3.89.88 \(\int \frac {1}{3} (3+e^{-e^{3/x} x+x^2} (1+e^{3/x} (3-x)+2 x^2)-\log (x)) \, dx\)

Optimal. Leaf size=34 \[ x+\frac {1}{3} \left (-x+\left (2+e^{x \left (-e^{3/x}+x\right )}\right ) x-x \log (x)\right ) \]

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Rubi [B]  time = 0.10, antiderivative size = 77, normalized size of antiderivative = 2.26, number of steps used = 4, number of rules used = 3, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {12, 2288, 2295} \begin {gather*} -\frac {e^{x^2-e^{3/x} x} \left (2 x^2+e^{3/x} (3-x)\right )}{3 \left (-2 x+e^{3/x}-\frac {3 e^{3/x}}{x}\right )}+\frac {4 x}{3}-\frac {1}{3} x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 26, normalized size = 0.76 \begin {gather*} \frac {1}{3} x \left (4+e^{x \left (-e^{3/x}+x\right )}-\log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 26, normalized size = 0.76 \begin {gather*} \frac {1}{3} \, x e^{\left (x^{2} - x e^{\frac {3}{x}}\right )} - \frac {1}{3} \, x \log \relax (x) + \frac {4}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.25, size = 39, normalized size = 1.15 \begin {gather*} \frac {1}{3} \, x e^{\left (\frac {x^{3} - x^{2} e^{\frac {3}{x}} + 3}{x} - \frac {3}{x}\right )} - \frac {1}{3} \, x \log \relax (x) + \frac {4}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 26, normalized size = 0.76

method	result	size
risch	\(\frac {4 x}{3}+\frac {x \,{\mathrm e}^{\left (x -{\mathrm e}^{\frac {3}{x}}\right ) x}}{3}-\frac {x \ln \relax (x )}{3}\)	\(26\)
default	\(\frac {4 x}{3}+\frac {x \,{\mathrm e}^{-x \,{\mathrm e}^{\frac {3}{x}}+x^{2}}}{3}-\frac {x \ln \relax (x )}{3}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 26, normalized size = 0.76 \begin {gather*} \frac {1}{3} \, x e^{\left (x^{2} - x e^{\frac {3}{x}}\right )} - \frac {1}{3} \, x \log \relax (x) + \frac {4}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.61, size = 23, normalized size = 0.68 \begin {gather*} \frac {x\,\left ({\mathrm {e}}^{x^2-x\,{\mathrm {e}}^{3/x}}-\ln \relax (x)+4\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.95, size = 26, normalized size = 0.76 \begin {gather*} \frac {x e^{x^{2} - x e^{\frac {3}{x}}}}{3} - \frac {x \log {\relax (x )}}{3} + \frac {4 x}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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