Integrand size = 125, antiderivative size = 26 \[ \int \frac {220 x^3+\left (27500 x-33000 e^4 x+9900 e^8 x+1090 x^2\right ) \log (3)+\left (-1250+1500 e^4-450 e^8-50 x\right ) \log ^2(3)+\left (2 x^3+\left (250 x-300 e^4 x+90 e^8 x+10 x^2\right ) \log (3)\right ) \log \left (25-30 e^4+9 e^8+x\right )}{625 x^3-750 e^4 x^3+225 e^8 x^3+25 x^4} \, dx=\left (-22+\frac {\log (3)}{x}-\frac {1}{5} \log \left (\left (5-3 e^4\right )^2+x\right )\right )^2 \] Output:
(ln(3)/x-22-1/5*ln((5-3*exp(4))^2+x))^2
Leaf count is larger than twice the leaf count of optimal. \(116\) vs. \(2(26)=52\).
Time = 0.13 (sec) , antiderivative size = 116, normalized size of antiderivative = 4.46 \[ \int \frac {220 x^3+\left (27500 x-33000 e^4 x+9900 e^8 x+1090 x^2\right ) \log (3)+\left (-1250+1500 e^4-450 e^8-50 x\right ) \log ^2(3)+\left (2 x^3+\left (250 x-300 e^4 x+90 e^8 x+10 x^2\right ) \log (3)\right ) \log \left (25-30 e^4+9 e^8+x\right )}{625 x^3-750 e^4 x^3+225 e^8 x^3+25 x^4} \, dx=\frac {2}{25} \left (\frac {25 \log ^2(3)}{2 x^2}-\frac {5 \log (3) \log \left (\left (5-3 e^4\right )^2+x\right )}{\left (5-3 e^4\right )^2}-\frac {5 \log (3) \left (110+\log \left (\left (5-3 e^4\right )^2+x\right )\right )}{x}+\frac {\left (5 \left (550-660 e^4+198 e^8+\log (3)\right )+\left (5-3 e^4\right )^2 \log \left (\left (5-3 e^4\right )^2+x\right )\right )^2}{2 \left (5-3 e^4\right )^4}\right ) \] Input:
Integrate[(220*x^3 + (27500*x - 33000*E^4*x + 9900*E^8*x + 1090*x^2)*Log[3 ] + (-1250 + 1500*E^4 - 450*E^8 - 50*x)*Log[3]^2 + (2*x^3 + (250*x - 300*E ^4*x + 90*E^8*x + 10*x^2)*Log[3])*Log[25 - 30*E^4 + 9*E^8 + x])/(625*x^3 - 750*E^4*x^3 + 225*E^8*x^3 + 25*x^4),x]
Output:
(2*((25*Log[3]^2)/(2*x^2) - (5*Log[3]*Log[(5 - 3*E^4)^2 + x])/(5 - 3*E^4)^ 2 - (5*Log[3]*(110 + Log[(5 - 3*E^4)^2 + x]))/x + (5*(550 - 660*E^4 + 198* E^8 + Log[3]) + (5 - 3*E^4)^2*Log[(5 - 3*E^4)^2 + x])^2/(2*(5 - 3*E^4)^4)) )/25
Leaf count is larger than twice the leaf count of optimal. \(122\) vs. \(2(26)=52\).
Time = 1.45 (sec) , antiderivative size = 122, normalized size of antiderivative = 4.69, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {6, 6, 2026, 7239, 27, 7293, 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.
