Integrand size = 75, antiderivative size = 25 \[ \int \frac {2562890625 e^{16+\log ^2\left (\frac {15 x}{20+5 x+\log (4)}\right )} x^8 \left (160+8 \log (4)+(40+2 \log (4)) \log \left (\frac {15 x}{20+5 x+\log (4)}\right )\right )}{(20+5 x+\log (4))^8 \left (100 x+25 x^2+5 x \log (4)\right )} \, dx=\frac {1}{5} e^{\left (4+\log \left (\frac {3 x}{4+x+\frac {\log (4)}{5}}\right )\right )^2} \] Output:
1/5*exp((ln(3/(x+4+2/5*ln(2))*x)+4)^2)
Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {2562890625 e^{16+\log ^2\left (\frac {15 x}{20+5 x+\log (4)}\right )} x^8 \left (160+8 \log (4)+(40+2 \log (4)) \log \left (\frac {15 x}{20+5 x+\log (4)}\right )\right )}{(20+5 x+\log (4))^8 \left (100 x+25 x^2+5 x \log (4)\right )} \, dx=\frac {512578125 e^{16+\log ^2\left (\frac {15 x}{20+5 x+\log (4)}\right )} x^8}{(20+5 x+\log (4))^8} \] Input:
Integrate[(2562890625*E^(16 + Log[(15*x)/(20 + 5*x + Log[4])]^2)*x^8*(160 + 8*Log[4] + (40 + 2*Log[4])*Log[(15*x)/(20 + 5*x + Log[4])]))/((20 + 5*x + Log[4])^8*(100*x + 25*x^2 + 5*x*Log[4])),x]
Output:
(512578125*E^(16 + Log[(15*x)/(20 + 5*x + Log[4])]^2)*x^8)/(20 + 5*x + Log [4])^8
Leaf count is larger than twice the leaf count of optimal. \(63\) vs. \(2(25)=50\).
Time = 0.39 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.52, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {6, 9, 27, 2726}
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.
\(\displaystyle \int \frac {2562890625 x^8 e^{\log ^2\left (\frac {15 x}{5 x+20+\log (4)}\right )+16} \left ((40+2 \log (4)) \log \left (\frac {15 x}{5 x+20+\log (4)}\right )+160+8 \log (4)\right )}{(5 x+20+\log (4))^8 \left (25 x^2+100 x+5 x \log (4)\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {2562890625 x^8 e^{\log ^2\left (\frac {15 x}{5 x+20+\log (4)}\right )+16} \left ((40+2 \log (4)) \log \left (\frac {15 x}{5 x+20+\log (4)}\right )+160+8 \log (4)\right )}{(5 x+20+\log (4))^8 \left (25 x^2+x (100+5 \log (4))\right )}dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {1025156250 x^7 e^{\log ^2\left (\frac {15 x}{5 x+20+\log (4)}\right )+16} \left ((20+\log (4)) \log \left (\frac {15 x}{5 x+20+\log (4)}\right )+4 (20+\log (4))\right )}{(5 x+20+\log (4))^9}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 1025156250 \int \frac {e^{\log ^2\left (\frac {15 x}{5 x+\log (4)+20}\right )+16} x^7 \left ((20+\log (4)) \log \left (\frac {15 x}{5 x+\log (4)+20}\right )+\log (256)+80\right )}{(5 x+\log (4)+20)^9}dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle -\frac {512578125 x^8 (20+\log (4)) e^{\log ^2\left (\frac {15 x}{5 x+20+\log (4)}\right )+16}}{(5 x+20+\log (4))^{10} \left (\frac {5 x}{(5 x+20+\log (4))^2}-\frac {1}{5 x+20+\log (4)}\right )}\) |
Input:
Int[(2562890625*E^(16 + Log[(15*x)/(20 + 5*x + Log[4])]^2)*x^8*(160 + 8*Lo g[4] + (40 + 2*Log[4])*Log[(15*x)/(20 + 5*x + Log[4])]))/((20 + 5*x + Log[ 4])^8*(100*x + 25*x^2 + 5*x*Log[4])),x]
