\(\int \frac {36+48 x+16 x^2+4 e^3 x^2+e^{3/2} (24 x+16 x^2)+(2 x^2+e^{3/2} x^2) \log (3)}{9 x^2+12 x^3+4 x^4+e^3 x^4+e^{3/2} (6 x^3+4 x^4)} \, dx\) [193]

Optimal result

Integrand size = 94, antiderivative size = 25 \[ \int \frac {36+48 x+16 x^2+4 e^3 x^2+e^{3/2} \left (24 x+16 x^2\right )+\left (2 x^2+e^{3/2} x^2\right ) \log (3)}{9 x^2+12 x^3+4 x^4+e^3 x^4+e^{3/2} \left (6 x^3+4 x^4\right )} \, dx=-2-\frac {4}{x}-\frac {\log (3)}{3+2 x+e^{3/2} x} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {36+48 x+16 x^2+4 e^3 x^2+e^{3/2} \left (24 x+16 x^2\right )+\left (2 x^2+e^{3/2} x^2\right ) \log (3)}{9 x^2+12 x^3+4 x^4+e^3 x^4+e^{3/2} \left (6 x^3+4 x^4\right )} \, dx=-\frac {4}{x}-\frac {e^{3/2} \log (3)+\log (9)}{\left (2+e^{3/2}\right ) \left (3+\left (2+e^{3/2}\right ) x\right )} \] Input:

Integrate[(36 + 48*x + 16*x^2 + 4*E^3*x^2 + E^(3/2)*(24*x + 16*x^2) + (2*x 
^2 + E^(3/2)*x^2)*Log[3])/(9*x^2 + 12*x^3 + 4*x^4 + E^3*x^4 + E^(3/2)*(6*x 
^3 + 4*x^4)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {6, 6, 2026, 2007, 2082, 1195, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {4 e^3 x^2+16 x^2+e^{3/2} \left (16 x^2+24 x\right )+\left (e^{3/2} x^2+2 x^2\right ) \log (3)+48 x+36}{e^3 x^4+4 x^4+12 x^3+9 x^2+e^{3/2} \left (4 x^4+6 x^3\right )} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {4 e^3 x^2+16 x^2+e^{3/2} \left (16 x^2+24 x\right )+\left (e^{3/2} x^2+2 x^2\right ) \log (3)+48 x+36}{\left (4+e^3\right ) x^4+12 x^3+9 x^2+e^{3/2} \left (4 x^4+6 x^3\right )}dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {\left (16+4 e^3\right ) x^2+e^{3/2} \left (16 x^2+24 x\right )+\left (e^{3/2} x^2+2 x^2\right ) \log (3)+48 x+36}{\left (4+e^3\right ) x^4+12 x^3+9 x^2+e^{3/2} \left (4 x^4+6 x^3\right )}dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {\left (16+4 e^3\right ) x^2+e^{3/2} \left (16 x^2+24 x\right )+\left (e^{3/2} x^2+2 x^2\right ) \log (3)+48 x+36}{x^2 \left (\left (2+e^{3/2}\right )^2 x^2+6 \left (2+e^{3/2}\right ) x+9\right )}dx\)

\(\Big \downarrow \) 2007

\(\displaystyle \int \frac {\left (16+4 e^3\right ) x^2+e^{3/2} \left (16 x^2+24 x\right )+\left (e^{3/2} x^2+2 x^2\right ) \log (3)+48 x+36}{x^2 \left (\left (2+e^{3/2}\right ) x+3\right )^2}dx\)

\(\Big \downarrow \) 2082

\(\displaystyle \int \frac {\left (2+e^{3/2}\right ) x^2 \left (8+4 e^{3/2}+\log (3)\right )+24 \left (2+e^{3/2}\right ) x+36}{x^2 \left (\left (2+e^{3/2}\right ) x+3\right )^2}dx\)

\(\Big \downarrow \) 1195

\(\displaystyle \int \left (\frac {4}{x^2}+\frac {\left (2+e^{3/2}\right ) \log (3)}{\left (\left (2+e^{3/2}\right ) x+3\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {4}{x}-\frac {\log (3)}{\left (2+e^{3/2}\right ) x+3}\)

Input:

Int[(36 + 48*x + 16*x^2 + 4*E^3*x^2 + E^(3/2)*(24*x + 16*x^2) + (2*x^2 + E 
^(3/2)*x^2)*Log[3])/(9*x^2 + 12*x^3 + 4*x^4 + E^3*x^4 + E^(3/2)*(6*x^3 + 4 
*x^4)),x]
 

Output:

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

Defintions of rubi rules used

rule 6

Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v 
+ (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] &&  !FreeQ[Fx, x]
 

rule 1195

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x 
_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + 
 g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x 
] && IGtQ[p, 0]
 

rule 2007

Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, 
x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex 
pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol 
yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2026

Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p 
*r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ 
erQ[p] &&  !MonomialQ[Px, x] && (ILtQ[p, 0] ||  !PolyQ[u, x])
 

rule 2082

Int[(u_)^(m_.)*(v_)^(n_.)*(w_)^(p_.), x_Symbol] :> Int[ExpandToSum[u, x]^m* 
ExpandToSum[v, x]^n*ExpandToSum[w, x]^p, x] /; FreeQ[{m, n, p}, x] && Linea 
rQ[{u, v}, x] && QuadraticQ[w, x] &&  !(LinearMatchQ[{u, v}, x] && Quadrati 
cMatchQ[w, x])
 

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20

method	result	size
risch	\(\frac {\left (-4 \,{\mathrm e}^{\frac {3}{2}}-\ln \left (3\right )-8\right ) x -12}{x \left (3+x \,{\mathrm e}^{\frac {3}{2}}+2 x \right )}\)	\(30\)
gosper	\(-\frac {x \ln \left (3\right )+4 x \,{\mathrm e}^{\frac {3}{2}}+8 x +12}{x \left (3+x \,{\mathrm e}^{\frac {3}{2}}+2 x \right )}\)	\(31\)
norman	\(\frac {-12+\left (\frac {\ln \left (3\right ) {\mathrm e}^{\frac {3}{2}}}{3}+\frac {4 \,{\mathrm e}^{3}}{3}+\frac {2 \ln \left (3\right )}{3}+\frac {16 \,{\mathrm e}^{\frac {3}{2}}}{3}+\frac {16}{3}\right ) x^{2}}{x \left (3+x \,{\mathrm e}^{\frac {3}{2}}+2 x \right )}\)	\(44\)
parallelrisch	\(\frac {\ln \left (3\right ) {\mathrm e}^{\frac {3}{2}} x^{2}+4 x^{2} {\mathrm e}^{3}-36+2 x^{2} \ln \left (3\right )+16 x^{2} {\mathrm e}^{\frac {3}{2}}+16 x^{2}}{3 x \left (3+x \,{\mathrm e}^{\frac {3}{2}}+2 x \right )}\)	\(55\)

Input:

int(((x^2*exp(3/2)+2*x^2)*ln(3)+4*x^2*exp(3/2)^2+(16*x^2+24*x)*exp(3/2)+16 
*x^2+48*x+36)/(x^4*exp(3/2)^2+(4*x^4+6*x^3)*exp(3/2)+4*x^4+12*x^3+9*x^2),x 
,method=_RETURNVERBOSE)
 

Output:

((-4*exp(3/2)-ln(3)-8)*x-12)/x/(3+x*exp(3/2)+2*x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {36+48 x+16 x^2+4 e^3 x^2+e^{3/2} \left (24 x+16 x^2\right )+\left (2 x^2+e^{3/2} x^2\right ) \log (3)}{9 x^2+12 x^3+4 x^4+e^3 x^4+e^{3/2} \left (6 x^3+4 x^4\right )} \, dx=-\frac {4 \, x e^{\frac {3}{2}} + x \log \left (3\right ) + 8 \, x + 12}{x^{2} e^{\frac {3}{2}} + 2 \, x^{2} + 3 \, x} \] Input:

integrate(((x^2*exp(3/2)+2*x^2)*log(3)+4*x^2*exp(3/2)^2+(16*x^2+24*x)*exp( 
3/2)+16*x^2+48*x+36)/(x^4*exp(3/2)^2+(4*x^4+6*x^3)*exp(3/2)+4*x^4+12*x^3+9 
*x^2),x, algorithm="fricas")
 

Output:

-(4*x*e^(3/2) + x*log(3) + 8*x + 12)/(x^2*e^(3/2) + 2*x^2 + 3*x)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1023 vs. \(2 (20) = 40\).