| \(\displaystyle \int \frac {220 x^3+\left (1090 x^2+9900 e^8 x-33000 e^4 x+27500 x\right ) \log (3)+\left (2 x^3+\left (10 x^2+90 e^8 x-300 e^4 x+250 x\right ) \log (3)\right ) \log \left (x+9 e^8-30 e^4+25\right )+\left (-50 x-450 e^8+1500 e^4-1250\right ) \log ^2(3)}{25 x^4+225 e^8 x^3-750 e^4 x^3+625 x^3} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {220 x^3+\left (1090 x^2+9900 e^8 x-33000 e^4 x+27500 x\right ) \log (3)+\left (2 x^3+\left (10 x^2+90 e^8 x-300 e^4 x+250 x\right ) \log (3)\right ) \log \left (x+9 e^8-30 e^4+25\right )+\left (-50 x-450 e^8+1500 e^4-1250\right ) \log ^2(3)}{25 x^4+\left (625-750 e^4\right ) x^3+225 e^8 x^3}dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {220 x^3+\left (1090 x^2+9900 e^8 x-33000 e^4 x+27500 x\right ) \log (3)+\left (2 x^3+\left (10 x^2+90 e^8 x-300 e^4 x+250 x\right ) \log (3)\right ) \log \left (x+9 e^8-30 e^4+25\right )+\left (-50 x-450 e^8+1500 e^4-1250\right ) \log ^2(3)}{25 x^4+\left (625-750 e^4+225 e^8\right ) x^3}dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {220 x^3+\left (1090 x^2+9900 e^8 x-33000 e^4 x+27500 x\right ) \log (3)+\left (2 x^3+\left (10 x^2+90 e^8 x-300 e^4 x+250 x\right ) \log (3)\right ) \log \left (x+9 e^8-30 e^4+25\right )+\left (-50 x-450 e^8+1500 e^4-1250\right ) \log ^2(3)}{x^3 \left (25 x+25 \left (5-3 e^4\right )^2\right )}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {2 \left (x^2+5 x \log (3)+5 \left (5-3 e^4\right )^2 \log (3)\right ) \left (110 x+x \log \left (x+9 e^8-30 e^4+25\right )-5 \log (3)\right )}{25 x^3 \left (x+9 e^8-30 e^4+25\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2}{25} \int \frac {\left (x^2+5 \log (3) x+5 \left (5-3 e^4\right )^2 \log (3)\right ) \left (\log \left (x+\left (5-3 e^4\right )^2\right ) x+110 x-5 \log (3)\right )}{x^3 \left (x+\left (5-3 e^4\right )^2\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {2}{25} \int \left (\frac {5 (22 x-\log (3)) \left (x^2+5 \log (3) x+5 \left (5-3 e^4\right )^2 \log (3)\right )}{x^3 \left (x+9 e^8-30 e^4+25\right )}+\frac {\log \left (x+\left (5-3 e^4\right )^2\right ) \left (x^2+5 \log (3) x+5 \left (5-3 e^4\right )^2 \log (3)\right )}{x^2 \left (x+9 e^8-30 e^4+25\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2}{25} \left (\frac {25 \log ^2(3)}{2 x^2}+\frac {1}{2} \log ^2\left (x+\left (5-3 e^4\right )^2\right )+\frac {5 \left (550-660 e^4+198 e^8+\log (3)\right ) \log \left (x+\left (5-3 e^4\right )^2\right )}{\left (5-3 e^4\right )^2}-\frac {5 \log (3) \log \left (x+\left (5-3 e^4\right )^2\right )}{x}-\frac {5 \log (3) \log \left (x+\left (5-3 e^4\right )^2\right )}{\left (5-3 e^4\right )^2}-\frac {550 \log (3)}{x}\right )\) |
Input:
Int[(220*x^3 + (27500*x - 33000*E^4*x + 9900*E^8*x + 1090*x^2)*Log[3] + (- 1250 + 1500*E^4 - 450*E^8 - 50*x)*Log[3]^2 + (2*x^3 + (250*x - 300*E^4*x + 90*E^8*x + 10*x^2)*Log[3])*Log[25 - 30*E^4 + 9*E^8 + x])/(625*x^3 - 750*E ^4*x^3 + 225*E^8*x^3 + 25*x^4),x]
Output:
(2*((-550*Log[3])/x + (25*Log[3]^2)/(2*x^2) - (5*Log[3]*Log[(5 - 3*E^4)^2 + x])/(5 - 3*E^4)^2 - (5*Log[3]*Log[(5 - 3*E^4)^2 + x])/x + (5*(550 - 660* E^4 + 198*E^8 + Log[3])*Log[(5 - 3*E^4)^2 + x])/(5 - 3*E^4)^2 + Log[(5 - 3 *E^4)^2 + x]^2/2))/25
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[(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_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Leaf count of result is larger than twice the leaf count of optimal. \(70\) vs. \(2(23)=46\).
