\(\int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+(-2+x^4 \log (5)-x^8 \log ^2(5)) \log (x)}{(8 x^2+8 x^6 \log (5)+2 x^{10} \log ^2(5)) \log (x)} \, dx\) [708]

Optimal result

Integrand size = 69, antiderivative size = 26 \[ \int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+\left (-2+x^4 \log (5)-x^8 \log ^2(5)\right ) \log (x)}{\left (8 x^2+8 x^6 \log (5)+2 x^{10} \log ^2(5)\right ) \log (x)} \, dx=\frac {1}{2 \left (x+\frac {x}{1+x^4 \log (5)}\right )}+\log \left (\log ^2(x)\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.145, Rules used = {1608, 28, 6820, 12, 14, 1498, 21, 30, 2339, 29} \[ \int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+\left (-2+x^4 \log (5)-x^8 \log ^2(5)\right ) \log (x)}{\left (8 x^2+8 x^6 \log (5)+2 x^{10} \log ^2(5)\right ) \log (x)} \, dx=\frac {x^3 \log (5)}{4 \left (x^4 \log (5)+2\right )}+\frac {1}{4 x}+2 \log (\log (x)) \]

[In]

Int[(16*x + 16*x^5*Log[5] + 4*x^9*Log[5]^2 + (-2 + x^4*Log[5] - x^8*Log[5]^2)*Log[x])/((8*x^2 + 8*x^6*Log[5] +
2*x^10*Log[5]^2)*Log[x]),x]

[Out]

1/(4*x) + (x^3*Log[5])/(4*(2 + x^4*Log[5])) + 2*Log[Log[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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 1498

Int[(x_)^(m_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_), x_Symbol] :> S
imp[(-d)^((m - Mod[m, n])/n - 1)*(c*d^2 - b*d*e + a*e^2)^p*x^(Mod[m, n] + 1)*((d + e*x^n)^(q + 1)/(n*e^(2*p +
(m - Mod[m, n])/n)*(q + 1))), x] + Dist[(-d)^((m - Mod[m, n])/n - 1)/(n*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^n)^
(q + 1)*ExpandToSum[Together[(1/(d + e*x^n))*(n*(-d)^(-(m - Mod[m, n])/n + 1)*e^(2*p)*(q + 1)*(a + b*x^n + c*x
^(2*n))^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^((m - Mod[m, n])/n)*x^(m - Mod[m, n])))*(d*(Mod[m, n] + 1) + e*(Mod[
m, n] + n*(q + 1) + 1)*x^n))], x], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0]
&& IGtQ[n, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m, 0]

Rule 1608

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

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+\left (-2+x^4 \log (5)-x^8 \log ^2(5)\right ) \log (x)}{x^2 \left (8+8 x^4 \log (5)+2 x^8 \log ^2(5)\right ) \log (x)} \, dx \\ & = \left (2 \log ^2(5)\right ) \int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+\left (-2+x^4 \log (5)-x^8 \log ^2(5)\right ) \log (x)}{x^2 \left (4 \log (5)+2 x^4 \log ^2(5)\right )^2 \log (x)} \, dx \\ & = \left (2 \log ^2(5)\right ) \int \frac {\frac {-2+x^4 \log (5)-x^8 \log ^2(5)}{\left (2+x^4 \log (5)\right )^2}+\frac {4 x}{\log (x)}}{4 x^2 \log ^2(5)} \, dx \\ & = \frac {1}{2} \int \frac {\frac {-2+x^4 \log (5)-x^8 \log ^2(5)}{\left (2+x^4 \log (5)\right )^2}+\frac {4 x}{\log (x)}}{x^2} \, dx \\ & = \frac {1}{2} \int \left (\frac {-2+x^4 \log (5)-x^8 \log ^2(5)}{x^2 \left (2+x^4 \log (5)\right )^2}+\frac {4}{x \log (x)}\right ) \, dx \\ & = \frac {1}{2} \int \frac {-2+x^4 \log (5)-x^8 \log ^2(5)}{x^2 \left (2+x^4 \log (5)\right )^2} \, dx+2 \int \frac {1}{x \log (x)} \, dx \\ & = \frac {x^3 \log (5)}{4 \left (2+x^4 \log (5)\right )}+2 \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )-\frac {\int \frac {16 \log ^2(5)+8 x^4 \log ^3(5)}{x^2 \left (2+x^4 \log (5)\right )} \, dx}{32 \log ^2(5)} \\ & = \frac {x^3 \log (5)}{4 \left (2+x^4 \log (5)\right )}+2 \log (\log (x))-\frac {1}{4} \int \frac {1}{x^2} \, dx \\ & = \frac {1}{4 x}+\frac {x^3 \log (5)}{4 \left (2+x^4 \log (5)\right )}+2 \log (\log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 5.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+\left (-2+x^4 \log (5)-x^8 \log ^2(5)\right ) \log (x)}{\left (8 x^2+8 x^6 \log (5)+2 x^{10} \log ^2(5)\right ) \log (x)} \, dx=\frac {1}{4} \left (\frac {1}{x}+\frac {x^3 \log (5)}{2+x^4 \log (5)}+8 \log (\log (x))\right ) \]

[In]

Integrate[(16*x + 16*x^5*Log[5] + 4*x^9*Log[5]^2 + (-2 + x^4*Log[5] - x^8*Log[5]^2)*Log[x])/((8*x^2 + 8*x^6*Lo
g[5] + 2*x^10*Log[5]^2)*Log[x]),x]

