Integrand size = 94, antiderivative size = 25 \[ \int \frac {36+48 x+16 x^2+4 e^3 x^2+e^{3/2} \left (24 x+16 x^2\right )+\left (2 x^2+e^{3/2} x^2\right ) \log (3)}{9 x^2+12 x^3+4 x^4+e^3 x^4+e^{3/2} \left (6 x^3+4 x^4\right )} \, dx=-2-\frac {4}{x}-\frac {\log (3)}{3+2 x+e^{3/2} x} \]
Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {36+48 x+16 x^2+4 e^3 x^2+e^{3/2} \left (24 x+16 x^2\right )+\left (2 x^2+e^{3/2} x^2\right ) \log (3)}{9 x^2+12 x^3+4 x^4+e^3 x^4+e^{3/2} \left (6 x^3+4 x^4\right )} \, dx=-\frac {4}{x}-\frac {e^{3/2} \log (3)+\log (9)}{\left (2+e^{3/2}\right ) \left (3+\left (2+e^{3/2}\right ) x\right )} \]
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]
Time = 0.36 (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 |
\(\displaystyle -\frac {4}{x}-\frac {\log (3)}{\left (2+e^{3/2}\right ) x+3}\) |
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]
3.2.93.3.1 Defintions of rubi rules used
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 = 0.24 (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 | \(\frac {\ln \left (3\right ) {\mathrm e}^{\frac {3}{2}} x^{2}+4 x^{2} {\mathrm e}^{3}-36+2 x^{2} \ln \left (3\right )+16 x^{2} {\mathrm e}^{\frac {3}{2}}+16 x^{2}}{3 x \left (3+x \,{\mathrm e}^{\frac {3}{2}}+2 x \right )}\) | \(55\) |
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)
Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {36+48 x+16 x^2+4 e^3 x^2+e^{3/2} \left (24 x+16 x^2\right )+\left (2 x^2+e^{3/2} x^2\right ) \log (3)}{9 x^2+12 x^3+4 x^4+e^3 x^4+e^{3/2} \left (6 x^3+4 x^4\right )} \, dx=-\frac {4 \, x e^{\frac {3}{2}} + x \log \left (3\right ) + 8 \, x + 12}{x^{2} e^{\frac {3}{2}} + 2 \, x^{2} + 3 \, x} \]
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=\
Leaf count of result is larger than twice the leaf count of optimal. 1023 vs. \(2 (20) = 40\).
Time = 3.99 (sec) , antiderivative size = 1023, normalized size of antiderivative = 40.92 \[ \int \frac {36+48 x+16 x^2+4 e^3 x^2+e^{3/2} \left (24 x+16 x^2\right )+\left (2 x^2+e^{3/2} x^2\right ) \log (3)}{9 x^2+12 x^3+4 x^4+e^3 x^4+e^{3/2} \left (6 x^3+4 x^4\right )} \, dx=\text {Too large to display} \]
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)
(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 = 0.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {36+48 x+16 x^2+4 e^3 x^2+e^{3/2} \left (24 x+16 x^2\right )+\left (2 x^2+e^{3/2} x^2\right ) \log (3)}{9 x^2+12 x^3+4 x^4+e^3 x^4+e^{3/2} \left (6 x^3+4 x^4\right )} \, dx=-\frac {{\left (e^{\frac {3}{2}} + 2\right )} \log \left (3\right )}{x {\left (e^{3} + 4 \, e^{\frac {3}{2}} + 4\right )} + 3 \, e^{\frac {3}{2}} + 6} - \frac {4}{x} \]
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=\
Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {36+48 x+16 x^2+4 e^3 x^2+e^{3/2} \left (24 x+16 x^2\right )+\left (2 x^2+e^{3/2} x^2\right ) \log (3)}{9 x^2+12 x^3+4 x^4+e^3 x^4+e^{3/2} \left (6 x^3+4 x^4\right )} \, dx=-\frac {4 \, x e^{\frac {3}{2}} + x \log \left (3\right ) + 8 \, x + 12}{x^{2} e^{\frac {3}{2}} + 2 \, x^{2} + 3 \, x} \]
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=\
Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {36+48 x+16 x^2+4 e^3 x^2+e^{3/2} \left (24 x+16 x^2\right )+\left (2 x^2+e^{3/2} x^2\right ) \log (3)}{9 x^2+12 x^3+4 x^4+e^3 x^4+e^{3/2} \left (6 x^3+4 x^4\right )} \, dx=-\frac {\ln \left (3\right )}{x\,\left ({\mathrm {e}}^{3/2}+2\right )+3}-\frac {4}{x} \]