\(\int \frac {e^{\frac {2 (1+i \pi +\log (-3+2 \log (5)))}{x}} (-2-2 (i \pi +\log (-3+2 \log (5))))}{x^2} \, dx\) [1325]

Optimal result

Integrand size = 42, antiderivative size = 21 \[ \int \frac {e^{\frac {2 (1+i \pi +\log (-3+2 \log (5)))}{x}} (-2-2 (i \pi +\log (-3+2 \log (5))))}{x^2} \, dx=e^{\frac {2 (1+i \pi +\log (-3+2 \log (5)))}{x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {12, 2240} \[ \int \frac {e^{\frac {2 (1+i \pi +\log (-3+2 \log (5)))}{x}} (-2-2 (i \pi +\log (-3+2 \log (5))))}{x^2} \, dx=e^{\frac {2 (1+i \pi )}{x}} (\log (25)-3)^{2/x} \]

[In]

Int[(E^((2*(1 + I*Pi + Log[-3 + 2*Log[5]]))/x)*(-2 - 2*(I*Pi + Log[-3 + 2*Log[5]])))/x^2,x]

[Out]

E^((2*(1 + I*Pi))/x)*(-3 + Log[25])^(2/x)

Rule 12

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

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps \begin{align*} \text {integral}& = -\left ((2 (1+i \pi +\log (-3+\log (25)))) \int \frac {e^{\frac {2 (1+i \pi +\log (-3+2 \log (5)))}{x}}}{x^2} \, dx\right ) \\ & = e^{\frac {2 (1+i \pi )}{x}} (-3+\log (25))^{2/x} \\ \end{align*}

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(52\) vs. \(2(21)=42\).

Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.48 \[ \int \frac {e^{\frac {2 (1+i \pi +\log (-3+2 \log (5)))}{x}} (-2-2 (i \pi +\log (-3+2 \log (5))))}{x^2} \, dx=-\frac {e^{\frac {2 (1+i \pi +\log (-3+\log (25)))}{x}} (-2 i \pi -2 (1+\log (-3+\log (25))))}{2 (1+i \pi +\log (-3+\log (25)))} \]

[In]

Integrate[(E^((2*(1 + I*Pi + Log[-3 + 2*Log[5]]))/x)*(-2 - 2*(I*Pi + Log[-3 + 2*Log[5]])))/x^2,x]

[Out]

-1/2*(E^((2*(1 + I*Pi + Log[-3 + Log[25]]))/x)*((-2*I)*Pi - 2*(1 + Log[-3 + Log[25]])))/(1 + I*Pi + Log[-3 + L
og[25]])

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81

method	result	size
norman	\({\mathrm e}^{\frac {2 \ln \left (-2 \ln \left (5\right )+3\right )+2}{x}}\)	\(17\)
derivativedivides	\(-\frac {\left (-2 \ln \left (-2 \ln \left (5\right )+3\right )-2\right ) {\mathrm e}^{\frac {2 \ln \left (-2 \ln \left (5\right )+3\right )+2}{x}}}{2 \left (\ln \left (-2 \ln \left (5\right )+3\right )+1\right )}\)	\(41\)
default	\(-\frac {\left (-2 \ln \left (-2 \ln \left (5\right )+3\right )-2\right ) {\mathrm e}^{\frac {2 \ln \left (-2 \ln \left (5\right )+3\right )+2}{x}}}{2 \left (\ln \left (-2 \ln \left (5\right )+3\right )+1\right )}\)	\(41\)
risch	\(-\frac {\left (-2 \ln \left (-2 \ln \left (5\right )+3\right )-2\right ) \left (-2 \ln \left (5\right )+3\right )^{\frac {2}{x}} {\mathrm e}^{\frac {2}{x}}}{2 \left (\ln \left (-2 \ln \left (5\right )+3\right )+1\right )}\)	\(43\)
meijerg	\(\frac {\ln \left (-2 \ln \left (5\right )+3\right ) \left (-2 \ln \left (-2 \ln \left (5\right )+3\right )-2\right ) \left (1-{\mathrm e}^{-\frac {2 \left (-\ln \left (-2 \ln \left (5\right )+3\right )-1\right )}{x}}\right )}{2 \left (-\ln \left (-2 \ln \left (5\right )+3\right )-1\right )^{2}}+\frac {\left (-2 \ln \left (-2 \ln \left (5\right )+3\right )-2\right ) \left (1-{\mathrm e}^{-\frac {2 \left (-\ln \left (-2 \ln \left (5\right )+3\right )-1\right )}{x}}\right )}{2 \left (-\ln \left (-2 \ln \left (5\right )+3\right )-1\right )^{2}}\)	\(103\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\frac {2 (1+i \pi +\log (-3+2 \log (5)))}{x}} (-2-2 (i \pi +\log (-3+2 \log (5))))}{x^2} \, dx=e^{\left (\frac {2 \, {\left (\log \left (-2 \, \log \left (5\right ) + 3\right ) + 1\right )}}{x}\right )} \]

[In]

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

[Out]

e^(2*(log(-2*log(5) + 3) + 1)/x)

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (19) = 38\).

Time = 4.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 3.00 \[ \int \frac {e^{\frac {2 (1+i \pi +\log (-3+2 \log (5)))}{x}} (-2-2 (i \pi +\log (-3+2 \log (5))))}{x^2} \, dx=- \frac {\left (-2 - 2 \log {\left (-3 + 2 \log {\left (5 \right )} \right )} - 2 i \pi \right ) e^{\frac {2}{x}} e^{\frac {2 \log {\left (-3 + 2 \log {\left (5 \right )} \right )}}{x}} e^{\frac {2 i \pi }{x}}}{2 \log {\left (-3 + 2 \log {\left (5 \right )} \right )} + 2 + 2 i \pi } \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\frac {2 (1+i \pi +\log (-3+2 \log (5)))}{x}} (-2-2 (i \pi +\log (-3+2 \log (5))))}{x^2} \, dx=e^{\left (\frac {2 \, {\left (\log \left (-2 \, \log \left (5\right ) + 3\right ) + 1\right )}}{x}\right )} \]

[In]

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

[Out]

e^(2*(log(-2*log(5) + 3) + 1)/x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {2 (1+i \pi +\log (-3+2 \log (5)))}{x}} (-2-2 (i \pi +\log (-3+2 \log (5))))}{x^2} \, dx=e^{\left (\frac {2 \, \log \left (-2 \, \log \left (5\right ) + 3\right )}{x} + \frac {2}{x}\right )} \]

[In]

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

[Out]

e^(2*log(-2*log(5) + 3)/x + 2/x)

Mupad [B] (verification not implemented)

Time = 8.58 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {2 (1+i \pi +\log (-3+2 \log (5)))}{x}} (-2-2 (i \pi +\log (-3+2 \log (5))))}{x^2} \, dx={\mathrm {e}}^{2/x}\,{\left (3-\ln \left (25\right )\right )}^{2/x} \]

[In]

int(-(exp((2*(log(3 - 2*log(5)) + 1))/x)*(2*log(3 - 2*log(5)) + 2))/x^2,x)

[Out]

exp(2/x)*(3 - log(25))^(2/x)