3.15.11 \(\int \frac {-18 e^4+36 (i \pi +\log (4))+e^{e^{\frac {x^2}{9}}+\frac {x^2}{9}} (-2 e^4 x+4 x (i \pi +\log (4)))}{9 e^{2 e^{\frac {x^2}{9}}}+36 e^{e^{\frac {x^2}{9}}} x+36 x^2} \, dx\)

Optimal. Leaf size=32 \[ \frac {e^4-2 (i \pi +\log (4))}{e^{e^{\frac {x^2}{9}}}+2 x} \]

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Rubi [A]  time = 0.27, antiderivative size = 31, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, integrand size = 92, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {6688, 12, 6686} \begin {gather*} \frac {e^4-2 i \pi -2 \log (4)}{e^{e^{\frac {x^2}{9}}}+2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-18*E^4 + 36*(I*Pi + Log[4]) + E^(E^(x^2/9) + x^2/9)*(-2*E^4*x + 4*x*(I*Pi + Log[4])))/(9*E^(2*E^(x^2/9))
+ 36*E^E^(x^2/9)*x + 36*x^2),x]

[Out]

(E^4 - (2*I)*Pi - 2*Log[4])/(E^E^(x^2/9) + 2*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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (9+e^{e^{\frac {x^2}{9}}+\frac {x^2}{9}} x\right ) \left (-e^4+2 i \pi +2 \log (4)\right )}{9 \left (e^{e^{\frac {x^2}{9}}}+2 x\right )^2} \, dx\\ &=-\left (\frac {1}{9} \left (2 \left (e^4-2 i \pi -\log (16)\right )\right ) \int \frac {9+e^{e^{\frac {x^2}{9}}+\frac {x^2}{9}} x}{\left (e^{e^{\frac {x^2}{9}}}+2 x\right )^2} \, dx\right )\\ &=\frac {e^4-2 i \pi -\log (16)}{e^{e^{\frac {x^2}{9}}}+2 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 31, normalized size = 0.97 \begin {gather*} \frac {e^4-2 i \pi -2 \log (4)}{e^{e^{\frac {x^2}{9}}}+2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-18*E^4 + 36*(I*Pi + Log[4]) + E^(E^(x^2/9) + x^2/9)*(-2*E^4*x + 4*x*(I*Pi + Log[4])))/(9*E^(2*E^(x
^2/9)) + 36*E^E^(x^2/9)*x + 36*x^2),x]

[Out]

(E^4 - (2*I)*Pi - 2*Log[4])/(E^E^(x^2/9) + 2*x)

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fricas [A]  time = 0.83, size = 42, normalized size = 1.31 \begin {gather*} \frac {{\left (-2 i \, \pi + e^{4} - 4 \, \log \relax (2)\right )} e^{\left (\frac {1}{9} \, x^{2}\right )}}{2 \, x e^{\left (\frac {1}{9} \, x^{2}\right )} + e^{\left (\frac {1}{9} \, x^{2} + e^{\left (\frac {1}{9} \, x^{2}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x*(2*log(2)+I*pi)-2*x*exp(2)^2)*exp(1/9*x^2)*exp(exp(1/9*x^2))+72*log(2)+36*I*pi-18*exp(2)^2)/(9
*exp(exp(1/9*x^2))^2+36*x*exp(exp(1/9*x^2))+36*x^2),x, algorithm="fricas")

[Out]

(-2*I*pi + e^4 - 4*log(2))*e^(1/9*x^2)/(2*x*e^(1/9*x^2) + e^(1/9*x^2 + e^(1/9*x^2)))

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giac [A]  time = 0.45, size = 27, normalized size = 0.84 \begin {gather*} -\frac {2 i \, \pi - e^{4} + 4 \, \log \relax (2)}{2 \, x + e^{\left (e^{\left (\frac {1}{9} \, x^{2}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x*(2*log(2)+I*pi)-2*x*exp(2)^2)*exp(1/9*x^2)*exp(exp(1/9*x^2))+72*log(2)+36*I*pi-18*exp(2)^2)/(9
*exp(exp(1/9*x^2))^2+36*x*exp(exp(1/9*x^2))+36*x^2),x, algorithm="giac")

