3.168 \(\int \frac{e^{\frac{1}{x}} (1+x)}{x^4} \, dx\)

Optimal. Leaf size=27 \[ -\frac{e^{\frac{1}{x}}}{x^2}-e^{\frac{1}{x}}+\frac{e^{\frac{1}{x}}}{x} \]

[Out]

-E^x^(-1) - E^x^(-1)/x^2 + E^x^(-1)/x

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Rubi [A]  time = 0.105, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6742, 2212, 2209} \[ -\frac{e^{\frac{1}{x}}}{x^2}-e^{\frac{1}{x}}+\frac{e^{\frac{1}{x}}}{x} \]

Antiderivative was successfully verified.

[In]

Int[(E^x^(-1)*(1 + x))/x^4,x]

[Out]

-E^x^(-1) - E^x^(-1)/x^2 + E^x^(-1)/x

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps

\begin{align*} \int \frac{e^{\frac{1}{x}} (1+x)}{x^4} \, dx &=\int \left (\frac{e^{\frac{1}{x}}}{x^4}+\frac{e^{\frac{1}{x}}}{x^3}\right ) \, dx\\ &=\int \frac{e^{\frac{1}{x}}}{x^4} \, dx+\int \frac{e^{\frac{1}{x}}}{x^3} \, dx\\ &=-\frac{e^{\frac{1}{x}}}{x^2}-\frac{e^{\frac{1}{x}}}{x}-2 \int \frac{e^{\frac{1}{x}}}{x^3} \, dx-\int \frac{e^{\frac{1}{x}}}{x^2} \, dx\\ &=e^{\frac{1}{x}}-\frac{e^{\frac{1}{x}}}{x^2}+\frac{e^{\frac{1}{x}}}{x}+2 \int \frac{e^{\frac{1}{x}}}{x^2} \, dx\\ &=-e^{\frac{1}{x}}-\frac{e^{\frac{1}{x}}}{x^2}+\frac{e^{\frac{1}{x}}}{x}\\ \end{align*}

Mathematica [A]  time = 0.0115183, size = 17, normalized size = 0.63 \[ \frac{e^{\frac{1}{x}} \left (-x^2+x-1\right )}{x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^x^(-1)*(1 + x))/x^4,x]

[Out]

(E^x^(-1)*(-1 + x - x^2))/x^2

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Maple [A]  time = 0.002, size = 18, normalized size = 0.7 \begin{align*} -{\frac{ \left ({x}^{2}-x+1 \right ){{\rm e}^{{x}^{-1}}}}{{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x^2-x+1)*exp(1/x)/x^2

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Maxima [C]  time = 1.02522, size = 23, normalized size = 0.85 \begin{align*} -\Gamma \left (3, -\frac{1}{x}\right ) + \Gamma \left (2, -\frac{1}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(3, -1/x) + gamma(2, -1/x)

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Fricas [A]  time = 1.50407, size = 38, normalized size = 1.41 \begin{align*} -\frac{{\left (x^{2} - x + 1\right )} e^{\frac{1}{x}}}{x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x^2 - x + 1)*e^(1/x)/x^2

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Sympy [A]  time = 0.095526, size = 14, normalized size = 0.52 \begin{align*} \frac{\left (- x^{2} + x - 1\right ) e^{\frac{1}{x}}}{x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(1/x)*(1+x)/x**4,x)

[Out]

(-x**2 + x - 1)*exp(1/x)/x**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (x + 1\right )} e^{\frac{1}{x}}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x + 1)*e^(1/x)/x^4, x)