3.50.51 \(\int \frac {e^{\frac {16+16 e^x+3 x^2+16 \log ^2(3)-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}} (6 x+e^x (6 x-3 x^2)+6 x \log ^2(3)+(-6 x-6 e^x x-6 x \log ^2(3)) \log (x)+(-6 x+e^x (-6 x+3 x^2)-6 x \log ^2(3)) \log ^2(x))}{1+e^{2 x}+2 \log ^2(3)+\log ^4(3)+e^x (2+2 \log ^2(3))} \, dx\)

Optimal. Leaf size=28 \[ e^{16-\frac {x^2 \left (-3+3 \log ^2(x)\right )}{1+e^x+\log ^2(3)}} \]

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Rubi [F]  time = 28.97, 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 {16+16 e^x+3 x^2+16 \log ^2(3)-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) \left (6 x+e^x \left (6 x-3 x^2\right )+6 x \log ^2(3)+\left (-6 x-6 e^x x-6 x \log ^2(3)\right ) \log (x)+\left (-6 x+e^x \left (-6 x+3 x^2\right )-6 x \log ^2(3)\right ) \log ^2(x)\right )}{1+e^{2 x}+2 \log ^2(3)+\log ^4(3)+e^x \left (2+2 \log ^2(3)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((16 + 16*E^x + 3*x^2 + 16*Log[3]^2 - 3*x^2*Log[x]^2)/(1 + E^x + Log[3]^2))*(6*x + E^x*(6*x - 3*x^2) +
6*x*Log[3]^2 + (-6*x - 6*E^x*x - 6*x*Log[3]^2)*Log[x] + (-6*x + E^x*(-6*x + 3*x^2) - 6*x*Log[3]^2)*Log[x]^2))/
(1 + E^(2*x) + 2*Log[3]^2 + Log[3]^4 + E^x*(2 + 2*Log[3]^2)),x]

[Out]

