\(\int \frac {2+4 e^e+e^x (1-x)}{5 x^2} \, dx\) [2213]

Optimal result

Integrand size = 23, antiderivative size = 25 \[ \int \frac {2+4 e^e+e^x (1-x)}{5 x^2} \, dx=\frac {1}{5} \left (4+\frac {-2-4 e^e-e^x-x}{x}\right ) \]

[Out]

1/5*(-2-4*exp(exp(1))-x-exp(x))/x+4/5

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {12, 14, 2228} \[ \int \frac {2+4 e^e+e^x (1-x)}{5 x^2} \, dx=-\frac {e^x}{5 x}-\frac {2 \left (1+2 e^e\right )}{5 x} \]

[In]

Int[(2 + 4*E^E + E^x*(1 - x))/(5*x^2),x]

[Out]

-1/5*E^x/x - (2*(1 + 2*E^E))/(5*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 2228

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[g*u^(m + 1)*(F^(c*v)/(b*c*
e*Log[F])), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {2+4 e^e+e^x (1-x)}{x^2} \, dx \\ & = \frac {1}{5} \int \left (\frac {2 \left (1+2 e^e\right )}{x^2}-\frac {e^x (-1+x)}{x^2}\right ) \, dx \\ & = -\frac {2 \left (1+2 e^e\right )}{5 x}-\frac {1}{5} \int \frac {e^x (-1+x)}{x^2} \, dx \\ & = -\frac {e^x}{5 x}-\frac {2 \left (1+2 e^e\right )}{5 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {2+4 e^e+e^x (1-x)}{5 x^2} \, dx=-\frac {2+4 e^e+e^x}{5 x} \]

[In]

Integrate[(2 + 4*E^E + E^x*(1 - x))/(5*x^2),x]

[Out]

-1/5*(2 + 4*E^E + E^x)/x

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.60

method	result	size
parallelrisch	\(-\frac {4 \,{\mathrm e}^{{\mathrm e}}+2+{\mathrm e}^{x}}{5 x}\)	\(15\)
norman	\(\frac {-\frac {{\mathrm e}^{x}}{5}-\frac {4 \,{\mathrm e}^{{\mathrm e}}}{5}-\frac {2}{5}}{x}\)	\(16\)
parts	\(-\frac {\frac {4 \,{\mathrm e}^{{\mathrm e}}}{5}+\frac {2}{5}}{x}-\frac {{\mathrm e}^{x}}{5 x}\)	\(21\)
default	\(-\frac {{\mathrm e}^{x}}{5 x}-\frac {2}{5 x}-\frac {4 \,{\mathrm e}^{{\mathrm e}}}{5 x}\)	\(22\)
risch	\(-\frac {{\mathrm e}^{x}}{5 x}-\frac {2}{5 x}-\frac {4 \,{\mathrm e}^{{\mathrm e}}}{5 x}\)	\(22\)

[In]

int(1/5*(4*exp(exp(1))+(1-x)*exp(x)+2)/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/5*(4*exp(exp(1))+2+exp(x))/x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.56 \[ \int \frac {2+4 e^e+e^x (1-x)}{5 x^2} \, dx=-\frac {e^{x} + 4 \, e^{e} + 2}{5 \, x} \]

[In]

integrate(1/5*(4*exp(exp(1))+(1-x)*exp(x)+2)/x^2,x, algorithm="fricas")

[Out]

-1/5*(e^x + 4*e^e + 2)/x

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {2+4 e^e+e^x (1-x)}{5 x^2} \, dx=- \frac {e^{x}}{5 x} - \frac {\frac {2}{5} + \frac {4 e^{e}}{5}}{x} \]

[In]

integrate(1/5*(4*exp(exp(1))+(1-x)*exp(x)+2)/x**2,x)

[Out]

-exp(x)/(5*x) - (2/5 + 4*exp(E)/5)/x

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {2+4 e^e+e^x (1-x)}{5 x^2} \, dx=-\frac {4 \, e^{e}}{5 \, x} - \frac {2}{5 \, x} - \frac {1}{5} \, {\rm Ei}\left (x\right ) + \frac {1}{5} \, \Gamma \left (-1, -x\right ) \]

[In]

integrate(1/5*(4*exp(exp(1))+(1-x)*exp(x)+2)/x^2,x, algorithm="maxima")

[Out]

-4/5*e^e/x - 2/5/x - 1/5*Ei(x) + 1/5*gamma(-1, -x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.56 \[ \int \frac {2+4 e^e+e^x (1-x)}{5 x^2} \, dx=-\frac {e^{x} + 4 \, e^{e} + 2}{5 \, x} \]

[In]

integrate(1/5*(4*exp(exp(1))+(1-x)*exp(x)+2)/x^2,x, algorithm="giac")

[Out]

-1/5*(e^x + 4*e^e + 2)/x

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.56 \[ \int \frac {2+4 e^e+e^x (1-x)}{5 x^2} \, dx=-\frac {4\,{\mathrm {e}}^{\mathrm {e}}+{\mathrm {e}}^x+2}{5\,x} \]

[In]

int(((4*exp(exp(1)))/5 - (exp(x)*(x - 1))/5 + 2/5)/x^2,x)

[Out]

-(4*exp(exp(1)) + exp(x) + 2)/(5*x)