3.91.90 \(\int \frac {e^{e^{\frac {1}{2} (18+8 x+x^2+(-144-36 x) \log (3)+324 \log ^2(3))}} (-1+e^{\frac {1}{2} (18+8 x+x^2+(-144-36 x) \log (3)+324 \log ^2(3))} (4 x+x^2-18 x \log (3)))}{x^2} \, dx\)

Optimal. Leaf size=25 \[ \frac {e^{e^{1+\frac {1}{2} (-4-x+18 \log (3))^2}}}{x} \]

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Rubi [B]  time = 0.63, antiderivative size = 124, normalized size of antiderivative = 4.96, number of steps used = 1, number of rules used = 1, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {2288} \begin {gather*} \frac {\left (x^2+4 x-18 x \log (3)\right ) \exp \left (\frac {1}{2} \left (-x^2-8 x+36 (x+4) \log (3)-18 \left (1+18 \log ^2(3)\right )\right )+3^{\frac {1}{2} (-36 x-144)} e^{\frac {1}{2} \left (x^2+8 x+18 \left (1+18 \log ^2(3)\right )\right )}+\frac {1}{2} \left (x^2+8 x-36 (x+4) \log (3)+18 \left (1+18 \log ^2(3)\right )\right )\right )}{x^2 (x+2 (2-9 \log (3)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^E^((18 + 8*x + x^2 + (-144 - 36*x)*Log[3] + 324*Log[3]^2)/2)*(-1 + E^((18 + 8*x + x^2 + (-144 - 36*x)*L
og[3] + 324*Log[3]^2)/2)*(4*x + x^2 - 18*x*Log[3])))/x^2,x]

[Out]

(E^(3^((-144 - 36*x)/2)*E^((8*x + x^2 + 18*(1 + 18*Log[3]^2))/2) + (-8*x - x^2 + 36*(4 + x)*Log[3] - 18*(1 + 1
8*Log[3]^2))/2 + (8*x + x^2 - 36*(4 + x)*Log[3] + 18*(1 + 18*Log[3]^2))/2)*(4*x + x^2 - 18*x*Log[3]))/(x^2*(x
+ 2*(2 - 9*Log[3])))

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\exp \left (3^{\frac {1}{2} (-144-36 x)} e^{\frac {1}{2} \left (8 x+x^2+18 \left (1+18 \log ^2(3)\right )\right )}+\frac {1}{2} \left (-8 x-x^2+36 (4+x) \log (3)-18 \left (1+18 \log ^2(3)\right )\right )+\frac {1}{2} \left (8 x+x^2-36 (4+x) \log (3)+18 \left (1+18 \log ^2(3)\right )\right )\right ) \left (4 x+x^2-18 x \log (3)\right )}{x^2 (x+2 (2-9 \log (3)))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 34, normalized size = 1.36 \begin {gather*} \frac {e^{3^{-72-18 x} e^{9+4 x+\frac {x^2}{2}+162 \log ^2(3)}}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^((18 + 8*x + x^2 + (-144 - 36*x)*Log[3] + 324*Log[3]^2)/2)*(-1 + E^((18 + 8*x + x^2 + (-144 - 3
6*x)*Log[3] + 324*Log[3]^2)/2)*(4*x + x^2 - 18*x*Log[3])))/x^2,x]

[Out]

E^(3^(-72 - 18*x)*E^(9 + 4*x + x^2/2 + 162*Log[3]^2))/x

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fricas [A]  time = 1.04, size = 29, normalized size = 1.16 \begin {gather*} \frac {e^{\left (e^{\left (\frac {1}{2} \, x^{2} - 18 \, {\left (x + 4\right )} \log \relax (3) + 162 \, \log \relax (3)^{2} + 4 \, x + 9\right )}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*x*log(3)+x^2+4*x)*exp(162*log(3)^2+1/2*(-36*x-144)*log(3)+1/2*x^2+4*x+9)-1)*exp(exp(162*log(3)
^2+1/2*(-36*x-144)*log(3)+1/2*x^2+4*x+9))/x^2,x, algorithm="fricas")

