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

Optimal. Leaf size=32 \[ 3+\left (e^{-5+\frac {x \left (-4+\frac {1}{2} \left (-5 e^3+x\right )\right )}{\log (\log (x))}}+x^2\right )^2 \]

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Rubi [F]  time = 3.06, 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 {4 x^3 \log (x) \log ^2(\log (x))+\exp \left (\frac {-8 x-5 e^3 x+x^2-10 \log (\log (x))}{\log (\log (x))}\right ) \left (8+5 e^3-x+\left (-8-5 e^3+2 x\right ) \log (x) \log (\log (x))\right )+\exp \left (\frac {-8 x-5 e^3 x+x^2-10 \log (\log (x))}{2 \log (\log (x))}\right ) \left (8 x^2+5 e^3 x^2-x^3+\left (-8 x^2-5 e^3 x^2+2 x^3\right ) \log (x) \log (\log (x))+4 x \log (x) \log ^2(\log (x))\right )}{\log (x) \log ^2(\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.41, size = 62, normalized size = 1.94 \begin {gather*} e^{-10-\frac {\left (8+5 e^3\right ) x}{\log (\log (x))}} \left (e^{\frac {x^2}{2 \log (\log (x))}}+e^{5+\frac {\left (8+5 e^3\right ) x}{2 \log (\log (x))}} x^2\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-10 - ((8 + 5*E^3)*x)/Log[Log[x]])*(E^(x^2/(2*Log[Log[x]])) + E^(5 + ((8 + 5*E^3)*x)/(2*Log[Log[x]]))*x^2)^
2

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fricas [B]  time = 0.63, size = 58, normalized size = 1.81 \begin {gather*} x^{4} + 2 \, x^{2} e^{\left (\frac {x^{2} - 5 \, x e^{3} - 8 \, x - 10 \, \log \left (\log \relax (x)\right )}{2 \, \log \left (\log \relax (x)\right )}\right )} + e^{\left (\frac {x^{2} - 5 \, x e^{3} - 8 \, x - 10 \, \log \left (\log \relax (x)\right )}{\log \left (\log \relax (x)\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^4 + 2*x^2*e^(1/2*(x^2 - 5*x*e^3 - 8*x - 10*log(log(x)))/log(log(x))) + e^((x^2 - 5*x*e^3 - 8*x - 10*log(log(
x)))/log(log(x)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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maple [B]  time = 0.19, size = 64, normalized size = 2.00

method	result	size
risch	\(x^{4}+2 x^{2} {\mathrm e}^{-\frac {5 x \,{\mathrm e}^{3}-x^{2}+10 \ln \left (\ln \relax (x )\right )+8 x}{2 \ln \left (\ln \relax (x )\right )}}+{\mathrm e}^{-\frac {5 x \,{\mathrm e}^{3}-x^{2}+10 \ln \left (\ln \relax (x )\right )+8 x}{\ln \left (\ln \relax (x )\right )}}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-5*exp(3)+2*x-8)*ln(x)*ln(ln(x))+5*exp(3)+8-x)*exp(1/2*(-10*ln(ln(x))-5*x*exp(3)+x^2-8*x)/ln(ln(x)))^2+
(4*x*ln(x)*ln(ln(x))^2+(-5*x^2*exp(3)+2*x^3-8*x^2)*ln(x)*ln(ln(x))+5*x^2*exp(3)-x^3+8*x^2)*exp(1/2*(-10*ln(ln(
x))-5*x*exp(3)+x^2-8*x)/ln(ln(x)))+4*x^3*ln(x)*ln(ln(x))^2)/ln(x)/ln(ln(x))^2,x,method=_RETURNVERBOSE)

[Out]

x^4+2*x^2*exp(-1/2*(5*x*exp(3)-x^2+10*ln(ln(x))+8*x)/ln(ln(x)))+exp(-(5*x*exp(3)-x^2+10*ln(ln(x))+8*x)/ln(ln(x
)))

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

[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 = 1.05, size = 75, normalized size = 2.34 \begin {gather*} x^4+{\mathrm {e}}^{\frac {x^2}{\ln \left (\ln \relax (x)\right )}}\,{\mathrm {e}}^{-\frac {5\,x\,{\mathrm {e}}^3}{\ln \left (\ln \relax (x)\right )}}\,{\mathrm {e}}^{-10}\,{\mathrm {e}}^{-\frac {8\,x}{\ln \left (\ln \relax (x)\right )}}+2\,x^2\,{\mathrm {e}}^{\frac {x^2}{2\,\ln \left (\ln \relax (x)\right )}}\,{\mathrm {e}}^{-\frac {5\,x\,{\mathrm {e}}^3}{2\,\ln \left (\ln \relax (x)\right )}}\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{-\frac {4\,x}{\ln \left (\ln \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(4*x + 5*log(log(x)) + (5*x*exp(3))/2 - x^2/2)/log(log(x)))*(5*x^2*exp(3) + 8*x^2 - x^3 + 4*x*log(lo
g(x))^2*log(x) - log(log(x))*log(x)*(5*x^2*exp(3) + 8*x^2 - 2*x^3)) - exp(-(2*(4*x + 5*log(log(x)) + (5*x*exp(
3))/2 - x^2/2))/log(log(x)))*(x - 5*exp(3) + log(log(x))*log(x)*(5*exp(3) - 2*x + 8) - 8) + 4*x^3*log(log(x))^
2*log(x))/(log(log(x))^2*log(x)),x)

[Out]

x^4 + exp(x^2/log(log(x)))*exp(-(5*x*exp(3))/log(log(x)))*exp(-10)*exp(-(8*x)/log(log(x))) + 2*x^2*exp(x^2/(2*
log(log(x))))*exp(-(5*x*exp(3))/(2*log(log(x))))*exp(-5)*exp(-(4*x)/log(log(x)))

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sympy [B]  time = 10.18, size = 65, normalized size = 2.03 \begin {gather*} x^{4} + 2 x^{2} e^{\frac {\frac {x^{2}}{2} - \frac {5 x e^{3}}{2} - 4 x - 5 \log {\left (\log {\relax (x )} \right )}}{\log {\left (\log {\relax (x )} \right )}}} + e^{\frac {2 \left (\frac {x^{2}}{2} - \frac {5 x e^{3}}{2} - 4 x - 5 \log {\left (\log {\relax (x )} \right )}\right )}{\log {\left (\log {\relax (x )} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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