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

Optimal. Leaf size=17 \[ \log \left (\frac {4}{3} \left (3+\log (4)+\log \left (\left (-\frac {1}{4}+x\right )^2\right )\right )\right ) \]

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Rubi [A]  time = 0.07, antiderivative size = 15, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {12, 6688, 2390, 2302, 29} \begin {gather*} \log \left (\log \left (\frac {1}{4} (1-4 x)^2\right )+3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[3 + Log[(1 - 4*x)^2/4]]

Rule 12

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

Rule 29

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

Rule 2302

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 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=8 \int \frac {1}{-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\\ &=8 \int \frac {1}{(-1+4 x) \left (3+\log \left (\frac {1}{4} (1-4 x)^2\right )\right )} \, dx\\ &=2 \operatorname {Subst}\left (\int \frac {1}{x \left (3+\log \left (\frac {x^2}{4}\right )\right )} \, dx,x,1-4 x\right )\\ &=\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,3+\log \left (\frac {1}{4} (1-4 x)^2\right )\right )\\ &=\log \left (3+\log \left (\frac {1}{4} (1-4 x)^2\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 15, normalized size = 0.88 \begin {gather*} \log \left (3+\log \left (\frac {1}{4} (1-4 x)^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[3 + Log[(1 - 4*x)^2/4]]

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fricas [A]  time = 0.60, size = 16, normalized size = 0.94 \begin {gather*} \log \left (2 \, \log \relax (2) + \log \left (x^{2} - \frac {1}{2} \, x + \frac {1}{16}\right ) + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8/((4*x-1)*log(x^2-1/2*x+1/16)+2*(4*x-1)*log(2)+12*x-3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 18, normalized size = 1.06 \begin {gather*} \log \left (-2 \, \log \relax (2) + \log \left (16 \, x^{2} - 8 \, x + 1\right ) + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8/((4*x-1)*log(x^2-1/2*x+1/16)+2*(4*x-1)*log(2)+12*x-3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 17, normalized size = 1.00

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 13, normalized size = 0.76 \begin {gather*} \log \left (-\log \relax (2) + \log \left (4 \, x - 1\right ) + \frac {3}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8/((4*x-1)*log(x^2-1/2*x+1/16)+2*(4*x-1)*log(2)+12*x-3),x, algorithm="maxima")

[Out]

log(-log(2) + log(4*x - 1) + 3/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \begin {gather*} \int \frac {8}{12\,x+2\,\ln \relax (2)\,\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.16, size = 19, normalized size = 1.12 \begin {gather*} \log {\left (\log {\left (x^{2} - \frac {x}{2} + \frac {1}{16} \right )} + 2 \log {\relax (2 )} + 3 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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