Integrand size = 117, antiderivative size = 27 \[ \int \frac {405+810 x+270 \log (4)+(405+810 x+270 \log (4)) \log (5)}{2916+972 x+1053 x^2+162 x^3+81 x^4+\left (648 x+108 x^2+108 x^3\right ) \log (4)+\left (108+18 x+54 x^2\right ) \log ^2(4)+12 x \log ^3(4)+\log ^4(4)+\left (5832+972 x+972 x^2+648 x \log (4)+108 \log ^2(4)\right ) \log (5)+2916 \log ^2(5)} \, dx=3-\frac {5}{6+\frac {x+\left (x+\frac {\log (4)}{3}\right )^2}{1+\log (5)}} \] Output:
3-5/(6+((x+2/3*ln(2))^2+x)/(ln(5)+1))
Leaf count is larger than twice the leaf count of optimal. \(69\) vs. \(2(27)=54\).
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.56 \[ \int \frac {405+810 x+270 \log (4)+(405+810 x+270 \log (4)) \log (5)}{2916+972 x+1053 x^2+162 x^3+81 x^4+\left (648 x+108 x^2+108 x^3\right ) \log (4)+\left (108+18 x+54 x^2\right ) \log ^2(4)+12 x \log ^3(4)+\log ^4(4)+\left (5832+972 x+972 x^2+648 x \log (4)+108 \log ^2(4)\right ) \log (5)+2916 \log ^2(5)} \, dx=\frac {15 (1+\log (5)) \left (207-6 \log (4)+4 \log ^2(4)+216 \log (5)-3 \log (16)-2 \log (4) \log (16)\right )}{(-69+4 \log (4)-72 \log (5)) \left (54+9 x+9 x^2+6 x \log (4)+\log ^2(4)+54 \log (5)\right )} \] Input:
Integrate[(405 + 810*x + 270*Log[4] + (405 + 810*x + 270*Log[4])*Log[5])/( 2916 + 972*x + 1053*x^2 + 162*x^3 + 81*x^4 + (648*x + 108*x^2 + 108*x^3)*L og[4] + (108 + 18*x + 54*x^2)*Log[4]^2 + 12*x*Log[4]^3 + Log[4]^4 + (5832 + 972*x + 972*x^2 + 648*x*Log[4] + 108*Log[4]^2)*Log[5] + 2916*Log[5]^2),x ]
Output:
(15*(1 + Log[5])*(207 - 6*Log[4] + 4*Log[4]^2 + 216*Log[5] - 3*Log[16] - 2 *Log[4]*Log[16]))/((-69 + 4*Log[4] - 72*Log[5])*(54 + 9*x + 9*x^2 + 6*x*Lo g[4] + Log[4]^2 + 54*Log[5]))
Time = 0.31 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {6, 2459, 27, 27, 1380, 27, 241}
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.
| \(\displaystyle \int \frac {810 x+\log (5) (810 x+405+270 \log (4))+405+270 \log (4)}{81 x^4+162 x^3+1053 x^2+\log (5) \left (972 x^2+972 x+648 x \log (4)+5832+108 \log ^2(4)\right )+\left (54 x^2+18 x+108\right ) \log ^2(4)+\left (108 x^3+108 x^2+648 x\right ) \log (4)+972 x+12 x \log ^3(4)+2916+\log ^4(4)+2916 \log ^2(5)} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {810 x+\log (5) (810 x+405+270 \log (4))+405+270 \log (4)}{81 x^4+162 x^3+1053 x^2+\log (5) \left (972 x^2+972 x+648 x \log (4)+5832+108 \log ^2(4)\right )+\left (54 x^2+18 x+108\right ) \log ^2(4)+\left (108 x^3+108 x^2+648 x\right ) \log (4)+x \left (972+12 \log ^3(4)\right )+2916+\log ^4(4)+2916 \log ^2(5)}dx\) |
| \(\Big \downarrow \) 2459 |
