3.75.53 \(\int \frac {12+10 x+5 x \log (4 x^2)+x \log ^2(4 x^2)+3 (i \pi +\log (-\log (\log (2))))^2}{16 x+8 x \log (4 x^2)+x \log ^2(4 x^2)} \, dx\)

Optimal. Leaf size=35 \[ x-\frac {3 \left (2+x+\frac {1}{2} (i \pi +\log (-\log (\log (2))))^2\right )}{4+\log \left (4 x^2\right )} \]

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Rubi [A]  time = 0.49, antiderivative size = 55, normalized size of antiderivative = 1.57, number of steps used = 12, number of rules used = 8, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {6741, 6742, 2300, 2178, 2353, 2297, 2302, 30} \begin {gather*} -\frac {3 \left (4-\pi ^2+\log ^2(-\log (\log (2)))+2 i \pi \log (-\log (\log (2)))\right )}{2 \left (\log \left (4 x^2\right )+4\right )}-\frac {3 x}{\log \left (4 x^2\right )+4}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12 + 10*x + 5*x*Log[4*x^2] + x*Log[4*x^2]^2 + 3*(I*Pi + Log[-Log[Log[2]]])^2)/(16*x + 8*x*Log[4*x^2] + x*
Log[4*x^2]^2),x]

[Out]

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

Rule 30

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

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2300

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[E^(x/n)*(a +
b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, 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 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10 x+5 x \log \left (4 x^2\right )+x \log ^2\left (4 x^2\right )+12 \left (1+\frac {1}{4} (i \pi +\log (-\log (\log (2))))^2\right )}{x \left (4+\log \left (4 x^2\right )\right )^2} \, dx\\ &=\int \left (1-\frac {3}{4+\log \left (4 x^2\right )}+\frac {3 \left (4-\pi ^2+2 x+2 i \pi \log (-\log (\log (2)))+\log ^2(-\log (\log (2)))\right )}{x \left (4+\log \left (4 x^2\right )\right )^2}\right ) \, dx\\ &=x-3 \int \frac {1}{4+\log \left (4 x^2\right )} \, dx+3 \int \frac {4-\pi ^2+2 x+2 i \pi \log (-\log (\log (2)))+\log ^2(-\log (\log (2)))}{x \left (4+\log \left (4 x^2\right )\right )^2} \, dx\\ &=x+3 \int \left (\frac {2}{\left (4+\log \left (4 x^2\right )\right )^2}+\frac {4-\pi ^2+2 i \pi \log (-\log (\log (2)))+\log ^2(-\log (\log (2)))}{x \left (4+\log \left (4 x^2\right )\right )^2}\right ) \, dx-\frac {(3 x) \operatorname {Subst}\left (\int \frac {e^{x/2}}{4+x} \, dx,x,\log \left (4 x^2\right )\right )}{4 \sqrt {x^2}}\\ &=x-\frac {3 x \text {Ei}\left (\frac {1}{2} \left (4+\log \left (4 x^2\right )\right )\right )}{4 e^2 \sqrt {x^2}}+6 \int \frac {1}{\left (4+\log \left (4 x^2\right )\right )^2} \, dx+\left (3 \left (4-\pi ^2+2 i \pi \log (-\log (\log (2)))+\log ^2(-\log (\log (2)))\right )\right ) \int \frac {1}{x \left (4+\log \left (4 x^2\right )\right )^2} \, dx\\ &=x-\frac {3 x \text {Ei}\left (\frac {1}{2} \left (4+\log \left (4 x^2\right )\right )\right )}{4 e^2 \sqrt {x^2}}-\frac {3 x}{4+\log \left (4 x^2\right )}+3 \int \frac {1}{4+\log \left (4 x^2\right )} \, dx+\frac {1}{2} \left (3 \left (4-\pi ^2+2 i \pi \log (-\log (\log (2)))+\log ^2(-\log (\log (2)))\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,4+\log \left (4 x^2\right )\right )\\ &=x-\frac {3 x \text {Ei}\left (\frac {1}{2} \left (4+\log \left (4 x^2\right )\right )\right )}{4 e^2 \sqrt {x^2}}-\frac {3 x}{4+\log \left (4 x^2\right )}-\frac {3 \left (4-\pi ^2+2 i \pi \log (-\log (\log (2)))+\log ^2(-\log (\log (2)))\right )}{2 \left (4+\log \left (4 x^2\right )\right )}+\frac {(3 x) \operatorname {Subst}\left (\int \frac {e^{x/2}}{4+x} \, dx,x,\log \left (4 x^2\right )\right )}{4 \sqrt {x^2}}\\ &=x-\frac {3 x}{4+\log \left (4 x^2\right )}-\frac {3 \left (4-\pi ^2+2 i \pi \log (-\log (\log (2)))+\log ^2(-\log (\log (2)))\right )}{2 \left (4+\log \left (4 x^2\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 45, normalized size = 1.29 \begin {gather*} x+\frac {3 \left (-4+\pi ^2-2 x-2 i \pi \log (-\log (\log (2)))-\log ^2(-\log (\log (2)))\right )}{2 \left (4+\log \left (4 x^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12 + 10*x + 5*x*Log[4*x^2] + x*Log[4*x^2]^2 + 3*(I*Pi + Log[-Log[Log[2]]])^2)/(16*x + 8*x*Log[4*x^2
] + x*Log[4*x^2]^2),x]

