3.19.42 \(\int \frac {2+2 \log (x)+2 \log ^2(x)}{256 x \log (x)+224 x \log ^2(x)-31 x \log ^3(x)+x \log ^4(x)+(32 x \log (x)+30 x \log ^2(x)-2 x \log ^3(x)) \log (\frac {3+3 \log (x)}{\log (x)})+(x \log (x)+x \log ^2(x)) \log ^2(\frac {3+3 \log (x)}{\log (x)})} \, dx\)

Optimal. Leaf size=19 \[ \frac {2}{16-\log (x)+\log \left (3+\frac {3}{\log (x)}\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 0.22, antiderivative size = 21, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 3, integrand size = 100, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {12, 1593, 6686} \begin {gather*} \frac {2}{-\log (x)+\log \left (\frac {3 (\log (x)+1)}{\log (x)}\right )+16} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 2*Log[x] + 2*Log[x]^2)/(256*x*Log[x] + 224*x*Log[x]^2 - 31*x*Log[x]^3 + x*Log[x]^4 + (32*x*Log[x] + 3
0*x*Log[x]^2 - 2*x*Log[x]^3)*Log[(3 + 3*Log[x])/Log[x]] + (x*Log[x] + x*Log[x]^2)*Log[(3 + 3*Log[x])/Log[x]]^2
),x]

[Out]

2/(16 - Log[x] + Log[(3*(1 + Log[x]))/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 1593

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\operatorname {Subst}\left (\int \frac {2 \left (1+x+x^2\right )}{\left (x+x^2\right ) \left (-16+x-\log \left (\frac {3+3 x}{x}\right )\right )^2} \, dx,x,\log (x)\right )\\ &=2 \operatorname {Subst}\left (\int \frac {1+x+x^2}{\left (x+x^2\right ) \left (-16+x-\log \left (\frac {3+3 x}{x}\right )\right )^2} \, dx,x,\log (x)\right )\\ &=2 \operatorname {Subst}\left (\int \frac {1+x+x^2}{x (1+x) \left (-16+x-\log \left (\frac {3+3 x}{x}\right )\right )^2} \, dx,x,\log (x)\right )\\ &=\frac {2}{16-\log (x)+\log \left (\frac {3 (1+\log (x))}{\log (x)}\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 19, normalized size = 1.00 \begin {gather*} \frac {2}{16-\log (x)+\log \left (3+\frac {3}{\log (x)}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 2*Log[x] + 2*Log[x]^2)/(256*x*Log[x] + 224*x*Log[x]^2 - 31*x*Log[x]^3 + x*Log[x]^4 + (32*x*Log[
x] + 30*x*Log[x]^2 - 2*x*Log[x]^3)*Log[(3 + 3*Log[x])/Log[x]] + (x*Log[x] + x*Log[x]^2)*Log[(3 + 3*Log[x])/Log
[x]]^2),x]

[Out]

2/(16 - Log[x] + Log[3 + 3/Log[x]])

________________________________________________________________________________________

fricas [A]  time = 0.85, size = 21, normalized size = 1.11 \begin {gather*} -\frac {2}{\log \relax (x) - \log \left (\frac {3 \, {\left (\log \relax (x) + 1\right )}}{\log \relax (x)}\right ) - 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)^2+2*log(x)+2)/((x*log(x)^2+x*log(x))*log((3*log(x)+3)/log(x))^2+(-2*x*log(x)^3+30*x*log(x)
^2+32*x*log(x))*log((3*log(x)+3)/log(x))+x*log(x)^4-31*x*log(x)^3+224*x*log(x)^2+256*x*log(x)),x, algorithm="f
ricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.28, size = 60, normalized size = 3.16 \begin {gather*} \frac {2 \, {\left (\frac {\log \relax (x) + 1}{\log \relax (x)} - 1\right )}}{\frac {{\left (\log \relax (x) + 1\right )} \log \left (\frac {3 \, {\left (\log \relax (x) + 1\right )}}{\log \relax (x)}\right )}{\log \relax (x)} + \frac {16 \, {\left (\log \relax (x) + 1\right )}}{\log \relax (x)} - \log \left (\frac {3 \, {\left (\log \relax (x) + 1\right )}}{\log \relax (x)}\right ) - 17} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)^2+2*log(x)+2)/((x*log(x)^2+x*log(x))*log((3*log(x)+3)/log(x))^2+(-2*x*log(x)^3+30*x*log(x)
^2+32*x*log(x))*log((3*log(x)+3)/log(x))+x*log(x)^4-31*x*log(x)^3+224*x*log(x)^2+256*x*log(x)),x, algorithm="g
iac")

