\(\int \frac {20 e^{\frac {2 (3 e^2-3 x+\log (i \pi +\log (2)))}{e^2-x}} \log (i \pi +\log (2))}{e^4-2 e^2 x+x^2} \, dx\) [7597]

Optimal result

Integrand size = 57, antiderivative size = 28 \[ \int \frac {20 e^{\frac {2 \left (3 e^2-3 x+\log (i \pi +\log (2))\right )}{e^2-x}} \log (i \pi +\log (2))}{e^4-2 e^2 x+x^2} \, dx=10 \left (5+e^{6-\frac {2 \log (i \pi +\log (2))}{-e^2+x}}\right ) \]

[Out]

10*exp(3-ln(ln(2)+I*Pi)/(x-exp(2)))^2+50

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {12, 27, 2262, 2240} \[ \int \frac {20 e^{\frac {2 \left (3 e^2-3 x+\log (i \pi +\log (2))\right )}{e^2-x}} \log (i \pi +\log (2))}{e^4-2 e^2 x+x^2} \, dx=10 e^6 (\log (2)+i \pi )^{\frac {2}{e^2-x}} \]

[In]

Int[(20*E^((2*(3*E^2 - 3*x + Log[I*Pi + Log[2]]))/(E^2 - x))*Log[I*Pi + Log[2]])/(E^4 - 2*E^2*x + x^2),x]

[Out]

10*E^6*(I*Pi + Log[2])^(2/(E^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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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]

Rule 2262

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

Rubi steps \begin{align*} \text {integral}& = (20 \log (i \pi +\log (2))) \int \frac {\exp \left (\frac {2 \left (3 e^2-3 x+\log (i \pi +\log (2))\right )}{e^2-x}\right )}{e^4-2 e^2 x+x^2} \, dx \\ & = (20 \log (i \pi +\log (2))) \int \frac {\exp \left (\frac {2 \left (3 e^2-3 x+\log (i \pi +\log (2))\right )}{e^2-x}\right )}{\left (-e^2+x\right )^2} \, dx \\ & = (20 \log (i \pi +\log (2))) \int \frac {e^{6-\frac {2 \log (i \pi +\log (2))}{-e^2+x}}}{\left (-e^2+x\right )^2} \, dx \\ & = 10 e^6 (i \pi +\log (2))^{\frac {2}{e^2-x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {20 e^{\frac {2 \left (3 e^2-3 x+\log (i \pi +\log (2))\right )}{e^2-x}} \log (i \pi +\log (2))}{e^4-2 e^2 x+x^2} \, dx=10 e^6 (i \pi +\log (2))^{\frac {2}{e^2-x}} \]

[In]

Integrate[(20*E^((2*(3*E^2 - 3*x + Log[I*Pi + Log[2]]))/(E^2 - x))*Log[I*Pi + Log[2]])/(E^4 - 2*E^2*x + x^2),x
]

[Out]

10*E^6*(I*Pi + Log[2])^(2/(E^2 - x))

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82

method	result	size
risch	\(10 \left (\ln \left (2\right )+i \pi \right )^{\frac {2}{{\mathrm e}^{2}-x}} {\mathrm e}^{6}\)	\(23\)
derivativedivides	\(10 \,{\mathrm e}^{6-\frac {2 \ln \left (\ln \left (2\right )+i \pi \right )}{x -{\mathrm e}^{2}}}\)	\(26\)
default	\(10 \,{\mathrm e}^{6-\frac {2 \ln \left (\ln \left (2\right )+i \pi \right )}{x -{\mathrm e}^{2}}}\)	\(26\)
norman	\(\frac {-10 x \,{\mathrm e}^{\frac {2 \ln \left (\ln \left (2\right )+i \pi \right )+6 \,{\mathrm e}^{2}-6 x}{{\mathrm e}^{2}-x}}+10 \,{\mathrm e}^{2} {\mathrm e}^{\frac {2 \ln \left (\ln \left (2\right )+i \pi \right )+6 \,{\mathrm e}^{2}-6 x}{{\mathrm e}^{2}-x}}}{{\mathrm e}^{2}-x}\)	\(74\)

[In]

int(20*ln(ln(2)+I*Pi)*exp((ln(ln(2)+I*Pi)+3*exp(2)-3*x)/(exp(2)-x))^2/(exp(2)^2-2*exp(2)*x+x^2),x,method=_RETU
RNVERBOSE)

[Out]

10*((ln(2)+I*Pi)^(1/(exp(2)-x)))^2*exp(6)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 3.18 \[ \int \frac {20 e^{\frac {2 \left (3 e^2-3 x+\log (i \pi +\log (2))\right )}{e^2-x}} \log (i \pi +\log (2))}{e^4-2 e^2 x+x^2} \, dx=10 \, {\left (\cosh \left (-\frac {3 \, x}{x - e^{2}} + \frac {3 \, e^{2}}{x - e^{2}} + \frac {\log \left (i \, \pi + \log \left (2\right )\right )}{x - e^{2}}\right ) - \sinh \left (-\frac {3 \, x}{x - e^{2}} + \frac {3 \, e^{2}}{x - e^{2}} + \frac {\log \left (i \, \pi + \log \left (2\right )\right )}{x - e^{2}}\right )\right )}^{2} \]

