\(\int \frac {6+2 \log (4)+(3+3 x+(1+x) \log (4)) \log (x^2)}{x \log (x^2)} \, dx\) [6854]

Optimal result

Integrand size = 32, antiderivative size = 15 \[ \int \frac {6+2 \log (4)+(3+3 x+(1+x) \log (4)) \log \left (x^2\right )}{x \log \left (x^2\right )} \, dx=(3+\log (4)) \left (x+\log \left (9 x \log \left (x^2\right )\right )\right ) \]

[Out]

(2*ln(2)+3)*(x+ln(9*x*ln(x^2)))

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.60, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {6820, 12, 6874, 45, 2339, 29} \[ \int \frac {6+2 \log (4)+(3+3 x+(1+x) \log (4)) \log \left (x^2\right )}{x \log \left (x^2\right )} \, dx=(3+\log (4)) \log \left (\log \left (x^2\right )\right )+x (3+\log (4))+(3+\log (4)) \log (x) \]

[In]

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

[Out]

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

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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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]]

Rule 6874

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

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(32\) vs. \(2(15)=30\).

Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 2.13 \[ \int \frac {6+2 \log (4)+(3+3 x+(1+x) \log (4)) \log \left (x^2\right )}{x \log \left (x^2\right )} \, dx=3 x+x \log (4)+3 \log (x)+\log (4) \log (x)+3 \log \left (\log \left (x^2\right )\right )+\log (4) \log \left (\log \left (x^2\right )\right ) \]

[In]

Integrate[(6 + 2*Log[4] + (3 + 3*x + (1 + x)*Log[4])*Log[x^2])/(x*Log[x^2]),x]

[Out]

3*x + x*Log[4] + 3*Log[x] + Log[4]*Log[x] + 3*Log[Log[x^2]] + Log[4]*Log[Log[x^2]]

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.73

method	result	size
parts	\(\left (2 \ln \left (2\right )+3\right ) \left (x +\ln \left (x \right )\right )+\frac {\left (4 \ln \left (2\right )+6\right ) \ln \left (\ln \left (x^{2}\right )\right )}{2}\)	\(26\)
default	\(2 \ln \left (2\right ) \left (x +\ln \left (x \right )+\ln \left (\ln \left (x^{2}\right )\right )\right )+3 \ln \left (\ln \left (x^{2}\right )\right )+3 x +3 \ln \left (x \right )\)	\(29\)
norman	\(x \left (2 \ln \left (2\right )+3\right )+\left (\frac {3}{2}+\ln \left (2\right )\right ) \ln \left (x^{2}\right )+\left (2 \ln \left (2\right )+3\right ) \ln \left (\ln \left (x^{2}\right )\right )\)	\(31\)
risch	\(2 \ln \left (2\right ) \ln \left (x \right )+2 x \ln \left (2\right )+3 \ln \left (x \right )+3 x +2 \ln \left (\ln \left (x^{2}\right )\right ) \ln \left (2\right )+3 \ln \left (\ln \left (x^{2}\right )\right )\)	\(36\)
parallelrisch	\(2 x \ln \left (2\right )+\ln \left (2\right ) \ln \left (x^{2}\right )+3 x +\frac {3 \ln \left (x^{2}\right )}{2}+2 \ln \left (\ln \left (x^{2}\right )\right ) \ln \left (2\right )+3 \ln \left (\ln \left (x^{2}\right )\right )\)	\(39\)

[In]

int(((2*ln(2)*(1+x)+3*x+3)*ln(x^2)+4*ln(2)+6)/x/ln(x^2),x,method=_RETURNVERBOSE)

[Out]

(2*ln(2)+3)*(x+ln(x))+1/2*(4*ln(2)+6)*ln(ln(x^2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 2.20 \[ \int \frac {6+2 \log (4)+(3+3 x+(1+x) \log (4)) \log \left (x^2\right )}{x \log \left (x^2\right )} \, dx=2 \, x \log \left (2\right ) + \frac {1}{2} \, {\left (2 \, \log \left (2\right ) + 3\right )} \log \left (x^{2}\right ) + {\left (2 \, \log \left (2\right ) + 3\right )} \log \left (\log \left (x^{2}\right )\right ) + 3 \, x \]

[In]

integrate(((2*log(2)*(1+x)+3*x+3)*log(x^2)+4*log(2)+6)/x/log(x^2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 2.07 \[ \int \frac {6+2 \log (4)+(3+3 x+(1+x) \log (4)) \log \left (x^2\right )}{x \log \left (x^2\right )} \, dx=x \left (2 \log {\left (2 \right )} + 3\right ) + \left (2 \log {\left (2 \right )} + 3\right ) \log {\left (x \right )} + \left (2 \log {\left (2 \right )} + 3\right ) \log {\left (\log {\left (x^{2} \right )} \right )} \]

[In]

integrate(((2*ln(2)*(1+x)+3*x+3)*ln(x**2)+4*ln(2)+6)/x/ln(x**2),x)

[Out]

x*(2*log(2) + 3) + (2*log(2) + 3)*log(x) + (2*log(2) + 3)*log(log(x**2))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (17) = 34\).

Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.33 \[ \int \frac {6+2 \log (4)+(3+3 x+(1+x) \log (4)) \log \left (x^2\right )}{x \log \left (x^2\right )} \, dx=2 \, x \log \left (2\right ) + 2 \, \log \left (2\right ) \log \left (x\right ) + 2 \, \log \left (2\right ) \log \left (\log \left (x^{2}\right )\right ) + 3 \, x + 3 \, \log \left (x\right ) + 3 \, \log \left (\log \left (x^{2}\right )\right ) \]

[In]

integrate(((2*log(2)*(1+x)+3*x+3)*log(x^2)+4*log(2)+6)/x/log(x^2),x, algorithm="maxima")

[Out]

2*x*log(2) + 2*log(2)*log(x) + 2*log(2)*log(log(x^2)) + 3*x + 3*log(x) + 3*log(log(x^2))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 2.00 \[ \int \frac {6+2 \log (4)+(3+3 x+(1+x) \log (4)) \log \left (x^2\right )}{x \log \left (x^2\right )} \, dx=x {\left (2 \, \log \left (2\right ) + 3\right )} + {\left (2 \, \log \left (2\right ) + 3\right )} \log \left (x\right ) + {\left (2 \, \log \left (2\right ) + 3\right )} \log \left (\log \left (x^{2}\right )\right ) \]

[In]

integrate(((2*log(2)*(1+x)+3*x+3)*log(x^2)+4*log(2)+6)/x/log(x^2),x, algorithm="giac")

[Out]

x*(2*log(2) + 3) + (2*log(2) + 3)*log(x) + (2*log(2) + 3)*log(log(x^2))

Mupad [B] (verification not implemented)

Time = 11.47 (sec) , antiderivative size = 36, normalized size of antiderivative = 2.40 \[ \int \frac {6+2 \log (4)+(3+3 x+(1+x) \log (4)) \log \left (x^2\right )}{x \log \left (x^2\right )} \, dx=\ln \left (\ln \left (x^2\right )\right )\,\left (\ln \left (4\right )+3\right )+\frac {x^3\,\left (\ln \left (4\right )+3\right )+x^2\,\ln \left (x^2\right )\,\left (\ln \left (2\right )+\frac {3}{2}\right )}{x^2} \]

[In]

int((4*log(2) + log(x^2)*(3*x + 2*log(2)*(x + 1) + 3) + 6)/(x*log(x^2)),x)

[Out]

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