Integrand size = 52, antiderivative size = 20 \[ \int \frac {e^{\log ^2\left (x^2 \log (x)\right )} \left ((6-3 \log (3)) \log (x)+(-12+6 \log (3)+(-24+12 \log (3)) \log (x)) \log \left (x^2 \log (x)\right )\right )}{x^2 \log (x)} \, dx=\frac {3 e^{\log ^2\left (x^2 \log (x)\right )} (-2+\log (3))}{x} \] Output:
3*exp(ln(x^2*ln(x))^2)*(ln(3)-2)/x
Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\log ^2\left (x^2 \log (x)\right )} \left ((6-3 \log (3)) \log (x)+(-12+6 \log (3)+(-24+12 \log (3)) \log (x)) \log \left (x^2 \log (x)\right )\right )}{x^2 \log (x)} \, dx=\frac {3 e^{\log ^2\left (x^2 \log (x)\right )} (-2+\log (3))}{x} \] Input:
Integrate[(E^Log[x^2*Log[x]]^2*((6 - 3*Log[3])*Log[x] + (-12 + 6*Log[3] + (-24 + 12*Log[3])*Log[x])*Log[x^2*Log[x]]))/(x^2*Log[x]),x]
Output:
(3*E^Log[x^2*Log[x]]^2*(-2 + Log[3]))/x
Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.90, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, 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.
\(\displaystyle \int \frac {e^{\log ^2\left (x^2 \log (x)\right )} \left (((12 \log (3)-24) \log (x)-12+6 \log (3)) \log \left (x^2 \log (x)\right )+(6-3 \log (3)) \log (x)\right )}{x^2 \log (x)} \, dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle -\frac {3 e^{\log ^2\left (x^2 \log (x)\right )} (2 (2-\log (3)) \log (x)+2-\log (3))}{x+2 x \log (x)}\) |
Input:
Int[(E^Log[x^2*Log[x]]^2*((6 - 3*Log[3])*Log[x] + (-12 + 6*Log[3] + (-24 + 12*Log[3])*Log[x])*Log[x^2*Log[x]]))/(x^2*Log[x]),x]
Output:
(-3*E^Log[x^2*Log[x]]^2*(2 - Log[3] + 2*(2 - Log[3])*Log[x]))/(x + 2*x*Log [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 = 0.69 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60
method | result | size |
parallelrisch | \(\frac {3 \,{\mathrm e}^{\ln \left (x^{2} \ln \left (x \right )\right )^{2}} \ln \left (3\right )-6 \,{\mathrm e}^{\ln \left (x^{2} \ln \left (x \right )\right )^{2}}}{x}\) | \(32\) |
risch | \(\frac {3 \left (\ln \left (3\right )-2\right ) {\mathrm e}^{\frac {{\left (-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+i \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{2} \ln \left (x \right )\right )^{2} \pi -i \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{2} \ln \left (x \right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )\right ) \pi -i \operatorname {csgn}\left (i x^{2} \ln \left (x \right )\right )^{3} \pi +i \operatorname {csgn}\left (i x^{2} \ln \left (x \right )\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right ) \pi +4 \ln \left (x \right )+2 \ln \left (\ln \left (x \right )\right )\right )}^{2}}{4}}}{x}\) | \(158\) |
Input:
int((((12*ln(3)-24)*ln(x)+6*ln(3)-12)*ln(x^2*ln(x))+(-3*ln(3)+6)*ln(x))*ex p(ln(x^2*ln(x))^2)/x^2/ln(x),x,method=_RETURNVERBOSE)
Output:
1/x*(3*exp(ln(x^2*ln(x))^2)*ln(3)-6*exp(ln(x^2*ln(x))^2))
Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\log ^2\left (x^2 \log (x)\right )} \left ((6-3 \log (3)) \log (x)+(-12+6 \log (3)+(-24+12 \log (3)) \log (x)) \log \left (x^2 \log (x)\right )\right )}{x^2 \log (x)} \, dx=\frac {3 \, {\left (\log \left (3\right ) - 2\right )} e^{\left (\log \left (x^{2} \log \left (x\right )\right )^{2}\right )}}{x} \] Input:
integrate((((12*log(3)-24)*log(x)+6*log(3)-12)*log(x^2*log(x))+(-3*log(3)+ 6)*log(x))*exp(log(x^2*log(x))^2)/x^2/log(x),x, algorithm="fricas")
Output:
3*(log(3) - 2)*e^(log(x^2*log(x))^2)/x
