3.6.71 \(\int \frac {-11 e x+e^{1+\frac {2+x \log (\frac {e^x}{x})}{x}} (2+2 x-x^2)}{5 x^3} \, dx\)

Optimal. Leaf size=24 \[ \frac {e \left (11-\frac {e^{\frac {2}{x}+x}}{x}\right )}{5 x} \]

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 42, normalized size of antiderivative = 1.75, number of steps used = 4, number of rules used = 3, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {12, 14, 2288} \begin {gather*} \frac {e^{x+\frac {2}{x}+1} \left (2-x^2\right )}{5 \left (1-\frac {2}{x^2}\right ) x^4}+\frac {11 e}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-11*E*x + E^(1 + (2 + x*Log[E^x/x])/x)*(2 + 2*x - x^2))/(5*x^3),x]

[Out]

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

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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {-11 e x+e^{1+\frac {2+x \log \left (\frac {e^x}{x}\right )}{x}} \left (2+2 x-x^2\right )}{x^3} \, dx\\ &=\frac {1}{5} \int \left (-\frac {11 e}{x^2}-\frac {e^{1+\frac {2}{x}+x} \left (-2-2 x+x^2\right )}{x^4}\right ) \, dx\\ &=\frac {11 e}{5 x}-\frac {1}{5} \int \frac {e^{1+\frac {2}{x}+x} \left (-2-2 x+x^2\right )}{x^4} \, dx\\ &=\frac {11 e}{5 x}+\frac {e^{1+\frac {2}{x}+x} \left (2-x^2\right )}{5 \left (1-\frac {2}{x^2}\right ) x^4}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 21, normalized size = 0.88 \begin {gather*} -\frac {e \left (e^{\frac {2}{x}+x}-11 x\right )}{5 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-11*E*x + E^(1 + (2 + x*Log[E^x/x])/x)*(2 + 2*x - x^2))/(5*x^3),x]

[Out]

-1/5*(E*(E^(2/x + x) - 11*x))/x^2

________________________________________________________________________________________

fricas [A]  time = 0.60, size = 29, normalized size = 1.21 \begin {gather*} \frac {11 \, e - e^{\left (\frac {x \log \left (\frac {e^{x}}{x}\right ) + x + 2}{x}\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-x^2+2*x+2)*exp(1)*exp((x*log(exp(x)/x)+2)/x)-11*x*exp(1))/x^3,x, algorithm="fricas")

[Out]

1/5*(11*e - e^((x*log(e^x/x) + x + 2)/x))/x

________________________________________________________________________________________

giac [A]  time = 0.67, size = 24, normalized size = 1.00 \begin {gather*} \frac {11 \, x e - e^{\left (\frac {x^{2} + x + 2}{x}\right )}}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-x^2+2*x+2)*exp(1)*exp((x*log(exp(x)/x)+2)/x)-11*x*exp(1))/x^3,x, algorithm="giac")

[Out]

1/5*(11*x*e - e^((x^2 + x + 2)/x))/x^2

________________________________________________________________________________________

maple [A]  time = 0.24, size = 32, normalized size = 1.33

method	result	size
default	\(\frac {11 \,{\mathrm e}}{5 x}-\frac {{\mathrm e} \,{\mathrm e}^{\frac {x \ln \left (\frac {{\mathrm e}^{x}}{x}\right )+2}{x}}}{5 x}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((-x^2+2*x+2)*exp(1)*exp((x*ln(exp(x)/x)+2)/x)-11*x*exp(1))/x^3,x,method=_RETURNVERBOSE)

[Out]

11/5*exp(1)/x-1/5*exp(1)/x*exp((x*ln(exp(x)/x)+2)/x)

________________________________________________________________________________________

maxima [A]  time = 0.76, size = 22, normalized size = 0.92 \begin {gather*} \frac {11 \, e}{5 \, x} - \frac {e^{\left (x + \frac {2}{x} + 1\right )}}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-x^2+2*x+2)*exp(1)*exp((x*log(exp(x)/x)+2)/x)-11*x*exp(1))/x^3,x, algorithm="maxima")

[Out]

11/5*e/x - 1/5*e^(x + 2/x + 1)/x^2

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {\frac {11\,x\,\mathrm {e}}{5}-\frac {{\mathrm {e}}^{\frac {x\,\ln \left (\frac {{\mathrm {e}}^x}{x}\right )+2}{x}}\,\mathrm {e}\,\left (-x^2+2\,x+2\right )}{5}}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((11*x*exp(1))/5 - (exp((x*log(exp(x)/x) + 2)/x)*exp(1)*(2*x - x^2 + 2))/5)/x^3,x)

[Out]

int(-((11*x*exp(1))/5 - (exp((x*log(exp(x)/x) + 2)/x)*exp(1)*(2*x - x^2 + 2))/5)/x^3, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {e \left (\int \frac {11}{x^{2}}\, dx + \int \left (- \frac {2 e^{\frac {2}{x}} e^{x}}{x^{4}}\right )\, dx + \int \left (- \frac {2 e^{\frac {2}{x}} e^{x}}{x^{3}}\right )\, dx + \int \frac {e^{\frac {2}{x}} e^{x}}{x^{2}}\, dx\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-x**2+2*x+2)*exp(1)*exp((x*ln(exp(x)/x)+2)/x)-11*x*exp(1))/x**3,x)

[Out]

-E*(Integral(11/x**2, x) + Integral(-2*exp(2/x)*exp(x)/x**4, x) + Integral(-2*exp(2/x)*exp(x)/x**3, x) + Integ
ral(exp(2/x)*exp(x)/x**2, x))/5

________________________________________________________________________________________