\(\int \frac {2562890625 e^{16+\log ^2(\frac {15 x}{20+5 x+\log (4)})} x^8 (160+8 \log (4)+(40+2 \log (4)) \log (\frac {15 x}{20+5 x+\log (4)}))}{(20+5 x+\log (4))^8 (100 x+25 x^2+5 x \log (4))} \, dx\) [6913]

Optimal result

Integrand size = 75, antiderivative size = 25 \[ \int \frac {2562890625 e^{16+\log ^2\left (\frac {15 x}{20+5 x+\log (4)}\right )} x^8 \left (160+8 \log (4)+(40+2 \log (4)) \log \left (\frac {15 x}{20+5 x+\log (4)}\right )\right )}{(20+5 x+\log (4))^8 \left (100 x+25 x^2+5 x \log (4)\right )} \, dx=\frac {1}{5} e^{\left (4+\log \left (\frac {3 x}{4+x+\frac {\log (4)}{5}}\right )\right )^2} \]

[Out]

1/5*exp((ln(3/(x+4+2/5*ln(2))*x)+4)^2)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(63\) vs. \(2(25)=50\).

Time = 0.14 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.52, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {6, 12, 1598, 21, 2326} \[ \int \frac {2562890625 e^{16+\log ^2\left (\frac {15 x}{20+5 x+\log (4)}\right )} x^8 \left (160+8 \log (4)+(40+2 \log (4)) \log \left (\frac {15 x}{20+5 x+\log (4)}\right )\right )}{(20+5 x+\log (4))^8 \left (100 x+25 x^2+5 x \log (4)\right )} \, dx=-\frac {512578125 x^8 (20+\log (4)) e^{\log ^2\left (\frac {15 x}{5 x+20+\log (4)}\right )+16}}{(5 x+20+\log (4))^{10} \left (\frac {5 x}{(5 x+20+\log (4))^2}-\frac {1}{5 x+20+\log (4)}\right )} \]

[In]

Int[(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)/(2
0 + 5*x + Log[4])]))/((20 + 5*x + Log[4])^8*(100*x + 25*x^2 + 5*x*Log[4])),x]

[Out]

(-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)))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2326

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]

Rubi steps \begin{align*} \text {integral}& = \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 (25 x^2+x (100+5 \log (4))\right )} \, dx \\ & = 2562890625 \int \frac {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 (25 x^2+x (100+5 \log (4))\right )} \, dx \\ & = 2562890625 \int \frac {e^{16+\log ^2\left (\frac {15 x}{20+5 x+\log (4)}\right )} x^7 \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 (100+25 x+5 \log (4))} \, dx \\ & = 512578125 \int \frac {e^{16+\log ^2\left (\frac {15 x}{20+5 x+\log (4)}\right )} x^7 \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))^9} \, dx \\ & = -\frac {512578125 e^{16+\log ^2\left (\frac {15 x}{20+5 x+\log (4)}\right )} x^8 (20+\log (4))}{(20+5 x+\log (4))^{10} \left (\frac {5 x}{(20+5 x+\log (4))^2}-\frac {1}{20+5 x+\log (4)}\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {2562890625 e^{16+\log ^2\left (\frac {15 x}{20+5 x+\log (4)}\right )} x^8 \left (160+8 \log (4)+(40+2 \log (4)) \log \left (\frac {15 x}{20+5 x+\log (4)}\right )\right )}{(20+5 x+\log (4))^8 \left (100 x+25 x^2+5 x \log (4)\right )} \, dx=\frac {512578125 e^{16+\log ^2\left (\frac {15 x}{20+5 x+\log (4)}\right )} x^8}{(20+5 x+\log (4))^8} \]

[In]

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]

[Out]

(512578125*E^(16 + Log[(15*x)/(20 + 5*x + Log[4])]^2)*x^8)/(20 + 5*x + Log[4])^8

Maple [A] (verified)

Time = 0.54 (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	\(\frac {{\mathrm e}^{\ln \left (\frac {15 x}{2 \ln \left (2\right )+20+5 x}\right )^{2}+8 \ln \left (\frac {15 x}{2 \ln \left (2\right )+20+5 x}\right )+16}}{5}\)	\(40\)