[Out]

2*((log(x) + 1)/log(x) - 1)/((log(x) + 1)*log(3*(log(x) + 1)/log(x))/log(x) + 16*(log(x) + 1)/log(x) - log(3*(
log(x) + 1)/log(x)) - 17)

________________________________________________________________________________________

maple [C]  time = 0.24, size = 129, normalized size = 6.79

method	result	size
risch	\(\frac {4 i}{\pi \,\mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) \mathrm {csgn}\left (i \left (\ln \relax (x )+1\right )\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )+1\right )}{\ln \relax (x )}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )+1\right )}{\ln \relax (x )}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (\ln \relax (x )+1\right )\right ) \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )+1\right )}{\ln \relax (x )}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (\ln \relax (x )+1\right )}{\ln \relax (x )}\right )^{3}+2 i \ln \relax (3)-2 i \ln \relax (x )-2 i \ln \left (\ln \relax (x )\right )+2 i \ln \left (\ln \relax (x )+1\right )+32 i}\)	\(129\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*ln(x)^2+2*ln(x)+2)/((x*ln(x)^2+x*ln(x))*ln((3*ln(x)+3)/ln(x))^2+(-2*x*ln(x)^3+30*x*ln(x)^2+32*x*ln(x))*
ln((3*ln(x)+3)/ln(x))+x*ln(x)^4-31*x*ln(x)^3+224*x*ln(x)^2+256*x*ln(x)),x,method=_RETURNVERBOSE)

[Out]

4*I/(Pi*csgn(I/ln(x))*csgn(I*(ln(x)+1))*csgn(I/ln(x)*(ln(x)+1))-Pi*csgn(I/ln(x))*csgn(I/ln(x)*(ln(x)+1))^2-Pi*
csgn(I*(ln(x)+1))*csgn(I/ln(x)*(ln(x)+1))^2+Pi*csgn(I/ln(x)*(ln(x)+1))^3+2*I*ln(3)-2*I*ln(x)-2*I*ln(ln(x))+2*I
*ln(ln(x)+1)+32*I)

________________________________________________________________________________________

maxima [A]  time = 0.86, size = 22, normalized size = 1.16 \begin {gather*} \frac {2}{\log \relax (3) - \log \relax (x) + \log \left (\log \relax (x) + 1\right ) - \log \left (\log \relax (x)\right ) + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)^2+2*log(x)+2)/((x*log(x)^2+x*log(x))*log((3*log(x)+3)/log(x))^2+(-2*x*log(x)^3+30*x*log(x)
^2+32*x*log(x))*log((3*log(x)+3)/log(x))+x*log(x)^4-31*x*log(x)^3+224*x*log(x)^2+256*x*log(x)),x, algorithm="m
axima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 2.00, size = 22, normalized size = 1.16 \begin {gather*} \frac {2}{\ln \left (\frac {3\,\ln \relax (x)+3}{\ln \relax (x)}\right )-\ln \relax (x)+16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*log(x) + 2*log(x)^2 + 2)/(224*x*log(x)^2 - 31*x*log(x)^3 + x*log(x)^4 + log((3*log(x) + 3)/log(x))*(30*
x*log(x)^2 - 2*x*log(x)^3 + 32*x*log(x)) + log((3*log(x) + 3)/log(x))^2*(x*log(x)^2 + x*log(x)) + 256*x*log(x)
),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.32, size = 17, normalized size = 0.89 \begin {gather*} \frac {2}{- \log {\relax (x )} + \log {\left (\frac {3 \log {\relax (x )} + 3}{\log {\relax (x )}} \right )} + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*ln(x)**2+2*ln(x)+2)/((x*ln(x)**2+x*ln(x))*ln((3*ln(x)+3)/ln(x))**2+(-2*x*ln(x)**3+30*x*ln(x)**2+3
2*x*ln(x))*ln((3*ln(x)+3)/ln(x))+x*ln(x)**4-31*x*ln(x)**3+224*x*ln(x)**2+256*x*ln(x)),x)

[Out]

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

________________________________________________________________________________________