Time = 25.29 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.73
| method | result | size |
| risch | \(\frac {\ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )^{2}}{25}-\frac {2 \ln \left (3\right ) \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )}{5 x}+\frac {44 x^{2} \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )+5 \ln \left (3\right )^{2}-220 x \ln \left (3\right )}{5 x^{2}}\) | \(71\) |
| norman | \(\frac {\ln \left (3\right )^{2}+\frac {44 x^{2} \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )}{5}-44 x \ln \left (3\right )+\frac {\ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )^{2} x^{2}}{25}-\frac {2 \ln \left (3\right ) \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right ) x}{5}}{x^{2}}\) | \(74\) |
| parallelrisch | \(\frac {\ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )^{2} x^{2}-10 \ln \left (3\right ) \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right ) x +220 x^{2} \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )+25 \ln \left (3\right )^{2}-1100 x \ln \left (3\right )}{25 x^{2}}\) | \(76\) |
| derivativedivides | \(\frac {2 \ln \left (3\right ) \left (\frac {\ln \left (-x \right )}{9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25}-\frac {\ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right ) \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )}{\left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25\right ) x}+\frac {\ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )}{9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25}-\frac {110}{x}-\frac {\ln \left (x \right )}{9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25}\right )}{5}+\frac {44 \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )}{5}+\frac {\ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )^{2}}{25}+\frac {\ln \left (3\right )^{2}}{x^{2}}\) | \(203\) |
| default | \(\frac {2 \ln \left (3\right ) \left (\frac {\ln \left (-x \right )}{9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25}-\frac {\ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right ) \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )}{\left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25\right ) x}+\frac {\ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )}{9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25}-\frac {110}{x}-\frac {\ln \left (x \right )}{9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25}\right )}{5}+\frac {44 \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )}{5}+\frac {\ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )^{2}}{25}+\frac {\ln \left (3\right )^{2}}{x^{2}}\) | \(203\) |
| parts | \(-\frac {2 \ln \left (3\right ) \ln \left (x \right )}{5 \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25\right )}-\frac {44 \ln \left (3\right )}{x}+\frac {\ln \left (3\right )^{2}}{x^{2}}-\frac {2 \left (-198 \,{\mathrm e}^{8}+660 \,{\mathrm e}^{4}-\ln \left (3\right )-550\right ) \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )}{5 \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25\right )}+\frac {2 \left (-540 \,{\mathrm e}^{8} {\mathrm e}^{4}+540 \,{\mathrm e}^{12}\right ) \left (\left (\ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )-\ln \left (\frac {9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25}{9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25}\right )\right ) \ln \left (-\frac {x}{9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25}\right )-\operatorname {dilog}\left (\frac {9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25}{9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25}\right )\right )}{25 \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25\right )^{2}}+\frac {2 \left (540 \,{\mathrm e}^{8} {\mathrm e}^{4}+45 \,{\mathrm e}^{8} \ln \left (3\right )-150 \,{\mathrm e}^{4} \ln \left (3\right )-540 \,{\mathrm e}^{12}+125 \ln \left (3\right )\right ) \left (\frac {\ln \left (-x \right )}{9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25}-\frac {\ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right ) \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )}{\left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25\right ) x}\right )}{25 \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25\right )}+\frac {\left (81 \,{\mathrm e}^{16}-540 \,{\mathrm e}^{12}+1350 \,{\mathrm e}^{8}-1500 \,{\mathrm e}^{4}+625\right ) \ln \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+x +25\right )^{2}}{25 \left (9 \,{\mathrm e}^{8}-30 \,{\mathrm e}^{4}+25\right )^{2}}\) | \(417\) |
Input:
int((((90*x*exp(4)^2-300*x*exp(4)+10*x^2+250*x)*ln(3)+2*x^3)*ln(9*exp(4)^2 -30*exp(4)+x+25)+(-450*exp(4)^2+1500*exp(4)-50*x-1250)*ln(3)^2+(9900*x*exp (4)^2-33000*x*exp(4)+1090*x^2+27500*x)*ln(3)+220*x^3)/(225*x^3*exp(4)^2-75 0*x^3*exp(4)+25*x^4+625*x^3),x,method=_RETURNVERBOSE)
Output:
1/25*ln(9*exp(8)-30*exp(4)+x+25)^2-2/5*ln(3)/x*ln(9*exp(8)-30*exp(4)+x+25) +1/5*(44*x^2*ln(9*exp(8)-30*exp(4)+x+25)+5*ln(3)^2-220*x*ln(3))/x^2
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (26) = 52\).
