\(\int \frac {e^{\frac {4 e^{e^2}}{\log (\frac {27+12 x}{x})}} (36 e^{e^2} \log (x)+(9+4 x) \log ^2(\frac {27+12 x}{x}))}{(9 x+4 x^2) \log ^2(\frac {27+12 x}{x})} \, dx\) [2555]

Optimal result

Integrand size = 73, antiderivative size = 27 \[ \int \frac {e^{\frac {4 e^{e^2}}{\log \left (\frac {27+12 x}{x}\right )}} \left (36 e^{e^2} \log (x)+(9+4 x) \log ^2\left (\frac {27+12 x}{x}\right )\right )}{\left (9 x+4 x^2\right ) \log ^2\left (\frac {27+12 x}{x}\right )} \, dx=e^{\frac {4 e^{e^2}}{\log \left (\frac {9 \left (3+\frac {4 x}{3}\right )}{x}\right )}} \log (x) \]

[Out]

exp(exp(2*ln(2)+exp(2))/ln(9*(3+4/3*x)/x))*ln(x)

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {1607, 2326} \[ \int \frac {e^{\frac {4 e^{e^2}}{\log \left (\frac {27+12 x}{x}\right )}} \left (36 e^{e^2} \log (x)+(9+4 x) \log ^2\left (\frac {27+12 x}{x}\right )\right )}{\left (9 x+4 x^2\right ) \log ^2\left (\frac {27+12 x}{x}\right )} \, dx=-\frac {9 e^{\frac {4 e^{e^2}}{\log \left (\frac {3 (4 x+9)}{x}\right )}} \log (x)}{x^2 \left (\frac {4}{x}-\frac {4 x+9}{x^2}\right )} \]

[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 1607

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 2326

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^{\frac {4 e^{e^2}}{\log \left (\frac {27+12 x}{x}\right )}} \left (36 e^{e^2} \log (x)+(9+4 x) \log ^2\left (\frac {27+12 x}{x}\right )\right )}{\left (9 x+4 x^2\right ) \log ^2\left (\frac {27+12 x}{x}\right )} \, dx=e^{\frac {4 e^{e^2}}{\log \left (12+\frac {27}{x}\right )}} \log (x) \]

[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]

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 9.79 (sec) , antiderivative size = 121, normalized size of antiderivative = 4.48

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

[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(x)+4*ln(2)+2*ln(3)+2*ln(x+9/4)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {4 e^{e^2}}{\log \left (\frac {27+12 x}{x}\right )}} \left (36 e^{e^2} \log (x)+(9+4 x) \log ^2\left (\frac {27+12 x}{x}\right )\right )}{\left (9 x+4 x^2\right ) \log ^2\left (\frac {27+12 x}{x}\right )} \, dx=e^{\left (\frac {e^{\left (e^{2} + 2 \, \log \left (2\right )\right )}}{\log \left (\frac {3 \, {\left (4 \, x + 9\right )}}{x}\right )}\right )} \log \left (x\right ) \]

[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)

Sympy [A] (verification not implemented)

Time = 5.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {e^{\frac {4 e^{e^2}}{\log \left (\frac {27+12 x}{x}\right )}} \left (36 e^{e^2} \log (x)+(9+4 x) \log ^2\left (\frac {27+12 x}{x}\right )\right )}{\left (9 x+4 x^2\right ) \log ^2\left (\frac {27+12 x}{x}\right )} \, dx=e^{\frac {4 e^{e^{2}}}{\log {\left (\frac {12 x + 27}{x} \right )}}} \log {\left (x \right )} \]

[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)

Maxima [F]

\[ \int \frac {e^{\frac {4 e^{e^2}}{\log \left (\frac {27+12 x}{x}\right )}} \left (36 e^{e^2} \log (x)+(9+4 x) \log ^2\left (\frac {27+12 x}{x}\right )\right )}{\left (9 x+4 x^2\right ) \log ^2\left (\frac {27+12 x}{x}\right )} \, dx=\int { \frac {{\left ({\left (4 \, x + 9\right )} \log \left (\frac {3 \, {\left (4 \, x + 9\right )}}{x}\right )^{2} + 9 \, e^{\left (e^{2} + 2 \, \log \left (2\right )\right )} \log \left (x\right )\right )} e^{\left (\frac {e^{\left (e^{2} + 2 \, \log \left (2\right )\right )}}{\log \left (\frac {3 \, {\left (4 \, x + 9\right )}}{x}\right )}\right )}}{{\left (4 \, x^{2} + 9 \, x\right )} \log \left (\frac {3 \, {\left (4 \, x + 9\right )}}{x}\right )^{2}} \,d x } \]

[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)

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {4 e^{e^2}}{\log \left (\frac {27+12 x}{x}\right )}} \left (36 e^{e^2} \log (x)+(9+4 x) \log ^2\left (\frac {27+12 x}{x}\right )\right )}{\left (9 x+4 x^2\right ) \log ^2\left (\frac {27+12 x}{x}\right )} \, dx=\text {Exception raised: TypeError} \]

[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:Unable to divide, perhaps due to rounding error%%%{186624,[0,0,1,3]%%%}+%%%{419904,[0,0,0,3]%%%} / %%%{1866
24,[0,0,2,3

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {4 e^{e^2}}{\log \left (\frac {27+12 x}{x}\right )}} \left (36 e^{e^2} \log (x)+(9+4 x) \log ^2\left (\frac {27+12 x}{x}\right )\right )}{\left (9 x+4 x^2\right ) \log ^2\left (\frac {27+12 x}{x}\right )} \, dx=\int \frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^{{\mathrm {e}}^2+2\,\ln \left (2\right )}}{\ln \left (\frac {12\,x+27}{x}\right )}}\,\left (\left (4\,x+9\right )\,{\ln \left (\frac {12\,x+27}{x}\right )}^2+9\,{\mathrm {e}}^{{\mathrm {e}}^2+2\,\ln \left (2\right )}\,\ln \left (x\right )\right )}{{\ln \left (\frac {12\,x+27}{x}\right )}^2\,\left (4\,x^2+9\,x\right )} \,d x \]

[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)