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

Optimal. Leaf size=33 \[ 5+x+\frac {e^{2 e^{2+\frac {4}{x}-x}+2 x (x+\log (3))}}{\log ^2(\log (5))} \]

________________________________________________________________________________________

Rubi [A]  time = 0.66, antiderivative size = 44, normalized size of antiderivative = 1.33, number of steps used = 4, number of rules used = 3, integrand size = 88, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {12, 14, 2288} \begin {gather*} \frac {2\ 9^x \log (3) \exp \left (-((1-2 x) x)+x+2 e^{-x+\frac {4}{x}+2}\right )}{\log (9) \log ^2(\log (5))}+x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x + (2*9^x*E^(2*E^(2 + 4/x - x) + x - (1 - 2*x)*x)*Log[3])/(Log[9]*Log[Log[5]]^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 1.94, size = 31, normalized size = 0.94 \begin {gather*} x+\frac {9^x e^{2 \left (e^{2+\frac {4}{x}-x}+x^2\right )}}{\log ^2(\log (5))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x + (9^x*E^(2*(E^(2 + 4/x - x) + x^2)))/Log[Log[5]]^2

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 42, normalized size = 1.27 \begin {gather*} \frac {x \log \left (\log \relax (5)\right )^{2} + e^{\left (2 \, x^{2} + 2 \, x \log \relax (3) + 2 \, e^{\left (-\frac {x^{2} - 2 \, x - 4}{x}\right )}\right )}}{\log \left (\log \relax (5)\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \log \left (\log \relax (5)\right )^{2} + 2 \, {\left (2 \, x^{3} + x^{2} \log \relax (3) - {\left (x^{2} + 4\right )} e^{\left (-\frac {x^{2} - 2 \, x - 4}{x}\right )}\right )} e^{\left (2 \, x^{2} + 2 \, x \log \relax (3) + 2 \, e^{\left (-\frac {x^{2} - 2 \, x - 4}{x}\right )}\right )}}{x^{2} \log \left (\log \relax (5)\right )^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.28, size = 37, normalized size = 1.12

method	result	size
risch	\(x +\frac {3^{2 x} {\mathrm e}^{2 \,{\mathrm e}^{-\frac {x^{2}-2 x -4}{x}}+2 x^{2}}}{\ln \left (\ln \relax (5)\right )^{2}}\)	\(37\)
norman	\(\frac {\ln \left (\ln \relax (5)\right ) x^{2}+\frac {x \,{\mathrm e}^{2 \,{\mathrm e}^{\frac {-x^{2}+2 x +4}{x}}+2 x \ln \relax (3)+2 x^{2}}}{\ln \left (\ln \relax (5)\right )}}{x \ln \left (\ln \relax (5)\right )}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.60, size = 39, normalized size = 1.18 \begin {gather*} \frac {x \log \left (\log \relax (5)\right )^{2} + e^{\left (2 \, x^{2} + 2 \, x \log \relax (3) + 2 \, e^{\left (-x + \frac {4}{x} + 2\right )}\right )}}{\log \left (\log \relax (5)\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 4.46, size = 34, normalized size = 1.03 \begin {gather*} x+\frac {3^{2\,x}\,{\mathrm {e}}^{2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^2\,{\mathrm {e}}^{4/x}}\,{\mathrm {e}}^{2\,x^2}}{{\ln \left (\ln \relax (5)\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.45, size = 34, normalized size = 1.03 \begin {gather*} x + \frac {e^{2 x^{2} + 2 x \log {\relax (3 )} + 2 e^{\frac {- x^{2} + 2 x + 4}{x}}}}{\log {\left (\log {\relax (5 )} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________