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

Optimal. Leaf size=10 \[ -\frac {e^x}{x}+x \]

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Rubi [A]  time = 0.03, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {14, 2197} \begin {gather*} x-\frac {e^x}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(1 - x) + x^2)/x^2,x]

[Out]

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

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Mathematica [A]  time = 0.01, size = 10, normalized size = 1.00 \begin {gather*} -\frac {e^x}{x}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(1 - x) + x^2)/x^2,x]

[Out]

-(E^x/x) + x

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fricas [A]  time = 0.47, size = 12, normalized size = 1.20 \begin {gather*} \frac {x^{2} - e^{x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^2 - e^x)/x

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giac [A]  time = 0.15, size = 12, normalized size = 1.20 \begin {gather*} \frac {x^{2} - e^{x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^2 - e^x)/x

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maple [A]  time = 0.08, size = 10, normalized size = 1.00

method	result	size
default	\(x -\frac {{\mathrm e}^{x}}{x}\)	\(10\)
risch	\(x -\frac {{\mathrm e}^{x}}{x}\)	\(10\)
norman	\(\frac {-{\mathrm e}^{x}+x^{2}}{x}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x-exp(x)/x

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maxima [C]  time = 0.38, size = 11, normalized size = 1.10 \begin {gather*} 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)+x^2)/x^2,x, algorithm="maxima")

[Out]

x - Ei(x) + gamma(-1, -x)

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mupad [B]  time = 5.60, size = 9, normalized size = 0.90 \begin {gather*} x-\frac {{\mathrm {e}}^x}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(x - 1) - x^2)/x^2,x)

[Out]

x - exp(x)/x

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sympy [A]  time = 0.08, size = 5, normalized size = 0.50 \begin {gather*} x - \frac {e^{x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - exp(x)/x

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