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

Optimal. Leaf size=27 \[ e^{-2+4 \left (x+\frac {2 x}{e^2}+\frac {x}{e}-\frac {2 \log (\log (x))}{x}\right )} \]

________________________________________________________________________________________

Rubi [F]  time = 2.36, 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 (-3+\frac {8 e x^2+e^2 \left (4 x^2+e \left (-2 x+4 x^2\right )\right )-8 e^3 \log (\log (x))}{e^3 x}\right ) \left (-8 e^3+\left (8 e x^2+e^2 \left (4 x^2+4 e x^2\right )\right ) \log (x)+8 e^3 \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

-8*Defer[Int][(E^(-2 + (4*(2 + E + E^2)*x)/E^2)*Log[x]^(-1 - 8/x))/x^2, x] + 4*(2 + E + E^2)*Defer[Int][E^(-4
+ (4*(2 + E + E^2)*x)/E^2)/Log[x]^(8/x), x] + 8*Defer[Int][(E^(-2 + (4*(2 + E + E^2)*x)/E^2)*Log[Log[x]])/(x^2
*Log[x]^(8/x)), x]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.32, size = 25, normalized size = 0.93 \begin {gather*} e^{-2+\frac {4 \left (2+e+e^2\right ) x}{e^2}} \log ^{-\frac {8}{x}}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-2 + (4*(2 + E + E^2)*x)/E^2)/Log[x]^(8/x)

________________________________________________________________________________________

fricas [A]  time = 0.90, size = 41, normalized size = 1.52 \begin {gather*} e^{\left (\frac {{\left (4 \, x^{2} e + 8 \, x^{2} + {\left (4 \, x^{2} - 5 \, x\right )} e^{2} - 8 \, e^{2} \log \left (\log \relax (x)\right )\right )} e^{\left (-2\right )}}{x} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.58, size = 24, normalized size = 0.89 \begin {gather*} e^{\left (4 \, x e^{\left (-1\right )} + 8 \, x e^{\left (-2\right )} + 4 \, x - \frac {8 \, \log \left (\log \relax (x)\right )}{x} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.06, size = 43, normalized size = 1.59

method	result	size
risch	\({\mathrm e}^{\frac {2 \left (2 x^{2} {\mathrm e}^{3}-x \,{\mathrm e}^{3}+2 x^{2} {\mathrm e}^{2}+4 x^{2} {\mathrm e}-4 \ln \left (\ln \relax (x )\right ) {\mathrm e}^{3}\right ) {\mathrm e}^{-3}}{x}}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.53, size = 24, normalized size = 0.89 \begin {gather*} e^{\left (4 \, x e^{\left (-1\right )} + 8 \, x e^{\left (-2\right )} + 4 \, x - \frac {8 \, \log \left (\log \relax (x)\right )}{x} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 8.13, size = 29, normalized size = 1.07 \begin {gather*} \frac {{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^{4\,x\,{\mathrm {e}}^{-1}}\,{\mathrm {e}}^{8\,x\,{\mathrm {e}}^{-2}}}{{\ln \relax (x)}^{8/x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(4*x)*exp(-2)*exp(4*x*exp(-1))*exp(8*x*exp(-2)))/log(x)^(8/x)

________________________________________________________________________________________

sympy [A]  time = 0.62, size = 44, normalized size = 1.63 \begin {gather*} e^{\frac {8 e x^{2} + \left (4 x^{2} + e \left (4 x^{2} - 2 x\right )\right ) e^{2} - 8 e^{3} \log {\left (\log {\relax (x )} \right )}}{x e^{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp((8*E*x**2 + (4*x**2 + E*(4*x**2 - 2*x))*exp(2) - 8*exp(3)*log(log(x)))*exp(-3)/x)

________________________________________________________________________________________