Time = 3.52 (sec) , antiderivative size = 1023, normalized size of antiderivative = 40.92 \[ \int \frac {36+48 x+16 x^2+4 e^3 x^2+e^{3/2} \left (24 x+16 x^2\right )+\left (2 x^2+e^{3/2} x^2\right ) \log (3)}{9 x^2+12 x^3+4 x^4+e^3 x^4+e^{3/2} \left (6 x^3+4 x^4\right )} \, dx=\text {Too large to display} \] Input:

integrate(((x**2*exp(3/2)+2*x**2)*ln(3)+4*x**2*exp(3/2)**2+(16*x**2+24*x)* 
exp(3/2)+16*x**2+48*x+36)/(x**4*exp(3/2)**2+(4*x**4+6*x**3)*exp(3/2)+4*x** 
4+12*x**3+9*x**2),x)
 

Output:

(x*(-41287833600*exp(33) - 181666467840*exp(63/2) - 8078054400*exp(69/2) - 
 693635604480*exp(30) - 1346342400*exp(36) - 2312118681600*exp(57/2) - 188 
487936*exp(75/2) - 6758500761600*exp(27) - 17379001958400*exp(51/2) - 2174 
8608*exp(39) - 39392404439040*exp(24) - 2013760*exp(81/2) - 78784808878080 
*exp(45/2) - 7325260800*exp(63/2)*log(3) - 30766095360*exp(30)*log(3) - 14 
98348800*exp(33)*log(3) - 111876710400*exp(57/2)*log(3) - 260582400*exp(69 
/2)*log(3) - 143840*exp(42) - 354276249600*exp(27)*log(3) - 13903201566720 
0*exp(21) - 38001600*exp(36)*log(3) - 981072691200*exp(51/2)*log(3) - 4560 
192*exp(75/2)*log(3) - 2382605107200*exp(24)*log(3) - 216272024371200*exp( 
39/2) - 7440*exp(87/2) - 438480*exp(39)*log(3) - 5082890895360*exp(45/2)*l 
og(3) - 295951191244800*exp(18) - 32480*exp(81/2)*log(3) - 9530420428800*e 
xp(21)*log(3) - 248*exp(45) - 355141429493760*exp(33/2) - 15697163059200*e 
xp(39/2)*log(3) - 1740*exp(42)*log(3) - 22673679974400*exp(18)*log(3) - 37 
2052926136320*exp(15) - 4*exp(93/2) - 60*exp(87/2)*log(3) - 28640437862400 
*exp(33/2)*log(3) - 338229932851200*exp(27/2) - 31504481648640*exp(15)*log 
(3) - 264701686579200*exp(12) - exp(45)*log(3) - 30004268236800*exp(27/2)* 
log(3) - 176467791052800*exp(21/2) - 24548946739200*exp(12)*log(3) - 98821 
962989568*exp(9) - 17077528166400*exp(21/2)*log(3) - 9961891430400*exp(9)* 
log(3) - 45610136764416*exp(15/2) - 4781707886592*exp(15/2)*log(3) - 16892 
643246080*exp(6) - 1839118417920*exp(6)*log(3) - 4826469498880*exp(9/2)...
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {36+48 x+16 x^2+4 e^3 x^2+e^{3/2} \left (24 x+16 x^2\right )+\left (2 x^2+e^{3/2} x^2\right ) \log (3)}{9 x^2+12 x^3+4 x^4+e^3 x^4+e^{3/2} \left (6 x^3+4 x^4\right )} \, dx=-\frac {{\left (e^{\frac {3}{2}} + 2\right )} \log \left (3\right )}{x {\left (e^{3} + 4 \, e^{\frac {3}{2}} + 4\right )} + 3 \, e^{\frac {3}{2}} + 6} - \frac {4}{x} \] Input:

integrate(((x^2*exp(3/2)+2*x^2)*log(3)+4*x^2*exp(3/2)^2+(16*x^2+24*x)*exp( 
3/2)+16*x^2+48*x+36)/(x^4*exp(3/2)^2+(4*x^4+6*x^3)*exp(3/2)+4*x^4+12*x^3+9 
*x^2),x, algorithm="maxima")
 

