3.22.97 \(\int \frac {e^{-\frac {e^{2 x+x^2}}{5+e^{x^2} (-1-\log (3))+e^{x^2} \log (x)}} (e^{2 x+x^2} (-40 x-40 x^2)+e^{2 x+2 x^2} (4+8 x+8 x \log (3))-8 e^{2 x+2 x^2} x \log (x))}{25 x+e^{x^2} (-10 x-10 x \log (3))+e^{2 x^2} (x+2 x \log (3)+x \log ^2(3))+(10 e^{x^2} x+e^{2 x^2} (-2 x-2 x \log (3))) \log (x)+e^{2 x^2} x \log ^2(x)} \, dx\)

Optimal. Leaf size=29 \[ 4 e^{\frac {e^{2 x}}{1-5 e^{-x^2}+\log (3)-\log (x)}} \]

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Rubi [F]  time = 59.41, 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 {\exp \left (-\frac {e^{2 x+x^2}}{5+e^{x^2} (-1-\log (3))+e^{x^2} \log (x)}\right ) \left (e^{2 x+x^2} \left (-40 x-40 x^2\right )+e^{2 x+2 x^2} (4+8 x+8 x \log (3))-8 e^{2 x+2 x^2} x \log (x)\right )}{25 x+e^{x^2} (-10 x-10 x \log (3))+e^{2 x^2} \left (x+2 x \log (3)+x \log ^2(3)\right )+\left (10 e^{x^2} x+e^{2 x^2} (-2 x-2 x \log (3))\right ) \log (x)+e^{2 x^2} x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2*x + x^2)*(-40*x - 40*x^2) + E^(2*x + 2*x^2)*(4 + 8*x + 8*x*Log[3]) - 8*E^(2*x + 2*x^2)*x*Log[x])/(E^
(E^(2*x + x^2)/(5 + E^x^2*(-1 - Log[3]) + E^x^2*Log[x]))*(25*x + E^x^2*(-10*x - 10*x*Log[3]) + E^(2*x^2)*(x +
2*x*Log[3] + x*Log[3]^2) + (10*E^x^2*x + E^(2*x^2)*(-2*x - 2*x*Log[3]))*Log[x] + E^(2*x^2)*x*Log[x]^2)),x]

[Out]

