \log \left (2 x^2\right )\right ) \log ^2\left (-e+x-\log \left (2 x^2\right )\right )\right )}{(e-x) \log ^2(2)+\log ^2(2) \log \left (2 x^2\right )} \, dx=e^{\frac {x^4 \log ^2\left (-e+x-\log \left (2 x^2\right )\right )}{\log ^2(2)}} \] Input:
Integrate[(E^((x^4*Log[-E + x - Log[2*x^2]]^2)/Log[2]^2)*((4*x^3 - 2*x^4)* Log[-E + x - Log[2*x^2]] + (4*E*x^3 - 4*x^4 + 4*x^3*Log[2*x^2])*Log[-E + x - Log[2*x^2]]^2))/((E - x)*Log[2]^2 + Log[2]^2*Log[2*x^2]),x]
Output:
E^((x^4*Log[-E + x - Log[2*x^2]]^2)/Log[2]^2)
Time = 0.49 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.008, Rules used = {7257}
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.
\(\displaystyle \int \frac {e^{\frac {x^4 \log ^2\left (-\log \left (2 x^2\right )+x-e\right )}{\log ^2(2)}} \left (\left (-4 x^4+4 e x^3+4 x^3 \log \left (2 x^2\right )\right ) \log ^2\left (-\log \left (2 x^2\right )+x-e\right )+\left (4 x^3-2 x^4\right ) \log \left (-\log \left (2 x^2\right )+x-e\right )\right )}{\log ^2(2) \log \left (2 x^2\right )+(e-x) \log ^2(2)} \, dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle e^{\frac {x^4 \log ^2\left (-\log \left (2 x^2\right )+x-e\right )}{\log ^2(2)}}\) |
Input:
Int[(E^((x^4*Log[-E + x - Log[2*x^2]]^2)/Log[2]^2)*((4*x^3 - 2*x^4)*Log[-E + x - Log[2*x^2]] + (4*E*x^3 - 4*x^4 + 4*x^3*Log[2*x^2])*Log[-E + x - Log [2*x^2]]^2))/((E - x)*Log[2]^2 + Log[2]^2*Log[2*x^2]),x]
Output:
E^((x^4*Log[-E + x - Log[2*x^2]]^2)/Log[2]^2)
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 24.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {x^{4} {\ln \left (-\ln \left (2 x^{2}\right )+x -{\mathrm e}\right )}^{2}}{\ln \left (2\right )^{2}}}\) | \(27\) |
Input:
int(((4*x^3*ln(2*x^2)+4*x^3*exp(1)-4*x^4)*ln(-ln(2*x^2)+x-exp(1))^2+(-2*x^ 4+4*x^3)*ln(-ln(2*x^2)+x-exp(1)))*exp(x^4*ln(-ln(2*x^2)+x-exp(1))^2/ln(2)^ 2)/(ln(2)^2*ln(2*x^2)+(exp(1)-x)*ln(2)^2),x,method=_RETURNVERBOSE)
Output:
exp(x^4*ln(-ln(2*x^2)+x-exp(1))^2/ln(2)^2)
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x^4 \log ^2\left (-e+x-\log \left (2 x^2\right )\right )}{\log ^2(2)}} \left (\left (4 x^3-2 x^4\right ) \log \left (-e+x-\log \left (2 x^2\right )\right )+\left (4 e x^3-4 x^4+4 x^3 \log \left (2 x^2\right )\right ) \log ^2\left (-e+x-\log \left (2 x^2\right )\right )\right )}{(e-x) \log ^2(2)+\log ^2(2) \log \left (2 x^2\right )} \, dx=e^{\left (\frac {x^{4} \log \left (x - e - \log \left (2 \, x^{2}\right )\right )^{2}}{\log \left (2\right )^{2}}\right )} \] Input:
integrate(((4*x^3*log(2*x^2)+4*x^3*exp(1)-4*x^4)*log(-log(2*x^2)+x-exp(1)) ^2+(-2*x^4+4*x^3)*log(-log(2*x^2)+x-exp(1)))*exp(x^4*log(-log(2*x^2)+x-exp (1))^2/log(2)^2)/(log(2)^2*log(2*x^2)+(exp(1)-x)*log(2)^2),x, algorithm="f ricas")
Output:
e^(x^4*log(x - e - log(2*x^2))^2/log(2)^2)
Time = 0.93 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {x^4 \log ^2\left (-e+x-\log \left (2 x^2\right )\right )}{\log ^2(2)}} \left (\left (4 x^3-2 x^4\right ) \log \left (-e+x-\log \left (2 x^2\right )\right )+\left (4 e x^3-4 x^4+4 x^3 \log \left (2 x^2\right )\right ) \log ^2\left (-e+x-\log \left (2 x^2\right )\right )\right )}{(e-x) \log ^2(2)+\log ^2(2) \log \left (2 x^2\right )} \, dx=e^{\frac {x^{4} \log {\left (x - \log {\left (2 x^{2} \right )} - e \right )}^{2}}{\log {\left (2 \right )}^{2}}} \] Input:
integrate(((4*x**3*ln(2*x**2)+4*x**3*exp(1)-4*x**4)*ln(-ln(2*x**2)+x-exp(1 ))**2+(-2*x**4+4*x**3)*ln(-ln(2*x**2)+x-exp(1)))*exp(x**4*ln(-ln(2*x**2)+x -exp(1))**2/ln(2)**2)/(ln(2)**2*ln(2*x**2)+(exp(1)-x)*ln(2)**2),x)