Time = 0.09 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.31 \[ \int \frac {220 x^3+\left (27500 x-33000 e^4 x+9900 e^8 x+1090 x^2\right ) \log (3)+\left (-1250+1500 e^4-450 e^8-50 x\right ) \log ^2(3)+\left (2 x^3+\left (250 x-300 e^4 x+90 e^8 x+10 x^2\right ) \log (3)\right ) \log \left (25-30 e^4+9 e^8+x\right )}{625 x^3-750 e^4 x^3+225 e^8 x^3+25 x^4} \, dx=\frac {x^{2} \log \left (x + 9 \, e^{8} - 30 \, e^{4} + 25\right )^{2} - 1100 \, x \log \left (3\right ) + 25 \, \log \left (3\right )^{2} + 10 \, {\left (22 \, x^{2} - x \log \left (3\right )\right )} \log \left (x + 9 \, e^{8} - 30 \, e^{4} + 25\right )}{25 \, x^{2}} \] Input:
integrate((((90*x*exp(4)^2-300*x*exp(4)+10*x^2+250*x)*log(3)+2*x^3)*log(9* exp(4)^2-30*exp(4)+x+25)+(-450*exp(4)^2+1500*exp(4)-50*x-1250)*log(3)^2+(9 900*x*exp(4)^2-33000*x*exp(4)+1090*x^2+27500*x)*log(3)+220*x^3)/(225*x^3*e xp(4)^2-750*x^3*exp(4)+25*x^4+625*x^3),x, algorithm="fricas")
Output:
1/25*(x^2*log(x + 9*e^8 - 30*e^4 + 25)^2 - 1100*x*log(3) + 25*log(3)^2 + 1 0*(22*x^2 - x*log(3))*log(x + 9*e^8 - 30*e^4 + 25))/x^2
Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (20) = 40\).
Time = 4.79 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.88 \[ \int \frac {220 x^3+\left (27500 x-33000 e^4 x+9900 e^8 x+1090 x^2\right ) \log (3)+\left (-1250+1500 e^4-450 e^8-50 x\right ) \log ^2(3)+\left (2 x^3+\left (250 x-300 e^4 x+90 e^8 x+10 x^2\right ) \log (3)\right ) \log \left (25-30 e^4+9 e^8+x\right )}{625 x^3-750 e^4 x^3+225 e^8 x^3+25 x^4} \, dx=\frac {\log {\left (x - 30 e^{4} + 25 + 9 e^{8} \right )}^{2}}{25} + \frac {44 \log {\left (x - 30 e^{4} + 25 + 9 e^{8} \right )}}{5} - \frac {2 \log {\left (3 \right )} \log {\left (x - 30 e^{4} + 25 + 9 e^{8} \right )}}{5 x} + \frac {- 44 x \log {\left (3 \right )} + \log {\left (3 \right )}^{2}}{x^{2}} \] Input:
integrate((((90*x*exp(4)**2-300*x*exp(4)+10*x**2+250*x)*ln(3)+2*x**3)*ln(9 *exp(4)**2-30*exp(4)+x+25)+(-450*exp(4)**2+1500*exp(4)-50*x-1250)*ln(3)**2 +(9900*x*exp(4)**2-33000*x*exp(4)+1090*x**2+27500*x)*ln(3)+220*x**3)/(225* x**3*exp(4)**2-750*x**3*exp(4)+25*x**4+625*x**3),x)
Output:
log(x - 30*exp(4) + 25 + 9*exp(8))**2/25 + 44*log(x - 30*exp(4) + 25 + 9*e xp(8))/5 - 2*log(3)*log(x - 30*exp(4) + 25 + 9*exp(8))/(5*x) + (-44*x*log( 3) + log(3)**2)/x**2
Leaf count of result is larger than twice the leaf count of optimal. 841 vs. \(2 (26) = 52\).
