3.100.69 \(\int \frac {e^x (2+6 x-x^2)+2 e^x x \log (x)}{2 e^3 x} \, dx\)

Optimal. Leaf size=18 \[ e^{-3+x} \left (\frac {7-x}{2}+\log (x)\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 28, normalized size of antiderivative = 1.56, number of steps used = 9, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 14, 2194, 2178, 2176, 2554} \begin {gather*} -\frac {1}{2} e^{x-3} x+\frac {7 e^{x-3}}{2}+e^{x-3} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(2 + 6*x - x^2) + 2*E^x*x*Log[x])/(2*E^3*x),x]

[Out]

(7*E^(-3 + x))/2 - (E^(-3 + x)*x)/2 + E^(-3 + 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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$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 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^x \left (2+6 x-x^2\right )+2 e^x x \log (x)}{x} \, dx}{2 e^3}\\ &=\frac {\int \left (6 e^x+\frac {2 e^x}{x}-e^x x+2 e^x \log (x)\right ) \, dx}{2 e^3}\\ &=-\frac {\int e^x x \, dx}{2 e^3}+\frac {\int \frac {e^x}{x} \, dx}{e^3}+\frac {\int e^x \log (x) \, dx}{e^3}+\frac {3 \int e^x \, dx}{e^3}\\ &=3 e^{-3+x}-\frac {1}{2} e^{-3+x} x+\frac {\text {Ei}(x)}{e^3}+e^{-3+x} \log (x)+\frac {\int e^x \, dx}{2 e^3}-\frac {\int \frac {e^x}{x} \, dx}{e^3}\\ &=\frac {7 e^{-3+x}}{2}-\frac {1}{2} e^{-3+x} x+e^{-3+x} \log (x)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 18, normalized size = 1.00 \begin {gather*} \frac {1}{2} e^{-3+x} (7-x+2 \log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(2 + 6*x - x^2) + 2*E^x*x*Log[x])/(2*E^3*x),x]

[Out]

(E^(-3 + x)*(7 - x + 2*Log[x]))/2

________________________________________________________________________________________

fricas [A]  time = 0.60, size = 17, normalized size = 0.94 \begin {gather*} -\frac {1}{2} \, {\left ({\left (x - 7\right )} e^{x} - 2 \, e^{x} \log \relax (x)\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*x*exp(x)*log(x)+(-x^2+6*x+2)*exp(x))/x/exp(3),x, algorithm="fricas")

[Out]

-1/2*((x - 7)*e^x - 2*e^x*log(x))*e^(-3)

________________________________________________________________________________________

giac [A]  time = 0.14, size = 19, normalized size = 1.06 \begin {gather*} -\frac {1}{2} \, {\left (x e^{x} - 2 \, e^{x} \log \relax (x) - 7 \, e^{x}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*x*exp(x)*log(x)+(-x^2+6*x+2)*exp(x))/x/exp(3),x, algorithm="giac")

[Out]

-1/2*(x*e^x - 2*e^x*log(x) - 7*e^x)*e^(-3)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 18, normalized size = 1.00

method	result	size
risch	\(\ln \relax (x ) {\mathrm e}^{x -3}-\frac {\left (x -7\right ) {\mathrm e}^{x -3}}{2}\)	\(18\)
norman	\({\mathrm e}^{x} {\mathrm e}^{-3} \ln \relax (x )+\frac {7 \,{\mathrm e}^{x} {\mathrm e}^{-3}}{2}-\frac {{\mathrm e}^{-3} {\mathrm e}^{x} x}{2}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(2*x*exp(x)*ln(x)+(-x^2+6*x+2)*exp(x))/x/exp(3),x,method=_RETURNVERBOSE)

[Out]

ln(x)*exp(x-3)-1/2*(x-7)*exp(x-3)

________________________________________________________________________________________

maxima [A]  time = 0.37, size = 21, normalized size = 1.17 \begin {gather*} -\frac {1}{2} \, {\left ({\left (x - 1\right )} e^{x} - 2 \, e^{x} \log \relax (x) - 6 \, e^{x}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*x*exp(x)*log(x)+(-x^2+6*x+2)*exp(x))/x/exp(3),x, algorithm="maxima")

[Out]

-1/2*((x - 1)*e^x - 2*e^x*log(x) - 6*e^x)*e^(-3)

________________________________________________________________________________________

mupad [B]  time = 8.51, size = 12, normalized size = 0.67 \begin {gather*} {\mathrm {e}}^{x-3}\,\left (\ln \relax (x)-\frac {x}{2}+\frac {7}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-3)*((exp(x)*(6*x - x^2 + 2))/2 + x*exp(x)*log(x)))/x,x)

[Out]

exp(x - 3)*(log(x) - x/2 + 7/2)

________________________________________________________________________________________

sympy [A]  time = 0.32, size = 15, normalized size = 0.83 \begin {gather*} \frac {\left (- x + 2 \log {\relax (x )} + 7\right ) e^{x}}{2 e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*x*exp(x)*ln(x)+(-x**2+6*x+2)*exp(x))/x/exp(3),x)

[Out]

(-x + 2*log(x) + 7)*exp(-3)*exp(x)/2

________________________________________________________________________________________