\(\int e^{\sqrt {x}} \, dx\) [257]

Optimal result

Integrand size = 7, antiderivative size = 24 \[ \int e^{\sqrt {x}} \, dx=-2 e^{\sqrt {x}}+2 e^{\sqrt {x}} \sqrt {x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2238, 2207, 2225} \[ \int e^{\sqrt {x}} \, dx=2 e^{\sqrt {x}} \sqrt {x}-2 e^{\sqrt {x}} \]

[In]

Int[E^Sqrt[x],x]

[Out]

-2*E^Sqrt[x] + 2*E^Sqrt[x]*Sqrt[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]

Rule 2238

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> With[{k = Denominator[n]}, Dist[k/d, Subst[In
t[x^(k - 1)*F^(a + b*x^(k*n)), x], x, (c + d*x)^(1/k)], x]] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n] &&
!IntegerQ[n]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int e^{\sqrt {x}} \, dx=2 e^{\sqrt {x}} \left (-1+\sqrt {x}\right ) \]

[In]

Integrate[E^Sqrt[x],x]

[Out]

2*E^Sqrt[x]*(-1 + Sqrt[x])

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67

method	result	size
meijerg	\(2-\left (-2 \sqrt {x}+2\right ) {\mathrm e}^{\sqrt {x}}\)	\(16\)
derivativedivides	\(-2 \,{\mathrm e}^{\sqrt {x}}+2 \,{\mathrm e}^{\sqrt {x}} \sqrt {x}\)	\(17\)
default	\(-2 \,{\mathrm e}^{\sqrt {x}}+2 \,{\mathrm e}^{\sqrt {x}} \sqrt {x}\)	\(17\)

[In]

int(exp(x^(1/2)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.46 \[ \int e^{\sqrt {x}} \, dx=2 \, {\left (\sqrt {x} - 1\right )} e^{\sqrt {x}} \]

[In]

integrate(exp(x^(1/2)),x, algorithm="fricas")

[Out]

2*(sqrt(x) - 1)*e^sqrt(x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int e^{\sqrt {x}} \, dx=2 \sqrt {x} e^{\sqrt {x}} - 2 e^{\sqrt {x}} \]

[In]

integrate(exp(x**(1/2)),x)

[Out]

2*sqrt(x)*exp(sqrt(x)) - 2*exp(sqrt(x))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.46 \[ \int e^{\sqrt {x}} \, dx=2 \, {\left (\sqrt {x} - 1\right )} e^{\sqrt {x}} \]

[In]

integrate(exp(x^(1/2)),x, algorithm="maxima")

[Out]

2*(sqrt(x) - 1)*e^sqrt(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.46 \[ \int e^{\sqrt {x}} \, dx=2 \, {\left (\sqrt {x} - 1\right )} e^{\sqrt {x}} \]

[In]

integrate(exp(x^(1/2)),x, algorithm="giac")

[Out]

2*(sqrt(x) - 1)*e^sqrt(x)

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.46 \[ \int e^{\sqrt {x}} \, dx=2\,{\mathrm {e}}^{\sqrt {x}}\,\left (\sqrt {x}-1\right ) \]

[In]

int(exp(x^(1/2)),x)

[Out]

2*exp(x^(1/2))*(x^(1/2) - 1)