3*(1 + Log[3]^2)*Defer[Int][(E^((16*E^x + 3*x^2 + 16*(1 + Log[3]^2) - 3*x^2*Log[x]^2)/(1 + E^x + Log[3]^2))*x^
2)/(1 + E^x + Log[3]^2)^2, x] + 6*Defer[Int][(E^((16*E^x + 3*x^2 + 16*(1 + Log[3]^2) - 3*x^2*Log[x]^2)/(1 + E^
x + Log[3]^2))*x)/(1 + E^x + Log[3]^2), x] - 3*Defer[Int][(E^((16*E^x + 3*x^2 + 16*(1 + Log[3]^2) - 3*x^2*Log[
x]^2)/(1 + E^x + Log[3]^2))*x^2)/(1 + E^x + Log[3]^2), x] - 6*Defer[Int][(E^((16*E^x + 3*x^2 + 16*(1 + Log[3]^
2) - 3*x^2*Log[x]^2)/(1 + E^x + Log[3]^2))*x*Log[x])/(1 + E^x + Log[3]^2), x] - 3*(1 + Log[3]^2)*Defer[Int][(E
^((16*E^x + 3*x^2 + 16*(1 + Log[3]^2) - 3*x^2*Log[x]^2)/(1 + E^x + Log[3]^2))*x^2*Log[x]^2)/(1 + E^x + Log[3]^
2)^2, x] - 6*Defer[Int][(E^((16*E^x + 3*x^2 + 16*(1 + Log[3]^2) - 3*x^2*Log[x]^2)/(1 + E^x + Log[3]^2))*x*Log[
x]^2)/(1 + E^x + Log[3]^2), x] + 3*Defer[Int][(E^((16*E^x + 3*x^2 + 16*(1 + Log[3]^2) - 3*x^2*Log[x]^2)/(1 + E
^x + Log[3]^2))*x^2*Log[x]^2)/(1 + E^x + Log[3]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {16+16 e^x+3 x^2+16 \log ^2(3)-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) \left (e^x \left (6 x-3 x^2\right )+x \left (6+6 \log ^2(3)\right )+\left (-6 x-6 e^x x-6 x \log ^2(3)\right ) \log (x)+\left (-6 x+e^x \left (-6 x+3 x^2\right )-6 x \log ^2(3)\right ) \log ^2(x)\right )}{1+e^{2 x}+2 \log ^2(3)+\log ^4(3)+e^x \left (2+2 \log ^2(3)\right )} \, dx\\ &=\int \frac {3 \exp \left (\frac {16 e^x+3 x^2+16 \left (1+\log ^2(3)\right )-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) x \left (-e^x (-2+x)+2 \left (1+\log ^2(3)\right )-2 \left (1+e^x+\log ^2(3)\right ) \log (x)+\left (e^x (-2+x)-2 \left (1+\log ^2(3)\right )\right ) \log ^2(x)\right )}{\left (1+e^x+\log ^2(3)\right )^2} \, dx\\ &=3 \int \frac {\exp \left (\frac {16 e^x+3 x^2+16 \left (1+\log ^2(3)\right )-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) x \left (-e^x (-2+x)+2 \left (1+\log ^2(3)\right )-2 \left (1+e^x+\log ^2(3)\right ) \log (x)+\left (e^x (-2+x)-2 \left (1+\log ^2(3)\right )\right ) \log ^2(x)\right )}{\left (1+e^x+\log ^2(3)\right )^2} \, dx\\ &=3 \int \left (-\frac {\exp \left (\frac {16 e^x+3 x^2+16 \left (1+\log ^2(3)\right )-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) x^2 \left (1+\log ^2(3)\right ) \left (-1+\log ^2(x)\right )}{\left (1+e^x+\log ^2(3)\right )^2}+\frac {\exp \left (\frac {16 e^x+3 x^2+16 \left (1+\log ^2(3)\right )-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) x \left (2-x-2 \log (x)-2 \log ^2(x)+x \log ^2(x)\right )}{1+e^x+\log ^2(3)}\right ) \, dx\\ &=3 \int \frac {\exp \left (\frac {16 e^x+3 x^2+16 \left (1+\log ^2(3)\right )-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) x \left (2-x-2 \log (x)-2 \log ^2(x)+x \log ^2(x)\right )}{1+e^x+\log ^2(3)} \, dx-\left (3 \left (1+\log ^2(3)\right )\right ) \int \frac {\exp \left (\frac {16 e^x+3 x^2+16 \left (1+\log ^2(3)\right )-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) x^2 \left (-1+\log ^2(x)\right )}{\left (1+e^x+\log ^2(3)\right )^2} \, dx\\ &=3 \int \left (\frac {2 \exp \left (\frac {16 e^x+3 x^2+16 \left (1+\log ^2(3)\right )-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) x}{1+e^x+\log ^2(3)}-\frac {\exp \left (\frac {16 e^x+3 x^2+16 \left (1+\log ^2(3)\right )-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) x^2}{1+e^x+\log ^2(3)}-\frac {2 \exp \left (\frac {16 e^x+3 x^2+16 \left (1+\log ^2(3)\right )-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) x \log (x)}{1+e^x+\log ^2(3)}-\frac {2 \exp \left (\frac {16 e^x+3 x^2+16 \left (1+\log ^2(3)\right )-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) x \log ^2(x)}{1+e^x+\log ^2(3)}+\frac {\exp \left (\frac {16 e^x+3 x^2+16 \left (1+\log ^2(3)\right )-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) \, dx-\left (3 \left (1+\log ^2(3)\right )\right ) \int \left (-\frac {\exp \left (\frac {16 e^x+3 x^2+16 \left (1+\log ^2(3)\right )-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) x^2}{\left (1+e^x+\log ^2(3)\right )^2}+\frac {\exp \left (\frac {16 e^x+3 x^2+16 \left (1+\log ^2(3)\right )-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) x^2 \log ^2(x)}{\left (1+e^x+\log ^2(3)\right )^2}\right ) \, dx\\ &=-\left (3 \int \frac {\exp \left (\frac {16 e^x+3 x^2+16 \left (1+\log ^2(3)\right )-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) x^2}{1+e^x+\log ^2(3)} \, dx\right )+3 \int \frac {\exp \left (\frac {16 e^x+3 x^2+16 \left (1+\log ^2(3)\right )-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) x^2 \log ^2(x)}{1+e^x+\log ^2(3)} \, dx+6 \int \frac {\exp \left (\frac {16 e^x+3 x^2+16 \left (1+\log ^2(3)\right )-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) x}{1+e^x+\log ^2(3)} \, dx-6 \int \frac {\exp \left (\frac {16 e^x+3 x^2+16 \left (1+\log ^2(3)\right )-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) x \log (x)}{1+e^x+\log ^2(3)} \, dx-6 \int \frac {\exp \left (\frac {16 e^x+3 x^2+16 \left (1+\log ^2(3)\right )-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) x \log ^2(x)}{1+e^x+\log ^2(3)} \, dx+\left (3 \left (1+\log ^2(3)\right )\right ) \int \frac {\exp \left (\frac {16 e^x+3 x^2+16 \left (1+\log ^2(3)\right )-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) x^2}{\left (1+e^x+\log ^2(3)\right )^2} \, dx-\left (3 \left (1+\log ^2(3)\right )\right ) \int \frac {\exp \left (\frac {16 e^x+3 x^2+16 \left (1+\log ^2(3)\right )-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}\right ) x^2 \log ^2(x)}{\left (1+e^x+\log ^2(3)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.31, size = 40, normalized size = 1.43 \begin {gather*} e^{16+\frac {3 x^2}{1+e^x+\log ^2(3)}-\frac {3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((16 + 16*E^x + 3*x^2 + 16*Log[3]^2 - 3*x^2*Log[x]^2)/(1 + E^x + Log[3]^2))*(6*x + E^x*(6*x - 3*x
^2) + 6*x*Log[3]^2 + (-6*x - 6*E^x*x - 6*x*Log[3]^2)*Log[x] + (-6*x + E^x*(-6*x + 3*x^2) - 6*x*Log[3]^2)*Log[x
]^2))/(1 + E^(2*x) + 2*Log[3]^2 + Log[3]^4 + E^x*(2 + 2*Log[3]^2)),x]

