3.8.36 \(\int \frac {8}{-3+12 x+(-1+4 x) \log (4)+(-1+4 x) \log (\frac {1}{16} (1-8 x+16 x^2))} \, dx\) [736]

3.8.36.1 Optimal result

Integrand size = 38, antiderivative size = 17 \begin {dmath*} \int \frac {8}{-3+12 x+(-1+4 x) \log (4)+(-1+4 x) \log \left (\frac {1}{16} \left (1-8 x+16 x^2\right )\right )} \, dx=\log \left (\frac {4}{3} \left (3+\log (4)+\log \left (\left (-\frac {1}{4}+x\right )^2\right )\right )\right ) \end {dmath*}

output

ln(4+8/3*ln(2)+4/3*ln((x-1/4)^2))
 

3.8.36.2 Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \begin {dmath*} \int \frac {8}{-3+12 x+(-1+4 x) \log (4)+(-1+4 x) \log \left (\frac {1}{16} \left (1-8 x+16 x^2\right )\right )} \, dx=\log \left (3+\log \left (\frac {1}{4} (1-4 x)^2\right )\right ) \end {dmath*}

input

Integrate[8/(-3 + 12*x + (-1 + 4*x)*Log[4] + (-1 + 4*x)*Log[(1 - 8*x + 16* 
x^2)/16]),x]
 

output

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

3.8.36.3 Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {27, 7292, 2837, 25, 2739, 14}

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.

\(\displaystyle \int \frac {8}{(4 x-1) \log \left (\frac {1}{16} \left (16 x^2-8 x+1\right )\right )+12 x+(4 x-1) \log (4)-3} \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle 8 \int \frac {1}{-\log \left (\frac {1}{16} \left (16 x^2-8 x+1\right )\right ) (1-4 x)-\log (4) (1-4 x)+12 x-3}dx\)

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 2837

\(\displaystyle -2 \int -\frac {1}{(1-4 x) \left (\log \left (\frac {1}{4} (1-4 x)^2\right )+3\right )}d(1-4 x)\)

\(\Big \downarrow \) 25

\(\displaystyle 2 \int \frac {1}{(1-4 x) \left (\log \left (\frac {1}{4} (1-4 x)^2\right )+3\right )}d(1-4 x)\)

\(\Big \downarrow \) 2739

\(\displaystyle \int \frac {1}{\log \left (\frac {1}{4} (1-4 x)^2\right )+3}d\left (\log \left (\frac {1}{4} (1-4 x)^2\right )+3\right )\)

\(\Big \downarrow \) 14

\(\displaystyle \log \left (\log \left (\frac {1}{4} (1-4 x)^2\right )+3\right )\)

input

Int[8/(-3 + 12*x + (-1 + 4*x)*Log[4] + (-1 + 4*x)*Log[(1 - 8*x + 16*x^2)/1 
6]),x]
 

output

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

3.8.36.3.1 Defintions of rubi rules used

rule 14

Int[(a_.)/(x_), x_Symbol] :> Simp[a*Log[x], x] /; FreeQ[a, x]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2739

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

rule 2837

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_. 
)*(x_))^(q_.), x_Symbol] :> Simp[1/e   Subst[Int[(f*(x/d))^q*(a + b*Log[c*x 
^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x] && 
EqQ[e*f - d*g, 0]
 

rule 7292

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

3.8.36.4 Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00

method	result	size
norman	\(\ln \left (2 \ln \left (2\right )+\ln \left (x^{2}-\frac {1}{2} x +\frac {1}{16}\right )+3\right )\)	\(17\)
risch	\(\ln \left (2 \ln \left (2\right )+\ln \left (x^{2}-\frac {1}{2} x +\frac {1}{16}\right )+3\right )\)	\(17\)
parallelrisch	\(\ln \left (2 \ln \left (2\right )+\ln \left (x^{2}-\frac {1}{2} x +\frac {1}{16}\right )+3\right )\)	\(17\)
default	\(\ln \left (2 \ln \left (2\right )-\ln \left (16 x^{2}-8 x +1\right )-3\right )\)	\(21\)

input

int(8/((-1+4*x)*ln(x^2-1/2*x+1/16)+2*(-1+4*x)*ln(2)+12*x-3),x,method=_RETU 
RNVERBOSE)
 

output

ln(2*ln(2)+ln(x^2-1/2*x+1/16)+3)
 

