3.36.4 \(\int \frac {-4-e^{3+x}+36 e^{3+e^{18 x^2}+18 x^2} x}{e^3} \, dx\)

Optimal. Leaf size=24 \[ -\frac {5}{2}+e^{e^{18 x^2}}-e^x-\frac {4 x}{e^3} \]

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Rubi [A]  time = 0.06, antiderivative size = 21, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 2194, 6715, 2282} \begin {gather*} e^{e^{18 x^2}}-\frac {4 x}{e^3}-e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 - E^(3 + x) + 36*E^(3 + E^(18*x^2) + 18*x^2)*x)/E^3,x]

[Out]

E^E^(18*x^2) - E^x - (4*x)/E^3

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]]]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (-4-e^{3+x}+36 e^{3+e^{18 x^2}+18 x^2} x\right ) \, dx}{e^3}\\ &=-\frac {4 x}{e^3}-\frac {\int e^{3+x} \, dx}{e^3}+\frac {36 \int e^{3+e^{18 x^2}+18 x^2} x \, dx}{e^3}\\ &=-e^x-\frac {4 x}{e^3}+\frac {18 \operatorname {Subst}\left (\int e^{3+e^{18 x}+18 x} \, dx,x,x^2\right )}{e^3}\\ &=-e^x-\frac {4 x}{e^3}+\frac {\operatorname {Subst}\left (\int e^{3+x} \, dx,x,e^{18 x^2}\right )}{e^3}\\ &=e^{e^{18 x^2}}-e^x-\frac {4 x}{e^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 21, normalized size = 0.88 \begin {gather*} e^{e^{18 x^2}}-e^x-\frac {4 x}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 - E^(3 + x) + 36*E^(3 + E^(18*x^2) + 18*x^2)*x)/E^3,x]

[Out]

E^E^(18*x^2) - E^x - (4*x)/E^3

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fricas [B]  time = 0.61, size = 42, normalized size = 1.75 \begin {gather*} -{\left ({\left (4 \, x + e^{\left (x + 3\right )}\right )} e^{\left (18 \, x^{2}\right )} - e^{\left (18 \, x^{2} + e^{\left (18 \, x^{2}\right )} + 3\right )}\right )} e^{\left (-18 \, x^{2} - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((36*x*exp(3)*exp(9*x^2)^2*exp(exp(9*x^2)^2)-exp(x)*exp(3)-4)/exp(3),x, algorithm="fricas")

[Out]

-((4*x + e^(x + 3))*e^(18*x^2) - e^(18*x^2 + e^(18*x^2) + 3))*e^(-18*x^2 - 3)

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giac [A]  time = 0.13, size = 23, normalized size = 0.96 \begin {gather*} -{\left (4 \, x + e^{\left (x + 3\right )} - e^{\left (e^{\left (18 \, x^{2}\right )} + 3\right )}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((36*x*exp(3)*exp(9*x^2)^2*exp(exp(9*x^2)^2)-exp(x)*exp(3)-4)/exp(3),x, algorithm="giac")

[Out]

-(4*x + e^(x + 3) - e^(e^(18*x^2) + 3))*e^(-3)

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maple [A]  time = 0.07, size = 18, normalized size = 0.75

method	result	size
risch	\(-4 x \,{\mathrm e}^{-3}-{\mathrm e}^{x}+{\mathrm e}^{{\mathrm e}^{18 x^{2}}}\)	\(18\)
norman	\(-4 x \,{\mathrm e}^{-3}-{\mathrm e}^{x}+{\mathrm e}^{{\mathrm e}^{18 x^{2}}}\)	\(22\)
default	\({\mathrm e}^{-3} \left (-4 x -{\mathrm e}^{x} {\mathrm e}^{3}+{\mathrm e}^{3} {\mathrm e}^{{\mathrm e}^{18 x^{2}}}\right )\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((36*x*exp(3)*exp(9*x^2)^2*exp(exp(9*x^2)^2)-exp(x)*exp(3)-4)/exp(3),x,method=_RETURNVERBOSE)

[Out]

-4*x*exp(-3)-exp(x)+exp(exp(18*x^2))

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maxima [A]  time = 0.48, size = 23, normalized size = 0.96 \begin {gather*} -{\left (4 \, x + e^{\left (x + 3\right )} - e^{\left (e^{\left (18 \, x^{2}\right )} + 3\right )}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((36*x*exp(3)*exp(9*x^2)^2*exp(exp(9*x^2)^2)-exp(x)*exp(3)-4)/exp(3),x, algorithm="maxima")

[Out]

-(4*x + e^(x + 3) - e^(e^(18*x^2) + 3))*e^(-3)

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mupad [B]  time = 0.16, size = 17, normalized size = 0.71 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^{18\,x^2}}-{\mathrm {e}}^x-4\,x\,{\mathrm {e}}^{-3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-3)*(exp(3)*exp(x) - 36*x*exp(exp(18*x^2))*exp(3)*exp(18*x^2) + 4),x)

[Out]

exp(exp(18*x^2)) - exp(x) - 4*x*exp(-3)

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sympy [A]  time = 0.24, size = 17, normalized size = 0.71 \begin {gather*} - \frac {4 x}{e^{3}} - e^{x} + e^{e^{18 x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((36*x*exp(3)*exp(9*x**2)**2*exp(exp(9*x**2)**2)-exp(x)*exp(3)-4)/exp(3),x)

[Out]

-4*x*exp(-3) - exp(x) + exp(exp(18*x**2))

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