20*Defer[Int][E^(2*x + x^2 + E^(x*(2 + x))/(-5 + E^x^2*(1 + Log[3]) - E^x^2*Log[x]))/(x*(1 + Log[3] - Log[x])*
(5 - E^x^2*(1 + Log[3]) + E^x^2*Log[x])^2), x] - 40*(1 + Log[3])*Defer[Int][(E^(2*x + x^2 + E^(x*(2 + x))/(-5
+ E^x^2*(1 + Log[3]) - E^x^2*Log[x]))*x)/((1 + Log[3] - Log[x])*(5 - E^x^2*(1 + Log[3]) + E^x^2*Log[x])^2), x]
+ 40*Defer[Int][(E^(2*x + x^2 + E^(x*(2 + x))/(-5 + E^x^2*(1 + Log[3]) - E^x^2*Log[x]))*x*Log[x])/((1 + Log[3
] - Log[x])*(5 - E^x^2*(1 + Log[3]) + E^x^2*Log[x])^2), x] - 4*(2 + Log[9])*Defer[Int][E^(2*x + x^2 + E^(x*(2
+ x))/(-5 + E^x^2*(1 + Log[3]) - E^x^2*Log[x]))/((1 + Log[3] - Log[x])*(5 - E^x^2*(1 + Log[3]) + E^x^2*Log[x])
), x] + 8*Defer[Int][(E^(2*x + x^2 + E^(x*(2 + x))/(-5 + E^x^2*(1 + Log[3]) - E^x^2*Log[x]))*Log[x])/((1 + Log
[3] - Log[x])*(5 - E^x^2*(1 + Log[3]) + E^x^2*Log[x])), x] + 4*Defer[Int][E^(2*x + x^2 + E^(x*(2 + x))/(-5 + E
^x^2*(1 + Log[3]) - E^x^2*Log[x]))/(x*(-1 - Log[3] + Log[x])*(5 - E^x^2*(1 + Log[3]) + E^x^2*Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 \exp \left (2 x+x^2+\frac {e^{x (2+x)}}{-5+e^{x^2} (1+\log (3))-e^{x^2} \log (x)}\right ) \left (-10 x (1+x)+e^{x^2} (1+x (2+\log (9)))-2 e^{x^2} x \log (x)\right )}{x \left (5-e^{x^2} (1+\log (3))+e^{x^2} \log (x)\right )^2} \, dx\\ &=4 \int \frac {\exp \left (2 x+x^2+\frac {e^{x (2+x)}}{-5+e^{x^2} (1+\log (3))-e^{x^2} \log (x)}\right ) \left (-10 x (1+x)+e^{x^2} (1+x (2+\log (9)))-2 e^{x^2} x \log (x)\right )}{x \left (5-e^{x^2} (1+\log (3))+e^{x^2} \log (x)\right )^2} \, dx\\ &=4 \int \left (\frac {\exp \left (2 x+x^2+\frac {e^{x (2+x)}}{-5+e^{x^2} (1+\log (3))-e^{x^2} \log (x)}\right ) (-1-2 x (1+\log (3))+2 x \log (x))}{x (1+\log (3)-\log (x)) \left (5-e^{x^2} (1+\log (3))+e^{x^2} \log (x)\right )}+\frac {5 \exp \left (2 x+x^2+\frac {e^{x (2+x)}}{-5+e^{x^2} (1+\log (3))-e^{x^2} \log (x)}\right ) \left (1-2 x^2 (1+\log (3))+2 x^2 \log (x)\right )}{x (1+\log (3)-\log (x)) \left (5-e^{x^2} (1+\log (3))+e^{x^2} \log (x)\right )^2}\right ) \, dx\\ &=4 \int \frac {\exp \left (2 x+x^2+\frac {e^{x (2+x)}}{-5+e^{x^2} (1+\log (3))-e^{x^2} \log (x)}\right ) (-1-2 x (1+\log (3))+2 x \log (x))}{x (1+\log (3)-\log (x)) \left (5-e^{x^2} (1+\log (3))+e^{x^2} \log (x)\right )} \, dx+20 \int \frac {\exp \left (2 x+x^2+\frac {e^{x (2+x)}}{-5+e^{x^2} (1+\log (3))-e^{x^2} \log (x)}\right ) \left (1-2 x^2 (1+\log (3))+2 x^2 \log (x)\right )}{x (1+\log (3)-\log (x)) \left (5-e^{x^2} (1+\log (3))+e^{x^2} \log (x)\right )^2} \, dx\\ &=4 \int \left (\frac {\exp \left (2 x+x^2+\frac {e^{x (2+x)}}{-5+e^{x^2} (1+\log (3))-e^{x^2} \log (x)}\right ) (-2-\log (9))}{(1+\log (3)-\log (x)) \left (5-e^{x^2} (1+\log (3))+e^{x^2} \log (x)\right )}+\frac {2 \exp \left (2 x+x^2+\frac {e^{x (2+x)}}{-5+e^{x^2} (1+\log (3))-e^{x^2} \log (x)}\right ) \log (x)}{(1+\log (3)-\log (x)) \left (5-e^{x^2} (1+\log (3))+e^{x^2} \log (x)\right )}+\frac {\exp \left (2 x+x^2+\frac {e^{x (2+x)}}{-5+e^{x^2} (1+\log (3))-e^{x^2} \log (x)}\right )}{x (-1-\log (3)+\log (x)) \left (5-e^{x^2} (1+\log (3))+e^{x^2} \log (x)\right )}\right ) \, dx+20 \int \left (\frac {\exp \left (2 x+x^2+\frac {e^{x (2+x)}}{-5+e^{x^2} (1+\log (3))-e^{x^2} \log (x)}\right )}{x (1+\log (3)-\log (x)) \left (5-e^{x^2} (1+\log (3))+e^{x^2} \log (x)\right )^2}+\frac {2 \exp \left (2 x+x^2+\frac {e^{x (2+x)}}{-5+e^{x^2} (1+\log (3))-e^{x^2} \log (x)}\right ) x (-1-\log (3))}{(1+\log (3)-\log (x)) \left (5-e^{x^2} (1+\log (3))+e^{x^2} \log (x)\right )^2}+\frac {2 \exp \left (2 x+x^2+\frac {e^{x (2+x)}}{-5+e^{x^2} (1+\log (3))-e^{x^2} \log (x)}\right ) x \log (x)}{(1+\log (3)-\log (x)) \left (5-e^{x^2} (1+\log (3))+e^{x^2} \log (x)\right )^2}\right ) \, dx\\ &=4 \int \frac {\exp \left (2 x+x^2+\frac {e^{x (2+x)}}{-5+e^{x^2} (1+\log (3))-e^{x^2} \log (x)}\right )}{x (-1-\log (3)+\log (x)) \left (5-e^{x^2} (1+\log (3))+e^{x^2} \log (x)\right )} \, dx+8 \int \frac {\exp \left (2 x+x^2+\frac {e^{x (2+x)}}{-5+e^{x^2} (1+\log (3))-e^{x^2} \log (x)}\right ) \log (x)}{(1+\log (3)-\log (x)) \left (5-e^{x^2} (1+\log (3))+e^{x^2} \log (x)\right )} \, dx+20 \int \frac {\exp \left (2 x+x^2+\frac {e^{x (2+x)}}{-5+e^{x^2} (1+\log (3))-e^{x^2} \log (x)}\right )}{x (1+\log (3)-\log (x)) \left (5-e^{x^2} (1+\log (3))+e^{x^2} \log (x)\right )^2} \, dx+40 \int \frac {\exp \left (2 x+x^2+\frac {e^{x (2+x)}}{-5+e^{x^2} (1+\log (3))-e^{x^2} \log (x)}\right ) x \log (x)}{(1+\log (3)-\log (x)) \left (5-e^{x^2} (1+\log (3))+e^{x^2} \log (x)\right )^2} \, dx-(40 (1+\log (3))) \int \frac {\exp \left (2 x+x^2+\frac {e^{x (2+x)}}{-5+e^{x^2} (1+\log (3))-e^{x^2} \log (x)}\right ) x}{(1+\log (3)-\log (x)) \left (5-e^{x^2} (1+\log (3))+e^{x^2} \log (x)\right )^2} \, dx-(4 (2+\log (9))) \int \frac {\exp \left (2 x+x^2+\frac {e^{x (2+x)}}{-5+e^{x^2} (1+\log (3))-e^{x^2} \log (x)}\right )}{(1+\log (3)-\log (x)) \left (5-e^{x^2} (1+\log (3))+e^{x^2} \log (x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 3.04, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{-\frac {e^{2 x+x^2}}{5+e^{x^2} (-1-\log (3))+e^{x^2} \log (x)}} \left (e^{2 x+x^2} \left (-40 x-40 x^2\right )+e^{2 x+2 x^2} (4+8 x+8 x \log (3))-8 e^{2 x+2 x^2} x \log (x)\right )}{25 x+e^{x^2} (-10 x-10 x \log (3))+e^{2 x^2} \left (x+2 x \log (3)+x \log ^2(3)\right )+\left (10 e^{x^2} x+e^{2 x^2} (-2 x-2 x \log (3))\right ) \log (x)+e^{2 x^2} x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^(2*x + x^2)*(-40*x - 40*x^2) + E^(2*x + 2*x^2)*(4 + 8*x + 8*x*Log[3]) - 8*E^(2*x + 2*x^2)*x*Log[x
])/(E^(E^(2*x + x^2)/(5 + E^x^2*(-1 - Log[3]) + E^x^2*Log[x]))*(25*x + E^x^2*(-10*x - 10*x*Log[3]) + E^(2*x^2)
*(x + 2*x*Log[3] + x*Log[3]^2) + (10*E^x^2*x + E^(2*x^2)*(-2*x - 2*x*Log[3]))*Log[x] + E^(2*x^2)*x*Log[x]^2)),
x]

