3.215 \(\int e^{\sqrt {5+3 x}} \, dx\)

Optimal. Leaf size=40 \[ \frac {2}{3} e^{\sqrt {3 x+5}} \sqrt {3 x+5}-\frac {2}{3} e^{\sqrt {3 x+5}} \]

[Out]

-2/3*exp((5+3*x)^(1/2))+2/3*exp((5+3*x)^(1/2))*(5+3*x)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2207, 2176, 2194} \[ \frac {2}{3} e^{\sqrt {3 x+5}} \sqrt {3 x+5}-\frac {2}{3} e^{\sqrt {3 x+5}} \]

Antiderivative was successfully verified.

[In]

Int[E^Sqrt[5 + 3*x],x]

[Out]

(-2*E^Sqrt[5 + 3*x])/3 + (2*E^Sqrt[5 + 3*x]*Sqrt[5 + 3*x])/3

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]

Rule 2207

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*} \int e^{\sqrt {5+3 x}} \, dx &=\frac {2}{3} \operatorname {Subst}\left (\int e^x x \, dx,x,\sqrt {5+3 x}\right )\\ &=\frac {2}{3} e^{\sqrt {5+3 x}} \sqrt {5+3 x}-\frac {2}{3} \operatorname {Subst}\left (\int e^x \, dx,x,\sqrt {5+3 x}\right )\\ &=-\frac {2}{3} e^{\sqrt {5+3 x}}+\frac {2}{3} e^{\sqrt {5+3 x}} \sqrt {5+3 x}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 26, normalized size = 0.65 \[ \frac {2}{3} e^{\sqrt {3 x+5}} \left (\sqrt {3 x+5}-1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^Sqrt[5 + 3*x],x]

[Out]

(2*E^Sqrt[5 + 3*x]*(-1 + Sqrt[5 + 3*x]))/3

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fricas [A]  time = 0.41, size = 19, normalized size = 0.48 \[ \frac {2}{3} \, {\left (\sqrt {3 \, x + 5} - 1\right )} e^{\left (\sqrt {3 \, x + 5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp((5+3*x)^(1/2)),x, algorithm="fricas")

[Out]

2/3*(sqrt(3*x + 5) - 1)*e^(sqrt(3*x + 5))

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giac [A]  time = 0.21, size = 19, normalized size = 0.48 \[ \frac {2}{3} \, {\left (\sqrt {3 \, x + 5} - 1\right )} e^{\left (\sqrt {3 \, x + 5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp((5+3*x)^(1/2)),x, algorithm="giac")

[Out]

2/3*(sqrt(3*x + 5) - 1)*e^(sqrt(3*x + 5))

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maple [A]  time = 0.01, size = 29, normalized size = 0.72 \[ -\frac {2 \,{\mathrm e}^{\sqrt {3 x +5}}}{3}+\frac {2 \sqrt {3 x +5}\, {\mathrm e}^{\sqrt {3 x +5}}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp((3*x+5)^(1/2)),x)

[Out]

-2/3*exp((3*x+5)^(1/2))+2/3*exp((3*x+5)^(1/2))*(3*x+5)^(1/2)

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maxima [A]  time = 0.74, size = 19, normalized size = 0.48 \[ \frac {2}{3} \, {\left (\sqrt {3 \, x + 5} - 1\right )} e^{\left (\sqrt {3 \, x + 5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp((5+3*x)^(1/2)),x, algorithm="maxima")

[Out]

2/3*(sqrt(3*x + 5) - 1)*e^(sqrt(3*x + 5))

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mupad [B]  time = 0.09, size = 19, normalized size = 0.48 \[ \frac {2\,{\mathrm {e}}^{\sqrt {3\,x+5}}\,\left (\sqrt {3\,x+5}-1\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp((3*x + 5)^(1/2)),x)

[Out]

(2*exp((3*x + 5)^(1/2))*((3*x + 5)^(1/2) - 1))/3

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sympy [A]  time = 0.21, size = 34, normalized size = 0.85 \[ \frac {2 \sqrt {3 x + 5} e^{\sqrt {3 x + 5}}}{3} - \frac {2 e^{\sqrt {3 x + 5}}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp((5+3*x)**(1/2)),x)

[Out]

2*sqrt(3*x + 5)*exp(sqrt(3*x + 5))/3 - 2*exp(sqrt(3*x + 5))/3

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