\(\int \frac {e^{-\frac {3 e^4}{x}} (3 e^7+e^{\frac {3 e^4}{x}} (2 x+e^{-3+x} (2 x^2+2 x^3)))}{2 x^2} \, dx\) [6426]

Optimal result

Integrand size = 55, antiderivative size = 27 \[ \int \frac {e^{-\frac {3 e^4}{x}} \left (3 e^7+e^{\frac {3 e^4}{x}} \left (2 x+e^{-3+x} \left (2 x^2+2 x^3\right )\right )\right )}{2 x^2} \, dx=4+\frac {1}{2} e^{3-\frac {3 e^4}{x}}+e^{-3+x} x+\log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 6820, 2240, 2207, 2225} \[ \int \frac {e^{-\frac {3 e^4}{x}} \left (3 e^7+e^{\frac {3 e^4}{x}} \left (2 x+e^{-3+x} \left (2 x^2+2 x^3\right )\right )\right )}{2 x^2} \, dx=e^{x-3} (x+1)+\frac {1}{2} e^{3-\frac {3 e^4}{x}}-e^{x-3}+\log (x) \]

[In]

Int[(3*E^7 + E^((3*E^4)/x)*(2*x + E^(-3 + x)*(2*x^2 + 2*x^3)))/(2*E^((3*E^4)/x)*x^2),x]

[Out]

E^(3 - (3*E^4)/x)/2 - E^(-3 + x) + E^(-3 + x)*(1 + x) + Log[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 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, 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]

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {e^{-\frac {3 e^4}{x}} \left (3 e^7+e^{\frac {3 e^4}{x}} \left (2 x+e^{-3+x} \left (2 x^2+2 x^3\right )\right )\right )}{x^2} \, dx \\ & = \frac {1}{2} \int \left (\frac {3 e^{7-\frac {3 e^4}{x}}}{x^2}+\frac {2}{x}+2 e^{-3+x} (1+x)\right ) \, dx \\ & = \log (x)+\frac {3}{2} \int \frac {e^{7-\frac {3 e^4}{x}}}{x^2} \, dx+\int e^{-3+x} (1+x) \, dx \\ & = \frac {1}{2} e^{3-\frac {3 e^4}{x}}+e^{-3+x} (1+x)+\log (x)-\int e^{-3+x} \, dx \\ & = \frac {1}{2} e^{3-\frac {3 e^4}{x}}-e^{-3+x}+e^{-3+x} (1+x)+\log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-\frac {3 e^4}{x}} \left (3 e^7+e^{\frac {3 e^4}{x}} \left (2 x+e^{-3+x} \left (2 x^2+2 x^3\right )\right )\right )}{2 x^2} \, dx=\frac {1}{2} e^{3-\frac {3 e^4}{x}}+e^{-3+x} x+\log (x) \]

[In]

Integrate[(3*E^7 + E^((3*E^4)/x)*(2*x + E^(-3 + x)*(2*x^2 + 2*x^3)))/(2*E^((3*E^4)/x)*x^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89

method	result	size
risch	\(\ln \left (x \right )+x \,{\mathrm e}^{-3+x}+\frac {{\mathrm e}^{-\frac {3 \left ({\mathrm e}^{4}-x \right )}{x}}}{2}\)	\(24\)
parts	\(\ln \left (x \right )+{\mathrm e}^{-3+x} \left (-3+x \right )+3 \,{\mathrm e}^{-3+x}+\frac {{\mathrm e}^{3} {\mathrm e}^{-\frac {3 \,{\mathrm e}^{4}}{x}}}{2}\)	\(32\)
default	\(\ln \left (x \right )+{\mathrm e}^{-3} {\mathrm e}^{x}+{\mathrm e}^{-3} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )+\frac {{\mathrm e}^{3} {\mathrm e}^{-\frac {3 \,{\mathrm e}^{4}}{x}}}{2}\)	\(34\)
norman	\(\frac {\left ({\mathrm e}^{-3+x} {\mathrm e}^{\frac {3 \,{\mathrm e}^{4}}{x}} x^{2}+\frac {x \,{\mathrm e}^{3}}{2}\right ) {\mathrm e}^{-\frac {3 \,{\mathrm e}^{4}}{x}}}{x}+\ln \left (x \right )\)	\(40\)
parallelrisch	\(\frac {\left (2 \,{\mathrm e}^{-3+x} {\mathrm e}^{\frac {3 \,{\mathrm e}^{4}}{x}} x^{2}+2 \ln \left (x \right ) {\mathrm e}^{\frac {3 \,{\mathrm e}^{4}}{x}} x +x \,{\mathrm e}^{3}\right ) {\mathrm e}^{-\frac {3 \,{\mathrm e}^{4}}{x}}}{2 x}\)	\(51\)