[Out]

Integrate[(E^(2*x + x^2)*(-40*x - 40*x^2) + E^(2*x + 2*x^2)*(4 + 8*x + 8*x*Log[3]) - 8*E^(2*x + 2*x^2)*x*Log[x
])/(E^(E^(2*x + x^2)/(5 + E^x^2*(-1 - Log[3]) + E^x^2*Log[x]))*(25*x + E^x^2*(-10*x - 10*x*Log[3]) + E^(2*x^2)
*(x + 2*x*Log[3] + x*Log[3]^2) + (10*E^x^2*x + E^(2*x^2)*(-2*x - 2*x*Log[3]))*Log[x] + E^(2*x^2)*x*Log[x]^2)),
x]

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fricas [B]  time = 0.90, size = 56, normalized size = 1.93 \begin {gather*} 4 \, e^{\left (\frac {e^{\left (2 \, x^{2} + 4 \, x\right )}}{{\left (\log \relax (3) + 1\right )} e^{\left (2 \, x^{2} + 2 \, x\right )} - e^{\left (2 \, x^{2} + 2 \, x\right )} \log \relax (x) - 5 \, e^{\left (x^{2} + 2 \, x\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x*exp(x)^2*exp(x^2)^2*log(x)+(8*x*log(3)+8*x+4)*exp(x)^2*exp(x^2)^2+(-40*x^2-40*x)*exp(x)^2*exp(
x^2))*exp(-exp(x)^2*exp(x^2)/(exp(x^2)*log(x)+(-log(3)-1)*exp(x^2)+5))/(x*exp(x^2)^2*log(x)^2+((-2*x*log(3)-2*
x)*exp(x^2)^2+10*exp(x^2)*x)*log(x)+(x*log(3)^2+2*x*log(3)+x)*exp(x^2)^2+(-10*x*log(3)-10*x)*exp(x^2)+25*x),x,
algorithm="fricas")

[Out]

4*e^(e^(2*x^2 + 4*x)/((log(3) + 1)*e^(2*x^2 + 2*x) - e^(2*x^2 + 2*x)*log(x) - 5*e^(x^2 + 2*x)))

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giac [A]  time = 0.53, size = 35, normalized size = 1.21 \begin {gather*} 4 \, e^{\left (\frac {e^{\left (x^{2} + 2 \, x\right )}}{e^{\left (x^{2}\right )} \log \relax (3) - e^{\left (x^{2}\right )} \log \relax (x) + e^{\left (x^{2}\right )} - 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x*exp(x)^2*exp(x^2)^2*log(x)+(8*x*log(3)+8*x+4)*exp(x)^2*exp(x^2)^2+(-40*x^2-40*x)*exp(x)^2*exp(
x^2))*exp(-exp(x)^2*exp(x^2)/(exp(x^2)*log(x)+(-log(3)-1)*exp(x^2)+5))/(x*exp(x^2)^2*log(x)^2+((-2*x*log(3)-2*
x)*exp(x^2)^2+10*exp(x^2)*x)*log(x)+(x*log(3)^2+2*x*log(3)+x)*exp(x^2)^2+(-10*x*log(3)-10*x)*exp(x^2)+25*x),x,
algorithm="giac")

[Out]

4*e^(e^(x^2 + 2*x)/(e^(x^2)*log(3) - e^(x^2)*log(x) + e^(x^2) - 5))

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maple [A]  time = 0.06, size = 34, normalized size = 1.17

method	result	size
risch	\(4 \,{\mathrm e}^{\frac {{\mathrm e}^{x \left (2+x \right )}}{-{\mathrm e}^{x^{2}} \ln \relax (x )+{\mathrm e}^{x^{2}} \ln \relax (3)+{\mathrm e}^{x^{2}}-5}}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*x*exp(x)^2*exp(x^2)^2*ln(x)+(8*x*ln(3)+8*x+4)*exp(x)^2*exp(x^2)^2+(-40*x^2-40*x)*exp(x)^2*exp(x^2))*ex
p(-exp(x)^2*exp(x^2)/(exp(x^2)*ln(x)+(-ln(3)-1)*exp(x^2)+5))/(x*exp(x^2)^2*ln(x)^2+((-2*x*ln(3)-2*x)*exp(x^2)^
2+10*exp(x^2)*x)*ln(x)+(x*ln(3)^2+2*x*ln(3)+x)*exp(x^2)^2+(-10*x*ln(3)-10*x)*exp(x^2)+25*x),x,method=_RETURNVE
RBOSE)

[Out]

4*exp(exp(x*(2+x))/(-exp(x^2)*ln(x)+exp(x^2)*ln(3)+exp(x^2)-5))

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maxima [A]  time = 1.01, size = 33, normalized size = 1.14 \begin {gather*} 4 \, e^{\left (\frac {e^{\left (x^{2} + 2 \, x\right )}}{{\left (\log \relax (3) + 1\right )} e^{\left (x^{2}\right )} - e^{\left (x^{2}\right )} \log \relax (x) - 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x*exp(x)^2*exp(x^2)^2*log(x)+(8*x*log(3)+8*x+4)*exp(x)^2*exp(x^2)^2+(-40*x^2-40*x)*exp(x)^2*exp(
x^2))*exp(-exp(x)^2*exp(x^2)/(exp(x^2)*log(x)+(-log(3)-1)*exp(x^2)+5))/(x*exp(x^2)^2*log(x)^2+((-2*x*log(3)-2*
x)*exp(x^2)^2+10*exp(x^2)*x)*log(x)+(x*log(3)^2+2*x*log(3)+x)*exp(x^2)^2+(-10*x*log(3)-10*x)*exp(x^2)+25*x),x,
algorithm="maxima")