[Out]

(x^(-1) + (x^3*Log[5])/(2 + x^4*Log[5]) + 8*Log[Log[x]])/4

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12

method	result	size
default	\(2 \ln \left (\ln \left (x \right )\right )+\frac {1}{4 x}+\frac {x^{3} \ln \left (5\right )}{4 x^{4} \ln \left (5\right )+8}\)	\(29\)
parts	\(2 \ln \left (\ln \left (x \right )\right )+\frac {1}{4 x}+\frac {x^{3} \ln \left (5\right )}{4 x^{4} \ln \left (5\right )+8}\)	\(29\)
norman	\(\frac {\frac {1}{2}+\frac {x^{4} \ln \left (5\right )}{2}}{x \left (x^{4} \ln \left (5\right )+2\right )}+2 \ln \left (\ln \left (x \right )\right )\)	\(30\)
risch	\(\frac {x^{4} \ln \left (5\right )+1}{2 \left (x^{4} \ln \left (5\right )+2\right ) x}+2 \ln \left (\ln \left (x \right )\right )\)	\(30\)
parallelrisch	\(\frac {8 \ln \left (5\right ) \ln \left (\ln \left (x \right )\right ) x^{5}+2+2 x^{4} \ln \left (5\right )+16 x \ln \left (\ln \left (x \right )\right )}{4 x \left (x^{4} \ln \left (5\right )+2\right )}\)	\(41\)

[In]

int(((-x^8*ln(5)^2+x^4*ln(5)-2)*ln(x)+4*x^9*ln(5)^2+16*x^5*ln(5)+16*x)/(2*x^10*ln(5)^2+8*x^6*ln(5)+8*x^2)/ln(x
),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+\left (-2+x^4 \log (5)-x^8 \log ^2(5)\right ) \log (x)}{\left (8 x^2+8 x^6 \log (5)+2 x^{10} \log ^2(5)\right ) \log (x)} \, dx=\frac {x^{4} \log \left (5\right ) + 4 \, {\left (x^{5} \log \left (5\right ) + 2 \, x\right )} \log \left (\log \left (x\right )\right ) + 1}{2 \, {\left (x^{5} \log \left (5\right ) + 2 \, x\right )}} \]

[In]

integrate(((-x^8*log(5)^2+x^4*log(5)-2)*log(x)+4*x^9*log(5)^2+16*x^5*log(5)+16*x)/(2*x^10*log(5)^2+8*x^6*log(5
)+8*x^2)/log(x),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+\left (-2+x^4 \log (5)-x^8 \log ^2(5)\right ) \log (x)}{\left (8 x^2+8 x^6 \log (5)+2 x^{10} \log ^2(5)\right ) \log (x)} \, dx=- \frac {- x^{4} \log {\left (5 \right )} - 1}{2 x^{5} \log {\left (5 \right )} + 4 x} + 2 \log {\left (\log {\left (x \right )} \right )} \]

[In]

integrate(((-x**8*ln(5)**2+x**4*ln(5)-2)*ln(x)+4*x**9*ln(5)**2+16*x**5*ln(5)+16*x)/(2*x**10*ln(5)**2+8*x**6*ln
(5)+8*x**2)/ln(x),x)

[Out]

-(-x**4*log(5) - 1)/(2*x**5*log(5) + 4*x) + 2*log(log(x))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+\left (-2+x^4 \log (5)-x^8 \log ^2(5)\right ) \log (x)}{\left (8 x^2+8 x^6 \log (5)+2 x^{10} \log ^2(5)\right ) \log (x)} \, dx=\frac {x^{4} \log \left (5\right ) + 1}{2 \, {\left (x^{5} \log \left (5\right ) + 2 \, x\right )}} + 2 \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate(((-x^8*log(5)^2+x^4*log(5)-2)*log(x)+4*x^9*log(5)^2+16*x^5*log(5)+16*x)/(2*x^10*log(5)^2+8*x^6*log(5
)+8*x^2)/log(x),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+\left (-2+x^4 \log (5)-x^8 \log ^2(5)\right ) \log (x)}{\left (8 x^2+8 x^6 \log (5)+2 x^{10} \log ^2(5)\right ) \log (x)} \, dx=\frac {x^{3} \log \left (5\right )}{4 \, {\left (x^{4} \log \left (5\right ) + 2\right )}} + \frac {1}{4 \, x} + 2 \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate(((-x^8*log(5)^2+x^4*log(5)-2)*log(x)+4*x^9*log(5)^2+16*x^5*log(5)+16*x)/(2*x^10*log(5)^2+8*x^6*log(5
)+8*x^2)/log(x),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.32 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {16 x+16 x^5 \log (5)+4 x^9 \log ^2(5)+\left (-2+x^4 \log (5)-x^8 \log ^2(5)\right ) \log (x)}{\left (8 x^2+8 x^6 \log (5)+2 x^{10} \log ^2(5)\right ) \log (x)} \, dx=2\,\ln \left (\ln \left (x\right )\right )+\frac {\ln \left (5\right )\,x^4+1}{2\,\ln \left (5\right )\,x^5+4\,x} \]

[In]

int((16*x + 4*x^9*log(5)^2 - log(x)*(x^8*log(5)^2 - x^4*log(5) + 2) + 16*x^5*log(5))/(log(x)*(2*x^10*log(5)^2
+ 8*x^6*log(5) + 8*x^2)),x)

[Out]

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