3.6.36 \(\int \frac {16 x^2+4 x^4+4 x \log (5)+e^{8 e} (16 x^2+8 x^4+x^6+(8 x+2 x^3) \log (5)+\log ^2(5))+(-8 x^4+4 x \log (5)) \log (x) \log (\log (x))}{(16 x^3+8 x^5+x^7+(8 x^2+2 x^4) \log (5)+x \log ^2(5)) \log (x)} \, dx\) [536]

3.6.36.1 Optimal result

Integrand size = 110, antiderivative size = 25 \begin {dmath*} \int \frac {16 x^2+4 x^4+4 x \log (5)+e^{8 e} \left (16 x^2+8 x^4+x^6+\left (8 x+2 x^3\right ) \log (5)+\log ^2(5)\right )+\left (-8 x^4+4 x \log (5)\right ) \log (x) \log (\log (x))}{\left (16 x^3+8 x^5+x^7+\left (8 x^2+2 x^4\right ) \log (5)+x \log ^2(5)\right ) \log (x)} \, dx=\left (e^{8 e}+\frac {4}{4+x^2+\frac {\log (5)}{x}}\right ) \log (\log (x)) \end {dmath*}

output

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

3.6.36.2 Mathematica [A] (verified)

Time = 5.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \begin {dmath*} \int \frac {16 x^2+4 x^4+4 x \log (5)+e^{8 e} \left (16 x^2+8 x^4+x^6+\left (8 x+2 x^3\right ) \log (5)+\log ^2(5)\right )+\left (-8 x^4+4 x \log (5)\right ) \log (x) \log (\log (x))}{\left (16 x^3+8 x^5+x^7+\left (8 x^2+2 x^4\right ) \log (5)+x \log ^2(5)\right ) \log (x)} \, dx=\left (e^{8 e}+\frac {4 x}{4 x+x^3+\log (5)}\right ) \log (\log (x)) \end {dmath*}

input

Integrate[(16*x^2 + 4*x^4 + 4*x*Log[5] + E^(8*E)*(16*x^2 + 8*x^4 + x^6 + ( 
8*x + 2*x^3)*Log[5] + Log[5]^2) + (-8*x^4 + 4*x*Log[5])*Log[x]*Log[Log[x]] 
)/((16*x^3 + 8*x^5 + x^7 + (8*x^2 + 2*x^4)*Log[5] + x*Log[5]^2)*Log[x]),x]
 

output

(E^(8*E) + (4*x)/(4*x + x^3 + Log[5]))*Log[Log[x]]
 

3.6.36.3 Rubi [F]

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.

\(\displaystyle \int \frac {4 x^4+\left (4 x \log (5)-8 x^4\right ) \log (x) \log (\log (x))+16 x^2+e^{8 e} \left (x^6+8 x^4+\left (2 x^3+8 x\right ) \log (5)+16 x^2+\log ^2(5)\right )+4 x \log (5)}{\left (x^7+8 x^5+16 x^3+\left (2 x^4+8 x^2\right ) \log (5)+x \log ^2(5)\right ) \log (x)} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {4 x^4+\left (4 x \log (5)-8 x^4\right ) \log (x) \log (\log (x))+16 x^2+e^{8 e} \left (x^6+8 x^4+\left (2 x^3+8 x\right ) \log (5)+16 x^2+\log ^2(5)\right )+4 x \log (5)}{x \left (x^6+8 x^4+2 x^3 \log (5)+16 x^2+8 x \log (5)+\log ^2(5)\right ) \log (x)}dx\)

