3.26.55 \(\int \frac {e^{\frac {4 e^{e^2}}{\log (\frac {27+12 x}{x})}} (36 e^{e^2} \log (x)+(9+4 x) \log ^2(\frac {27+12 x}{x}))}{(9 x+4 x^2) \log ^2(\frac {27+12 x}{x})} \, dx\)

Optimal. Leaf size=27 \[ e^{\frac {4 e^{e^2}}{\log \left (\frac {9 \left (3+\frac {4 x}{3}\right )}{x}\right )}} \log (x) \]

________________________________________________________________________________________

Rubi [A]  time = 0.45, antiderivative size = 47, normalized size of antiderivative = 1.74, number of steps used = 2, number of rules used = 2, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {1593, 2288} \begin {gather*} -\frac {9 e^{\frac {4 e^{e^2}}{\log \left (\frac {3 (4 x+9)}{x}\right )}} \log (x)}{x^2 \left (\frac {4}{x}-\frac {4 x+9}{x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((4*E^E^2)/Log[(27 + 12*x)/x])*(36*E^E^2*Log[x] + (9 + 4*x)*Log[(27 + 12*x)/x]^2))/((9*x + 4*x^2)*Log[(
27 + 12*x)/x]^2),x]

[Out]

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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} &=\int \frac {e^{\frac {4 e^{e^2}}{\log \left (\frac {27+12 x}{x}\right )}} \left (36 e^{e^2} \log (x)+(9+4 x) \log ^2\left (\frac {27+12 x}{x}\right )\right )}{x (9+4 x) \log ^2\left (\frac {27+12 x}{x}\right )} \, dx\\ &=-\frac {9 e^{\frac {4 e^{e^2}}{\log \left (\frac {3 (9+4 x)}{x}\right )}} \log (x)}{x^2 \left (\frac {4}{x}-\frac {9+4 x}{x^2}\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 22, normalized size = 0.81 \begin {gather*} e^{\frac {4 e^{e^2}}{\log \left (12+\frac {27}{x}\right )}} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((4*E^E^2)/Log[(27 + 12*x)/x])*(36*E^E^2*Log[x] + (9 + 4*x)*Log[(27 + 12*x)/x]^2))/((9*x + 4*x^2)
*Log[(27 + 12*x)/x]^2),x]

[Out]

E^((4*E^E^2)/Log[12 + 27/x])*Log[x]

________________________________________________________________________________________

fricas [A]  time = 0.62, size = 26, normalized size = 0.96 \begin {gather*} e^{\left (\frac {e^{\left (e^{2} + 2 \, \log \relax (2)\right )}}{\log \left (\frac {3 \, {\left (4 \, x + 9\right )}}{x}\right )}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*exp(2*log(2)+exp(2))*log(x)+(4*x+9)*log((12*x+27)/x)^2)*exp(exp(2*log(2)+exp(2))/log((12*x+27)/x)
)/(4*x^2+9*x)/log((12*x+27)/x)^2,x, algorithm="fricas")

[Out]

e^(e^(e^2 + 2*log(2))/log(3*(4*x + 9)/x))*log(x)

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*exp(2*log(2)+exp(2))*log(x)+(4*x+9)*log((12*x+27)/x)^2)*exp(exp(2*log(2)+exp(2))/log((12*x+27)/x)
)/(4*x^2+9*x)/log((12*x+27)/x)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Simplification assuming t_nostep near 0Simplification assuming t_nostep near 0Simplification assuming sageV
ARx near 0E