[Out]

4*e^(e^(x^2 + 2*x)/((log(3) + 1)*e^(x^2) - e^(x^2)*log(x) - 5))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {{\mathrm {e}}^{-\frac {{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{x^2}}{{\mathrm {e}}^{x^2}\,\ln \relax (x)-{\mathrm {e}}^{x^2}\,\left (\ln \relax (3)+1\right )+5}}\,\left ({\mathrm {e}}^{x^2+2\,x}\,\left (40\,x^2+40\,x\right )-{\mathrm {e}}^{2\,x^2+2\,x}\,\left (8\,x+8\,x\,\ln \relax (3)+4\right )+8\,x\,{\mathrm {e}}^{2\,x^2+2\,x}\,\ln \relax (x)\right )}{x\,{\mathrm {e}}^{2\,x^2}\,{\ln \relax (x)}^2+\left (10\,x\,{\mathrm {e}}^{x^2}-{\mathrm {e}}^{2\,x^2}\,\left (2\,x+2\,x\,\ln \relax (3)\right )\right )\,\ln \relax (x)+25\,x-{\mathrm {e}}^{x^2}\,\left (10\,x+10\,x\,\ln \relax (3)\right )+{\mathrm {e}}^{2\,x^2}\,\left (x+2\,x\,\ln \relax (3)+x\,{\ln \relax (3)}^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(exp(2*x)*exp(x^2))/(exp(x^2)*log(x) - exp(x^2)*(log(3) + 1) + 5))*(exp(2*x)*exp(x^2)*(40*x + 40*x^
2) - exp(2*x)*exp(2*x^2)*(8*x + 8*x*log(3) + 4) + 8*x*exp(2*x)*exp(2*x^2)*log(x)))/(25*x - exp(x^2)*(10*x + 10
*x*log(3)) + log(x)*(10*x*exp(x^2) - exp(2*x^2)*(2*x + 2*x*log(3))) + exp(2*x^2)*(x + 2*x*log(3) + x*log(3)^2)
+ x*exp(2*x^2)*log(x)^2),x)

[Out]

-int((exp(-(exp(2*x)*exp(x^2))/(exp(x^2)*log(x) - exp(x^2)*(log(3) + 1) + 5))*(exp(2*x + x^2)*(40*x + 40*x^2)
- exp(2*x + 2*x^2)*(8*x + 8*x*log(3) + 4) + 8*x*exp(2*x + 2*x^2)*log(x)))/(25*x - exp(x^2)*(10*x + 10*x*log(3)
) + log(x)*(10*x*exp(x^2) - exp(2*x^2)*(2*x + 2*x*log(3))) + exp(2*x^2)*(x + 2*x*log(3) + x*log(3)^2) + x*exp(
2*x^2)*log(x)^2), x)

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sympy [A]  time = 23.09, size = 34, normalized size = 1.17 \begin {gather*} 4 e^{- \frac {e^{2 x} e^{x^{2}}}{e^{x^{2}} \log {\relax (x )} + \left (- \log {\relax (3 )} - 1\right ) e^{x^{2}} + 5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x*exp(x)**2*exp(x**2)**2*ln(x)+(8*x*ln(3)+8*x+4)*exp(x)**2*exp(x**2)**2+(-40*x**2-40*x)*exp(x)**
2*exp(x**2))*exp(-exp(x)**2*exp(x**2)/(exp(x**2)*ln(x)+(-ln(3)-1)*exp(x**2)+5))/(x*exp(x**2)**2*ln(x)**2+((-2*
x*ln(3)-2*x)*exp(x**2)**2+10*exp(x**2)*x)*ln(x)+(x*ln(3)**2+2*x*ln(3)+x)*exp(x**2)**2+(-10*x*ln(3)-10*x)*exp(x
**2)+25*x),x)

[Out]

4*exp(-exp(2*x)*exp(x**2)/(exp(x**2)*log(x) + (-log(3) - 1)*exp(x**2) + 5))

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