[Out]

-(2*I*pi - e^4 + 4*log(2))/(2*x + e^(e^(1/9*x^2)))

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maple [A]  time = 0.15, size = 28, normalized size = 0.88

method	result	size
norman	\(\frac {{\mathrm e}^{4}-4 \ln \relax (2)-2 i \pi }{2 x +{\mathrm e}^{{\mathrm e}^{\frac {x^{2}}{9}}}}\)	\(28\)
risch	\(-\frac {2 i \pi }{2 x +{\mathrm e}^{{\mathrm e}^{\frac {x^{2}}{9}}}}+\frac {{\mathrm e}^{4}}{2 x +{\mathrm e}^{{\mathrm e}^{\frac {x^{2}}{9}}}}-\frac {4 \ln \relax (2)}{2 x +{\mathrm e}^{{\mathrm e}^{\frac {x^{2}}{9}}}}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x*(2*ln(2)+I*Pi)-2*x*exp(2)^2)*exp(1/9*x^2)*exp(exp(1/9*x^2))+72*ln(2)+36*I*Pi-18*exp(2)^2)/(9*exp(exp
(1/9*x^2))^2+36*x*exp(exp(1/9*x^2))+36*x^2),x,method=_RETURNVERBOSE)

[Out]

(exp(2)^2-4*ln(2)-2*I*Pi)/(2*x+exp(exp(1/9*x^2)))

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maxima [A]  time = 0.69, size = 27, normalized size = 0.84 \begin {gather*} -\frac {2 i \, \pi - e^{4} + 4 \, \log \relax (2)}{2 \, x + e^{\left (e^{\left (\frac {1}{9} \, x^{2}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x*(2*log(2)+I*pi)-2*x*exp(2)^2)*exp(1/9*x^2)*exp(exp(1/9*x^2))+72*log(2)+36*I*pi-18*exp(2)^2)/(9
*exp(exp(1/9*x^2))^2+36*x*exp(exp(1/9*x^2))+36*x^2),x, algorithm="maxima")

[Out]

-(2*I*pi - e^4 + 4*log(2))/(2*x + e^(e^(1/9*x^2)))

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mupad [B]  time = 0.24, size = 26, normalized size = 0.81 \begin {gather*} -\frac {\ln \left (16\right )-{\mathrm {e}}^4+\Pi \,2{}\mathrm {i}}{2\,x+{\mathrm {e}}^{{\mathrm {e}}^{\frac {x^2}{9}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((Pi*36i - 18*exp(4) + 72*log(2) - exp(exp(x^2/9))*exp(x^2/9)*(2*x*exp(4) - 4*x*(Pi*1i + 2*log(2))))/(9*exp
(2*exp(x^2/9)) + 36*x^2 + 36*x*exp(exp(x^2/9))),x)

[Out]

-(Pi*2i - exp(4) + log(16))/(2*x + exp(exp(x^2/9)))

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sympy [A]  time = 0.47, size = 26, normalized size = 0.81 \begin {gather*} \frac {- e^{4} + 4 \log {\relax (2 )} + 2 i \pi }{- 2 x - e^{e^{\frac {x^{2}}{9}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x*(2*ln(2)+I*pi)-2*x*exp(2)**2)*exp(1/9*x**2)*exp(exp(1/9*x**2))+72*ln(2)+36*I*pi-18*exp(2)**2)/
(9*exp(exp(1/9*x**2))**2+36*x*exp(exp(1/9*x**2))+36*x**2),x)

[Out]

(-exp(4) + 4*log(2) + 2*I*pi)/(-2*x - exp(exp(x**2/9)))

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