3.8.36.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \begin {dmath*} \int \frac {8}{-3+12 x+(-1+4 x) \log (4)+(-1+4 x) \log \left (\frac {1}{16} \left (1-8 x+16 x^2\right )\right )} \, dx=\log \left (2 \, \log \left (2\right ) + \log \left (x^{2} - \frac {1}{2} \, x + \frac {1}{16}\right ) + 3\right ) \end {dmath*}

input

integrate(8/((-1+4*x)*log(x^2-1/2*x+1/16)+2*(-1+4*x)*log(2)+12*x-3),x, alg 
orithm=\
 

output

log(2*log(2) + log(x^2 - 1/2*x + 1/16) + 3)
 

3.8.36.6 Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \begin {dmath*} \int \frac {8}{-3+12 x+(-1+4 x) \log (4)+(-1+4 x) \log \left (\frac {1}{16} \left (1-8 x+16 x^2\right )\right )} \, dx=\log {\left (\log {\left (x^{2} - \frac {x}{2} + \frac {1}{16} \right )} + 2 \log {\left (2 \right )} + 3 \right )} \end {dmath*}

input

integrate(8/((-1+4*x)*ln(x**2-1/2*x+1/16)+2*(-1+4*x)*ln(2)+12*x-3),x)
 

output

log(log(x**2 - x/2 + 1/16) + 2*log(2) + 3)
 

3.8.36.7 Maxima [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \begin {dmath*} \int \frac {8}{-3+12 x+(-1+4 x) \log (4)+(-1+4 x) \log \left (\frac {1}{16} \left (1-8 x+16 x^2\right )\right )} \, dx=\log \left (-\log \left (2\right ) + \log \left (4 \, x - 1\right ) + \frac {3}{2}\right ) \end {dmath*}

input

integrate(8/((-1+4*x)*log(x^2-1/2*x+1/16)+2*(-1+4*x)*log(2)+12*x-3),x, alg 
orithm=\
 

output

log(-log(2) + log(4*x - 1) + 3/2)
 

3.8.36.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \begin {dmath*} \int \frac {8}{-3+12 x+(-1+4 x) \log (4)+(-1+4 x) \log \left (\frac {1}{16} \left (1-8 x+16 x^2\right )\right )} \, dx=\log \left (-2 \, \log \left (2\right ) + \log \left (16 \, x^{2} - 8 \, x + 1\right ) + 3\right ) \end {dmath*}

input

integrate(8/((-1+4*x)*log(x^2-1/2*x+1/16)+2*(-1+4*x)*log(2)+12*x-3),x, alg 
orithm=\
 

output

log(-2*log(2) + log(16*x^2 - 8*x + 1) + 3)
 

3.8.36.9 Mupad [F(-1)]

Timed out. \begin {dmath*} \int \frac {8}{-3+12 x+(-1+4 x) \log (4)+(-1+4 x) \log \left (\frac {1}{16} \left (1-8 x+16 x^2\right )\right )} \, dx=\int \frac {8}{12\,x+2\,\ln \left (2\right )\,\left (4\,x-1\right )+\ln \left (x^2-\frac {x}{2}+\frac {1}{16}\right )\,\left (4\,x-1\right )-3} \,d x \end {dmath*}

input

int(8/(12*x + 2*log(2)*(4*x - 1) + log(x^2 - x/2 + 1/16)*(4*x - 1) - 3),x)
 

output

int(8/(12*x + 2*log(2)*(4*x - 1) + log(x^2 - x/2 + 1/16)*(4*x - 1) - 3), x 
)