[Out]

E^(16 + (3*x^2)/(1 + E^x + Log[3]^2) - (3*x^2*Log[x]^2)/(1 + E^x + Log[3]^2))

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fricas [A]  time = 0.90, size = 39, normalized size = 1.39 \begin {gather*} e^{\left (-\frac {3 \, x^{2} \log \relax (x)^{2} - 3 \, x^{2} - 16 \, \log \relax (3)^{2} - 16 \, e^{x} - 16}{\log \relax (3)^{2} + e^{x} + 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2-6*x)*exp(x)-6*x*log(3)^2-6*x)*log(x)^2+(-6*exp(x)*x-6*x*log(3)^2-6*x)*log(x)+(-3*x^2+6*x)*e
xp(x)+6*x*log(3)^2+6*x)*exp((-3*x^2*log(x)^2+16*exp(x)+16*log(3)^2+3*x^2+16)/(exp(x)+log(3)^2+1))/(exp(x)^2+(2
*log(3)^2+2)*exp(x)+log(3)^4+2*log(3)^2+1),x, algorithm="fricas")

[Out]

e^(-(3*x^2*log(x)^2 - 3*x^2 - 16*log(3)^2 - 16*e^x - 16)/(log(3)^2 + e^x + 1))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {3 \, {\left (2 \, x \log \relax (3)^{2} - {\left (2 \, x \log \relax (3)^{2} - {\left (x^{2} - 2 \, x\right )} e^{x} + 2 \, x\right )} \log \relax (x)^{2} - {\left (x^{2} - 2 \, x\right )} e^{x} - 2 \, {\left (x \log \relax (3)^{2} + x e^{x} + x\right )} \log \relax (x) + 2 \, x\right )} e^{\left (-\frac {3 \, x^{2} \log \relax (x)^{2} - 3 \, x^{2} - 16 \, \log \relax (3)^{2} - 16 \, e^{x} - 16}{\log \relax (3)^{2} + e^{x} + 1}\right )}}{\log \relax (3)^{4} + 2 \, {\left (\log \relax (3)^{2} + 1\right )} e^{x} + 2 \, \log \relax (3)^{2} + e^{\left (2 \, x\right )} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2-6*x)*exp(x)-6*x*log(3)^2-6*x)*log(x)^2+(-6*exp(x)*x-6*x*log(3)^2-6*x)*log(x)+(-3*x^2+6*x)*e
xp(x)+6*x*log(3)^2+6*x)*exp((-3*x^2*log(x)^2+16*exp(x)+16*log(3)^2+3*x^2+16)/(exp(x)+log(3)^2+1))/(exp(x)^2+(2
*log(3)^2+2)*exp(x)+log(3)^4+2*log(3)^2+1),x, algorithm="giac")

[Out]

integrate(3*(2*x*log(3)^2 - (2*x*log(3)^2 - (x^2 - 2*x)*e^x + 2*x)*log(x)^2 - (x^2 - 2*x)*e^x - 2*(x*log(3)^2
+ x*e^x + x)*log(x) + 2*x)*e^(-(3*x^2*log(x)^2 - 3*x^2 - 16*log(3)^2 - 16*e^x - 16)/(log(3)^2 + e^x + 1))/(log
(3)^4 + 2*(log(3)^2 + 1)*e^x + 2*log(3)^2 + e^(2*x) + 1), x)

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maple [A]  time = 0.06, size = 39, normalized size = 1.39

method	result	size
risch	\({\mathrm e}^{\frac {-3 x^{2} \ln \relax (x )^{2}+16 \,{\mathrm e}^{x}+16 \ln \relax (3)^{2}+3 x^{2}+16}{{\mathrm e}^{x}+\ln \relax (3)^{2}+1}}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((3*x^2-6*x)*exp(x)-6*x*ln(3)^2-6*x)*ln(x)^2+(-6*exp(x)*x-6*x*ln(3)^2-6*x)*ln(x)+(-3*x^2+6*x)*exp(x)+6*x*
ln(3)^2+6*x)*exp((-3*x^2*ln(x)^2+16*exp(x)+16*ln(3)^2+3*x^2+16)/(exp(x)+ln(3)^2+1))/(exp(x)^2+(2*ln(3)^2+2)*ex
p(x)+ln(3)^4+2*ln(3)^2+1),x,method=_RETURNVERBOSE)