\(\Big \downarrow \) 2463

\(\displaystyle \int \frac {4 x^4+\left (4 x \log (5)-8 x^4\right ) \log (x) \log (\log (x))+16 x^2+e^{8 e} \left (x^6+8 x^4+\left (2 x^3+8 x\right ) \log (5)+16 x^2+\log ^2(5)\right )+4 x \log (5)}{x \left (x^3+4 x+\log (5)\right )^2 \log (x)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {e^{8 e} x^3+4 \left (1+e^{8 e}\right ) x+e^{8 e} \log (5)}{x \left (x^3+4 x+\log (5)\right ) \log (x)}-\frac {4 \left (2 x^3-\log (5)\right ) \log (\log (x))}{\left (x^3+4 x+\log (5)\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \int \frac {e^{8 e} x^3+4 \left (1+e^{8 e}\right ) x+e^{8 e} \log (5)}{x \left (x^3+4 x+\log (5)\right ) \log (x)}dx+4 \log (125) \int \frac {\log (\log (x))}{\left (x^3+4 x+\log (5)\right )^2}dx+32 \int \frac {x \log (\log (x))}{\left (x^3+4 x+\log (5)\right )^2}dx-8 \int \frac {\log (\log (x))}{x^3+4 x+\log (5)}dx\)

input

Int[(16*x^2 + 4*x^4 + 4*x*Log[5] + E^(8*E)*(16*x^2 + 8*x^4 + x^6 + (8*x + 
2*x^3)*Log[5] + Log[5]^2) + (-8*x^4 + 4*x*Log[5])*Log[x]*Log[Log[x]])/((16 
*x^3 + 8*x^5 + x^7 + (8*x^2 + 2*x^4)*Log[5] + x*Log[5]^2)*Log[x]),x]
 

output

$Aborted
 

3.6.36.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2026

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])
 

rule 2463

Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr 
and[u, Qx^p, x], x] /;  !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt 
Q[Expon[Px, x], 2] &&  !BinomialQ[Px, x] &&  !TrinomialQ[Px, x] && ILtQ[p, 
0]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.6.36.4 Maple [A] (verified)

Time = 61.59 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12

method	result	size
risch	\(\frac {4 x \ln \left (\ln \left (x \right )\right )}{x^{3}+\ln \left (5\right )+4 x}+{\mathrm e}^{8 \,{\mathrm e}} \ln \left (\ln \left (x \right )\right )\)	\(28\)
parallelrisch	\(\frac {{\mathrm e}^{8 \,{\mathrm e}} x^{3} \ln \left (\ln \left (x \right )\right )+\ln \left (5\right ) {\mathrm e}^{8 \,{\mathrm e}} \ln \left (\ln \left (x \right )\right )+4 \,{\mathrm e}^{8 \,{\mathrm e}} x \ln \left (\ln \left (x \right )\right )+4 x \ln \left (\ln \left (x \right )\right )}{x^{3}+\ln \left (5\right )+4 x}\)	\(69\)

input

int(((4*x*ln(5)-8*x^4)*ln(x)*ln(ln(x))+(ln(5)^2+(2*x^3+8*x)*ln(5)+x^6+8*x^ 
4+16*x^2)*exp(exp(1+2*ln(2)))^2+4*x*ln(5)+4*x^4+16*x^2)/(x*ln(5)^2+(2*x^4+ 
8*x^2)*ln(5)+x^7+8*x^5+16*x^3)/ln(x),x,method=_RETURNVERBOSE)
 

output

4*x/(x^3+ln(5)+4*x)*ln(ln(x))+exp(8*exp(1))*ln(ln(x))
 

3.6.36.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \begin {dmath*} \int \frac {16 x^2+4 x^4+4 x \log (5)+e^{8 e} \left (16 x^2+8 x^4+x^6+\left (8 x+2 x^3\right ) \log (5)+\log ^2(5)\right )+\left (-8 x^4+4 x \log (5)\right ) \log (x) \log (\log (x))}{\left (16 x^3+8 x^5+x^7+\left (8 x^2+2 x^4\right ) \log (5)+x \log ^2(5)\right ) \log (x)} \, dx=\frac {{\left ({\left (x^{3} + 4 \, x + \log \left (5\right )\right )} e^{\left (2 \, e^{\left (2 \, \log \left (2\right ) + 1\right )}\right )} + 4 \, x\right )} \log \left (\log \left (x\right )\right )}{x^{3} + 4 \, x + \log \left (5\right )} \end {dmath*}

