3.99.86 \(\int 1250 e^{\frac {2}{3} (1+3 e^{x+(i \pi +\log (-2+e^5))^2})+x+(i \pi +\log (-2+e^5))^2} \, dx\)

Optimal. Leaf size=28 \[ 625 e^{\frac {2}{3}+2 e^{x+\left (i \pi +\log \left (-2+e^5\right )\right )^2}} \]

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Rubi [A]  time = 0.04, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {12, 2282, 2194} \begin {gather*} 625 e^{\frac {2}{3}+2 e^{x+\left (\log \left (e^5-2\right )+i \pi \right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1250*E^((2*(1 + 3*E^(x + (I*Pi + Log[-2 + E^5])^2)))/3 + x + (I*Pi + Log[-2 + E^5])^2),x]

[Out]

625*E^(2/3 + 2*E^(x + (I*Pi + Log[-2 + E^5])^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} &=1250 \int \exp \left (\frac {2}{3} \left (1+3 e^{x+\left (i \pi +\log \left (-2+e^5\right )\right )^2}\right )+x+\left (i \pi +\log \left (-2+e^5\right )\right )^2\right ) \, dx\\ &=1250 \operatorname {Subst}\left (\int e^{\frac {2}{3}+2 x} \, dx,x,e^{x+\left (i \pi +\log \left (-2+e^5\right )\right )^2}\right )\\ &=625 e^{\frac {2}{3}+2 e^{x+\left (i \pi +\log \left (-2+e^5\right )\right )^2}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 28, normalized size = 1.00 \begin {gather*} 625 e^{\frac {2}{3}+2 e^{x+\left (i \pi +\log \left (-2+e^5\right )\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1250*E^((2*(1 + 3*E^(x + (I*Pi + Log[-2 + E^5])^2)))/3 + x + (I*Pi + Log[-2 + E^5])^2),x]

[Out]

625*E^(2/3 + 2*E^(x + (I*Pi + Log[-2 + E^5])^2))

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fricas [A]  time = 1.15, size = 19, normalized size = 0.68 \begin {gather*} 625 \, e^{\left (2 \, e^{\left (\log \left (-e^{5} + 2\right )^{2} + x\right )} + \frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1250*exp(log(2-exp(5))^2+x)*exp(exp(log(2-exp(5))^2+x)+1/3)^2,x, algorithm="fricas")

[Out]

625*e^(2*e^(log(-e^5 + 2)^2 + x) + 2/3)

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giac [A]  time = 0.19, size = 19, normalized size = 0.68 \begin {gather*} 625 \, e^{\left (2 \, e^{\left (\log \left (-e^{5} + 2\right )^{2} + x\right )} + \frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1250*exp(log(2-exp(5))^2+x)*exp(exp(log(2-exp(5))^2+x)+1/3)^2,x, algorithm="giac")

[Out]

625*e^(2*e^(log(-e^5 + 2)^2 + x) + 2/3)

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maple [A]  time = 0.03, size = 20, normalized size = 0.71

method	result	size
derivativedivides	\(625 \,{\mathrm e}^{2 \,{\mathrm e}^{\ln \left (2-{\mathrm e}^{5}\right )^{2}+x}+\frac {2}{3}}\)	\(20\)
default	\(625 \,{\mathrm e}^{2 \,{\mathrm e}^{\ln \left (2-{\mathrm e}^{5}\right )^{2}+x}+\frac {2}{3}}\)	\(20\)
norman	\(625 \,{\mathrm e}^{2 \,{\mathrm e}^{\ln \left (2-{\mathrm e}^{5}\right )^{2}+x}+\frac {2}{3}}\)	\(20\)
risch	\(625 \,{\mathrm e}^{2 \,{\mathrm e}^{\ln \left (2-{\mathrm e}^{5}\right )^{2}+x}+\frac {2}{3}}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1250*exp(ln(2-exp(5))^2+x)*exp(exp(ln(2-exp(5))^2+x)+1/3)^2,x,method=_RETURNVERBOSE)

[Out]

625*exp(exp(ln(2-exp(5))^2+x)+1/3)^2

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maxima [A]  time = 0.44, size = 19, normalized size = 0.68 \begin {gather*} 625 \, e^{\left (2 \, e^{\left (\log \left (-e^{5} + 2\right )^{2} + x\right )} + \frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1250*exp(log(2-exp(5))^2+x)*exp(exp(log(2-exp(5))^2+x)+1/3)^2,x, algorithm="maxima")

[Out]

625*e^(2*e^(log(-e^5 + 2)^2 + x) + 2/3)

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mupad [B]  time = 0.06, size = 19, normalized size = 0.68 \begin {gather*} 625\,{\mathrm {e}}^{2/3}\,{\mathrm {e}}^{2\,{\mathrm {e}}^{{\ln \left (2-{\mathrm {e}}^5\right )}^2}\,{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1250*exp(x + log(2 - exp(5))^2)*exp(2*exp(x + log(2 - exp(5))^2) + 2/3),x)

[Out]

625*exp(2/3)*exp(2*exp(log(2 - exp(5))^2)*exp(x))

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sympy [A]  time = 0.27, size = 41, normalized size = 1.46 \begin {gather*} 625 e^{\frac {2}{3}} e^{\frac {2 e^{x} e^{2 i \pi \log {\left (-2 + e^{5} \right )}} e^{\log {\left (-2 + e^{5} \right )}^{2}}}{e^{\pi ^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1250*exp(ln(2-exp(5))**2+x)*exp(exp(ln(2-exp(5))**2+x)+1/3)**2,x)

[Out]

625*exp(2/3)*exp(2*exp(-pi**2)*exp(x)*exp(2*I*pi*log(-2 + exp(5)))*exp(log(-2 + exp(5))**2))

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