[Out]

exp((-3*x^2*ln(x)^2+16*exp(x)+16*ln(3)^2+3*x^2+16)/(exp(x)+ln(3)^2+1))

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maxima [B]  time = 0.70, size = 78, normalized size = 2.79 \begin {gather*} e^{\left (-\frac {3 \, x^{2} \log \relax (x)^{2}}{\log \relax (3)^{2} + e^{x} + 1} + \frac {3 \, x^{2}}{\log \relax (3)^{2} + e^{x} + 1} + \frac {16 \, \log \relax (3)^{2}}{\log \relax (3)^{2} + e^{x} + 1} + \frac {16 \, e^{x}}{\log \relax (3)^{2} + e^{x} + 1} + \frac {16}{\log \relax (3)^{2} + e^{x} + 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2-6*x)*exp(x)-6*x*log(3)^2-6*x)*log(x)^2+(-6*exp(x)*x-6*x*log(3)^2-6*x)*log(x)+(-3*x^2+6*x)*e
xp(x)+6*x*log(3)^2+6*x)*exp((-3*x^2*log(x)^2+16*exp(x)+16*log(3)^2+3*x^2+16)/(exp(x)+log(3)^2+1))/(exp(x)^2+(2
*log(3)^2+2)*exp(x)+log(3)^4+2*log(3)^2+1),x, algorithm="maxima")

[Out]

e^(-3*x^2*log(x)^2/(log(3)^2 + e^x + 1) + 3*x^2/(log(3)^2 + e^x + 1) + 16*log(3)^2/(log(3)^2 + e^x + 1) + 16*e
^x/(log(3)^2 + e^x + 1) + 16/(log(3)^2 + e^x + 1))

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mupad [B]  time = 3.58, size = 82, normalized size = 2.93 \begin {gather*} {\mathrm {e}}^{\frac {3\,x^2}{{\mathrm {e}}^x+{\ln \relax (3)}^2+1}}\,{\mathrm {e}}^{\frac {16\,{\mathrm {e}}^x}{{\mathrm {e}}^x+{\ln \relax (3)}^2+1}}\,{\mathrm {e}}^{\frac {16}{{\mathrm {e}}^x+{\ln \relax (3)}^2+1}}\,{\mathrm {e}}^{-\frac {3\,x^2\,{\ln \relax (x)}^2}{{\mathrm {e}}^x+{\ln \relax (3)}^2+1}}\,{\mathrm {e}}^{\frac {16\,{\ln \relax (3)}^2}{{\mathrm {e}}^x+{\ln \relax (3)}^2+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((16*exp(x) - 3*x^2*log(x)^2 + 16*log(3)^2 + 3*x^2 + 16)/(exp(x) + log(3)^2 + 1))*(6*x - log(x)^2*(6*x
+ 6*x*log(3)^2 + exp(x)*(6*x - 3*x^2)) + 6*x*log(3)^2 + exp(x)*(6*x - 3*x^2) - log(x)*(6*x + 6*x*log(3)^2 + 6
*x*exp(x))))/(exp(2*x) + exp(x)*(2*log(3)^2 + 2) + 2*log(3)^2 + log(3)^4 + 1),x)

[Out]

exp((3*x^2)/(exp(x) + log(3)^2 + 1))*exp((16*exp(x))/(exp(x) + log(3)^2 + 1))*exp(16/(exp(x) + log(3)^2 + 1))*
exp(-(3*x^2*log(x)^2)/(exp(x) + log(3)^2 + 1))*exp((16*log(3)^2)/(exp(x) + log(3)^2 + 1))

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sympy [A]  time = 1.30, size = 39, normalized size = 1.39 \begin {gather*} e^{\frac {- 3 x^{2} \log {\relax (x )}^{2} + 3 x^{2} + 16 e^{x} + 16 + 16 \log {\relax (3 )}^{2}}{e^{x} + 1 + \log {\relax (3 )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x**2-6*x)*exp(x)-6*x*ln(3)**2-6*x)*ln(x)**2+(-6*exp(x)*x-6*x*ln(3)**2-6*x)*ln(x)+(-3*x**2+6*x)*
exp(x)+6*x*ln(3)**2+6*x)*exp((-3*x**2*ln(x)**2+16*exp(x)+16*ln(3)**2+3*x**2+16)/(exp(x)+ln(3)**2+1))/(exp(x)**
2+(2*ln(3)**2+2)*exp(x)+ln(3)**4+2*ln(3)**2+1),x)

[Out]

exp((-3*x**2*log(x)**2 + 3*x**2 + 16*exp(x) + 16 + 16*log(3)**2)/(exp(x) + 1 + log(3)**2))

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