3.99.65 \(\int (2 e^{16+2 x}+e^{8+x} (2+2 x)) \, dx\)

Optimal. Leaf size=19 \[ -x^2+\left (e^{8+x}+x\right )^2+4 \log (2) \]

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Rubi [A]  time = 0.02, antiderivative size = 25, normalized size of antiderivative = 1.32, number of steps used = 4, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2194, 2176} \begin {gather*} 2 e^{x+8} (x+1)-2 e^{x+8}+e^{2 x+16} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-2*E^(8 + x) + E^(16 + 2*x) + 2*E^(8 + x)*(1 + x)

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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} &=2 \int e^{16+2 x} \, dx+\int e^{8+x} (2+2 x) \, dx\\ &=e^{16+2 x}+2 e^{8+x} (1+x)-2 \int e^{8+x} \, dx\\ &=-2 e^{8+x}+e^{16+2 x}+2 e^{8+x} (1+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 21, normalized size = 1.11 \begin {gather*} 2 \left (\frac {1}{2} e^{16+2 x}+e^{8+x} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2*(E^(16 + 2*x)/2 + E^(8 + x)*x)

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fricas [A]  time = 0.63, size = 14, normalized size = 0.74 \begin {gather*} 2 \, x e^{\left (x + 8\right )} + e^{\left (2 \, x + 16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(4)^4*exp(x)^2+(2*x+2)*exp(4)^2*exp(x),x, algorithm="fricas")

[Out]

2*x*e^(x + 8) + e^(2*x + 16)

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giac [A]  time = 0.19, size = 14, normalized size = 0.74 \begin {gather*} 2 \, x e^{\left (x + 8\right )} + e^{\left (2 \, x + 16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(4)^4*exp(x)^2+(2*x+2)*exp(4)^2*exp(x),x, algorithm="giac")

[Out]

2*x*e^(x + 8) + e^(2*x + 16)

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maple [A]  time = 0.03, size = 15, normalized size = 0.79




method result size



risch \(2 x \,{\mathrm e}^{x +8}+{\mathrm e}^{2 x +16}\) \(15\)
default \(2 \,{\mathrm e}^{8} x \,{\mathrm e}^{x}+{\mathrm e}^{16} {\mathrm e}^{2 x}\) \(20\)
norman \(2 \,{\mathrm e}^{8} x \,{\mathrm e}^{x}+{\mathrm e}^{16} {\mathrm e}^{2 x}\) \(20\)
meijerg \(-{\mathrm e}^{16} \left (1-{\mathrm e}^{2 x}\right )+2 \,{\mathrm e}^{8} \left (1-\frac {\left (-2 x +2\right ) {\mathrm e}^{x}}{2}\right )-2 \,{\mathrm e}^{8} \left (1-{\mathrm e}^{x}\right )\) \(39\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*exp(4)^4*exp(x)^2+(2*x+2)*exp(4)^2*exp(x),x,method=_RETURNVERBOSE)

[Out]

2*x*exp(x+8)+exp(2*x+16)

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maxima [A]  time = 0.36, size = 26, normalized size = 1.37 \begin {gather*} 2 \, {\left (x e^{8} - e^{8}\right )} e^{x} + e^{\left (2 \, x + 16\right )} + 2 \, e^{\left (x + 8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(4)^4*exp(x)^2+(2*x+2)*exp(4)^2*exp(x),x, algorithm="maxima")

[Out]

2*(x*e^8 - e^8)*e^x + e^(2*x + 16) + 2*e^(x + 8)

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mupad [B]  time = 0.05, size = 13, normalized size = 0.68 \begin {gather*} {\mathrm {e}}^{x+8}\,\left (2\,x+{\mathrm {e}}^{x+8}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*exp(2*x)*exp(16) + exp(8)*exp(x)*(2*x + 2),x)

[Out]

exp(x + 8)*(2*x + exp(x + 8))

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sympy [A]  time = 0.12, size = 17, normalized size = 0.89 \begin {gather*} 2 x e^{8} e^{x} + e^{16} e^{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(4)**4*exp(x)**2+(2*x+2)*exp(4)**2*exp(x),x)

[Out]

2*x*exp(8)*exp(x) + exp(16)*exp(2*x)

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