3.47.44 \(\int \frac {-11+e^x (-1+x)}{x^2} \, dx\)

Optimal. Leaf size=17 \[ \frac {11+e^x-x+e^3 x}{x} \]

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Rubi [A]  time = 0.06, antiderivative size = 13, normalized size of antiderivative = 0.76, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {14, 2197} \begin {gather*} \frac {e^x}{x}+\frac {11}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-11 + E^x*(-1 + x))/x^2,x]

[Out]

11/x + E^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 (-\frac {11}{x^2}+\frac {e^x (-1+x)}{x^2}\right ) \, dx\\ &=\frac {11}{x}+\int \frac {e^x (-1+x)}{x^2} \, dx\\ &=\frac {11}{x}+\frac {e^x}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 9, normalized size = 0.53 \begin {gather*} \frac {11+e^x}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-11 + E^x*(-1 + x))/x^2,x]

[Out]

(11 + E^x)/x

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fricas [A]  time = 0.67, size = 8, normalized size = 0.47 \begin {gather*} \frac {e^{x} + 11}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^x + 11)/x

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giac [A]  time = 0.22, size = 8, normalized size = 0.47 \begin {gather*} \frac {e^{x} + 11}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^x + 11)/x

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maple [A]  time = 0.02, size = 9, normalized size = 0.53

method	result	size
norman	\(\frac {11+{\mathrm e}^{x}}{x}\)	\(9\)
default	\(\frac {11}{x}+\frac {{\mathrm e}^{x}}{x}\)	\(13\)
risch	\(\frac {11}{x}+\frac {{\mathrm e}^{x}}{x}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(11+exp(x))/x

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maxima [C]  time = 0.38, size = 15, normalized size = 0.88 \begin {gather*} \frac {11}{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)-11)/x^2,x, algorithm="maxima")

[Out]

11/x + Ei(x) - gamma(-1, -x)

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mupad [B]  time = 0.04, size = 8, normalized size = 0.47 \begin {gather*} \frac {{\mathrm {e}}^x+11}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(x - 1) - 11)/x^2,x)

[Out]

(exp(x) + 11)/x

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sympy [A]  time = 0.08, size = 7, normalized size = 0.41 \begin {gather*} \frac {e^{x}}{x} + \frac {11}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-1)*exp(x)-11)/x**2,x)

[Out]

exp(x)/x + 11/x

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