Integrand size = 88, antiderivative size = 35 \[ \int \frac {\left (-2 x \log ^2(3)+e^{2+x-x^4} \left (-2+4 x-16 x^4\right ) \log ^2(3)\right ) \log \left (\frac {e^{4+2 x-2 x^4}-2 e^{2+x-x^4} x+x^2}{x}\right )}{e^{2+x-x^4} x-x^2} \, dx=\log ^2(3) \log ^2\left (\frac {\left (e^{x \left (-x^3+\frac {2+x}{x}\right )}-x\right )^2}{x}\right ) \] Output:
ln((exp(x*((2+x)/x-x^3))-x)^2/x)^2*ln(3)^2
Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-2 x \log ^2(3)+e^{2+x-x^4} \left (-2+4 x-16 x^4\right ) \log ^2(3)\right ) \log \left (\frac {e^{4+2 x-2 x^4}-2 e^{2+x-x^4} x+x^2}{x}\right )}{e^{2+x-x^4} x-x^2} \, dx=\log ^2(3) \log ^2\left (\frac {e^{-2 x^4} \left (e^{2+x}-e^{x^4} x\right )^2}{x}\right ) \] Input:
Integrate[((-2*x*Log[3]^2 + E^(2 + x - x^4)*(-2 + 4*x - 16*x^4)*Log[3]^2)* Log[(E^(4 + 2*x - 2*x^4) - 2*E^(2 + x - x^4)*x + x^2)/x])/(E^(2 + x - x^4) *x - x^2),x]
Output:
Log[3]^2*Log[(E^(2 + x) - E^x^4*x)^2/(E^(2*x^4)*x)]^2
Time = 3.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.17, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used = {7237}
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.
\(\displaystyle \int \frac {\left (e^{-x^4+x+2} \left (-16 x^4+4 x-2\right ) \log ^2(3)-2 x \log ^2(3)\right ) \log \left (\frac {-2 e^{-x^4+x+2} x+e^{-2 x^4+2 x+4}+x^2}{x}\right )}{e^{-x^4+x+2} x-x^2} \, dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \log ^2(3) \log ^2\left (\frac {-2 e^{-x^4+x+2} x+e^{-2 x^4+2 x+4}+x^2}{x}\right )\) |
Input:
Int[((-2*x*Log[3]^2 + E^(2 + x - x^4)*(-2 + 4*x - 16*x^4)*Log[3]^2)*Log[(E ^(4 + 2*x - 2*x^4) - 2*E^(2 + x - x^4)*x + x^2)/x])/(E^(2 + x - x^4)*x - x ^2),x]
Output:
Log[3]^2*Log[(E^(4 + 2*x - 2*x^4) - 2*E^(2 + x - x^4)*x + x^2)/x]^2
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Time = 1.43 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.14
method | result | size |
norman | \(\ln \left (3\right )^{2} \ln \left (\frac {{\mathrm e}^{-2 x^{4}+2 x +4}-2 x \,{\mathrm e}^{-x^{4}+x +2}+x^{2}}{x}\right )^{2}\) | \(40\) |
parallelrisch | \(\ln \left (3\right )^{2} \ln \left (\frac {{\mathrm e}^{-2 x^{4}+2 x +4}-2 x \,{\mathrm e}^{-x^{4}+x +2}+x^{2}}{x}\right )^{2}\) | \(40\) |
risch | \(\text {Expression too large to display}\) | \(1442\) |
Input:
int(((-16*x^4+4*x-2)*ln(3)^2*exp(-x^4+x+2)-2*x*ln(3)^2)*ln((exp(-x^4+x+2)^ 2-2*x*exp(-x^4+x+2)+x^2)/x)/(x*exp(-x^4+x+2)-x^2),x,method=_RETURNVERBOSE)
Output:
ln(3)^2*ln((exp(-x^4+x+2)^2-2*x*exp(-x^4+x+2)+x^2)/x)^2
Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.11 \[ \int \frac {\left (-2 x \log ^2(3)+e^{2+x-x^4} \left (-2+4 x-16 x^4\right ) \log ^2(3)\right ) \log \left (\frac {e^{4+2 x-2 x^4}-2 e^{2+x-x^4} x+x^2}{x}\right )}{e^{2+x-x^4} x-x^2} \, dx=\log \left (3\right )^{2} \log \left (\frac {x^{2} - 2 \, x e^{\left (-x^{4} + x + 2\right )} + e^{\left (-2 \, x^{4} + 2 \, x + 4\right )}}{x}\right )^{2} \] Input:
integrate(((-16*x^4+4*x-2)*log(3)^2*exp(-x^4+x+2)-2*x*log(3)^2)*log((exp(- x^4+x+2)^2-2*x*exp(-x^4+x+2)+x^2)/x)/(x*exp(-x^4+x+2)-x^2),x, algorithm="f ricas")
Output:
log(3)^2*log((x^2 - 2*x*e^(-x^4 + x + 2) + e^(-2*x^4 + 2*x + 4))/x)^2
Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \frac {\left (-2 x \log ^2(3)+e^{2+x-x^4} \left (-2+4 x-16 x^4\right ) \log ^2(3)\right ) \log \left (\frac {e^{4+2 x-2 x^4}-2 e^{2+x-x^4} x+x^2}{x}\right )}{e^{2+x-x^4} x-x^2} \, dx=\log {\left (3 \right )}^{2} \log {\left (\frac {x^{2} - 2 x e^{- x^{4} + x + 2} + e^{- 2 x^{4} + 2 x + 4}}{x} \right )}^{2} \] Input:
integrate(((-16*x**4+4*x-2)*ln(3)**2*exp(-x**4+x+2)-2*x*ln(3)**2)*ln((exp( -x**4+x+2)**2-2*x*exp(-x**4+x+2)+x**2)/x)/(x*exp(-x**4+x+2)-x**2),x)
Output:
log(3)**2*log((x**2 - 2*x*exp(-x**4 + x + 2) + exp(-2*x**4 + 2*x + 4))/x)* *2
\[ \int \frac {\left (-2 x \log ^2(3)+e^{2+x-x^4} \left (-2+4 x-16 x^4\right ) \log ^2(3)\right ) \log \left (\frac {e^{4+2 x-2 x^4}-2 e^{2+x-x^4} x+x^2}{x}\right )}{e^{2+x-x^4} x-x^2} \, dx=\int { \frac {2 \, {\left ({\left (8 \, x^{4} - 2 \, x + 1\right )} e^{\left (-x^{4} + x + 2\right )} \log \left (3\right )^{2} + x \log \left (3\right )^{2}\right )} \log \left (\frac {x^{2} - 2 \, x e^{\left (-x^{4} + x + 2\right )} + e^{\left (-2 \, x^{4} + 2 \, x + 4\right )}}{x}\right )}{x^{2} - x e^{\left (-x^{4} + x + 2\right )}} \,d x } \] Input:
integrate(((-16*x^4+4*x-2)*log(3)^2*exp(-x^4+x+2)-2*x*log(3)^2)*log((exp(- x^4+x+2)^2-2*x*exp(-x^4+x+2)+x^2)/x)/(x*exp(-x^4+x+2)-x^2),x, algorithm="m axima")
Output:
2*integrate(((8*x^4 - 2*x + 1)*e^(-x^4 + x + 2)*log(3)^2 + x*log(3)^2)*log ((x^2 - 2*x*e^(-x^4 + x + 2) + e^(-2*x^4 + 2*x + 4))/x)/(x^2 - x*e^(-x^4 + x + 2)), x)
\[ \int \frac {\left (-2 x \log ^2(3)+e^{2+x-x^4} \left (-2+4 x-16 x^4\right ) \log ^2(3)\right ) \log \left (\frac {e^{4+2 x-2 x^4}-2 e^{2+x-x^4} x+x^2}{x}\right )}{e^{2+x-x^4} x-x^2} \, dx=\int { \frac {2 \, {\left ({\left (8 \, x^{4} - 2 \, x + 1\right )} e^{\left (-x^{4} + x + 2\right )} \log \left (3\right )^{2} + x \log \left (3\right )^{2}\right )} \log \left (\frac {x^{2} - 2 \, x e^{\left (-x^{4} + x + 2\right )} + e^{\left (-2 \, x^{4} + 2 \, x + 4\right )}}{x}\right )}{x^{2} - x e^{\left (-x^{4} + x + 2\right )}} \,d x } \] Input:
integrate(((-16*x^4+4*x-2)*log(3)^2*exp(-x^4+x+2)-2*x*log(3)^2)*log((exp(- x^4+x+2)^2-2*x*exp(-x^4+x+2)+x^2)/x)/(x*exp(-x^4+x+2)-x^2),x, algorithm="g iac")
Output:
integrate(2*((8*x^4 - 2*x + 1)*e^(-x^4 + x + 2)*log(3)^2 + x*log(3)^2)*log ((x^2 - 2*x*e^(-x^4 + x + 2) + e^(-2*x^4 + 2*x + 4))/x)/(x^2 - x*e^(-x^4 + x + 2)), x)
Time = 4.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.09 \[ \int \frac {\left (-2 x \log ^2(3)+e^{2+x-x^4} \left (-2+4 x-16 x^4\right ) \log ^2(3)\right ) \log \left (\frac {e^{4+2 x-2 x^4}-2 e^{2+x-x^4} x+x^2}{x}\right )}{e^{2+x-x^4} x-x^2} \, dx={\ln \left (3\right )}^2\,{\ln \left (x-2\,{\mathrm {e}}^2\,{\mathrm {e}}^{-x^4}\,{\mathrm {e}}^x+\frac {{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^4\,{\mathrm {e}}^{-2\,x^4}}{x}\right )}^2 \] Input:
int(-(log((exp(2*x - 2*x^4 + 4) - 2*x*exp(x - x^4 + 2) + x^2)/x)*(2*x*log( 3)^2 + exp(x - x^4 + 2)*log(3)^2*(16*x^4 - 4*x + 2)))/(x*exp(x - x^4 + 2) - x^2),x)
Output:
log(3)^2*log(x - 2*exp(2)*exp(-x^4)*exp(x) + (exp(2*x)*exp(4)*exp(-2*x^4)) /x)^2
Time = 0.19 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.57 \[ \int \frac {\left (-2 x \log ^2(3)+e^{2+x-x^4} \left (-2+4 x-16 x^4\right ) \log ^2(3)\right ) \log \left (\frac {e^{4+2 x-2 x^4}-2 e^{2+x-x^4} x+x^2}{x}\right )}{e^{2+x-x^4} x-x^2} \, dx=\mathrm {log}\left (\frac {e^{2 x^{4}} x^{2}-2 e^{x^{4}+x} e^{2} x +e^{2 x} e^{4}}{e^{2 x^{4}} x}\right )^{2} \mathrm {log}\left (3\right )^{2} \] Input:
int(((-16*x^4+4*x-2)*log(3)^2*exp(-x^4+x+2)-2*x*log(3)^2)*log((exp(-x^4+x+ 2)^2-2*x*exp(-x^4+x+2)+x^2)/x)/(x*exp(-x^4+x+2)-x^2),x)
Output:
log((e**(2*x**4)*x**2 - 2*e**(x**4 + x)*e**2*x + e**(2*x)*e**4)/(e**(2*x** 4)*x))**2*log(3)**2