[In]

int(1/2*(((2*x^3+2*x^2)*exp(-3+x)+2*x)*exp(3*exp(4)/x)+3*exp(3)*exp(4))/x^2/exp(3*exp(4)/x),x,method=_RETURNVE
RBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {e^{-\frac {3 e^4}{x}} \left (3 e^7+e^{\frac {3 e^4}{x}} \left (2 x+e^{-3+x} \left (2 x^2+2 x^3\right )\right )\right )}{2 x^2} \, dx=\frac {1}{2} \, {\left (2 \, x e^{\left (x + \frac {3 \, e^{4}}{x} - 3\right )} + 2 \, e^{\left (\frac {3 \, e^{4}}{x}\right )} \log \left (x\right ) + e^{3}\right )} e^{\left (-\frac {3 \, e^{4}}{x}\right )} \]

[In]

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

[Out]

1/2*(2*x*e^(x + 3*e^4/x - 3) + 2*e^(3*e^4/x)*log(x) + e^3)*e^(-3*e^4/x)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-\frac {3 e^4}{x}} \left (3 e^7+e^{\frac {3 e^4}{x}} \left (2 x+e^{-3+x} \left (2 x^2+2 x^3\right )\right )\right )}{2 x^2} \, dx=x e^{x - 3} + \log {\left (x \right )} + \frac {e^{3} e^{- \frac {3 e^{4}}{x}}}{2} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {3 e^4}{x}} \left (3 e^7+e^{\frac {3 e^4}{x}} \left (2 x+e^{-3+x} \left (2 x^2+2 x^3\right )\right )\right )}{2 x^2} \, dx={\left (x - 1\right )} e^{\left (x - 3\right )} + e^{\left (x - 3\right )} + \frac {1}{2} \, e^{\left (-\frac {3 \, e^{4}}{x} + 3\right )} + \log \left (x\right ) \]

[In]

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

[Out]

(x - 1)*e^(x - 3) + e^(x - 3) + 1/2*e^(-3*e^4/x + 3) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-\frac {3 e^4}{x}} \left (3 e^7+e^{\frac {3 e^4}{x}} \left (2 x+e^{-3+x} \left (2 x^2+2 x^3\right )\right )\right )}{2 x^2} \, dx={\left (x e^{x} + e^{3} \log \left (x\right )\right )} e^{\left (-3\right )} + \frac {1}{2} \, e^{\left (-\frac {3 \, e^{4}}{x} + 3\right )} \]

[In]

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

[Out]

(x*e^x + e^3*log(x))*e^(-3) + 1/2*e^(-3*e^4/x + 3)

Mupad [B] (verification not implemented)

Time = 13.81 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {e^{-\frac {3 e^4}{x}} \left (3 e^7+e^{\frac {3 e^4}{x}} \left (2 x+e^{-3+x} \left (2 x^2+2 x^3\right )\right )\right )}{2 x^2} \, dx=\ln \left (x\right )+\frac {{\mathrm {e}}^{-\frac {3\,{\mathrm {e}}^4}{x}}\,{\mathrm {e}}^3}{2}+x\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^x \]

[In]

int((exp(-(3*exp(4))/x)*((3*exp(7))/2 + (exp((3*exp(4))/x)*(2*x + exp(x - 3)*(2*x^2 + 2*x^3)))/2))/x^2,x)

[Out]

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