Output:

-(e^(3/2) + 2)*log(3)/(x*(e^3 + 4*e^(3/2) + 4) + 3*e^(3/2) + 6) - 4/x
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {36+48 x+16 x^2+4 e^3 x^2+e^{3/2} \left (24 x+16 x^2\right )+\left (2 x^2+e^{3/2} x^2\right ) \log (3)}{9 x^2+12 x^3+4 x^4+e^3 x^4+e^{3/2} \left (6 x^3+4 x^4\right )} \, dx=-\frac {4 \, x e^{\frac {3}{2}} + x \log \left (3\right ) + 8 \, x + 12}{x^{2} e^{\frac {3}{2}} + 2 \, x^{2} + 3 \, x} \] Input:

integrate(((x^2*exp(3/2)+2*x^2)*log(3)+4*x^2*exp(3/2)^2+(16*x^2+24*x)*exp( 
3/2)+16*x^2+48*x+36)/(x^4*exp(3/2)^2+(4*x^4+6*x^3)*exp(3/2)+4*x^4+12*x^3+9 
*x^2),x, algorithm="giac")
 

Output:

-(4*x*e^(3/2) + x*log(3) + 8*x + 12)/(x^2*e^(3/2) + 2*x^2 + 3*x)
 

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {36+48 x+16 x^2+4 e^3 x^2+e^{3/2} \left (24 x+16 x^2\right )+\left (2 x^2+e^{3/2} x^2\right ) \log (3)}{9 x^2+12 x^3+4 x^4+e^3 x^4+e^{3/2} \left (6 x^3+4 x^4\right )} \, dx=-\frac {\ln \left (3\right )}{x\,\left ({\mathrm {e}}^{3/2}+2\right )+3}-\frac {4}{x} \] Input:

int((48*x + exp(3/2)*(24*x + 16*x^2) + 4*x^2*exp(3) + 16*x^2 + log(3)*(x^2 
*exp(3/2) + 2*x^2) + 36)/(exp(3/2)*(6*x^3 + 4*x^4) + x^4*exp(3) + 9*x^2 + 
12*x^3 + 4*x^4),x)
 

Output:

- log(3)/(x*(exp(3/2) + 2) + 3) - 4/x
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 108, normalized size of antiderivative = 4.32 \[ \int \frac {36+48 x+16 x^2+4 e^3 x^2+e^{3/2} \left (24 x+16 x^2\right )+\left (2 x^2+e^{3/2} x^2\right ) \log (3)}{9 x^2+12 x^3+4 x^4+e^3 x^4+e^{3/2} \left (6 x^3+4 x^4\right )} \, dx=\frac {-\sqrt {e}\, \mathrm {log}\left (3\right ) e^{4} x^{3}+4 \sqrt {e}\, \mathrm {log}\left (3\right ) e \,x^{3}+9 \sqrt {e}\, \mathrm {log}\left (3\right ) e x +2 \,\mathrm {log}\left (3\right ) e^{3} x^{3}-8 \,\mathrm {log}\left (3\right ) x^{3}+18 \,\mathrm {log}\left (3\right ) x -4 e^{6} x^{3}+32 e^{3} x^{3}+36 e^{3} x -64 x^{3}+432 x +432}{12 x \left (e^{3} x^{2}-4 x^{2}-12 x -9\right )} \] Input:

int(((x^2*exp(3/2)+2*x^2)*log(3)+4*x^2*exp(3/2)^2+(16*x^2+24*x)*exp(3/2)+1 
6*x^2+48*x+36)/(x^4*exp(3/2)^2+(4*x^4+6*x^3)*exp(3/2)+4*x^4+12*x^3+9*x^2), 
x)
 

Output:

( - sqrt(e)*log(3)*e**4*x**3 + 4*sqrt(e)*log(3)*e*x**3 + 9*sqrt(e)*log(3)* 
e*x + 2*log(3)*e**3*x**3 - 8*log(3)*x**3 + 18*log(3)*x - 4*e**6*x**3 + 32* 
e**3*x**3 + 36*e**3*x - 64*x**3 + 432*x + 432)/(12*x*(e**3*x**2 - 4*x**2 - 
 12*x - 9))