3.27.32 \(\int \frac {1}{4} (2+4 e^3-e^x+8 x) \, dx\)

Optimal. Leaf size=25 \[ \frac {1}{4} \left (-e^x-2 x\right )+x+x^2+e^3 (4+x) \]

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 2194} \begin {gather*} x^2+\frac {1}{2} \left (1+2 e^3\right ) x-\frac {e^x}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 4*E^3 - E^x + 8*x)/4,x]

[Out]

-1/4*E^x + ((1 + 2*E^3)*x)/2 + 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 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \left (2+4 e^3-e^x+8 x\right ) \, dx\\ &=\frac {1}{2} \left (1+2 e^3\right ) x+x^2-\frac {\int e^x \, dx}{4}\\ &=-\frac {e^x}{4}+\frac {1}{2} \left (1+2 e^3\right ) x+x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 21, normalized size = 0.84 \begin {gather*} -\frac {e^x}{4}+\frac {x}{2}+e^3 x+x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 4*E^3 - E^x + 8*x)/4,x]

[Out]

-1/4*E^x + x/2 + E^3*x + x^2

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fricas [A]  time = 0.51, size = 15, normalized size = 0.60 \begin {gather*} x^{2} + x e^{3} + \frac {1}{2} \, x - \frac {1}{4} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/4*exp(x)+exp(3)+2*x+1/2,x, algorithm="fricas")

[Out]

x^2 + x*e^3 + 1/2*x - 1/4*e^x

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giac [A]  time = 0.20, size = 15, normalized size = 0.60 \begin {gather*} x^{2} + x e^{3} + \frac {1}{2} \, x - \frac {1}{4} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/4*exp(x)+exp(3)+2*x+1/2,x, algorithm="giac")

[Out]

x^2 + x*e^3 + 1/2*x - 1/4*e^x

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maple [A]  time = 0.02, size = 15, normalized size = 0.60

method	result	size
norman	\(x^{2}+\left ({\mathrm e}^{3}+\frac {1}{2}\right ) x -\frac {{\mathrm e}^{x}}{4}\)	\(15\)
default	\(x^{2}+\frac {x}{2}-\frac {{\mathrm e}^{x}}{4}+x \,{\mathrm e}^{3}\)	\(16\)
risch	\(x^{2}+\frac {x}{2}-\frac {{\mathrm e}^{x}}{4}+x \,{\mathrm e}^{3}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/4*exp(x)+exp(3)+2*x+1/2,x,method=_RETURNVERBOSE)

[Out]

x^2+(exp(3)+1/2)*x-1/4*exp(x)

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maxima [A]  time = 0.41, size = 15, normalized size = 0.60 \begin {gather*} x^{2} + x e^{3} + \frac {1}{2} \, x - \frac {1}{4} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/4*exp(x)+exp(3)+2*x+1/2,x, algorithm="maxima")

[Out]

x^2 + x*e^3 + 1/2*x - 1/4*e^x

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mupad [B]  time = 0.06, size = 14, normalized size = 0.56 \begin {gather*} x\,\left ({\mathrm {e}}^3+\frac {1}{2}\right )-\frac {{\mathrm {e}}^x}{4}+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x + exp(3) - exp(x)/4 + 1/2,x)

[Out]

x*(exp(3) + 1/2) - exp(x)/4 + x^2

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sympy [A]  time = 0.08, size = 15, normalized size = 0.60 \begin {gather*} x^{2} + x \left (\frac {1}{2} + e^{3}\right ) - \frac {e^{x}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/4*exp(x)+exp(3)+2*x+1/2,x)

[Out]

x**2 + x*(1/2 + exp(3)) - exp(x)/4

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