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

Optimal. Leaf size=27 \[ 4+\frac {1}{2} e^{3-\frac {3 e^4}{x}}+e^{-3+x} x+\log (x) \]

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Rubi [A]  time = 0.25, antiderivative size = 35, normalized size of antiderivative = 1.30, number of steps used = 6, number of rules used = 5, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 6688, 2209, 2176, 2194} \begin {gather*} e^{x-3} (x+1)+\frac {1}{2} e^{3-\frac {3 e^4}{x}}-e^{x-3}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[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 2176

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] &&  !$UseGamma === True

Rule 2194

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 2209

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 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} &=\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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 26, normalized size = 0.96 \begin {gather*} \frac {1}{2} e^{3-\frac {3 e^4}{x}}+e^{-3+x} x+\log (x) \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.73, size = 39, normalized size = 1.44 \begin {gather*} \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 \relax (x) + e^{3}\right )} e^{\left (-\frac {3 \, e^{4}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [A]  time = 0.20, size = 26, normalized size = 0.96 \begin {gather*} {\left (x e^{x} + e^{3} \log \relax (x)\right )} e^{\left (-3\right )} + \frac {1}{2} \, e^{\left (-\frac {3 \, e^{4}}{x} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.09, size = 24, normalized size = 0.89

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.38, size = 27, normalized size = 1.00 \begin {gather*} {\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 \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 4.39, size = 21, normalized size = 0.78 \begin {gather*} \ln \relax (x)+\frac {{\mathrm {e}}^{-\frac {3\,{\mathrm {e}}^4}{x}}\,{\mathrm {e}}^3}{2}+x\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.36, size = 22, normalized size = 0.81 \begin {gather*} x e^{x - 3} + \log {\relax (x )} + \frac {e^{3} e^{- \frac {3 e^{4}}{x}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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