[Out]

e^(e^(1/2*x^2 - 18*(x + 4)*log(3) + 162*log(3)^2 + 4*x + 9))/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left ({\left (x^{2} - 18 \, x \log \relax (3) + 4 \, x\right )} e^{\left (\frac {1}{2} \, x^{2} - 18 \, {\left (x + 4\right )} \log \relax (3) + 162 \, \log \relax (3)^{2} + 4 \, x + 9\right )} - 1\right )} e^{\left (e^{\left (\frac {1}{2} \, x^{2} - 18 \, {\left (x + 4\right )} \log \relax (3) + 162 \, \log \relax (3)^{2} + 4 \, x + 9\right )}\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*x*log(3)+x^2+4*x)*exp(162*log(3)^2+1/2*(-36*x-144)*log(3)+1/2*x^2+4*x+9)-1)*exp(exp(162*log(3)
^2+1/2*(-36*x-144)*log(3)+1/2*x^2+4*x+9))/x^2,x, algorithm="giac")

[Out]

integrate(((x^2 - 18*x*log(3) + 4*x)*e^(1/2*x^2 - 18*(x + 4)*log(3) + 162*log(3)^2 + 4*x + 9) - 1)*e^(e^(1/2*x
^2 - 18*(x + 4)*log(3) + 162*log(3)^2 + 4*x + 9))/x^2, x)

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maple [A]  time = 0.10, size = 31, normalized size = 1.24

method	result	size
risch	\(\frac {{\mathrm e}^{3^{-72-18 x} {\mathrm e}^{162 \ln \relax (3)^{2}+9+\frac {x^{2}}{2}+4 x}}}{x}\)	\(31\)
norman	\(\frac {{\mathrm e}^{{\mathrm e}^{162 \ln \relax (3)^{2}+\frac {\left (-36 x -144\right ) \ln \relax (3)}{2}+\frac {x^{2}}{2}+4 x +9}}}{x}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-18*x*ln(3)+x^2+4*x)*exp(162*ln(3)^2+1/2*(-36*x-144)*ln(3)+1/2*x^2+4*x+9)-1)*exp(exp(162*ln(3)^2+1/2*(-3
6*x-144)*ln(3)+1/2*x^2+4*x+9))/x^2,x,method=_RETURNVERBOSE)

[Out]

exp(3^(-72-18*x)*exp(162*ln(3)^2+9+1/2*x^2+4*x))/x

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maxima [A]  time = 0.85, size = 29, normalized size = 1.16 \begin {gather*} \frac {e^{\left (\frac {1}{22528399544939174411840147874772641} \, e^{\left (\frac {1}{2} \, x^{2} - 18 \, x \log \relax (3) + 162 \, \log \relax (3)^{2} + 4 \, x + 9\right )}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*x*log(3)+x^2+4*x)*exp(162*log(3)^2+1/2*(-36*x-144)*log(3)+1/2*x^2+4*x+9)-1)*exp(exp(162*log(3)
^2+1/2*(-36*x-144)*log(3)+1/2*x^2+4*x+9))/x^2,x, algorithm="maxima")

[Out]

e^(1/22528399544939174411840147874772641*e^(1/2*x^2 - 18*x*log(3) + 162*log(3)^2 + 4*x + 9))/x

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mupad [B]  time = 5.82, size = 29, normalized size = 1.16 \begin {gather*} \frac {{\mathrm {e}}^{\frac {{\left (\frac {1}{387420489}\right )}^x\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^9\,{\mathrm {e}}^{162\,{\ln \relax (3)}^2}\,{\mathrm {e}}^{\frac {x^2}{2}}}{22528399544939174411840147874772641}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(4*x - (log(3)*(36*x + 144))/2 + 162*log(3)^2 + x^2/2 + 9))*(exp(4*x - (log(3)*(36*x + 144))/2 + 1
62*log(3)^2 + x^2/2 + 9)*(4*x - 18*x*log(3) + x^2) - 1))/x^2,x)

[Out]

exp(((1/387420489)^x*exp(4*x)*exp(9)*exp(162*log(3)^2)*exp(x^2/2))/22528399544939174411840147874772641)/x

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sympy [A]  time = 0.31, size = 31, normalized size = 1.24 \begin {gather*} \frac {e^{e^{\frac {x^{2}}{2} + 4 x + \left (- 18 x - 72\right ) \log {\relax (3 )} + 9 + 162 \log {\relax (3 )}^{2}}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*x*ln(3)+x**2+4*x)*exp(162*ln(3)**2+1/2*(-36*x-144)*ln(3)+1/2*x**2+4*x+9)-1)*exp(exp(162*ln(3)*
*2+1/2*(-36*x-144)*ln(3)+1/2*x**2+4*x+9))/x**2,x)

[Out]

exp(exp(x**2/2 + 4*x + (-18*x - 72)*log(3) + 9 + 162*log(3)**2))/x

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