3.95.100 \(\int (4+e^{-16-4 e^{3 x}+4 x} (4-12 e^{3 x})+6 x) \, dx\)

Optimal. Leaf size=30 \[ e^{4 \left (-4-e^{3 x}+x\right )}+x+\frac {x+3 x \left (x+x^2\right )}{x} \]

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Rubi [C]  time = 0.14, antiderivative size = 81, normalized size of antiderivative = 2.70, number of steps used = 7, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2282, 12, 2226, 2218} \begin {gather*} 3 x^2+4 x-\frac {e^{4 x-16} \Gamma \left (\frac {4}{3},4 e^{3 x}\right )}{3\ 2^{2/3} \left (e^{3 x}\right )^{4/3}}+\frac {e^{7 x-16} \Gamma \left (\frac {7}{3},4 e^{3 x}\right )}{4\ 2^{2/3} \left (e^{3 x}\right )^{7/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[4 + E^(-16 - 4*E^(3*x) + 4*x)*(4 - 12*E^(3*x)) + 6*x,x]

[Out]

4*x + 3*x^2 - (E^(-16 + 4*x)*Gamma[4/3, 4*E^(3*x)])/(3*2^(2/3)*(E^(3*x))^(4/3)) + (E^(-16 + 7*x)*Gamma[7/3, 4*
E^(3*x)])/(4*2^(2/3)*(E^(3*x))^(7/3))

Rule 12

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

Rule 2218

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Simp[(F^a*(e + f*
x)^(m + 1)*Gamma[(m + 1)/n, -(b*(c + d*x)^n*Log[F])])/(f*n*(-(b*(c + d*x)^n*Log[F]))^((m + 1)/n)), x] /; FreeQ
[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rule 2226

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, 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} &=4 x+3 x^2+\int e^{-16-4 e^{3 x}+4 x} \left (4-12 e^{3 x}\right ) \, dx\\ &=4 x+3 x^2+\operatorname {Subst}\left (\int 4 e^{-16-4 x^3} x^3 \left (1-3 x^3\right ) \, dx,x,e^x\right )\\ &=4 x+3 x^2+4 \operatorname {Subst}\left (\int e^{-16-4 x^3} x^3 \left (1-3 x^3\right ) \, dx,x,e^x\right )\\ &=4 x+3 x^2+4 \operatorname {Subst}\left (\int \left (e^{-16-4 x^3} x^3-3 e^{-16-4 x^3} x^6\right ) \, dx,x,e^x\right )\\ &=4 x+3 x^2+4 \operatorname {Subst}\left (\int e^{-16-4 x^3} x^3 \, dx,x,e^x\right )-12 \operatorname {Subst}\left (\int e^{-16-4 x^3} x^6 \, dx,x,e^x\right )\\ &=4 x+3 x^2-\frac {e^{-16+4 x} \Gamma \left (\frac {4}{3},4 e^{3 x}\right )}{3\ 2^{2/3} \left (e^{3 x}\right )^{4/3}}+\frac {e^{-16+7 x} \Gamma \left (\frac {7}{3},4 e^{3 x}\right )}{4\ 2^{2/3} \left (e^{3 x}\right )^{7/3}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 23, normalized size = 0.77 \begin {gather*} e^{-16-4 e^{3 x}+4 x}+4 x+3 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[4 + E^(-16 - 4*E^(3*x) + 4*x)*(4 - 12*E^(3*x)) + 6*x,x]

[Out]

E^(-16 - 4*E^(3*x) + 4*x) + 4*x + 3*x^2

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fricas [A]  time = 0.89, size = 21, normalized size = 0.70 \begin {gather*} 3 \, x^{2} + 4 \, x + e^{\left (4 \, x - 4 \, e^{\left (3 \, x\right )} - 16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*exp(3*x)+4)*exp(-4*exp(3*x)+4*x-16)+6*x+4,x, algorithm="fricas")

[Out]

3*x^2 + 4*x + e^(4*x - 4*e^(3*x) - 16)

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giac [A]  time = 0.43, size = 21, normalized size = 0.70 \begin {gather*} 3 \, x^{2} + 4 \, x + e^{\left (4 \, x - 4 \, e^{\left (3 \, x\right )} - 16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*exp(3*x)+4)*exp(-4*exp(3*x)+4*x-16)+6*x+4,x, algorithm="giac")

[Out]

3*x^2 + 4*x + e^(4*x - 4*e^(3*x) - 16)

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maple [A]  time = 0.08, size = 22, normalized size = 0.73

method	result	size
default	\(4 x +{\mathrm e}^{-4 \,{\mathrm e}^{3 x}+4 x -16}+3 x^{2}\)	\(22\)
norman	\(4 x +{\mathrm e}^{-4 \,{\mathrm e}^{3 x}+4 x -16}+3 x^{2}\)	\(22\)
risch	\(4 x +{\mathrm e}^{-4 \,{\mathrm e}^{3 x}+4 x -16}+3 x^{2}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-12*exp(3*x)+4)*exp(-4*exp(3*x)+4*x-16)+6*x+4,x,method=_RETURNVERBOSE)

[Out]

4*x+exp(-4*exp(3*x)+4*x-16)+3*x^2

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maxima [A]  time = 0.34, size = 21, normalized size = 0.70 \begin {gather*} 3 \, x^{2} + 4 \, x + e^{\left (4 \, x - 4 \, e^{\left (3 \, x\right )} - 16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*exp(3*x)+4)*exp(-4*exp(3*x)+4*x-16)+6*x+4,x, algorithm="maxima")

[Out]

3*x^2 + 4*x + e^(4*x - 4*e^(3*x) - 16)

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mupad [B]  time = 0.11, size = 23, normalized size = 0.77 \begin {gather*} 4\,x+3\,x^2+{\mathrm {e}}^{-4\,{\mathrm {e}}^{3\,x}}\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{-16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(6*x - exp(4*x - 4*exp(3*x) - 16)*(12*exp(3*x) - 4) + 4,x)

[Out]

4*x + 3*x^2 + exp(-4*exp(3*x))*exp(4*x)*exp(-16)

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sympy [A]  time = 0.13, size = 20, normalized size = 0.67 \begin {gather*} 3 x^{2} + 4 x + e^{4 x - 4 e^{3 x} - 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*exp(3*x)+4)*exp(-4*exp(3*x)+4*x-16)+6*x+4,x)

[Out]

3*x**2 + 4*x + exp(4*x - 4*exp(3*x) - 16)

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