Output:
(-512578125*E^(16 + Log[(15*x)/(20 + 5*x + Log[4])]^2)*x^8*(20 + Log[4]))/ ((20 + 5*x + Log[4])^10*((5*x)/(20 + 5*x + Log[4])^2 - (20 + 5*x + Log[4]) ^(-1)))
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[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Time = 1.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48
method | result | size |
risch | \(\frac {512578125 x^{8} {\mathrm e}^{\ln \left (\frac {15 x}{2 \ln \left (2\right )+20+5 x}\right )^{2}+16}}{\left (2 \ln \left (2\right )+20+5 x \right )^{8}}\) | \(37\) |
norman | \(\frac {{\mathrm e}^{\ln \left (\frac {15 x}{2 \ln \left (2\right )+20+5 x}\right )^{2}+8 \ln \left (\frac {15 x}{2 \ln \left (2\right )+20+5 x}\right )+16}}{5}\) | \(40\) |
parallelrisch | \(\frac {{\mathrm e}^{\ln \left (\frac {15 x}{2 \ln \left (2\right )+20+5 x}\right )^{2}+8 \ln \left (\frac {15 x}{2 \ln \left (2\right )+20+5 x}\right )+16}}{5}\) | \(40\) |
Input:
int(((4*ln(2)+40)*ln(15*x/(2*ln(2)+20+5*x))+16*ln(2)+160)*exp(ln(15*x/(2*l n(2)+20+5*x))^2+8*ln(15*x/(2*ln(2)+20+5*x))+16)/(10*x*ln(2)+25*x^2+100*x), x,method=_RETURNVERBOSE)
Output:
512578125*x^8/(2*ln(2)+20+5*x)^8*exp(ln(15*x/(2*ln(2)+20+5*x))^2+16)
Time = 0.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {2562890625 e^{16+\log ^2\left (\frac {15 x}{20+5 x+\log (4)}\right )} x^8 \left (160+8 \log (4)+(40+2 \log (4)) \log \left (\frac {15 x}{20+5 x+\log (4)}\right )\right )}{(20+5 x+\log (4))^8 \left (100 x+25 x^2+5 x \log (4)\right )} \, dx=\frac {1}{5} \, e^{\left (\log \left (\frac {15 \, x}{5 \, x + 2 \, \log \left (2\right ) + 20}\right )^{2} + 8 \, \log \left (\frac {15 \, x}{5 \, x + 2 \, \log \left (2\right ) + 20}\right ) + 16\right )} \] Input:
integrate(((4*log(2)+40)*log(15*x/(2*log(2)+20+5*x))+16*log(2)+160)*exp(lo g(15*x/(2*log(2)+20+5*x))^2+8*log(15*x/(2*log(2)+20+5*x))+16)/(10*x*log(2) +25*x^2+100*x),x, algorithm="fricas")
Output:
1/5*e^(log(15*x/(5*x + 2*log(2) + 20))^2 + 8*log(15*x/(5*x + 2*log(2) + 20 )) + 16)
Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (20) = 40\).
Time = 1.39 (sec) , antiderivative size = 379, normalized size of antiderivative = 15.16 \[ \int \frac {2562890625 e^{16+\log ^2\left (\frac {15 x}{20+5 x+\log (4)}\right )} x^8 \left (160+8 \log (4)+(40+2 \log (4)) \log \left (\frac {15 x}{20+5 x+\log (4)}\right )\right )}{(20+5 x+\log (4))^8 \left (100 x+25 x^2+5 x \log (4)\right )} \, dx=\frac {512578125 x^{8} e^{\log {\left (\frac {15 x}{5 x + 2 \log {\left (2 \right )} + 20} \right )}^{2} + 16}}{390625 x^{8} + 1250000 x^{7} \log {\left (2 \right )} + 12500000 x^{7} + 1750000 x^{6} \log {\left (2 \right )}^{2} + 35000000 x^{6} \log {\left (2 \right )} + 175000000 x^{6} + 1400000 x^{5} \log {\left (2 \right )}^{3} + 42000000 x^{5} \log {\left (2 \right )}^{2} + 420000000 x^{5} \log {\left (2 \right )} + 1400000000 x^{5} + 700000 x^{4} \log {\left (2 \right )}^{4} + 28000000 x^{4} \log {\left (2 \right )}^{3} + 420000000 x^{4} \log {\left (2 \right )}^{2} + 2800000000 x^{4} \log {\left (2 \right )} + 7000000000 x^{4} + 224000 x^{3} \log {\left (2 \right )}^{5} + 11200000 x^{3} \log {\left (2 \right )}^{4} + 224000000 x^{3} \log {\left (2 \right )}^{3} + 2240000000 x^{3} \log {\left (2 \right )}^{2} + 11200000000 x^{3} \log {\left (2 \right )} + 22400000000 x^{3} + 44800 x^{2} \log {\left (2 \right )}^{6} + 2688000 x^{2} \log {\left (2 \right )}^{5} + 67200000 x^{2} \log {\left (2 \right )}^{4} + 896000000 x^{2} \log {\left (2 \right )}^{3} + 6720000000 x^{2} \log {\left (2 \right )}^{2} + 26880000000 x^{2} \log {\left (2 \right )} + 44800000000 x^{2} + 5120 x \log {\left (2 \right )}^{7} + 358400 x \log {\left (2 \right )}^{6} + 10752000 x \log {\left (2 \right )}^{5} + 179200000 x \log {\left (2 \right )}^{4} + 1792000000 x \log {\left (2 \right )}^{3} + 10752000000 x \log {\left (2 \right )}^{2} + 35840000000 x \log {\left (2 \right )} + 51200000000 x + 256 \log {\left (2 \right )}^{8} + 20480 \log {\left (2 \right )}^{7} + 716800 \log {\left (2 \right )}^{6} + 14336000 \log {\left (2 \right )}^{5} + 179200000 \log {\left (2 \right )}^{4} + 1433600000 \log {\left (2 \right )}^{3} + 7168000000 \log {\left (2 \right )}^{2} + 20480000000 \log {\left (2 \right )} + 25600000000} \] Input:
integrate(((4*ln(2)+40)*ln(15*x/(2*ln(2)+20+5*x))+16*ln(2)+160)*exp(ln(15* x/(2*ln(2)+20+5*x))**2+8*ln(15*x/(2*ln(2)+20+5*x))+16)/(10*x*ln(2)+25*x**2 +100*x),x)
Output:
512578125*x**8*exp(log(15*x/(5*x + 2*log(2) + 20))**2 + 16)/(390625*x**8 + 1250000*x**7*log(2) + 12500000*x**7 + 1750000*x**6*log(2)**2 + 35000000*x **6*log(2) + 175000000*x**6 + 1400000*x**5*log(2)**3 + 42000000*x**5*log(2 )**2 + 420000000*x**5*log(2) + 1400000000*x**5 + 700000*x**4*log(2)**4 + 2 8000000*x**4*log(2)**3 + 420000000*x**4*log(2)**2 + 2800000000*x**4*log(2) + 7000000000*x**4 + 224000*x**3*log(2)**5 + 11200000*x**3*log(2)**4 + 224 000000*x**3*log(2)**3 + 2240000000*x**3*log(2)**2 + 11200000000*x**3*log(2 ) + 22400000000*x**3 + 44800*x**2*log(2)**6 + 2688000*x**2*log(2)**5 + 672 00000*x**2*log(2)**4 + 896000000*x**2*log(2)**3 + 6720000000*x**2*log(2)** 2 + 26880000000*x**2*log(2) + 44800000000*x**2 + 5120*x*log(2)**7 + 358400 *x*log(2)**6 + 10752000*x*log(2)**5 + 179200000*x*log(2)**4 + 1792000000*x *log(2)**3 + 10752000000*x*log(2)**2 + 35840000000*x*log(2) + 51200000000* x + 256*log(2)**8 + 20480*log(2)**7 + 716800*log(2)**6 + 14336000*log(2)** 5 + 179200000*log(2)**4 + 1433600000*log(2)**3 + 7168000000*log(2)**2 + 20 480000000*log(2) + 25600000000)
Leaf count of result is larger than twice the leaf count of optimal. 336 vs. \(2 (22) = 44\).