| \(\displaystyle \int \frac {810 (1+\log (5)) \left (x+\frac {1}{324} (162+108 \log (4))\right )}{81 \left (x+\frac {1}{324} (162+108 \log (4))\right )^4+\frac {27}{2} (69-4 \log (4)+72 \log (5)) \left (x+\frac {1}{324} (162+108 \log (4))\right )^2+\frac {9}{16} (69+72 \log (5)-\log (256))^2}d\left (x+\frac {1}{324} (162+108 \log (4))\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 810 (1+\log (5)) \int \frac {16 \left (x+\frac {1}{324} (162+108 \log (4))\right )}{9 \left (144 \left (x+\frac {1}{324} (162+108 \log (4))\right )^4+24 (69-4 \log (4)+72 \log (5)) \left (x+\frac {1}{324} (162+108 \log (4))\right )^2+(-69-72 \log (5)+\log (256))^2\right )}d\left (x+\frac {1}{324} (162+108 \log (4))\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 1440 (1+\log (5)) \int \frac {x+\frac {1}{324} (162+108 \log (4))}{144 \left (x+\frac {1}{324} (162+108 \log (4))\right )^4+24 (69-4 \log (4)+72 \log (5)) \left (x+\frac {1}{324} (162+108 \log (4))\right )^2+(-69-72 \log (5)+\log (256))^2}d\left (x+\frac {1}{324} (162+108 \log (4))\right )\) |
| \(\Big \downarrow \) 1380 |
| \(\displaystyle 207360 (1+\log (5)) \int \frac {x+\frac {1}{324} (162+108 \log (4))}{144 \left (12 \left (x+\frac {1}{324} (162+108 \log (4))\right )^2+72 \log (5)-4 \log (4)+69\right )^2}d\left (x+\frac {1}{324} (162+108 \log (4))\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 1440 (1+\log (5)) \int \frac {x+\frac {1}{324} (162+108 \log (4))}{\left (12 \left (x+\frac {1}{324} (162+108 \log (4))\right )^2+72 \log (5)-4 \log (4)+69\right )^2}d\left (x+\frac {1}{324} (162+108 \log (4))\right )\) |
| \(\Big \downarrow \) 241 |
| \(\displaystyle -\frac {60 (1+\log (5))}{12 \left (x+\frac {1}{324} (162+108 \log (4))\right )^2+69+72 \log (5)-4 \log (4)}\) |
Input:
Int[(405 + 810*x + 270*Log[4] + (405 + 810*x + 270*Log[4])*Log[5])/(2916 + 972*x + 1053*x^2 + 162*x^3 + 81*x^4 + (648*x + 108*x^2 + 108*x^3)*Log[4] + (108 + 18*x + 54*x^2)*Log[4]^2 + 12*x*Log[4]^3 + Log[4]^4 + (5832 + 972* x + 972*x^2 + 648*x*Log[4] + 108*Log[4]^2)*Log[5] + 2916*Log[5]^2),x]
Output:
(-60*(1 + Log[5]))/(69 - 4*Log[4] + 12*(x + (162 + 108*Log[4])/324)^2 + 72 *Log[5])
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[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ (2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -1]
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Int[(Pn_)^(p_.)*(Qx_), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1 ]/(Expon[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x - S, x]^p*ExpandToSum[Qx /. x -> x - S, x], x], x, x + S] /; Binomial Q[Pn /. x -> x - S, x] || (IntegerQ[Expon[Pn, x]/2] && TrinomialQ[Pn /. x - > x - S, x])] /; FreeQ[p, x] && PolyQ[Pn, x] && GtQ[Expon[Pn, x], 2] && NeQ [Coeff[Pn, x, Expon[Pn, x] - 1], 0] && PolyQ[Qx, x] && !(MonomialQ[Qx, x] && IGtQ[p, 0])
Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26
| method | result | size |
| gosper | \(-\frac {45 \left (\ln \left (5\right )+1\right )}{4 \ln \left (2\right )^{2}+12 x \ln \left (2\right )+9 x^{2}+54 \ln \left (5\right )+9 x +54}\) | \(34\) |
| norman | \(\frac {-45-45 \ln \left (5\right )}{4 \ln \left (2\right )^{2}+12 x \ln \left (2\right )+9 x^{2}+54 \ln \left (5\right )+9 x +54}\) | \(35\) |
