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

Optimal. Leaf size=24 \[ 5 e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \log (4) \]

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Rubi [F]  time = 9.00, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.64, size = 23, normalized size = 0.96 \begin {gather*} 5 e^{\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} \log (4) \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.74, size = 21, normalized size = 0.88 \begin {gather*} 10 \, e^{\left (\frac {3 \, e^{\left (x^{2}\right )}}{\log \left (-{\left (x - 4\right )} \log \relax (x)\right )}\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 1.31, size = 24, normalized size = 1.00 \begin {gather*} 10 \, e^{\left (\frac {3 \, e^{\left (x^{2}\right )}}{\log \left (-x \log \relax (x) + 4 \, \log \relax (x)\right )}\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [C]  time = 5.78, size = 129, normalized size = 5.38

method	result	size
risch	\(10 \ln \relax (2) {\mathrm e}^{\frac {6 \,{\mathrm e}^{x^{2}}}{i \pi \mathrm {csgn}\left (i \ln \relax (x ) \left (x -4\right )\right )^{3}+i \pi \mathrm {csgn}\left (i \ln \relax (x ) \left (x -4\right )\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )\right )+i \pi \mathrm {csgn}\left (i \ln \relax (x ) \left (x -4\right )\right )^{2} \mathrm {csgn}\left (i \left (x -4\right )\right )-i \pi \,\mathrm {csgn}\left (i \ln \relax (x ) \left (x -4\right )\right ) \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i \left (x -4\right )\right )-2 i \pi \mathrm {csgn}\left (i \ln \relax (x ) \left (x -4\right )\right )^{2}+2 i \pi +2 \ln \left (\ln \relax (x )\right )+2 \ln \left (x -4\right )}}\)	\(129\)

Verification of antiderivative is not currently implemented for this CAS.

[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,method=_RETURNVERBOSE)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 4.68, size = 24, normalized size = 1.00 \begin {gather*} 10\,{\mathrm {e}}^{\frac {3\,{\mathrm {e}}^{x^2}}{\ln \left (4\,\ln \relax (x)-x\,\ln \relax (x)\right )}}\,\ln \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 22.45, size = 20, normalized size = 0.83 \begin {gather*} 10 e^{\frac {3 e^{x^{2}}}{\log {\left (\left (4 - x\right ) \log {\relax (x )} \right )}}} \log {\relax (2 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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