\(\int \frac {e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} (e^{x^2} (60-15 x) \log (4)-15 e^{x^2} x \log (4) \log (x)+e^{x^2} (-120 x^2+30 x^3) \log (4) \log (x) \log ((4-x) \log (x)))}{(-4 x+x^2) \log (x) \log ^2((4-x) \log (x))} \, dx\) [6303]

Optimal result

Integrand size = 101, antiderivative size = 24 \[ \int \frac {e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (e^{x^2} (60-15 x) \log (4)-15 e^{x^2} x \log (4) \log (x)+e^{x^2} \left (-120 x^2+30 x^3\right ) \log (4) \log (x) \log ((4-x) \log (x))\right )}{\left (-4 x+x^2\right ) \log (x) \log ^2((4-x) \log (x))} \, dx=5 e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \log (4) \]

[Out]

10*ln(2)*exp(3*exp(x^2)/ln((-x+4)*ln(x)))

Rubi [F]

\[ \int \frac {e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (e^{x^2} (60-15 x) \log (4)-15 e^{x^2} x \log (4) \log (x)+e^{x^2} \left (-120 x^2+30 x^3\right ) \log (4) \log (x) \log ((4-x) \log (x))\right )}{\left (-4 x+x^2\right ) \log (x) \log ^2((4-x) \log (x))} \, dx=\int \frac {e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (e^{x^2} (60-15 x) \log (4)-15 e^{x^2} x \log (4) \log (x)+e^{x^2} \left (-120 x^2+30 x^3\right ) \log (4) \log (x) \log ((4-x) \log (x))\right )}{\left (-4 x+x^2\right ) \log (x) \log ^2((4-x) \log (x))} \, dx \]

[In]

Int[(E^((3*E^x^2)/Log[(4 - x)*Log[x]])*(E^x^2*(60 - 15*x)*Log[4] - 15*E^x^2*x*Log[4]*Log[x] + E^x^2*(-120*x^2
+ 30*x^3)*Log[4]*Log[x]*Log[(4 - x)*Log[x]]))/((-4*x + x^2)*Log[x]*Log[(4 - x)*Log[x]]^2),x]

[Out]