Output:
exp(x**4*log(x - log(2*x**2) - E)**2/log(2)**2)
Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x^4 \log ^2\left (-e+x-\log \left (2 x^2\right )\right )}{\log ^2(2)}} \left (\left (4 x^3-2 x^4\right ) \log \left (-e+x-\log \left (2 x^2\right )\right )+\left (4 e x^3-4 x^4+4 x^3 \log \left (2 x^2\right )\right ) \log ^2\left (-e+x-\log \left (2 x^2\right )\right )\right )}{(e-x) \log ^2(2)+\log ^2(2) \log \left (2 x^2\right )} \, dx=e^{\left (\frac {x^{4} \log \left (x - e - \log \left (2\right ) - 2 \, \log \left (x\right )\right )^{2}}{\log \left (2\right )^{2}}\right )} \] Input:
integrate(((4*x^3*log(2*x^2)+4*x^3*exp(1)-4*x^4)*log(-log(2*x^2)+x-exp(1)) ^2+(-2*x^4+4*x^3)*log(-log(2*x^2)+x-exp(1)))*exp(x^4*log(-log(2*x^2)+x-exp (1))^2/log(2)^2)/(log(2)^2*log(2*x^2)+(exp(1)-x)*log(2)^2),x, algorithm="m axima")
Output:
e^(x^4*log(x - e - log(2) - 2*log(x))^2/log(2)^2)
Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x^4 \log ^2\left (-e+x-\log \left (2 x^2\right )\right )}{\log ^2(2)}} \left (\left (4 x^3-2 x^4\right ) \log \left (-e+x-\log \left (2 x^2\right )\right )+\left (4 e x^3-4 x^4+4 x^3 \log \left (2 x^2\right )\right ) \log ^2\left (-e+x-\log \left (2 x^2\right )\right )\right )}{(e-x) \log ^2(2)+\log ^2(2) \log \left (2 x^2\right )} \, dx=e^{\left (\frac {x^{4} \log \left (x - e - \log \left (2 \, x^{2}\right )\right )^{2}}{\log \left (2\right )^{2}}\right )} \] Input:
integrate(((4*x^3*log(2*x^2)+4*x^3*exp(1)-4*x^4)*log(-log(2*x^2)+x-exp(1)) ^2+(-2*x^4+4*x^3)*log(-log(2*x^2)+x-exp(1)))*exp(x^4*log(-log(2*x^2)+x-exp (1))^2/log(2)^2)/(log(2)^2*log(2*x^2)+(exp(1)-x)*log(2)^2),x, algorithm="g iac")
Output:
e^(x^4*log(x - e - log(2*x^2))^2/log(2)^2)
Time = 4.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {x^4 \log ^2\left (-e+x-\log \left (2 x^2\right )\right )}{\log ^2(2)}} \left (\left (4 x^3-2 x^4\right ) \log \left (-e+x-\log \left (2 x^2\right )\right )+\left (4 e x^3-4 x^4+4 x^3 \log \left (2 x^2\right )\right ) \log ^2\left (-e+x-\log \left (2 x^2\right )\right )\right )}{(e-x) \log ^2(2)+\log ^2(2) \log \left (2 x^2\right )} \, dx={\mathrm {e}}^{\frac {x^4\,{\ln \left (x-\ln \left (x^2\right )-\mathrm {e}-\ln \left (2\right )\right )}^2}{{\ln \left (2\right )}^2}} \] Input:
int(-(exp((x^4*log(x - exp(1) - log(2*x^2))^2)/log(2)^2)*(log(x - exp(1) - log(2*x^2))*(4*x^3 - 2*x^4) + log(x - exp(1) - log(2*x^2))^2*(4*x^3*exp(1 ) - 4*x^4 + 4*x^3*log(2*x^2))))/(log(2)^2*(x - exp(1)) - log(2)^2*log(2*x^ 2)),x)
Output:
exp((x^4*log(x - log(x^2) - exp(1) - log(2))^2)/log(2)^2)
Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x^4 \log ^2\left (-e+x-\log \left (2 x^2\right )\right )}{\log ^2(2)}} \left (\left (4 x^3-2 x^4\right ) \log \left (-e+x-\log \left (2 x^2\right )\right )+\left (4 e x^3-4 x^4+4 x^3 \log \left (2 x^2\right )\right ) \log ^2\left (-e+x-\log \left (2 x^2\right )\right )\right )}{(e-x) \log ^2(2)+\log ^2(2) \log \left (2 x^2\right )} \, dx=e^{\frac {{\mathrm {log}\left (-\mathrm {log}\left (2 x^{2}\right )-e +x \right )}^{2} x^{4}}{\mathrm {log}\left (2\right )^{2}}} \] Input:
int(((4*x^3*log(2*x^2)+4*x^3*exp(1)-4*x^4)*log(-log(2*x^2)+x-exp(1))^2+(-2 *x^4+4*x^3)*log(-log(2*x^2)+x-exp(1)))*exp(x^4*log(-log(2*x^2)+x-exp(1))^2 /log(2)^2)/(log(2)^2*log(2*x^2)+(exp(1)-x)*log(2)^2),x)
Output:
e**((log( - log(2*x**2) - e + x)**2*x**4)/log(2)**2)