[Out]

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

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fricas [A]  time = 1.17, size = 34, normalized size = 0.97 \begin {gather*} \frac {2 \, x \log \left (4 \, x^{2}\right ) - 3 \, \log \left (\log \left (\log \relax (2)\right )\right )^{2} + 2 \, x - 12}{2 \, {\left (\log \left (4 \, x^{2}\right ) + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*log(log(log(2)))^2+x*log(4*x^2)^2+5*x*log(4*x^2)+10*x+12)/(x*log(4*x^2)^2+8*x*log(4*x^2)+16*x),x,
algorithm="fricas")

[Out]

1/2*(2*x*log(4*x^2) - 3*log(log(log(2)))^2 + 2*x - 12)/(log(4*x^2) + 4)

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giac [A]  time = 0.21, size = 27, normalized size = 0.77 \begin {gather*} x - \frac {3 \, {\left (\log \left (\log \left (\log \relax (2)\right )\right )^{2} + 2 \, x + 4\right )}}{2 \, {\left (2 \, \log \relax (2) + \log \left (x^{2}\right ) + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*log(log(log(2)))^2+x*log(4*x^2)^2+5*x*log(4*x^2)+10*x+12)/(x*log(4*x^2)^2+8*x*log(4*x^2)+16*x),x,
algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 26, normalized size = 0.74

method	result	size
risch	\(x -\frac {3 \left (\ln \left (\ln \left (\ln \relax (2)\right )\right )^{2}+2 x +4\right )}{2 \left (\ln \left (4 x^{2}\right )+4\right )}\)	\(26\)
norman	\(\frac {x +x \ln \left (4 x^{2}\right )-6-\frac {3 \ln \left (\ln \left (\ln \relax (2)\right )\right )^{2}}{2}}{\ln \left (4 x^{2}\right )+4}\)	\(31\)
default	\(\frac {x \ln \left (x^{2}\right )+\left (1+2 \ln \relax (2)\right ) x -6}{2 \ln \relax (2)+\ln \left (x^{2}\right )+4}-\frac {3 \ln \left (\ln \left (\ln \relax (2)\right )\right )^{2}}{2 \left (2 \ln \relax (2)+\ln \left (x^{2}\right )+4\right )}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*ln(ln(ln(2)))^2+x*ln(4*x^2)^2+5*x*ln(4*x^2)+10*x+12)/(x*ln(4*x^2)^2+8*x*ln(4*x^2)+16*x),x,method=_RETUR
NVERBOSE)

[Out]

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

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maxima [B]  time = 0.49, size = 51, normalized size = 1.46 \begin {gather*} -\frac {3 \, \log \left (\log \left (\log \relax (2)\right )\right )^{2}}{4 \, {\left (\log \relax (2) + \log \relax (x) + 2\right )}} + \frac {x {\left (2 \, \log \relax (2) + 1\right )} + 2 \, x \log \relax (x)}{2 \, {\left (\log \relax (2) + \log \relax (x) + 2\right )}} - \frac {3}{\log \relax (2) + \log \relax (x) + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*log(log(log(2)))^2+x*log(4*x^2)^2+5*x*log(4*x^2)+10*x+12)/(x*log(4*x^2)^2+8*x*log(4*x^2)+16*x),x,
algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.08, size = 48, normalized size = 1.37 \begin {gather*} \frac {x\,\left (\ln \left (4\,x^2\right )+1\right )}{\ln \left (4\,x^2\right )+4}+\frac {\ln \left (4\,x^2\right )\,\left (\frac {3\,{\ln \left (\ln \left (\ln \relax (2)\right )\right )}^2}{8}+\frac {3}{2}\right )}{\ln \left (4\,x^2\right )+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x + 5*x*log(4*x^2) + 3*log(log(log(2)))^2 + x*log(4*x^2)^2 + 12)/(16*x + 8*x*log(4*x^2) + x*log(4*x^2)
^2),x)

[Out]

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

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sympy [A]  time = 0.26, size = 49, normalized size = 1.40 \begin {gather*} x + \frac {- 6 x - 12 - 3 \log {\left (- \log {\left (\log {\relax (2 )} \right )} \right )}^{2} + 3 \pi ^{2} - 6 i \pi \log {\left (- \log {\left (\log {\relax (2 )} \right )} \right )}}{2 \log {\left (x^{2} \right )} + 4 \log {\relax (2 )} + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*ln(ln(ln(2)))**2+x*ln(4*x**2)**2+5*x*ln(4*x**2)+10*x+12)/(x*ln(4*x**2)**2+8*x*ln(4*x**2)+16*x),x)

[Out]

x + (-6*x - 12 - 3*log(-log(log(2)))**2 + 3*pi**2 - 6*I*pi*log(-log(log(2))))/(2*log(x**2) + 4*log(2) + 8)

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