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

Optimal. Leaf size=19 \[ \frac {3-e^2+e^x-\frac {2 x}{3}}{x} \]

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Rubi [A]  time = 0.03, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {14, 2197} \begin {gather*} \frac {e^x}{x}+\frac {3-e^2}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3 + E^2 + E^x*(-1 + x))/x^2,x]

[Out]

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

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Mathematica [A]  time = 0.01, size = 14, normalized size = 0.74 \begin {gather*} \frac {3-e^2+e^x}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3 - E^2 + E^x)/x

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fricas [A]  time = 0.59, size = 13, normalized size = 0.68 \begin {gather*} -\frac {e^{2} - e^{x} - 3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(e^2 - e^x - 3)/x

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giac [A]  time = 0.17, size = 13, normalized size = 0.68 \begin {gather*} -\frac {e^{2} - e^{x} - 3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(e^2 - e^x - 3)/x

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maple [A]  time = 0.02, size = 13, normalized size = 0.68

method	result	size
norman	\(\frac {-{\mathrm e}^{2}+3+{\mathrm e}^{x}}{x}\)	\(13\)
default	\(-\frac {{\mathrm e}^{2}}{x}+\frac {3}{x}+\frac {{\mathrm e}^{x}}{x}\)	\(20\)
risch	\(-\frac {{\mathrm e}^{2}}{x}+\frac {3}{x}+\frac {{\mathrm e}^{x}}{x}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-exp(2)+3+exp(x))/x

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maxima [C]  time = 0.37, size = 22, normalized size = 1.16 \begin {gather*} -\frac {e^{2}}{x} + \frac {3}{x} + {\rm Ei}\relax (x) - \Gamma \left (-1, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^2/x + 3/x + Ei(x) - gamma(-1, -x)

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mupad [B]  time = 0.06, size = 12, normalized size = 0.63 \begin {gather*} \frac {{\mathrm {e}}^x-{\mathrm {e}}^2+3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(x) - exp(2) + 3)/x

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sympy [A]  time = 0.09, size = 10, normalized size = 0.53 \begin {gather*} \frac {e^{x}}{x} - \frac {-3 + e^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x)/x - (-3 + exp(2))/x

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