Time = 0.19 (sec) , antiderivative size = 841, normalized size of antiderivative = 32.35 \[ \int \frac {220 x^3+\left (27500 x-33000 e^4 x+9900 e^8 x+1090 x^2\right ) \log (3)+\left (-1250+1500 e^4-450 e^8-50 x\right ) \log ^2(3)+\left (2 x^3+\left (250 x-300 e^4 x+90 e^8 x+10 x^2\right ) \log (3)\right ) \log \left (25-30 e^4+9 e^8+x\right )}{625 x^3-750 e^4 x^3+225 e^8 x^3+25 x^4} \, dx=\text {Too large to display} \] Input:
integrate((((90*x*exp(4)^2-300*x*exp(4)+10*x^2+250*x)*log(3)+2*x^3)*log(9* exp(4)^2-30*exp(4)+x+25)+(-450*exp(4)^2+1500*exp(4)-50*x-1250)*log(3)^2+(9 900*x*exp(4)^2-33000*x*exp(4)+1090*x^2+27500*x)*log(3)+220*x^3)/(225*x^3*e xp(4)^2-750*x^3*exp(4)+25*x^4+625*x^3),x, algorithm="maxima")
Output:
9*(2*log(x + 9*e^8 - 30*e^4 + 25)/(729*e^24 - 7290*e^20 + 30375*e^16 - 675 00*e^12 + 84375*e^8 - 56250*e^4 + 15625) - 2*log(x)/(729*e^24 - 7290*e^20 + 30375*e^16 - 67500*e^12 + 84375*e^8 - 56250*e^4 + 15625) - (2*x - 9*e^8 + 30*e^4 - 25)/(x^2*(81*e^16 - 540*e^12 + 1350*e^8 - 1500*e^4 + 625)))*e^8 *log(3)^2 - 30*(2*log(x + 9*e^8 - 30*e^4 + 25)/(729*e^24 - 7290*e^20 + 303 75*e^16 - 67500*e^12 + 84375*e^8 - 56250*e^4 + 15625) - 2*log(x)/(729*e^24 - 7290*e^20 + 30375*e^16 - 67500*e^12 + 84375*e^8 - 56250*e^4 + 15625) - (2*x - 9*e^8 + 30*e^4 - 25)/(x^2*(81*e^16 - 540*e^12 + 1350*e^8 - 1500*e^4 + 625)))*e^4*log(3)^2 + 396*(log(x + 9*e^8 - 30*e^4 + 25)/(81*e^16 - 540* e^12 + 1350*e^8 - 1500*e^4 + 625) - log(x)/(81*e^16 - 540*e^12 + 1350*e^8 - 1500*e^4 + 625) - 1/(x*(9*e^8 - 30*e^4 + 25)))*e^8*log(3) - 1320*(log(x + 9*e^8 - 30*e^4 + 25)/(81*e^16 - 540*e^12 + 1350*e^8 - 1500*e^4 + 625) - log(x)/(81*e^16 - 540*e^12 + 1350*e^8 - 1500*e^4 + 625) - 1/(x*(9*e^8 - 30 *e^4 + 25)))*e^4*log(3) + 25*(2*log(x + 9*e^8 - 30*e^4 + 25)/(729*e^24 - 7 290*e^20 + 30375*e^16 - 67500*e^12 + 84375*e^8 - 56250*e^4 + 15625) - 2*lo g(x)/(729*e^24 - 7290*e^20 + 30375*e^16 - 67500*e^12 + 84375*e^8 - 56250*e ^4 + 15625) - (2*x - 9*e^8 + 30*e^4 - 25)/(x^2*(81*e^16 - 540*e^12 + 1350* e^8 - 1500*e^4 + 625)))*log(3)^2 - 2*(log(x + 9*e^8 - 30*e^4 + 25)/(81*e^1 6 - 540*e^12 + 1350*e^8 - 1500*e^4 + 625) - log(x)/(81*e^16 - 540*e^12 + 1 350*e^8 - 1500*e^4 + 625) - 1/(x*(9*e^8 - 30*e^4 + 25)))*log(3)^2 + 110...
