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)
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))
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)
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]
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]
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]
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])
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])
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)
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)
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)...
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
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)
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
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))