Time = 0.36 (sec) , antiderivative size = 336, normalized size of antiderivative = 13.44 \[ \int \frac {2562890625 e^{16+\log ^2\left (\frac {15 x}{20+5 x+\log (4)}\right )} x^8 \left (160+8 \log (4)+(40+2 \log (4)) \log \left (\frac {15 x}{20+5 x+\log (4)}\right )\right )}{(20+5 x+\log (4))^8 \left (100 x+25 x^2+5 x \log (4)\right )} \, dx=\frac {78125 \cdot 3^{2 \, \log \left (5\right ) + 8} x^{8} e^{\left (\log \left (5\right )^{2} + \log \left (3\right )^{2} - 2 \, \log \left (5\right ) \log \left (5 \, x + 2 \, \log \left (2\right ) + 20\right ) - 2 \, \log \left (3\right ) \log \left (5 \, x + 2 \, \log \left (2\right ) + 20\right ) + \log \left (5 \, x + 2 \, \log \left (2\right ) + 20\right )^{2} + 2 \, \log \left (5\right ) \log \left (x\right ) + 2 \, \log \left (3\right ) \log \left (x\right ) - 2 \, \log \left (5 \, x + 2 \, \log \left (2\right ) + 20\right ) \log \left (x\right ) + \log \left (x\right )^{2} + 16\right )}}{390625 \, x^{8} + 1250000 \, x^{7} {\left (\log \left (2\right ) + 10\right )} + 256 \, \log \left (2\right )^{8} + 1750000 \, {\left (\log \left (2\right )^{2} + 20 \, \log \left (2\right ) + 100\right )} x^{6} + 20480 \, \log \left (2\right )^{7} + 1400000 \, {\left (\log \left (2\right )^{3} + 30 \, \log \left (2\right )^{2} + 300 \, \log \left (2\right ) + 1000\right )} x^{5} + 716800 \, \log \left (2\right )^{6} + 700000 \, {\left (\log \left (2\right )^{4} + 40 \, \log \left (2\right )^{3} + 600 \, \log \left (2\right )^{2} + 4000 \, \log \left (2\right ) + 10000\right )} x^{4} + 14336000 \, \log \left (2\right )^{5} + 224000 \, {\left (\log \left (2\right )^{5} + 50 \, \log \left (2\right )^{4} + 1000 \, \log \left (2\right )^{3} + 10000 \, \log \left (2\right )^{2} + 50000 \, \log \left (2\right ) + 100000\right )} x^{3} + 179200000 \, \log \left (2\right )^{4} + 44800 \, {\left (\log \left (2\right )^{6} + 60 \, \log \left (2\right )^{5} + 1500 \, \log \left (2\right )^{4} + 20000 \, \log \left (2\right )^{3} + 150000 \, \log \left (2\right )^{2} + 600000 \, \log \left (2\right ) + 1000000\right )} x^{2} + 1433600000 \, \log \left (2\right )^{3} + 5120 \, {\left (\log \left (2\right )^{7} + 70 \, \log \left (2\right )^{6} + 2100 \, \log \left (2\right )^{5} + 35000 \, \log \left (2\right )^{4} + 350000 \, \log \left (2\right )^{3} + 2100000 \, \log \left (2\right )^{2} + 7000000 \, \log \left (2\right ) + 10000000\right )} x + 7168000000 \, \log \left (2\right )^{2} + 20480000000 \, \log \left (2\right ) + 25600000000} \] Input:
integrate(((4*log(2)+40)*log(15*x/(2*log(2)+20+5*x))+16*log(2)+160)*exp(lo g(15*x/(2*log(2)+20+5*x))^2+8*log(15*x/(2*log(2)+20+5*x))+16)/(10*x*log(2) +25*x^2+100*x),x, algorithm="maxima")
Output:
78125*3^(2*log(5) + 8)*x^8*e^(log(5)^2 + log(3)^2 - 2*log(5)*log(5*x + 2*l og(2) + 20) - 2*log(3)*log(5*x + 2*log(2) + 20) + log(5*x + 2*log(2) + 20) ^2 + 2*log(5)*log(x) + 2*log(3)*log(x) - 2*log(5*x + 2*log(2) + 20)*log(x) + log(x)^2 + 16)/(390625*x^8 + 1250000*x^7*(log(2) + 10) + 256*log(2)^8 + 1750000*(log(2)^2 + 20*log(2) + 100)*x^6 + 20480*log(2)^7 + 1400000*(log( 2)^3 + 30*log(2)^2 + 300*log(2) + 1000)*x^5 + 716800*log(2)^6 + 700000*(lo g(2)^4 + 40*log(2)^3 + 600*log(2)^2 + 4000*log(2) + 10000)*x^4 + 14336000* log(2)^5 + 224000*(log(2)^5 + 50*log(2)^4 + 1000*log(2)^3 + 10000*log(2)^2 + 50000*log(2) + 100000)*x^3 + 179200000*log(2)^4 + 44800*(log(2)^6 + 60* log(2)^5 + 1500*log(2)^4 + 20000*log(2)^3 + 150000*log(2)^2 + 600000*log(2 ) + 1000000)*x^2 + 1433600000*log(2)^3 + 5120*(log(2)^7 + 70*log(2)^6 + 21 00*log(2)^5 + 35000*log(2)^4 + 350000*log(2)^3 + 2100000*log(2)^2 + 700000 0*log(2) + 10000000)*x + 7168000000*log(2)^2 + 20480000000*log(2) + 256000 00000)
Time = 0.18 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {2562890625 e^{16+\log ^2\left (\frac {15 x}{20+5 x+\log (4)}\right )} x^8 \left (160+8 \log (4)+(40+2 \log (4)) \log \left (\frac {15 x}{20+5 x+\log (4)}\right )\right )}{(20+5 x+\log (4))^8 \left (100 x+25 x^2+5 x \log (4)\right )} \, dx=\frac {1}{5} \, e^{\left (\log \left (\frac {15 \, x}{5 \, x + 2 \, \log \left (2\right ) + 20}\right )^{2} + 8 \, \log \left (\frac {15 \, x}{5 \, x + 2 \, \log \left (2\right ) + 20}\right ) + 16\right )} \] Input:
integrate(((4*log(2)+40)*log(15*x/(2*log(2)+20+5*x))+16*log(2)+160)*exp(lo g(15*x/(2*log(2)+20+5*x))^2+8*log(15*x/(2*log(2)+20+5*x))+16)/(10*x*log(2) +25*x^2+100*x),x, algorithm="giac")
Output:
1/5*e^(log(15*x/(5*x + 2*log(2) + 20))^2 + 8*log(15*x/(5*x + 2*log(2) + 20 )) + 16)
Time = 6.39 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.96 \[ \int \frac {2562890625 e^{16+\log ^2\left (\frac {15 x}{20+5 x+\log (4)}\right )} x^8 \left (160+8 \log (4)+(40+2 \log (4)) \log \left (\frac {15 x}{20+5 x+\log (4)}\right )\right )}{(20+5 x+\log (4))^8 \left (100 x+25 x^2+5 x \log (4)\right )} \, dx=\frac {512578125\,x^8\,{\mathrm {e}}^{{\ln \left (\frac {x}{5\,x+\ln \left (4\right )+20}\right )}^2+{\ln \left (15\right )}^2+16}\,{\left (\frac {x}{5\,x+\ln \left (4\right )+20}\right )}^{\ln \left (225\right )}}{{\left (5\,x+\ln \left (4\right )+20\right )}^8} \] Input:
int((exp(8*log((15*x)/(5*x + 2*log(2) + 20)) + log((15*x)/(5*x + 2*log(2) + 20))^2 + 16)*(16*log(2) + log((15*x)/(5*x + 2*log(2) + 20))*(4*log(2) + 40) + 160))/(100*x + 10*x*log(2) + 25*x^2),x)
Output:
(512578125*x^8*exp(log(x/(5*x + log(4) + 20))^2 + log(15)^2 + 16)*(x/(5*x + log(4) + 20))^log(225))/(5*x + log(4) + 20)^8
Time = 0.17 (sec) , antiderivative size = 339, normalized size of antiderivative = 13.56 \[ \int \frac {2562890625 e^{16+\log ^2\left (\frac {15 x}{20+5 x+\log (4)}\right )} x^8 \left (160+8 \log (4)+(40+2 \log (4)) \log \left (\frac {15 x}{20+5 x+\log (4)}\right )\right )}{(20+5 x+\log (4))^8 \left (100 x+25 x^2+5 x \log (4)\right )} \, dx=\frac {512578125 e^{\mathrm {log}\left (\frac {15 x}{2 \,\mathrm {log}\left (2\right )+5 x +20}\right )^{2}} e^{16} x^{8}}{25600000000+51200000000 x +2240000000 \mathrm {log}\left (2\right )^{2} x^{3}+26880000000 \,\mathrm {log}\left (2\right ) x^{2}+35840000000 \,\mathrm {log}\left (2\right ) x +42000000 \mathrm {log}\left (2\right )^{2} x^{5}+44800000000 x^{2}+7168000000 \mathrm {log}\left (2\right )^{2}+1400000000 x^{5}+12500000 x^{7}+22400000000 x^{3}+390625 x^{8}+175000000 x^{6}+256 \mathrm {log}\left (2\right )^{8}+20480 \mathrm {log}\left (2\right )^{7}+716800 \mathrm {log}\left (2\right )^{6}+14336000 \mathrm {log}\left (2\right )^{5}+11200000000 \,\mathrm {log}\left (2\right ) x^{3}+6720000000 \mathrm {log}\left (2\right )^{2} x^{2}+10752000000 \mathrm {log}\left (2\right )^{2} x +5120 \mathrm {log}\left (2\right )^{7} x +44800 \mathrm {log}\left (2\right )^{6} x^{2}+358400 \mathrm {log}\left (2\right )^{6} x +224000 \mathrm {log}\left (2\right )^{5} x^{3}+2688000 \mathrm {log}\left (2\right )^{5} x^{2}+10752000 \mathrm {log}\left (2\right )^{5} x +11200000 \mathrm {log}\left (2\right )^{4} x^{3}+179200000 \mathrm {log}\left (2\right )^{4} x +1400000 \mathrm {log}\left (2\right )^{3} x^{5}+28000000 \mathrm {log}\left (2\right )^{3} x^{4}+224000000 \mathrm {log}\left (2\right )^{3} x^{3}+1750000 \mathrm {log}\left (2\right )^{2} x^{6}+1250000 \,\mathrm {log}\left (2\right ) x^{7}+35000000 \,\mathrm {log}\left (2\right ) x^{6}+420000000 \,\mathrm {log}\left (2\right ) x^{5}+20480000000 \,\mathrm {log}\left (2\right )+7000000000 x^{4}+1433600000 \mathrm {log}\left (2\right )^{3}+179200000 \mathrm {log}\left (2\right )^{4}+700000 \mathrm {log}\left (2\right )^{4} x^{4}+67200000 \mathrm {log}\left (2\right )^{4} x^{2}+2800000000 \,\mathrm {log}\left (2\right ) x^{4}+896000000 \mathrm {log}\left (2\right )^{3} x^{2}+1792000000 \mathrm {log}\left (2\right )^{3} x +420000000 \mathrm {log}\left (2\right )^{2} x^{4}} \] Input:
int(((4*log(2)+40)*log(15*x/(2*log(2)+20+5*x))+16*log(2)+160)*exp(log(15*x /(2*log(2)+20+5*x))^2+8*log(15*x/(2*log(2)+20+5*x))+16)/(10*x*log(2)+25*x^ 2+100*x),x)
Output:
(512578125*e**(log((15*x)/(2*log(2) + 5*x + 20))**2)*e**16*x**8)/(256*log( 2)**8 + 5120*log(2)**7*x + 20480*log(2)**7 + 44800*log(2)**6*x**2 + 358400 *log(2)**6*x + 716800*log(2)**6 + 224000*log(2)**5*x**3 + 2688000*log(2)** 5*x**2 + 10752000*log(2)**5*x + 14336000*log(2)**5 + 700000*log(2)**4*x**4 + 11200000*log(2)**4*x**3 + 67200000*log(2)**4*x**2 + 179200000*log(2)**4 *x + 179200000*log(2)**4 + 1400000*log(2)**3*x**5 + 28000000*log(2)**3*x** 4 + 224000000*log(2)**3*x**3 + 896000000*log(2)**3*x**2 + 1792000000*log(2 )**3*x + 1433600000*log(2)**3 + 1750000*log(2)**2*x**6 + 42000000*log(2)** 2*x**5 + 420000000*log(2)**2*x**4 + 2240000000*log(2)**2*x**3 + 6720000000 *log(2)**2*x**2 + 10752000000*log(2)**2*x + 7168000000*log(2)**2 + 1250000 *log(2)*x**7 + 35000000*log(2)*x**6 + 420000000*log(2)*x**5 + 2800000000*l og(2)*x**4 + 11200000000*log(2)*x**3 + 26880000000*log(2)*x**2 + 358400000 00*log(2)*x + 20480000000*log(2) + 390625*x**8 + 12500000*x**7 + 175000000 *x**6 + 1400000000*x**5 + 7000000000*x**4 + 22400000000*x**3 + 44800000000 *x**2 + 51200000000*x + 25600000000)