\[ \int \frac {220 x^3+\left (27500 x-33000 e^4 x+9900 e^8 x+1090 x^2\right ) \log (3)+\left (-1250+1500 e^4-450 e^8-50 x\right ) \log ^2(3)+\left (2 x^3+\left (250 x-300 e^4 x+90 e^8 x+10 x^2\right ) \log (3)\right ) \log \left (25-30 e^4+9 e^8+x\right )}{625 x^3-750 e^4 x^3+225 e^8 x^3+25 x^4} \, dx=\int { \frac {2 \, {\left (110 \, x^{3} - 25 \, {\left (x + 9 \, e^{8} - 30 \, e^{4} + 25\right )} \log \left (3\right )^{2} + 5 \, {\left (109 \, x^{2} + 990 \, x e^{8} - 3300 \, x e^{4} + 2750 \, x\right )} \log \left (3\right ) + {\left (x^{3} + 5 \, {\left (x^{2} + 9 \, x e^{8} - 30 \, x e^{4} + 25 \, x\right )} \log \left (3\right )\right )} \log \left (x + 9 \, e^{8} - 30 \, e^{4} + 25\right )\right )}}{25 \, {\left (x^{4} + 9 \, x^{3} e^{8} - 30 \, x^{3} e^{4} + 25 \, x^{3}\right )}} \,d x } \] Input:
integrate((((90*x*exp(4)^2-300*x*exp(4)+10*x^2+250*x)*log(3)+2*x^3)*log(9* exp(4)^2-30*exp(4)+x+25)+(-450*exp(4)^2+1500*exp(4)-50*x-1250)*log(3)^2+(9 900*x*exp(4)^2-33000*x*exp(4)+1090*x^2+27500*x)*log(3)+220*x^3)/(225*x^3*e xp(4)^2-750*x^3*exp(4)+25*x^4+625*x^3),x, algorithm="giac")
Output:
integrate(2/25*(110*x^3 - 25*(x + 9*e^8 - 30*e^4 + 25)*log(3)^2 + 5*(109*x ^2 + 990*x*e^8 - 3300*x*e^4 + 2750*x)*log(3) + (x^3 + 5*(x^2 + 9*x*e^8 - 3 0*x*e^4 + 25*x)*log(3))*log(x + 9*e^8 - 30*e^4 + 25))/(x^4 + 9*x^3*e^8 - 3 0*x^3*e^4 + 25*x^3), x)
Time = 3.11 (sec) , antiderivative size = 88, normalized size of antiderivative = 3.38 \[ \int \frac {220 x^3+\left (27500 x-33000 e^4 x+9900 e^8 x+1090 x^2\right ) \log (3)+\left (-1250+1500 e^4-450 e^8-50 x\right ) \log ^2(3)+\left (2 x^3+\left (250 x-300 e^4 x+90 e^8 x+10 x^2\right ) \log (3)\right ) \log \left (25-30 e^4+9 e^8+x\right )}{625 x^3-750 e^4 x^3+225 e^8 x^3+25 x^4} \, dx=\frac {44\,\ln \left (x+{\left (3\,{\mathrm {e}}^4-5\right )}^2\right )}{5}-\frac {220\,x\,\ln \left (3\right )-5\,{\ln \left (3\right )}^2}{5\,x^2}+\frac {{\ln \left (x-30\,{\mathrm {e}}^4+9\,{\mathrm {e}}^8+25\right )}^2}{25}-\frac {\ln \left (x-30\,{\mathrm {e}}^4+9\,{\mathrm {e}}^8+25\right )\,\left (\frac {12\,{\mathrm {e}}^4}{5}-\frac {18\,{\mathrm {e}}^8}{25}+\frac {2\,\ln \left (3\right )}{5}+\frac {2\,{\left (3\,{\mathrm {e}}^4-5\right )}^2}{25}-2\right )}{x} \] Input:
int((log(x - 30*exp(4) + 9*exp(8) + 25)*(log(3)*(250*x - 300*x*exp(4) + 90 *x*exp(8) + 10*x^2) + 2*x^3) + log(3)*(27500*x - 33000*x*exp(4) + 9900*x*e xp(8) + 1090*x^2) + 220*x^3 - log(3)^2*(50*x - 1500*exp(4) + 450*exp(8) + 1250))/(225*x^3*exp(8) - 750*x^3*exp(4) + 625*x^3 + 25*x^4),x)
Output:
(44*log(x + (3*exp(4) - 5)^2))/5 - (220*x*log(3) - 5*log(3)^2)/(5*x^2) + l og(x - 30*exp(4) + 9*exp(8) + 25)^2/25 - (log(x - 30*exp(4) + 9*exp(8) + 2 5)*((12*exp(4))/5 - (18*exp(8))/25 + (2*log(3))/5 + (2*(3*exp(4) - 5)^2)/2 5 - 2))/x
Time = 0.21 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.88 \[ \int \frac {220 x^3+\left (27500 x-33000 e^4 x+9900 e^8 x+1090 x^2\right ) \log (3)+\left (-1250+1500 e^4-450 e^8-50 x\right ) \log ^2(3)+\left (2 x^3+\left (250 x-300 e^4 x+90 e^8 x+10 x^2\right ) \log (3)\right ) \log \left (25-30 e^4+9 e^8+x\right )}{625 x^3-750 e^4 x^3+225 e^8 x^3+25 x^4} \, dx=\frac {\mathrm {log}\left (9 e^{8}-30 e^{4}+x +25\right )^{2} x^{2}-10 \,\mathrm {log}\left (9 e^{8}-30 e^{4}+x +25\right ) \mathrm {log}\left (3\right ) x +220 \,\mathrm {log}\left (9 e^{8}-30 e^{4}+x +25\right ) x^{2}+25 \mathrm {log}\left (3\right )^{2}-1100 \,\mathrm {log}\left (3\right ) x}{25 x^{2}} \] Input:
int((((90*x*exp(4)^2-300*x*exp(4)+10*x^2+250*x)*log(3)+2*x^3)*log(9*exp(4) ^2-30*exp(4)+x+25)+(-450*exp(4)^2+1500*exp(4)-50*x-1250)*log(3)^2+(9900*x* exp(4)^2-33000*x*exp(4)+1090*x^2+27500*x)*log(3)+220*x^3)/(225*x^3*exp(4)^ 2-750*x^3*exp(4)+25*x^4+625*x^3),x)
Output:
(log(9*e**8 - 30*e**4 + x + 25)**2*x**2 - 10*log(9*e**8 - 30*e**4 + x + 25 )*log(3)*x + 220*log(9*e**8 - 30*e**4 + x + 25)*x**2 + 25*log(3)**2 - 1100 *log(3)*x)/(25*x**2)