\(\int e^{x+2 e^5 x} (1+2 e^5) \, dx\) [7434]

Optimal result

Integrand size = 18, antiderivative size = 12 \[ \int e^{x+2 e^5 x} \left (1+2 e^5\right ) \, dx=3+e^{x+2 e^5 x} \]

[Out]

3+exp(x+2*x*exp(5))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {12, 2259, 2225} \[ \int e^{x+2 e^5 x} \left (1+2 e^5\right ) \, dx=e^{\left (1+2 e^5\right ) x} \]

[In]

Int[E^(x + 2*E^5*x)*(1 + 2*E^5),x]

[Out]

E^((1 + 2*E^5)*x)

Rule 12

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

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]

Rule 2259

Int[(u_.)*(F_)^((a_.) + (b_.)*(v_)), x_Symbol] :> Int[u*F^(a + b*NormalizePowerOfLinear[v, x]), x] /; FreeQ[{F
, a, b}, x] && PolynomialQ[u, x] && PowerOfLinearQ[v, x] &&  !PowerOfLinearMatchQ[v, x]

Rubi steps \begin{align*} \text {integral}& = \left (1+2 e^5\right ) \int e^{x+2 e^5 x} \, dx \\ & = \left (1+2 e^5\right ) \int e^{\left (1+2 e^5\right ) x} \, dx \\ & = e^{\left (1+2 e^5\right ) x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int e^{x+2 e^5 x} \left (1+2 e^5\right ) \, dx=e^{\left (1+2 e^5\right ) x} \]

[In]

Integrate[E^(x + 2*E^5*x)*(1 + 2*E^5),x]

[Out]

E^((1 + 2*E^5)*x)

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75

method	result	size
gosper	\({\mathrm e}^{x +2 x \,{\mathrm e}^{5}}\)	\(9\)
derivativedivides	\({\mathrm e}^{x +2 x \,{\mathrm e}^{5}}\)	\(9\)
default	\({\mathrm e}^{x +2 x \,{\mathrm e}^{5}}\)	\(9\)
norman	\({\mathrm e}^{x +2 x \,{\mathrm e}^{5}}\)	\(9\)
parts	\({\mathrm e}^{x +2 x \,{\mathrm e}^{5}}\)	\(9\)
risch	\({\mathrm e}^{x \left (2 \,{\mathrm e}^{5}+1\right )}\)	\(10\)
parallelrisch	\({\mathrm e}^{x \left (2 \,{\mathrm e}^{5}+1\right )}\)	\(10\)
meijerg	\(\frac {2 \,{\mathrm e}^{5} \left (1-{\mathrm e}^{-x \left (-2 \,{\mathrm e}^{5}-1\right )}\right )}{-2 \,{\mathrm e}^{5}-1}+\frac {1-{\mathrm e}^{-x \left (-2 \,{\mathrm e}^{5}-1\right )}}{-2 \,{\mathrm e}^{5}-1}\)	\(51\)

[In]

int((2*exp(5)+1)*exp(x+2*x*exp(5)),x,method=_RETURNVERBOSE)

[Out]

exp(x+2*x*exp(5))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int e^{x+2 e^5 x} \left (1+2 e^5\right ) \, dx=e^{\left (2 \, x e^{5} + x\right )} \]

[In]

integrate((2*exp(5)+1)*exp(x+2*x*exp(5)),x, algorithm="fricas")

[Out]

e^(2*x*e^5 + x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int e^{x+2 e^5 x} \left (1+2 e^5\right ) \, dx=e^{x + 2 x e^{5}} \]

[In]

integrate((2*exp(5)+1)*exp(x+2*x*exp(5)),x)

[Out]

exp(x + 2*x*exp(5))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int e^{x+2 e^5 x} \left (1+2 e^5\right ) \, dx=e^{\left (2 \, x e^{5} + x\right )} \]

[In]

integrate((2*exp(5)+1)*exp(x+2*x*exp(5)),x, algorithm="maxima")

[Out]

e^(2*x*e^5 + x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int e^{x+2 e^5 x} \left (1+2 e^5\right ) \, dx=e^{\left (2 \, x e^{5} + x\right )} \]

[In]

integrate((2*exp(5)+1)*exp(x+2*x*exp(5)),x, algorithm="giac")

[Out]

e^(2*x*e^5 + x)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int e^{x+2 e^5 x} \left (1+2 e^5\right ) \, dx={\mathrm {e}}^{x\,\left (2\,{\mathrm {e}}^5+1\right )} \]

[In]

int(exp(x + 2*x*exp(5))*(2*exp(5) + 1),x)

[Out]

exp(x*(2*exp(5) + 1))