| parallelrisch | \(\frac {-405-405 \ln \left (5\right )}{36 \ln \left (2\right )^{2}+108 x \ln \left (2\right )+81 x^{2}+486 \ln \left (5\right )+81 x +486}\) | \(36\) |
| risch | \(-\frac {45 \ln \left (5\right )}{4 \left (\ln \left (2\right )^{2}+3 x \ln \left (2\right )+\frac {9 x^{2}}{4}+\frac {27 \ln \left (5\right )}{2}+\frac {9 x}{4}+\frac {27}{2}\right )}-\frac {45}{4 \left (\ln \left (2\right )^{2}+3 x \ln \left (2\right )+\frac {9 x^{2}}{4}+\frac {27 \ln \left (5\right )}{2}+\frac {9 x}{4}+\frac {27}{2}\right )}\) | \(58\) |
| default | \(\frac {\left (135 \ln \left (5\right )+135\right ) \left (-648+\left (4 \ln \left (2\right )+3\right ) \left (12 \ln \left (2\right )+9\right )-48 \ln \left (2\right )^{2}-648 \ln \left (5\right )\right )}{\left (-216 \ln \left (2\right )+1944 \ln \left (5\right )+1863\right ) \left (4 \ln \left (2\right )^{2}+12 x \ln \left (2\right )+9 x^{2}+54 \ln \left (5\right )+9 x +54\right )}\) | \(72\) |
Input:
int(((540*ln(2)+810*x+405)*ln(5)+540*ln(2)+810*x+405)/(2916*ln(5)^2+(432*l n(2)^2+1296*x*ln(2)+972*x^2+972*x+5832)*ln(5)+16*ln(2)^4+96*x*ln(2)^3+4*(5 4*x^2+18*x+108)*ln(2)^2+2*(108*x^3+108*x^2+648*x)*ln(2)+81*x^4+162*x^3+105 3*x^2+972*x+2916),x,method=_RETURNVERBOSE)
Output:
-45*(ln(5)+1)/(4*ln(2)^2+12*x*ln(2)+9*x^2+54*ln(5)+9*x+54)
Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {405+810 x+270 \log (4)+(405+810 x+270 \log (4)) \log (5)}{2916+972 x+1053 x^2+162 x^3+81 x^4+\left (648 x+108 x^2+108 x^3\right ) \log (4)+\left (108+18 x+54 x^2\right ) \log ^2(4)+12 x \log ^3(4)+\log ^4(4)+\left (5832+972 x+972 x^2+648 x \log (4)+108 \log ^2(4)\right ) \log (5)+2916 \log ^2(5)} \, dx=-\frac {45 \, {\left (\log \left (5\right ) + 1\right )}}{9 \, x^{2} + 12 \, x \log \left (2\right ) + 4 \, \log \left (2\right )^{2} + 9 \, x + 54 \, \log \left (5\right ) + 54} \] Input:
integrate(((540*log(2)+810*x+405)*log(5)+540*log(2)+810*x+405)/(2916*log(5 )^2+(432*log(2)^2+1296*x*log(2)+972*x^2+972*x+5832)*log(5)+16*log(2)^4+96* x*log(2)^3+4*(54*x^2+18*x+108)*log(2)^2+2*(108*x^3+108*x^2+648*x)*log(2)+8 1*x^4+162*x^3+1053*x^2+972*x+2916),x, algorithm="fricas")
Output:
-45*(log(5) + 1)/(9*x^2 + 12*x*log(2) + 4*log(2)^2 + 9*x + 54*log(5) + 54)
Time = 0.65 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {405+810 x+270 \log (4)+(405+810 x+270 \log (4)) \log (5)}{2916+972 x+1053 x^2+162 x^3+81 x^4+\left (648 x+108 x^2+108 x^3\right ) \log (4)+\left (108+18 x+54 x^2\right ) \log ^2(4)+12 x \log ^3(4)+\log ^4(4)+\left (5832+972 x+972 x^2+648 x \log (4)+108 \log ^2(4)\right ) \log (5)+2916 \log ^2(5)} \, dx=\frac {- 45 \log {\left (5 \right )} - 45}{9 x^{2} + x \left (12 \log {\left (2 \right )} + 9\right ) + 4 \log {\left (2 \right )}^{2} + 54 + 54 \log {\left (5 \right )}} \] Input:
integrate(((540*ln(2)+810*x+405)*ln(5)+540*ln(2)+810*x+405)/(2916*ln(5)**2 +(432*ln(2)**2+1296*x*ln(2)+972*x**2+972*x+5832)*ln(5)+16*ln(2)**4+96*x*ln (2)**3+4*(54*x**2+18*x+108)*ln(2)**2+2*(108*x**3+108*x**2+648*x)*ln(2)+81* x**4+162*x**3+1053*x**2+972*x+2916),x)
Output:
(-45*log(5) - 45)/(9*x**2 + x*(12*log(2) + 9) + 4*log(2)**2 + 54 + 54*log( 5))
Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {405+810 x+270 \log (4)+(405+810 x+270 \log (4)) \log (5)}{2916+972 x+1053 x^2+162 x^3+81 x^4+\left (648 x+108 x^2+108 x^3\right ) \log (4)+\left (108+18 x+54 x^2\right ) \log ^2(4)+12 x \log ^3(4)+\log ^4(4)+\left (5832+972 x+972 x^2+648 x \log (4)+108 \log ^2(4)\right ) \log (5)+2916 \log ^2(5)} \, dx=-\frac {45 \, {\left (\log \left (5\right ) + 1\right )}}{9 \, x^{2} + 3 \, x {\left (4 \, \log \left (2\right ) + 3\right )} + 4 \, \log \left (2\right )^{2} + 54 \, \log \left (5\right ) + 54} \] Input:
integrate(((540*log(2)+810*x+405)*log(5)+540*log(2)+810*x+405)/(2916*log(5 )^2+(432*log(2)^2+1296*x*log(2)+972*x^2+972*x+5832)*log(5)+16*log(2)^4+96* x*log(2)^3+4*(54*x^2+18*x+108)*log(2)^2+2*(108*x^3+108*x^2+648*x)*log(2)+8 1*x^4+162*x^3+1053*x^2+972*x+2916),x, algorithm="maxima")
Output:
-45*(log(5) + 1)/(9*x^2 + 3*x*(4*log(2) + 3) + 4*log(2)^2 + 54*log(5) + 54 )
Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (29) = 58\).
Time = 0.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.63 \[ \int \frac {405+810 x+270 \log (4)+(405+810 x+270 \log (4)) \log (5)}{2916+972 x+1053 x^2+162 x^3+81 x^4+\left (648 x+108 x^2+108 x^3\right ) \log (4)+\left (108+18 x+54 x^2\right ) \log ^2(4)+12 x \log ^3(4)+\log ^4(4)+\left (5832+972 x+972 x^2+648 x \log (4)+108 \log ^2(4)\right ) \log (5)+2916 \log ^2(5)} \, dx=-\frac {45 \, {\left (\log \left (5\right )^{2} + 2 \, \log \left (5\right ) + 1\right )}}{4 \, \log \left (5\right ) \log \left (2\right )^{2} + 9 \, x^{2} + 3 \, {\left (3 \, x^{2} + 4 \, x \log \left (2\right ) + 3 \, x\right )} \log \left (5\right ) + 54 \, \log \left (5\right )^{2} + 12 \, x \log \left (2\right ) + 4 \, \log \left (2\right )^{2} + 9 \, x + 108 \, \log \left (5\right ) + 54} \] Input:
integrate(((540*log(2)+810*x+405)*log(5)+540*log(2)+810*x+405)/(2916*log(5 )^2+(432*log(2)^2+1296*x*log(2)+972*x^2+972*x+5832)*log(5)+16*log(2)^4+96* x*log(2)^3+4*(54*x^2+18*x+108)*log(2)^2+2*(108*x^3+108*x^2+648*x)*log(2)+8 1*x^4+162*x^3+1053*x^2+972*x+2916),x, algorithm="giac")
Output:
-45*(log(5)^2 + 2*log(5) + 1)/(4*log(5)*log(2)^2 + 9*x^2 + 3*(3*x^2 + 4*x* log(2) + 3*x)*log(5) + 54*log(5)^2 + 12*x*log(2) + 4*log(2)^2 + 9*x + 108* log(5) + 54)
Time = 10.30 (sec) , antiderivative size = 2558, normalized size of antiderivative = 94.74 \[ \int \frac {405+810 x+270 \log (4)+(405+810 x+270 \log (4)) \log (5)}{2916+972 x+1053 x^2+162 x^3+81 x^4+\left (648 x+108 x^2+108 x^3\right ) \log (4)+\left (108+18 x+54 x^2\right ) \log ^2(4)+12 x \log ^3(4)+\log ^4(4)+\left (5832+972 x+972 x^2+648 x \log (4)+108 \log ^2(4)\right ) \log (5)+2916 \log ^2(5)} \, dx=\text {Too large to display} \] Input:
int((810*x + 540*log(2) + log(5)*(810*x + 540*log(2) + 405) + 405)/(972*x + 2*log(2)*(648*x + 108*x^2 + 108*x^3) + 4*log(2)^2*(18*x + 54*x^2 + 108) + 96*x*log(2)^3 + log(5)*(972*x + 1296*x*log(2) + 432*log(2)^2 + 972*x^2 + 5832) + 16*log(2)^4 + 2916*log(5)^2 + 1053*x^2 + 162*x^3 + 81*x^4 + 2916) ,x)
Output:
symsum(log(4009802061150*root(418066920000*log(2)*log(5)^3 - 9447840000*lo g(2)^2*log(5) + 210804930000*log(2)*log(5)^4 - 2361960000*log(2)^2*log(5)^ 4 + 42515280000*log(2)*log(5)^5 - 14171760000*log(2)^2*log(5)^2 - 94478400 00*log(2)^2*log(5)^3 + 414523980000*log(2)*log(5)^2 + 205490520000*log(2)* log(5) - 2790397400625*log(5)^4 - 1069525012500*log(5) - 1131969330000*log (5)^5 - 2361960000*log(2)^2 - 191318760000*log(5)^6 - 2712342003750*log(5) ^2 + 40743810000*log(2) - 3668271502500*log(5)^3 - 175707680625, z, k) + 3 486784401000*x + 2324522934000*log(2) + 5230176601500*log(5) + 48814981614 00*root(418066920000*log(2)*log(5)^3 - 9447840000*log(2)^2*log(5) + 210804 930000*log(2)*log(5)^4 - 2361960000*log(2)^2*log(5)^4 + 42515280000*log(2) *log(5)^5 - 14171760000*log(2)^2*log(5)^2 - 9447840000*log(2)^2*log(5)^3 + 414523980000*log(2)*log(5)^2 + 205490520000*log(2)*log(5) - 2790397400625 *log(5)^4 - 1069525012500*log(5) - 1131969330000*log(5)^5 - 2361960000*log (2)^2 - 191318760000*log(5)^6 - 2712342003750*log(5)^2 + 40743810000*log(2 ) - 3668271502500*log(5)^3 - 175707680625, z, k)*log(2) + 12203745403500*r oot(418066920000*log(2)*log(5)^3 - 9447840000*log(2)^2*log(5) + 2108049300 00*log(2)*log(5)^4 - 2361960000*log(2)^2*log(5)^4 + 42515280000*log(2)*log (5)^5 - 14171760000*log(2)^2*log(5)^2 - 9447840000*log(2)^2*log(5)^3 + 414 523980000*log(2)*log(5)^2 + 205490520000*log(2)*log(5) - 2790397400625*log (5)^4 - 1069525012500*log(5) - 1131969330000*log(5)^5 - 2361960000*log(...
Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {405+810 x+270 \log (4)+(405+810 x+270 \log (4)) \log (5)}{2916+972 x+1053 x^2+162 x^3+81 x^4+\left (648 x+108 x^2+108 x^3\right ) \log (4)+\left (108+18 x+54 x^2\right ) \log ^2(4)+12 x \log ^3(4)+\log ^4(4)+\left (5832+972 x+972 x^2+648 x \log (4)+108 \log ^2(4)\right ) \log (5)+2916 \log ^2(5)} \, dx=\frac {-45 \,\mathrm {log}\left (5\right )-45}{54 \,\mathrm {log}\left (5\right )+4 \mathrm {log}\left (2\right )^{2}+12 \,\mathrm {log}\left (2\right ) x +9 x^{2}+9 x +54} \] Input:
int(((540*log(2)+810*x+405)*log(5)+540*log(2)+810*x+405)/(2916*log(5)^2+(4 32*log(2)^2+1296*x*log(2)+972*x^2+972*x+5832)*log(5)+16*log(2)^4+96*x*log( 2)^3+4*(54*x^2+18*x+108)*log(2)^2+2*(108*x^3+108*x^2+648*x)*log(2)+81*x^4+ 162*x^3+1053*x^2+972*x+2916),x)
Output:
( - 45*(log(5) + 1))/(54*log(5) + 4*log(2)**2 + 12*log(2)*x + 9*x**2 + 9*x + 54)