input

integrate(((4*x*log(5)-8*x^4)*log(x)*log(log(x))+(log(5)^2+(2*x^3+8*x)*log 
(5)+x^6+8*x^4+16*x^2)*exp(exp(1+2*log(2)))^2+4*x*log(5)+4*x^4+16*x^2)/(x*l 
og(5)^2+(2*x^4+8*x^2)*log(5)+x^7+8*x^5+16*x^3)/log(x),x, algorithm=\
 

output

((x^3 + 4*x + log(5))*e^(2*e^(2*log(2) + 1)) + 4*x)*log(log(x))/(x^3 + 4*x 
 + log(5))
 

3.6.36.6 Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \begin {dmath*} \int \frac {16 x^2+4 x^4+4 x \log (5)+e^{8 e} \left (16 x^2+8 x^4+x^6+\left (8 x+2 x^3\right ) \log (5)+\log ^2(5)\right )+\left (-8 x^4+4 x \log (5)\right ) \log (x) \log (\log (x))}{\left (16 x^3+8 x^5+x^7+\left (8 x^2+2 x^4\right ) \log (5)+x \log ^2(5)\right ) \log (x)} \, dx=\frac {4 x \log {\left (\log {\left (x \right )} \right )}}{x^{3} + 4 x + \log {\left (5 \right )}} + e^{8 e} \log {\left (\log {\left (x \right )} \right )} \end {dmath*}

input

integrate(((4*x*ln(5)-8*x**4)*ln(x)*ln(ln(x))+(ln(5)**2+(2*x**3+8*x)*ln(5) 
+x**6+8*x**4+16*x**2)*exp(exp(1+2*ln(2)))**2+4*x*ln(5)+4*x**4+16*x**2)/(x* 
ln(5)**2+(2*x**4+8*x**2)*ln(5)+x**7+8*x**5+16*x**3)/ln(x),x)
 

output

4*x*log(log(x))/(x**3 + 4*x + log(5)) + exp(8*E)*log(log(x))
 

3.6.36.7 Maxima [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.72 \begin {dmath*} \int \frac {16 x^2+4 x^4+4 x \log (5)+e^{8 e} \left (16 x^2+8 x^4+x^6+\left (8 x+2 x^3\right ) \log (5)+\log ^2(5)\right )+\left (-8 x^4+4 x \log (5)\right ) \log (x) \log (\log (x))}{\left (16 x^3+8 x^5+x^7+\left (8 x^2+2 x^4\right ) \log (5)+x \log ^2(5)\right ) \log (x)} \, dx=\frac {{\left (x^{3} e^{\left (8 \, e\right )} + 4 \, x {\left (e^{\left (8 \, e\right )} + 1\right )} + e^{\left (8 \, e\right )} \log \left (5\right )\right )} \log \left (\log \left (x\right )\right )}{x^{3} + 4 \, x + \log \left (5\right )} \end {dmath*}

input

integrate(((4*x*log(5)-8*x^4)*log(x)*log(log(x))+(log(5)^2+(2*x^3+8*x)*log 
(5)+x^6+8*x^4+16*x^2)*exp(exp(1+2*log(2)))^2+4*x*log(5)+4*x^4+16*x^2)/(x*l 
og(5)^2+(2*x^4+8*x^2)*log(5)+x^7+8*x^5+16*x^3)/log(x),x, algorithm=\
 

output

(x^3*e^(8*e) + 4*x*(e^(8*e) + 1) + e^(8*e)*log(5))*log(log(x))/(x^3 + 4*x 
+ log(5))
 