________________________________________________________________________________________

maple [C]  time = 12.84, size = 121, normalized size = 4.48

method	result	size
risch	\(\ln \relax (x ) {\mathrm e}^{\frac {8 \,{\mathrm e}^{{\mathrm e}^{2}}}{-i \pi \mathrm {csgn}\left (\frac {i \left (x +\frac {9}{4}\right )}{x}\right )^{3}+i \pi \mathrm {csgn}\left (\frac {i \left (x +\frac {9}{4}\right )}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right )+i \pi \mathrm {csgn}\left (\frac {i \left (x +\frac {9}{4}\right )}{x}\right )^{2} \mathrm {csgn}\left (i \left (x +\frac {9}{4}\right )\right )-i \pi \,\mathrm {csgn}\left (\frac {i \left (x +\frac {9}{4}\right )}{x}\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (x +\frac {9}{4}\right )\right )+2 \ln \relax (3)+4 \ln \relax (2)-2 \ln \relax (x )+2 \ln \left (x +\frac {9}{4}\right )}}\)	\(121\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((9*exp(2*ln(2)+exp(2))*ln(x)+(4*x+9)*ln((12*x+27)/x)^2)*exp(exp(2*ln(2)+exp(2))/ln((12*x+27)/x))/(4*x^2+9*
x)/ln((12*x+27)/x)^2,x,method=_RETURNVERBOSE)

[Out]

ln(x)*exp(8*exp(exp(2))/(-I*Pi*csgn(I/x*(x+9/4))^3+I*Pi*csgn(I/x*(x+9/4))^2*csgn(I/x)+I*Pi*csgn(I/x*(x+9/4))^2
*csgn(I*(x+9/4))-I*Pi*csgn(I/x*(x+9/4))*csgn(I/x)*csgn(I*(x+9/4))+2*ln(3)+4*ln(2)-2*ln(x)+2*ln(x+9/4)))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left ({\left (4 \, x + 9\right )} \log \left (\frac {3 \, {\left (4 \, x + 9\right )}}{x}\right )^{2} + 36 \, e^{\left (e^{2}\right )} \log \relax (x)\right )} e^{\left (\frac {4 \, e^{\left (e^{2}\right )}}{\log \left (\frac {3 \, {\left (4 \, x + 9\right )}}{x}\right )}\right )}}{{\left (4 \, x^{2} + 9 \, x\right )} \log \left (\frac {3 \, {\left (4 \, x + 9\right )}}{x}\right )^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*exp(2*log(2)+exp(2))*log(x)+(4*x+9)*log((12*x+27)/x)^2)*exp(exp(2*log(2)+exp(2))/log((12*x+27)/x)
)/(4*x^2+9*x)/log((12*x+27)/x)^2,x, algorithm="maxima")

[Out]

integrate(((4*x + 9)*log(3*(4*x + 9)/x)^2 + 36*e^(e^2)*log(x))*e^(4*e^(e^2)/log(3*(4*x + 9)/x))/((4*x^2 + 9*x)
*log(3*(4*x + 9)/x)^2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^{{\mathrm {e}}^2+2\,\ln \relax (2)}}{\ln \left (\frac {12\,x+27}{x}\right )}}\,\left (\left (4\,x+9\right )\,{\ln \left (\frac {12\,x+27}{x}\right )}^2+9\,{\mathrm {e}}^{{\mathrm {e}}^2+2\,\ln \relax (2)}\,\ln \relax (x)\right )}{{\ln \left (\frac {12\,x+27}{x}\right )}^2\,\left (4\,x^2+9\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(exp(2) + 2*log(2))/log((12*x + 27)/x))*(9*exp(exp(2) + 2*log(2))*log(x) + log((12*x + 27)/x)^2*(4
*x + 9)))/(log((12*x + 27)/x)^2*(9*x + 4*x^2)),x)

[Out]

int((exp(exp(exp(2) + 2*log(2))/log((12*x + 27)/x))*(9*exp(exp(2) + 2*log(2))*log(x) + log((12*x + 27)/x)^2*(4
*x + 9)))/(log((12*x + 27)/x)^2*(9*x + 4*x^2)), x)

________________________________________________________________________________________

sympy [A]  time = 8.18, size = 19, normalized size = 0.70 \begin {gather*} e^{\frac {4 e^{e^{2}}}{\log {\left (\frac {12 x + 27}{x} \right )}}} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*exp(2*ln(2)+exp(2))*ln(x)+(4*x+9)*ln((12*x+27)/x)**2)*exp(exp(2*ln(2)+exp(2))/ln((12*x+27)/x))/(4
*x**2+9*x)/ln((12*x+27)/x)**2,x)

[Out]

exp(4*exp(exp(2))/log((12*x + 27)/x))*log(x)

________________________________________________________________________________________