\(\int (2 e^{16+2 x}+e^{8+x} (2+2 x)) \, dx\) [9865]

Optimal result

Integrand size = 21, antiderivative size = 19 \[ \int \left (2 e^{16+2 x}+e^{8+x} (2+2 x)\right ) \, dx=-x^2+\left (e^{8+x}+x\right )^2+4 \log (2) \]

[Out]

(x+exp(4)^2*exp(x))^2+4*ln(2)-x^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.32, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2225, 2207} \[ \int \left (2 e^{16+2 x}+e^{8+x} (2+2 x)\right ) \, dx=2 e^{x+8} (x+1)-2 e^{x+8}+e^{2 x+16} \]

[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 2207

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] &&  !TrueQ[$UseGamma]

Rule 2225

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \left (2 e^{16+2 x}+e^{8+x} (2+2 x)\right ) \, dx=2 \left (\frac {1}{2} e^{16+2 x}+e^{8+x} x\right ) \]

[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)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \left (2 e^{16+2 x}+e^{8+x} (2+2 x)\right ) \, dx=2 \, x e^{\left (x + 8\right )} + e^{\left (2 \, x + 16\right )} \]

[In]

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

[Out]

2*x*e^(x + 8) + e^(2*x + 16)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \left (2 e^{16+2 x}+e^{8+x} (2+2 x)\right ) \, dx=2 x e^{8} e^{x} + e^{16} e^{2 x} \]

[In]

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

[Out]

2*x*exp(8)*exp(x) + exp(16)*exp(2*x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \left (2 e^{16+2 x}+e^{8+x} (2+2 x)\right ) \, dx=2 \, {\left (x e^{8} - e^{8}\right )} e^{x} + e^{\left (2 \, x + 16\right )} + 2 \, e^{\left (x + 8\right )} \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \left (2 e^{16+2 x}+e^{8+x} (2+2 x)\right ) \, dx=2 \, x e^{\left (x + 8\right )} + e^{\left (2 \, x + 16\right )} \]

[In]

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

[Out]

2*x*e^(x + 8) + e^(2*x + 16)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \left (2 e^{16+2 x}+e^{8+x} (2+2 x)\right ) \, dx={\mathrm {e}}^{x+8}\,\left (2\,x+{\mathrm {e}}^{x+8}\right ) \]

[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))