Time = 0.15 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\log ^2\left (x^2 \log (x)\right )} \left ((6-3 \log (3)) \log (x)+(-12+6 \log (3)+(-24+12 \log (3)) \log (x)) \log \left (x^2 \log (x)\right )\right )}{x^2 \log (x)} \, dx=\frac {\left (-6 + 3 \log {\left (3 \right )}\right ) e^{\log {\left (x^{2} \log {\left (x \right )} \right )}^{2}}}{x} \] Input:
integrate((((12*ln(3)-24)*ln(x)+6*ln(3)-12)*ln(x**2*ln(x))+(-3*ln(3)+6)*ln (x))*exp(ln(x**2*ln(x))**2)/x**2/ln(x),x)
Output:
(-6 + 3*log(3))*exp(log(x**2*log(x))**2)/x
Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45 \[ \int \frac {e^{\log ^2\left (x^2 \log (x)\right )} \left ((6-3 \log (3)) \log (x)+(-12+6 \log (3)+(-24+12 \log (3)) \log (x)) \log \left (x^2 \log (x)\right )\right )}{x^2 \log (x)} \, dx=\frac {3 \, {\left (\log \left (3\right ) - 2\right )} e^{\left (4 \, \log \left (x\right )^{2} + 4 \, \log \left (x\right ) \log \left (\log \left (x\right )\right ) + \log \left (\log \left (x\right )\right )^{2}\right )}}{x} \] Input:
integrate((((12*log(3)-24)*log(x)+6*log(3)-12)*log(x^2*log(x))+(-3*log(3)+ 6)*log(x))*exp(log(x^2*log(x))^2)/x^2/log(x),x, algorithm="maxima")
Output:
3*(log(3) - 2)*e^(4*log(x)^2 + 4*log(x)*log(log(x)) + log(log(x))^2)/x
\[ \int \frac {e^{\log ^2\left (x^2 \log (x)\right )} \left ((6-3 \log (3)) \log (x)+(-12+6 \log (3)+(-24+12 \log (3)) \log (x)) \log \left (x^2 \log (x)\right )\right )}{x^2 \log (x)} \, dx=\int { \frac {3 \, {\left (2 \, {\left (2 \, {\left (\log \left (3\right ) - 2\right )} \log \left (x\right ) + \log \left (3\right ) - 2\right )} \log \left (x^{2} \log \left (x\right )\right ) - {\left (\log \left (3\right ) - 2\right )} \log \left (x\right )\right )} e^{\left (\log \left (x^{2} \log \left (x\right )\right )^{2}\right )}}{x^{2} \log \left (x\right )} \,d x } \] Input:
integrate((((12*log(3)-24)*log(x)+6*log(3)-12)*log(x^2*log(x))+(-3*log(3)+ 6)*log(x))*exp(log(x^2*log(x))^2)/x^2/log(x),x, algorithm="giac")
Output:
undef
Timed out. \[ \int \frac {e^{\log ^2\left (x^2 \log (x)\right )} \left ((6-3 \log (3)) \log (x)+(-12+6 \log (3)+(-24+12 \log (3)) \log (x)) \log \left (x^2 \log (x)\right )\right )}{x^2 \log (x)} \, dx=\int -\frac {{\mathrm {e}}^{{\ln \left (x^2\,\ln \left (x\right )\right )}^2}\,\left (\ln \left (x\right )\,\left (3\,\ln \left (3\right )-6\right )-\ln \left (x^2\,\ln \left (x\right )\right )\,\left (6\,\ln \left (3\right )+\ln \left (x\right )\,\left (12\,\ln \left (3\right )-24\right )-12\right )\right )}{x^2\,\ln \left (x\right )} \,d x \] Input:
int(-(exp(log(x^2*log(x))^2)*(log(x)*(3*log(3) - 6) - log(x^2*log(x))*(6*l og(3) + log(x)*(12*log(3) - 24) - 12)))/(x^2*log(x)),x)
Output:
int(-(exp(log(x^2*log(x))^2)*(log(x)*(3*log(3) - 6) - log(x^2*log(x))*(6*l og(3) + log(x)*(12*log(3) - 24) - 12)))/(x^2*log(x)), x)
Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\log ^2\left (x^2 \log (x)\right )} \left ((6-3 \log (3)) \log (x)+(-12+6 \log (3)+(-24+12 \log (3)) \log (x)) \log \left (x^2 \log (x)\right )\right )}{x^2 \log (x)} \, dx=\frac {3 e^{\mathrm {log}\left (\mathrm {log}\left (x \right ) x^{2}\right )^{2}} \left (\mathrm {log}\left (3\right )-2\right )}{x} \] Input:
int((((12*log(3)-24)*log(x)+6*log(3)-12)*log(x^2*log(x))+(-3*log(3)+6)*log (x))*exp(log(x^2*log(x))^2)/x^2/log(x),x)
Output:
(3*e**(log(log(x)*x**2)**2)*(log(3) - 2))/x