-15*Log[4]*Defer[Int][E^(x^2 + (3*E^x^2)/Log[-((-4 + x)*Log[x])])/((-4 + x)*Log[-((-4 + x)*Log[x])]^2), x] - 1
5*Log[4]*Defer[Int][E^(x^2 + (3*E^x^2)/Log[-((-4 + x)*Log[x])])/(x*Log[x]*Log[-((-4 + x)*Log[x])]^2), x] + 30*
Log[4]*Defer[Int][(E^(x^2 + (3*E^x^2)/Log[-((-4 + x)*Log[x])])*x)/Log[-((-4 + x)*Log[x])], x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (e^{x^2} (60-15 x) \log (4)-15 e^{x^2} x \log (4) \log (x)+e^{x^2} \left (-120 x^2+30 x^3\right ) \log (4) \log (x) \log ((4-x) \log (x))\right )}{(-4+x) x \log (x) \log ^2((4-x) \log (x))} \, dx \\ & = \int \frac {15 e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} \log (4) (-4+x-x \log (x) (-1+2 (-4+x) x \log (-((-4+x) \log (x)))))}{(4-x) x \log (x) \log ^2(-((-4+x) \log (x)))} \, dx \\ & = (15 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} (-4+x-x \log (x) (-1+2 (-4+x) x \log (-((-4+x) \log (x)))))}{(4-x) x \log (x) \log ^2(-((-4+x) \log (x)))} \, dx \\ & = (15 \log (4)) \int \left (\frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} (4-x-x \log (x))}{(-4+x) x \log (x) \log ^2(-((-4+x) \log (x)))}+\frac {2 e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} x}{\log (-((-4+x) \log (x)))}\right ) \, dx \\ & = (15 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} (4-x-x \log (x))}{(-4+x) x \log (x) \log ^2(-((-4+x) \log (x)))} \, dx+(30 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} x}{\log (-((-4+x) \log (x)))} \, dx \\ & = (15 \log (4)) \int \left (\frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} (4-x-x \log (x))}{4 (-4+x) \log (x) \log ^2(-((-4+x) \log (x)))}+\frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} (-4+x+x \log (x))}{4 x \log (x) \log ^2(-((-4+x) \log (x)))}\right ) \, dx+(30 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} x}{\log (-((-4+x) \log (x)))} \, dx \\ & = \frac {1}{4} (15 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} (4-x-x \log (x))}{(-4+x) \log (x) \log ^2(-((-4+x) \log (x)))} \, dx+\frac {1}{4} (15 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} (-4+x+x \log (x))}{x \log (x) \log ^2(-((-4+x) \log (x)))} \, dx+(30 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} x}{\log (-((-4+x) \log (x)))} \, dx \\ & = \frac {1}{4} (15 \log (4)) \int \left (\frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}}}{\log ^2(-((-4+x) \log (x)))}+\frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}}}{\log (x) \log ^2(-((-4+x) \log (x)))}-\frac {4 e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}}}{x \log (x) \log ^2(-((-4+x) \log (x)))}\right ) \, dx+\frac {1}{4} (15 \log (4)) \int \left (-\frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} x}{(-4+x) \log ^2(-((-4+x) \log (x)))}+\frac {4 e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}}}{(-4+x) \log (x) \log ^2(-((-4+x) \log (x)))}-\frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} x}{(-4+x) \log (x) \log ^2(-((-4+x) \log (x)))}\right ) \, dx+(30 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} x}{\log (-((-4+x) \log (x)))} \, dx \\ & = \frac {1}{4} (15 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}}}{\log ^2(-((-4+x) \log (x)))} \, dx-\frac {1}{4} (15 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} x}{(-4+x) \log ^2(-((-4+x) \log (x)))} \, dx+\frac {1}{4} (15 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}}}{\log (x) \log ^2(-((-4+x) \log (x)))} \, dx-\frac {1}{4} (15 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} x}{(-4+x) \log (x) \log ^2(-((-4+x) \log (x)))} \, dx+(15 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}}}{(-4+x) \log (x) \log ^2(-((-4+x) \log (x)))} \, dx-(15 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}}}{x \log (x) \log ^2(-((-4+x) \log (x)))} \, dx+(30 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} x}{\log (-((-4+x) \log (x)))} \, dx \\ & = -\left (\frac {1}{4} (15 \log (4)) \int \left (\frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}}}{\log ^2(-((-4+x) \log (x)))}+\frac {4 e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}}}{(-4+x) \log ^2(-((-4+x) \log (x)))}\right ) \, dx\right )-\frac {1}{4} (15 \log (4)) \int \left (\frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}}}{\log (x) \log ^2(-((-4+x) \log (x)))}+\frac {4 e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}}}{(-4+x) \log (x) \log ^2(-((-4+x) \log (x)))}\right ) \, dx+\frac {1}{4} (15 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}}}{\log ^2(-((-4+x) \log (x)))} \, dx+\frac {1}{4} (15 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}}}{\log (x) \log ^2(-((-4+x) \log (x)))} \, dx+(15 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}}}{(-4+x) \log (x) \log ^2(-((-4+x) \log (x)))} \, dx-(15 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}}}{x \log (x) \log ^2(-((-4+x) \log (x)))} \, dx+(30 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} x}{\log (-((-4+x) \log (x)))} \, dx \\ & = -\left ((15 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}}}{(-4+x) \log ^2(-((-4+x) \log (x)))} \, dx\right )-(15 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}}}{x \log (x) \log ^2(-((-4+x) \log (x)))} \, dx+(30 \log (4)) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} x}{\log (-((-4+x) \log (x)))} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 2.37 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (e^{x^2} (60-15 x) \log (4)-15 e^{x^2} x \log (4) \log (x)+e^{x^2} \left (-120 x^2+30 x^3\right ) \log (4) \log (x) \log ((4-x) \log (x))\right )}{\left (-4 x+x^2\right ) \log (x) \log ^2((4-x) \log (x))} \, dx=5 e^{\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} \log (4) \]

