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

Optimal. Leaf size=25 \[ \frac {1}{5} \left (4+\frac {-2-4 e^e-e^x-x}{x}\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {12, 14, 2197} \begin {gather*} -\frac {e^x}{5 x}-\frac {2 \left (1+2 e^e\right )}{5 x} \end {gather*}

Antiderivative was successfully verified.

[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 2197

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 17, normalized size = 0.68 \begin {gather*} -\frac {2+4 e^e+e^x}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(2 + 4*E^E + E^x)/x

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fricas [A]  time = 1.16, size = 14, normalized size = 0.56 \begin {gather*} -\frac {e^{x} + 4 \, e^{e} + 2}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(e^x + 4*e^e + 2)/x

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giac [A]  time = 0.18, size = 14, normalized size = 0.56 \begin {gather*} -\frac {e^{x} + 4 \, e^{e} + 2}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(e^x + 4*e^e + 2)/x

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maple [A]  time = 0.03, size = 16, normalized size = 0.64

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.55, size = 25, normalized size = 1.00 \begin {gather*} -\frac {4 \, e^{e}}{5 \, x} - \frac {2}{5 \, x} - \frac {1}{5} \, {\rm Ei}\relax (x) + \frac {1}{5} \, \Gamma \left (-1, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 0.06, size = 14, normalized size = 0.56 \begin {gather*} -\frac {4\,{\mathrm {e}}^{\mathrm {e}}+{\mathrm {e}}^x+2}{5\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.10, size = 20, normalized size = 0.80 \begin {gather*} - \frac {e^{x}}{5 x} - \frac {\frac {2}{5} + \frac {4 e^{e}}{5}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp(x)/(5*x) - (2/5 + 4*exp(E)/5)/x

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