3.14.47 \(\int \frac {1}{8} (-16+3 x^2+e^{2/x} (-8+8 x)) \, dx\)

Optimal. Leaf size=28 \[ -2 x+\frac {1}{2} \left (3+\frac {1}{4} x^2 \left (4 e^{2/x}+x\right )\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 25, normalized size of antiderivative = 0.89, number of steps used = 9, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {12, 2226, 2206, 2210, 2214} \begin {gather*} \frac {x^3}{8}+\frac {1}{2} e^{2/x} x^2-2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16 + 3*x^2 + E^(2/x)*(-8 + 8*x))/8,x]

[Out]

-2*x + (E^(2/x)*x^2)/2 + x^3/8

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}{8} \int \left (-16+3 x^2+e^{2/x} (-8+8 x)\right ) \, dx\\ &=-2 x+\frac {x^3}{8}+\frac {1}{8} \int e^{2/x} (-8+8 x) \, dx\\ &=-2 x+\frac {x^3}{8}+\frac {1}{8} \int \left (-8 e^{2/x}+8 e^{2/x} x\right ) \, dx\\ &=-2 x+\frac {x^3}{8}-\int e^{2/x} \, dx+\int e^{2/x} x \, dx\\ &=-2 x-e^{2/x} x+\frac {1}{2} e^{2/x} x^2+\frac {x^3}{8}-2 \int \frac {e^{2/x}}{x} \, dx+\int e^{2/x} \, dx\\ &=-2 x+\frac {1}{2} e^{2/x} x^2+\frac {x^3}{8}+2 \text {Ei}\left (\frac {2}{x}\right )+2 \int \frac {e^{2/x}}{x} \, dx\\ &=-2 x+\frac {1}{2} e^{2/x} x^2+\frac {x^3}{8}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 0.89 \begin {gather*} -2 x+\frac {1}{2} e^{2/x} x^2+\frac {x^3}{8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16 + 3*x^2 + E^(2/x)*(-8 + 8*x))/8,x]

[Out]

-2*x + (E^(2/x)*x^2)/2 + x^3/8

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fricas [A]  time = 1.09, size = 20, normalized size = 0.71 \begin {gather*} \frac {1}{8} \, x^{3} + \frac {1}{2} \, x^{2} e^{\frac {2}{x}} - 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(8*x-8)*exp(2/x)+3/8*x^2-2,x, algorithm="fricas")

[Out]

1/8*x^3 + 1/2*x^2*e^(2/x) - 2*x

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giac [A]  time = 0.27, size = 20, normalized size = 0.71 \begin {gather*} \frac {1}{8} \, x^{3} + \frac {1}{2} \, x^{2} e^{\frac {2}{x}} - 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(8*x-8)*exp(2/x)+3/8*x^2-2,x, algorithm="giac")

[Out]

1/8*x^3 + 1/2*x^2*e^(2/x) - 2*x

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maple [A]  time = 0.07, size = 21, normalized size = 0.75

method	result	size
derivativedivides	\(\frac {x^{3}}{8}-2 x +\frac {x^{2} {\mathrm e}^{\frac {2}{x}}}{2}\)	\(21\)
default	\(\frac {x^{3}}{8}-2 x +\frac {x^{2} {\mathrm e}^{\frac {2}{x}}}{2}\)	\(21\)
norman	\(\frac {x^{3}}{8}-2 x +\frac {x^{2} {\mathrm e}^{\frac {2}{x}}}{2}\)	\(21\)
risch	\(\frac {x^{3}}{8}-2 x +\frac {x^{2} {\mathrm e}^{\frac {2}{x}}}{2}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/8*(8*x-8)*exp(2/x)+3/8*x^2-2,x,method=_RETURNVERBOSE)

[Out]

1/8*x^3-2*x+1/2*x^2*exp(2/x)

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maxima [A]  time = 0.41, size = 20, normalized size = 0.71 \begin {gather*} \frac {1}{8} \, x^{3} + \frac {1}{2} \, x^{2} e^{\frac {2}{x}} - 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(8*x-8)*exp(2/x)+3/8*x^2-2,x, algorithm="maxima")

[Out]

1/8*x^3 + 1/2*x^2*e^(2/x) - 2*x

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mupad [B]  time = 0.92, size = 17, normalized size = 0.61 \begin {gather*} \frac {x\,\left (4\,x\,{\mathrm {e}}^{2/x}+x^2-16\right )}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2/x)*(8*x - 8))/8 + (3*x^2)/8 - 2,x)

[Out]

(x*(4*x*exp(2/x) + x^2 - 16))/8

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sympy [A]  time = 0.09, size = 17, normalized size = 0.61 \begin {gather*} \frac {x^{3}}{8} + \frac {x^{2} e^{\frac {2}{x}}}{2} - 2 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(8*x-8)*exp(2/x)+3/8*x**2-2,x)

[Out]

x**3/8 + x**2*exp(2/x)/2 - 2*x

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