[In]

Integrate[(E^((3*E^x^2)/Log[(4 - x)*Log[x]])*(E^x^2*(60 - 15*x)*Log[4] - 15*E^x^2*x*Log[4]*Log[x] + E^x^2*(-12
0*x^2 + 30*x^3)*Log[4]*Log[x]*Log[(4 - x)*Log[x]]))/((-4*x + x^2)*Log[x]*Log[(4 - x)*Log[x]]^2),x]

[Out]

5*E^((3*E^x^2)/Log[-((-4 + x)*Log[x])])*Log[4]

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.15 (sec) , antiderivative size = 129, normalized size of antiderivative = 5.38

\[10 \ln \left (2\right ) {\mathrm e}^{\frac {6 \,{\mathrm e}^{x^{2}}}{i \pi \operatorname {csgn}\left (i \ln \left (x \right ) \left (x -4\right )\right )^{3}+i \operatorname {csgn}\left (i \ln \left (x \right ) \left (x -4\right )\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right ) \pi +i \operatorname {csgn}\left (i \ln \left (x \right ) \left (x -4\right )\right )^{2} \operatorname {csgn}\left (i \left (x -4\right )\right ) \pi -i \operatorname {csgn}\left (i \ln \left (x \right ) \left (x -4\right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \left (x -4\right )\right ) \pi -2 i \pi \operatorname {csgn}\left (i \ln \left (x \right ) \left (x -4\right )\right )^{2}+2 i \pi +2 \ln \left (x -4\right )+2 \ln \left (\ln \left (x \right )\right )}}\]

[In]

int((2*(30*x^3-120*x^2)*ln(2)*exp(x^2)*ln(x)*ln((-x+4)*ln(x))-30*x*ln(2)*exp(x^2)*ln(x)+2*(-15*x+60)*ln(2)*exp
(x^2))*exp(3*exp(x^2)/ln((-x+4)*ln(x)))/(x^2-4*x)/ln(x)/ln((-x+4)*ln(x))^2,x)

[Out]

10*ln(2)*exp(6*exp(x^2)/(I*Pi*csgn(I*ln(x)*(x-4))^3+I*csgn(I*ln(x)*(x-4))^2*csgn(I*ln(x))*Pi+I*csgn(I*ln(x)*(x
-4))^2*csgn(I*(x-4))*Pi-I*csgn(I*ln(x)*(x-4))*csgn(I*ln(x))*csgn(I*(x-4))*Pi-2*I*Pi*csgn(I*ln(x)*(x-4))^2+2*I*
Pi+2*ln(x-4)+2*ln(ln(x))))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (e^{x^2} (60-15 x) \log (4)-15 e^{x^2} x \log (4) \log (x)+e^{x^2} \left (-120 x^2+30 x^3\right ) \log (4) \log (x) \log ((4-x) \log (x))\right )}{\left (-4 x+x^2\right ) \log (x) \log ^2((4-x) \log (x))} \, dx=10 \, e^{\left (\frac {3 \, e^{\left (x^{2}\right )}}{\log \left (-{\left (x - 4\right )} \log \left (x\right )\right )}\right )} \log \left (2\right ) \]

[In]

integrate((2*(30*x^3-120*x^2)*log(2)*exp(x^2)*log(x)*log((-x+4)*log(x))-30*x*log(2)*exp(x^2)*log(x)+2*(-15*x+6
0)*log(2)*exp(x^2))*exp(3*exp(x^2)/log((-x+4)*log(x)))/(x^2-4*x)/log(x)/log((-x+4)*log(x))^2,x, algorithm="fri
cas")

[Out]

10*e^(3*e^(x^2)/log(-(x - 4)*log(x)))*log(2)

Sympy [A] (verification not implemented)