[In]

int(((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)+2
0+5*x))+16)/(10*x*ln(2)+25*x^2+100*x),x,method=_RETURNVERBOSE)

[Out]

512578125*x^8/(2*ln(2)+20+5*x)^8*exp(ln(15*x/(2*ln(2)+20+5*x))^2+16)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {2562890625 e^{16+\log ^2\left (\frac {15 x}{20+5 x+\log (4)}\right )} x^8 \left (160+8 \log (4)+(40+2 \log (4)) \log \left (\frac {15 x}{20+5 x+\log (4)}\right )\right )}{(20+5 x+\log (4))^8 \left (100 x+25 x^2+5 x \log (4)\right )} \, dx=\frac {1}{5} \, e^{\left (\log \left (\frac {15 \, x}{5 \, x + 2 \, \log \left (2\right ) + 20}\right )^{2} + 8 \, \log \left (\frac {15 \, x}{5 \, x + 2 \, \log \left (2\right ) + 20}\right ) + 16\right )} \]

[In]

integrate(((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, algorithm="fricas")

[Out]

1/5*e^(log(15*x/(5*x + 2*log(2) + 20))^2 + 8*log(15*x/(5*x + 2*log(2) + 20)) + 16)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (20) = 40\).

Time = 1.36 (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} \]

[In]

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)

[Out]

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 + 28000000*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 +
224000000*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 + 67200000*x**2*log(2)**4 + 896000000*x**2*log(2)**3 + 6720000000*x**2*lo
g(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) + 5120
0000000*x + 256*log(2)**8 + 20480*log(2)**7 + 716800*log(2)**6 + 14336000*log(2)**5 + 179200000*log(2)**4 + 14
33600000*log(2)**3 + 7168000000*log(2)**2 + 20480000000*log(2) + 25600000000)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 336 vs. \(2 (22) = 44\).

Time = 0.56 (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} \]

[In]

integrate(((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, algorithm="maxima")

[Out]

78125*3^(2*log(5) + 8)*x^8*e^(log(5)^2 + log(3)^2 - 2*log(5)*log(5*x + 2*log(2) + 20) - 2*log(3)*log(5*x + 2*l
og(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) + 1
00)*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
*(log(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 + 2100*log(2)^5 + 35000*log(2)^4 + 350000*log(2)^3 + 2100000*log(2)^2
+ 7000000*log(2) + 10000000)*x + 7168000000*log(2)^2 + 20480000000*log(2) + 25600000000)

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {2562890625 e^{16+\log ^2\left (\frac {15 x}{20+5 x+\log (4)}\right )} x^8 \left (160+8 \log (4)+(40+2 \log (4)) \log \left (\frac {15 x}{20+5 x+\log (4)}\right )\right )}{(20+5 x+\log (4))^8 \left (100 x+25 x^2+5 x \log (4)\right )} \, dx=\frac {1}{5} \, e^{\left (\log \left (\frac {15 \, x}{5 \, x + 2 \, \log \left (2\right ) + 20}\right )^{2} + 8 \, \log \left (\frac {15 \, x}{5 \, x + 2 \, \log \left (2\right ) + 20}\right ) + 16\right )} \]

[In]

integrate(((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, algorithm="giac")

[Out]

1/5*e^(log(15*x/(5*x + 2*log(2) + 20))^2 + 8*log(15*x/(5*x + 2*log(2) + 20)) + 16)

Mupad [B] (verification not implemented)

Time = 17.94 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.96 \[ \int \frac {2562890625 e^{16+\log ^2\left (\frac {15 x}{20+5 x+\log (4)}\right )} x^8 \left (160+8 \log (4)+(40+2 \log (4)) \log \left (\frac {15 x}{20+5 x+\log (4)}\right )\right )}{(20+5 x+\log (4))^8 \left (100 x+25 x^2+5 x \log (4)\right )} \, dx=\frac {512578125\,x^8\,{\mathrm {e}}^{{\ln \left (\frac {x}{5\,x+\ln \left (4\right )+20}\right )}^2+{\ln \left (15\right )}^2+16}\,{\left (\frac {x}{5\,x+\ln \left (4\right )+20}\right )}^{\ln \left (225\right )}}{{\left (5\,x+\ln \left (4\right )+20\right )}^8} \]

[In]

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)

[Out]

(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