3.6.36.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.12 \begin {dmath*} \int \frac {16 x^2+4 x^4+4 x \log (5)+e^{8 e} \left (16 x^2+8 x^4+x^6+\left (8 x+2 x^3\right ) \log (5)+\log ^2(5)\right )+\left (-8 x^4+4 x \log (5)\right ) \log (x) \log (\log (x))}{\left (16 x^3+8 x^5+x^7+\left (8 x^2+2 x^4\right ) \log (5)+x \log ^2(5)\right ) \log (x)} \, dx=\frac {x^{3} e^{\left (8 \, e\right )} \log \left (\log \left (x\right )\right ) + 4 \, x e^{\left (8 \, e\right )} \log \left (\log \left (x\right )\right ) + e^{\left (8 \, e\right )} \log \left (5\right ) \log \left (\log \left (x\right )\right ) + 4 \, x \log \left (\log \left (x\right )\right )}{x^{3} + 4 \, x + \log \left (5\right )} \end {dmath*}

input

integrate(((4*x*log(5)-8*x^4)*log(x)*log(log(x))+(log(5)^2+(2*x^3+8*x)*log 
(5)+x^6+8*x^4+16*x^2)*exp(exp(1+2*log(2)))^2+4*x*log(5)+4*x^4+16*x^2)/(x*l 
og(5)^2+(2*x^4+8*x^2)*log(5)+x^7+8*x^5+16*x^3)/log(x),x, algorithm=\
 

output

(x^3*e^(8*e)*log(log(x)) + 4*x*e^(8*e)*log(log(x)) + e^(8*e)*log(5)*log(lo 
g(x)) + 4*x*log(log(x)))/(x^3 + 4*x + log(5))
 

3.6.36.9 Mupad [F(-1)]

Timed out. \begin {dmath*} \int \frac {16 x^2+4 x^4+4 x \log (5)+e^{8 e} \left (16 x^2+8 x^4+x^6+\left (8 x+2 x^3\right ) \log (5)+\log ^2(5)\right )+\left (-8 x^4+4 x \log (5)\right ) \log (x) \log (\log (x))}{\left (16 x^3+8 x^5+x^7+\left (8 x^2+2 x^4\right ) \log (5)+x \log ^2(5)\right ) \log (x)} \, dx=\int \frac {4\,x\,\ln \left (5\right )+{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,\ln \left (2\right )+1}}\,\left (\ln \left (5\right )\,\left (2\,x^3+8\,x\right )+{\ln \left (5\right )}^2+16\,x^2+8\,x^4+x^6\right )+16\,x^2+4\,x^4+\ln \left (\ln \left (x\right )\right )\,\ln \left (x\right )\,\left (4\,x\,\ln \left (5\right )-8\,x^4\right )}{\ln \left (x\right )\,\left (\ln \left (5\right )\,\left (2\,x^4+8\,x^2\right )+x\,{\ln \left (5\right )}^2+16\,x^3+8\,x^5+x^7\right )} \,d x \end {dmath*}

input

int((4*x*log(5) + exp(2*exp(2*log(2) + 1))*(log(5)*(8*x + 2*x^3) + log(5)^ 
2 + 16*x^2 + 8*x^4 + x^6) + 16*x^2 + 4*x^4 + log(log(x))*log(x)*(4*x*log(5 
) - 8*x^4))/(log(x)*(log(5)*(8*x^2 + 2*x^4) + x*log(5)^2 + 16*x^3 + 8*x^5 
+ x^7)),x)
 

output

int((4*x*log(5) + exp(2*exp(2*log(2) + 1))*(log(5)*(8*x + 2*x^3) + log(5)^ 
2 + 16*x^2 + 8*x^4 + x^6) + 16*x^2 + 4*x^4 + log(log(x))*log(x)*(4*x*log(5 
) - 8*x^4))/(log(x)*(log(5)*(8*x^2 + 2*x^4) + x*log(5)^2 + 16*x^3 + 8*x^5 
+ x^7)), x)