Time = 5.64 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (e^{x^2} (60-15 x) \log (4)-15 e^{x^2} x \log (4) \log (x)+e^{x^2} \left (-120 x^2+30 x^3\right ) \log (4) \log (x) \log ((4-x) \log (x))\right )}{\left (-4 x+x^2\right ) \log (x) \log ^2((4-x) \log (x))} \, dx=10 e^{\frac {3 e^{x^{2}}}{\log {\left (\left (4 - x\right ) \log {\left (x \right )} \right )}}} \log {\left (2 \right )} \]

[In]

integrate((2*(30*x**3-120*x**2)*ln(2)*exp(x**2)*ln(x)*ln((-x+4)*ln(x))-30*x*ln(2)*exp(x**2)*ln(x)+2*(-15*x+60)
*ln(2)*exp(x**2))*exp(3*exp(x**2)/ln((-x+4)*ln(x)))/(x**2-4*x)/ln(x)/ln((-x+4)*ln(x))**2,x)

[Out]

10*exp(3*exp(x**2)/log((4 - x)*log(x)))*log(2)

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (e^{x^2} (60-15 x) \log (4)-15 e^{x^2} x \log (4) \log (x)+e^{x^2} \left (-120 x^2+30 x^3\right ) \log (4) \log (x) \log ((4-x) \log (x))\right )}{\left (-4 x+x^2\right ) \log (x) \log ^2((4-x) \log (x))} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((2*(30*x^3-120*x^2)*log(2)*exp(x^2)*log(x)*log((-x+4)*log(x))-30*x*log(2)*exp(x^2)*log(x)+2*(-15*x+6
0)*log(2)*exp(x^2))*exp(3*exp(x^2)/log((-x+4)*log(x)))/(x^2-4*x)/log(x)/log((-x+4)*log(x))^2,x, algorithm="max
ima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

Giac [A] (verification not implemented)

none

Time = 0.92 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (e^{x^2} (60-15 x) \log (4)-15 e^{x^2} x \log (4) \log (x)+e^{x^2} \left (-120 x^2+30 x^3\right ) \log (4) \log (x) \log ((4-x) \log (x))\right )}{\left (-4 x+x^2\right ) \log (x) \log ^2((4-x) \log (x))} \, dx=10 \, e^{\left (\frac {3 \, e^{\left (x^{2}\right )}}{\log \left (-x \log \left (x\right ) + 4 \, \log \left (x\right )\right )}\right )} \log \left (2\right ) \]

[In]

integrate((2*(30*x^3-120*x^2)*log(2)*exp(x^2)*log(x)*log((-x+4)*log(x))-30*x*log(2)*exp(x^2)*log(x)+2*(-15*x+6
0)*log(2)*exp(x^2))*exp(3*exp(x^2)/log((-x+4)*log(x)))/(x^2-4*x)/log(x)/log((-x+4)*log(x))^2,x, algorithm="gia
c")

[Out]

10*e^(3*e^(x^2)/log(-x*log(x) + 4*log(x)))*log(2)

Mupad [B] (verification not implemented)

Time = 12.37 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (e^{x^2} (60-15 x) \log (4)-15 e^{x^2} x \log (4) \log (x)+e^{x^2} \left (-120 x^2+30 x^3\right ) \log (4) \log (x) \log ((4-x) \log (x))\right )}{\left (-4 x+x^2\right ) \log (x) \log ^2((4-x) \log (x))} \, dx=10\,{\mathrm {e}}^{\frac {3\,{\mathrm {e}}^{x^2}}{\ln \left (4\,\ln \left (x\right )-x\,\ln \left (x\right )\right )}}\,\ln \left (2\right ) \]

[In]

int((exp((3*exp(x^2))/log(-log(x)*(x - 4)))*(2*exp(x^2)*log(2)*(15*x - 60) + 30*x*exp(x^2)*log(2)*log(x) + 2*e
xp(x^2)*log(2)*log(-log(x)*(x - 4))*log(x)*(120*x^2 - 30*x^3)))/(log(-log(x)*(x - 4))^2*log(x)*(4*x - x^2)),x)

[Out]

10*exp((3*exp(x^2))/log(4*log(x) - x*log(x)))*log(2)