3.99.62 \(\int (1+59049 e^{-e^{-3+4 x}+2 x} (-2+4 e^{-3+4 x})) \, dx\)

Optimal. Leaf size=20 \[ -1-59049 e^{-e^{-3+4 x}+2 x}+x \]

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Rubi [A]  time = 0.10, antiderivative size = 19, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2282, 2226, 2205, 2212} \begin {gather*} x-59049 e^{2 x-e^{4 x-3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1 + 59049*E^(-E^(-3 + 4*x) + 2*x)*(-2 + 4*E^(-3 + 4*x)),x]

[Out]

-59049*E^(-E^(-3 + 4*x) + 2*x) + x

Rule 2205

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erf[(c + d*x)*Rt[-(b*Log[F]),
2]])/(2*d*Rt[-(b*Log[F]), 2]), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 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} &=x+59049 \int e^{-e^{-3+4 x}+2 x} \left (-2+4 e^{-3+4 x}\right ) \, dx\\ &=x+\frac {59049}{2} \operatorname {Subst}\left (\int e^{-\frac {x^2}{e^3}} \left (-2+\frac {4 x^2}{e^3}\right ) \, dx,x,e^{2 x}\right )\\ &=x+\frac {59049}{2} \operatorname {Subst}\left (\int \left (-2 e^{-\frac {x^2}{e^3}}+4 e^{-3-\frac {x^2}{e^3}} x^2\right ) \, dx,x,e^{2 x}\right )\\ &=x-59049 \operatorname {Subst}\left (\int e^{-\frac {x^2}{e^3}} \, dx,x,e^{2 x}\right )+118098 \operatorname {Subst}\left (\int e^{-3-\frac {x^2}{e^3}} x^2 \, dx,x,e^{2 x}\right )\\ &=-59049 e^{-e^{-3+4 x}+2 x}+x-\frac {59049}{2} e^{3/2} \sqrt {\pi } \text {erf}\left (e^{-\frac {3}{2}+2 x}\right )+\left (59049 e^3\right ) \operatorname {Subst}\left (\int e^{-3-\frac {x^2}{e^3}} \, dx,x,e^{2 x}\right )\\ &=-59049 e^{-e^{-3+4 x}+2 x}+x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 19, normalized size = 0.95 \begin {gather*} -59049 e^{-e^{-3+4 x}+2 x}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1 + 59049*E^(-E^(-3 + 4*x) + 2*x)*(-2 + 4*E^(-3 + 4*x)),x]

[Out]

-59049*E^(-E^(-3 + 4*x) + 2*x) + x

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fricas [A]  time = 0.76, size = 21, normalized size = 1.05 \begin {gather*} x - e^{\left (2 \, x - e^{\left (4 \, x - 3\right )} + 10 \, \log \relax (3)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(4*x-3)-2)*exp(-exp(4*x-3)+10*log(3)+2*x)+1,x, algorithm="fricas")

[Out]

x - e^(2*x - e^(4*x - 3) + 10*log(3))

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giac [A]  time = 0.14, size = 21, normalized size = 1.05 \begin {gather*} x - e^{\left (2 \, x - e^{\left (4 \, x - 3\right )} + 10 \, \log \relax (3)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(4*x-3)-2)*exp(-exp(4*x-3)+10*log(3)+2*x)+1,x, algorithm="giac")

[Out]

x - e^(2*x - e^(4*x - 3) + 10*log(3))

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maple [A]  time = 0.08, size = 18, normalized size = 0.90

method	result	size
risch	\(x -59049 \,{\mathrm e}^{-{\mathrm e}^{4 x -3}+2 x}\)	\(18\)
default	\(x -{\mathrm e}^{-{\mathrm e}^{4 x -3}+10 \ln \relax (3)+2 x}\)	\(22\)
norman	\(x -{\mathrm e}^{-{\mathrm e}^{4 x -3}+10 \ln \relax (3)+2 x}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*exp(4*x-3)-2)*exp(-exp(4*x-3)+10*ln(3)+2*x)+1,x,method=_RETURNVERBOSE)

[Out]

x-59049*exp(-exp(4*x-3)+2*x)

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maxima [A]  time = 0.37, size = 17, normalized size = 0.85 \begin {gather*} x - 59049 \, e^{\left (2 \, x - e^{\left (4 \, x - 3\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(4*x-3)-2)*exp(-exp(4*x-3)+10*log(3)+2*x)+1,x, algorithm="maxima")

[Out]

x - 59049*e^(2*x - e^(4*x - 3))

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mupad [B]  time = 0.08, size = 17, normalized size = 0.85 \begin {gather*} x-59049\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{-3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x + 10*log(3) - exp(4*x - 3))*(4*exp(4*x - 3) - 2) + 1,x)

[Out]

x - 59049*exp(2*x)*exp(-exp(4*x)*exp(-3))

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sympy [A]  time = 0.15, size = 14, normalized size = 0.70 \begin {gather*} x - 59049 e^{2 x - e^{4 x - 3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(4*x-3)-2)*exp(-exp(4*x-3)+10*ln(3)+2*x)+1,x)

[Out]

x - 59049*exp(2*x - exp(4*x - 3))

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