3.27.88 \(\int x^{-3+\frac {1-x}{x^2}} (e^{5 e^4+x-e^4 x \log (2)} (1-x+x^3-e^4 x^3 \log (2))+e^{5 e^4+x-e^4 x \log (2)} (-2+x) \log (x)) \, dx\)

Optimal. Leaf size=27 \[ e^{x-e^4 (-5+x \log (2))} x^{\frac {1-x}{x^2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.33, antiderivative size = 31, normalized size of antiderivative = 1.15, number of steps used = 3, number of rules used = 3, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {6688, 2287, 2288} \begin {gather*} x^{\frac {1}{x^2}-\frac {1}{x}} e^{x \left (1-e^4 \log (2)\right )+5 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-3 + (1 - x)/x^2)*(E^(5*E^4 + x - E^4*x*Log[2])*(1 - x + x^3 - E^4*x^3*Log[2]) + E^(5*E^4 + x - E^4*x*L
og[2])*(-2 + x)*Log[x]),x]

[Out]

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

Rule 2287

Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x],
x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]

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]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 27, normalized size = 1.00 \begin {gather*} e^{x+e^4 (5-x \log (2))} x^{\frac {1-x}{x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.54, size = 26, normalized size = 0.96 \begin {gather*} \frac {e^{\left (-x e^{4} \log \relax (2) + x + 5 \, e^{4}\right )}}{x^{\frac {x - 1}{x^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-2)*exp(-x*exp(4)*log(2)+5*exp(4)+x)*log(x)+(-x^3*exp(4)*log(2)+x^3-x+1)*exp(-x*exp(4)*log(2)+5*e
xp(4)+x))*exp((-x+1)*log(x)/x^2)/x^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 1.43, size = 35, normalized size = 1.30 \begin {gather*} e^{\left (-\frac {x^{3} e^{4} \log \relax (2) - x^{3} - 5 \, x^{2} e^{4} + x \log \relax (x) - \log \relax (x)}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-2)*exp(-x*exp(4)*log(2)+5*exp(4)+x)*log(x)+(-x^3*exp(4)*log(2)+x^3-x+1)*exp(-x*exp(4)*log(2)+5*e
xp(4)+x))*exp((-x+1)*log(x)/x^2)/x^3,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.03, size = 25, normalized size = 0.93

method	result	size
risch	\(\left (\frac {1}{2}\right )^{x \,{\mathrm e}^{4}} {\mathrm e}^{5 \,{\mathrm e}^{4}+x} x^{-\frac {x -1}{x^{2}}}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.02, size = 27, normalized size = 1.00 \begin {gather*} e^{\left (-x e^{4} \log \relax (2) + x - \frac {\log \relax (x)}{x} + \frac {\log \relax (x)}{x^{2}} + 5 \, e^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-2)*exp(-x*exp(4)*log(2)+5*exp(4)+x)*log(x)+(-x^3*exp(4)*log(2)+x^3-x+1)*exp(-x*exp(4)*log(2)+5*e
xp(4)+x))*exp((-x+1)*log(x)/x^2)/x^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.94, size = 25, normalized size = 0.93 \begin {gather*} {\left (\frac {1}{2}\right )}^{x\,{\mathrm {e}}^4}\,x^{\frac {1}{x^2}-\frac {1}{x}}\,{\mathrm {e}}^{5\,{\mathrm {e}}^4}\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(log(x)*(x - 1))/x^2)*(exp(x + 5*exp(4) - x*exp(4)*log(2))*(x - x^3 + x^3*exp(4)*log(2) - 1) - exp(
x + 5*exp(4) - x*exp(4)*log(2))*log(x)*(x - 2)))/x^3,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 129.38, size = 27, normalized size = 1.00 \begin {gather*} e^{\frac {\left (1 - x\right ) \log {\relax (x )}}{x^{2}}} e^{- x e^{4} \log {\relax (2 )} + x + 5 e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp((1 - x)*log(x)/x**2)*exp(-x*exp(4)*log(2) + x + 5*exp(4))

________________________________________________________________________________________