3.18.58 \(\int \frac {1}{3} e^{\frac {6-x}{3}} (-3 e^{-7+\frac {1}{3} (-6+x)+x}+e^{e^x} (-5+15 e^x)) \, dx\)

Optimal. Leaf size=22 \[ 5 e^{2+e^x-\frac {x}{3}}-e^{-7+x} \]

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Rubi [A]  time = 0.10, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {12, 2282, 14, 2288} \begin {gather*} 5 e^{-\frac {x}{3}+e^x+2}-e^{x-7} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((6 - x)/3)*(-3*E^(-7 + (-6 + x)/3 + x) + E^E^x*(-5 + 15*E^x)))/3,x]

[Out]

5*E^(2 + E^x - x/3) - E^(-7 + 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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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]]]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int e^{\frac {6-x}{3}} \left (-3 e^{-7+\frac {1}{3} (-6+x)+x}+e^{e^x} \left (-5+15 e^x\right )\right ) \, dx\\ &=\operatorname {Subst}\left (\int \frac {-\frac {3 x^3}{e^7}+\frac {5 e^{2+x^3} \left (-1+3 x^3\right )}{x}}{x} \, dx,x,e^{x/3}\right )\\ &=\operatorname {Subst}\left (\int \left (-\frac {3 x^2}{e^7}+\frac {5 e^{2+x^3} \left (-1+3 x^3\right )}{x^2}\right ) \, dx,x,e^{x/3}\right )\\ &=-e^{-7+x}+5 \operatorname {Subst}\left (\int \frac {e^{2+x^3} \left (-1+3 x^3\right )}{x^2} \, dx,x,e^{x/3}\right )\\ &=5 e^{2+e^x-\frac {x}{3}}-e^{-7+x}\\ \end {aligned} \end {gather*}

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Mathematica [C]  time = 0.05, size = 68, normalized size = 3.09 \begin {gather*} -e^{-7+x}+\frac {5}{3} e^{2-\frac {x}{3}} \sqrt [3]{-e^x} \Gamma \left (-\frac {1}{3},-e^x\right )+5 e^{2-\frac {x}{3}} \sqrt [3]{-e^x} \Gamma \left (\frac {2}{3},-e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((6 - x)/3)*(-3*E^(-7 + (-6 + x)/3 + x) + E^E^x*(-5 + 15*E^x)))/3,x]

[Out]

-E^(-7 + x) + (5*E^(2 - x/3)*(-E^x)^(1/3)*Gamma[-1/3, -E^x])/3 + 5*E^(2 - x/3)*(-E^x)^(1/3)*Gamma[2/3, -E^x]

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fricas [A]  time = 0.91, size = 17, normalized size = 0.77 \begin {gather*} {\left (5 \, e^{\left (-\frac {4}{3} \, x + e^{x} + 9\right )} - 1\right )} e^{\left (x - 7\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((15*exp(x)-5)*exp(exp(x))-3*exp(1/3*x-2)*exp(x-7))/exp(1/3*x-2),x, algorithm="fricas")

[Out]

(5*e^(-4/3*x + e^x + 9) - 1)*e^(x - 7)

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giac [A]  time = 0.19, size = 23, normalized size = 1.05 \begin {gather*} -{\left (e^{\left (2 \, x\right )} - 5 \, e^{\left (\frac {2}{3} \, x + e^{x} + 9\right )}\right )} e^{\left (-x - 7\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((15*exp(x)-5)*exp(exp(x))-3*exp(1/3*x-2)*exp(x-7))/exp(1/3*x-2),x, algorithm="giac")

[Out]

-(e^(2*x) - 5*e^(2/3*x + e^x + 9))*e^(-x - 7)

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maple [A]  time = 0.08, size = 18, normalized size = 0.82

method	result	size
risch	\(-{\mathrm e}^{x -7}+5 \,{\mathrm e}^{-\frac {x}{3}+2+{\mathrm e}^{x}}\)	\(18\)
norman	\(\left (5 \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{x}}-{\mathrm e}^{-7} {\mathrm e}^{\frac {4 x}{3}}\right ) {\mathrm e}^{-\frac {x}{3}}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((15*exp(x)-5)*exp(exp(x))-3*exp(1/3*x-2)*exp(x-7))/exp(1/3*x-2),x,method=_RETURNVERBOSE)

[Out]

-exp(x-7)+5*exp(-1/3*x+2+exp(x))

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maxima [A]  time = 0.35, size = 17, normalized size = 0.77 \begin {gather*} -e^{\left (x - 7\right )} + 5 \, e^{\left (-\frac {1}{3} \, x + e^{x} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((15*exp(x)-5)*exp(exp(x))-3*exp(1/3*x-2)*exp(x-7))/exp(1/3*x-2),x, algorithm="maxima")

[Out]

-e^(x - 7) + 5*e^(-1/3*x + e^x + 2)

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mupad [B]  time = 1.17, size = 22, normalized size = 1.00 \begin {gather*} -\frac {{\mathrm {e}}^{-\frac {x}{3}}\,\left (3\,{\mathrm {e}}^{\frac {4\,x}{3}-7}-15\,{\mathrm {e}}^{{\mathrm {e}}^x+2}\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(2 - x/3)*(exp(x - 7)*exp(x/3 - 2) - (exp(exp(x))*(15*exp(x) - 5))/3),x)

[Out]

-(exp(-x/3)*(3*exp((4*x)/3 - 7) - 15*exp(exp(x) + 2)))/3

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sympy [A]  time = 0.21, size = 22, normalized size = 1.00 \begin {gather*} - \frac {e^{x}}{e^{7}} + \frac {5 e^{2} e^{e^{x}}}{\sqrt [3]{e^{x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((15*exp(x)-5)*exp(exp(x))-3*exp(1/3*x-2)*exp(x-7))/exp(1/3*x-2),x)

[Out]

-exp(-7)*exp(x) + 5*exp(2)*exp(exp(x))/exp(x)**(1/3)

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