3.20.3 \(\int \frac {1}{5} (-1+e^{\frac {1}{x}} (-5+10 x)) \, dx\)

Optimal. Leaf size=17 \[ \left (e^{\frac {1}{x}}-\frac {1}{5 x}\right ) x^2 \]

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 15, normalized size of antiderivative = 0.88, number of steps used = 9, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {12, 2226, 2206, 2210, 2214} \begin {gather*} e^{\frac {1}{x}} x^2-\frac {x}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + E^x^(-1)*(-5 + 10*x))/5,x]

[Out]

-1/5*x + E^x^(-1)*x^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2206

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

Rule 2210

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

Rule 2214

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*F^(a + b*(c + d*x)^n))/(d*(m + 1)), x] - Dist[(b*n*Log[F])/(m + 1), 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[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rule 2226

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \left (-1+e^{\frac {1}{x}} (-5+10 x)\right ) \, dx\\ &=-\frac {x}{5}+\frac {1}{5} \int e^{\frac {1}{x}} (-5+10 x) \, dx\\ &=-\frac {x}{5}+\frac {1}{5} \int \left (-5 e^{\frac {1}{x}}+10 e^{\frac {1}{x}} x\right ) \, dx\\ &=-\frac {x}{5}+2 \int e^{\frac {1}{x}} x \, dx-\int e^{\frac {1}{x}} \, dx\\ &=-\frac {x}{5}-e^{\frac {1}{x}} x+e^{\frac {1}{x}} x^2+\int e^{\frac {1}{x}} \, dx-\int \frac {e^{\frac {1}{x}}}{x} \, dx\\ &=-\frac {x}{5}+e^{\frac {1}{x}} x^2+\text {Ei}\left (\frac {1}{x}\right )+\int \frac {e^{\frac {1}{x}}}{x} \, dx\\ &=-\frac {x}{5}+e^{\frac {1}{x}} x^2\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 15, normalized size = 0.88 \begin {gather*} -\frac {x}{5}+e^{\frac {1}{x}} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + E^x^(-1)*(-5 + 10*x))/5,x]

[Out]

-1/5*x + E^x^(-1)*x^2

________________________________________________________________________________________

fricas [A]  time = 0.85, size = 12, normalized size = 0.71 \begin {gather*} x^{2} e^{\frac {1}{x}} - \frac {1}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(10*x-5)*exp(1/x)-1/5,x, algorithm="fricas")

[Out]

x^2*e^(1/x) - 1/5*x

________________________________________________________________________________________

giac [A]  time = 0.38, size = 12, normalized size = 0.71 \begin {gather*} x^{2} e^{\frac {1}{x}} - \frac {1}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(10*x-5)*exp(1/x)-1/5,x, algorithm="giac")

[Out]

x^2*e^(1/x) - 1/5*x

________________________________________________________________________________________

maple [A]  time = 0.04, size = 13, normalized size = 0.76

method	result	size
derivativedivides	\(-\frac {x}{5}+x^{2} {\mathrm e}^{\frac {1}{x}}\)	\(13\)
default	\(-\frac {x}{5}+x^{2} {\mathrm e}^{\frac {1}{x}}\)	\(13\)
norman	\(-\frac {x}{5}+x^{2} {\mathrm e}^{\frac {1}{x}}\)	\(13\)
risch	\(-\frac {x}{5}+x^{2} {\mathrm e}^{\frac {1}{x}}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(10*x-5)*exp(1/x)-1/5,x,method=_RETURNVERBOSE)

[Out]

-1/5*x+x^2*exp(1/x)

________________________________________________________________________________________

maxima [C]  time = 0.73, size = 20, normalized size = 1.18 \begin {gather*} -\frac {1}{5} \, x + \Gamma \left (-1, -\frac {1}{x}\right ) + 2 \, \Gamma \left (-2, -\frac {1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(10*x-5)*exp(1/x)-1/5,x, algorithm="maxima")

[Out]

-1/5*x + gamma(-1, -1/x) + 2*gamma(-2, -1/x)

________________________________________________________________________________________

mupad [B]  time = 1.12, size = 12, normalized size = 0.71 \begin {gather*} x^2\,{\mathrm {e}}^{1/x}-\frac {x}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1/x)*(10*x - 5))/5 - 1/5,x)

[Out]

x^2*exp(1/x) - x/5

________________________________________________________________________________________

sympy [A]  time = 0.09, size = 10, normalized size = 0.59 \begin {gather*} x^{2} e^{\frac {1}{x}} - \frac {x}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(10*x-5)*exp(1/x)-1/5,x)

[Out]

x**2*exp(1/x) - x/5

________________________________________________________________________________________