[In]

integrate(20*log(log(2)+I*pi)*exp((log(log(2)+I*pi)+3*exp(2)-3*x)/(exp(2)-x))^2/(exp(2)^2-2*exp(2)*x+x^2),x, a
lgorithm="fricas")

[Out]

10*(cosh(-3*x/(x - e^2) + 3*e^2/(x - e^2) + log(I*pi + log(2))/(x - e^2)) - sinh(-3*x/(x - e^2) + 3*e^2/(x - e
^2) + log(I*pi + log(2))/(x - e^2)))^2

Sympy [A] (verification not implemented)

Time = 100.71 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {20 e^{\frac {2 \left (3 e^2-3 x+\log (i \pi +\log (2))\right )}{e^2-x}} \log (i \pi +\log (2))}{e^4-2 e^2 x+x^2} \, dx=10 e^{\frac {6 x}{x - e^{2}}} e^{- \frac {6 e^{2}}{x - e^{2}}} e^{- \frac {2 \log {\left (\log {\left (2 \right )} + i \pi \right )}}{x - e^{2}}} \]

[In]

integrate(20*ln(ln(2)+I*pi)*exp((ln(ln(2)+I*pi)+3*exp(2)-3*x)/(exp(2)-x))**2/(exp(2)**2-2*exp(2)*x+x**2),x)

[Out]

10*exp(6*x/(x - exp(2)))*exp(-6*exp(2)/(x - exp(2)))*exp(-2*log(log(2) + I*pi)/(x - exp(2)))

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.61 \[ \int \frac {20 e^{\frac {2 \left (3 e^2-3 x+\log (i \pi +\log (2))\right )}{e^2-x}} \log (i \pi +\log (2))}{e^4-2 e^2 x+x^2} \, dx=10 \, \cosh \left (\frac {2 \, \log \left (i \, \pi + \log \left (2\right )\right )}{x - e^{2}} - 6\right ) - 10 \, \sinh \left (\frac {2 \, \log \left (i \, \pi + \log \left (2\right )\right )}{x - e^{2}} - 6\right ) \]

[In]

integrate(20*log(log(2)+I*pi)*exp((log(log(2)+I*pi)+3*exp(2)-3*x)/(exp(2)-x))^2/(exp(2)^2-2*exp(2)*x+x^2),x, a
lgorithm="maxima")

[Out]

10*cosh(2*log(I*pi + log(2))/(x - e^2) - 6) - 10*sinh(2*log(I*pi + log(2))/(x - e^2) - 6)

Giac [F]

\[ \int \frac {20 e^{\frac {2 \left (3 e^2-3 x+\log (i \pi +\log (2))\right )}{e^2-x}} \log (i \pi +\log (2))}{e^4-2 e^2 x+x^2} \, dx=\int { \frac {20 \, {\left (\cosh \left (-\frac {3 \, x}{x - e^{2}} + \frac {3 \, e^{2}}{x - e^{2}} + \frac {\log \left (i \, \pi + \log \left (2\right )\right )}{x - e^{2}}\right ) - \sinh \left (-\frac {3 \, x}{x - e^{2}} + \frac {3 \, e^{2}}{x - e^{2}} + \frac {\log \left (i \, \pi + \log \left (2\right )\right )}{x - e^{2}}\right )\right )}^{2} \log \left (i \, \pi + \log \left (2\right )\right )}{x^{2} - 2 \, x e^{2} + e^{4}} \,d x } \]

[In]

integrate(20*log(log(2)+I*pi)*exp((log(log(2)+I*pi)+3*exp(2)-3*x)/(exp(2)-x))^2/(exp(2)^2-2*exp(2)*x+x^2),x, a
lgorithm="giac")

[Out]

integrate(20*(cosh(-3*x/(x - e^2) + 3*e^2/(x - e^2) + log(I*pi + log(2))/(x - e^2)) - sinh(-3*x/(x - e^2) + 3*
e^2/(x - e^2) + log(I*pi + log(2))/(x - e^2)))^2*log(I*pi + log(2))/(x^2 - 2*x*e^2 + e^4), x)

Mupad [B] (verification not implemented)

Time = 1.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {20 e^{\frac {2 \left (3 e^2-3 x+\log (i \pi +\log (2))\right )}{e^2-x}} \log (i \pi +\log (2))}{e^4-2 e^2 x+x^2} \, dx=\frac {10\,{\mathrm {e}}^6}{{\left (\ln \left (2\right )+\Pi \,1{}\mathrm {i}\right )}^{\frac {2}{x-{\mathrm {e}}^2}}} \]

[In]

int((20*exp(-(2*(3*exp(2) - 3*x + log(Pi*1i + log(2))))/(x - exp(2)))*log(Pi*1i + log(2)))/(exp(4) - 2*x*exp(2
) + x^2),x)

[Out]

(10*exp(6))/(Pi*1i + log(2))^(2/(x - exp(2)))