3.12.38 \(\int \frac {3 e^{e^{3 x}+3 x}+e^5 (-24+24 x)}{e^5} \, dx\)

Optimal. Leaf size=23 \[ e^{-5+e^{3 x}}+3 \left (e^4+(-2+2 x)^2\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 19, normalized size of antiderivative = 0.83, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 2282, 2194} \begin {gather*} 12 (1-x)^2+e^{e^{3 x}-5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3*E^(E^(3*x) + 3*x) + E^5*(-24 + 24*x))/E^5,x]

[Out]

E^(-5 + E^(3*x)) + 12*(1 - x)^2

Rule 12

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

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 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (3 e^{e^{3 x}+3 x}+e^5 (-24+24 x)\right ) \, dx}{e^5}\\ &=12 (1-x)^2+\frac {3 \int e^{e^{3 x}+3 x} \, dx}{e^5}\\ &=12 (1-x)^2+\frac {\operatorname {Subst}\left (\int e^x \, dx,x,e^{3 x}\right )}{e^5}\\ &=e^{-5+e^{3 x}}+12 (1-x)^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 24, normalized size = 1.04 \begin {gather*} 3 \left (\frac {1}{3} e^{-5+e^{3 x}}-8 x+4 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3*E^(E^(3*x) + 3*x) + E^5*(-24 + 24*x))/E^5,x]

[Out]

3*(E^(-5 + E^(3*x))/3 - 8*x + 4*x^2)

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fricas [A]  time = 0.69, size = 32, normalized size = 1.39 \begin {gather*} {\left (12 \, {\left (x^{2} - 2 \, x\right )} e^{\left (3 \, x + 5\right )} + e^{\left (3 \, x + e^{\left (3 \, x\right )}\right )}\right )} e^{\left (-3 \, x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*exp(x)*exp(2*x)*exp(exp(x)*exp(2*x))+(24*x-24)*exp(5))/exp(5),x, algorithm="fricas")

[Out]

(12*(x^2 - 2*x)*e^(3*x + 5) + e^(3*x + e^(3*x)))*e^(-3*x - 5)

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giac [A]  time = 0.42, size = 20, normalized size = 0.87 \begin {gather*} {\left (12 \, {\left (x^{2} - 2 \, x\right )} e^{5} + e^{\left (e^{\left (3 \, x\right )}\right )}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*exp(x)*exp(2*x)*exp(exp(x)*exp(2*x))+(24*x-24)*exp(5))/exp(5),x, algorithm="giac")

[Out]

(12*(x^2 - 2*x)*e^5 + e^(e^(3*x)))*e^(-5)

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maple [A]  time = 0.05, size = 17, normalized size = 0.74

method	result	size
risch	\({\mathrm e}^{-5+{\mathrm e}^{3 x}}-24 x +12 x^{2}\)	\(17\)
norman	\({\mathrm e}^{-5} {\mathrm e}^{{\mathrm e}^{3 x}}-24 x +12 x^{2}\)	\(20\)
default	\({\mathrm e}^{-5} \left ({\mathrm e}^{5} \left (12 x^{2}-24 x \right )+{\mathrm e}^{{\mathrm e}^{x} {\mathrm e}^{2 x}}\right )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*exp(x)*exp(2*x)*exp(exp(x)*exp(2*x))+(24*x-24)*exp(5))/exp(5),x,method=_RETURNVERBOSE)

[Out]

exp(-5+exp(3*x))-24*x+12*x^2

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maxima [A]  time = 0.51, size = 20, normalized size = 0.87 \begin {gather*} {\left (12 \, {\left (x^{2} - 2 \, x\right )} e^{5} + e^{\left (e^{\left (3 \, x\right )}\right )}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*exp(x)*exp(2*x)*exp(exp(x)*exp(2*x))+(24*x-24)*exp(5))/exp(5),x, algorithm="maxima")

[Out]

(12*(x^2 - 2*x)*e^5 + e^(e^(3*x)))*e^(-5)

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mupad [B]  time = 0.09, size = 17, normalized size = 0.74 \begin {gather*} {\mathrm {e}}^{-5}\,{\mathrm {e}}^{{\mathrm {e}}^{3\,x}}-24\,x+12\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-5)*(3*exp(3*x)*exp(exp(3*x)) + exp(5)*(24*x - 24)),x)

[Out]

exp(-5)*exp(exp(3*x)) - 24*x + 12*x^2

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sympy [A]  time = 0.15, size = 17, normalized size = 0.74 \begin {gather*} 12 x^{2} - 24 x + \frac {e^{e^{3 x}}}{e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*exp(x)*exp(2*x)*exp(exp(x)*exp(2*x))+(24*x-24)*exp(5))/exp(5),x)

[Out]

12*x**2 - 24*x + exp(-5)*exp(exp(3*x))

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