\(\int \frac {(4+4 x) \log (1+x)+(-4-8 x) \log ^2(1+x)+\log (3 x) (-4 x+(4+8 x) \log (1+x))}{(x+x^2) \log (3 x) \log (1+x)+(-x-x^2) \log ^2(1+x)} \, dx\) [3565]

Optimal result

Integrand size = 76, antiderivative size = 23 \[ \int \frac {(4+4 x) \log (1+x)+(-4-8 x) \log ^2(1+x)+\log (3 x) (-4 x+(4+8 x) \log (1+x))}{\left (x+x^2\right ) \log (3 x) \log (1+x)+\left (-x-x^2\right ) \log ^2(1+x)} \, dx=4 \log \left ((1+x) \left (-x+\frac {x \log (3 x)}{\log (1+x)}\right )\right ) \]

[Out]

4*ln((-x+x*ln(3*x)/ln(1+x))*(1+x))

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.092, Rules used = {6873, 6874, 78, 6816, 2437, 2339, 29} \[ \int \frac {(4+4 x) \log (1+x)+(-4-8 x) \log ^2(1+x)+\log (3 x) (-4 x+(4+8 x) \log (1+x))}{\left (x+x^2\right ) \log (3 x) \log (1+x)+\left (-x-x^2\right ) \log ^2(1+x)} \, dx=4 \log (x)+4 \log (x+1)+4 \log (\log (3 x)-\log (x+1))-4 \log (\log (x+1)) \]

[In]

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

[Out]

4*Log[x] + 4*Log[1 + x] + 4*Log[Log[3*x] - Log[1 + x]] - 4*Log[Log[1 + x]]

Rule 29

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 2437

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 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6873

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

Rule 6874

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

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {(4+4 x) \log (1+x)+(-4-8 x) \log ^2(1+x)+\log (3 x) (-4 x+(4+8 x) \log (1+x))}{\left (x+x^2\right ) \log (3 x) \log (1+x)+\left (-x-x^2\right ) \log ^2(1+x)} \, dx=4 (\log (x)+\log (1+x)+\log (\log (3 x)-\log (1+x))-\log (\log (1+x))) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.89 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35

method	result	size
risch	\(4 \ln \left (x^{2}+x \right )-4 \ln \left (\ln \left (1+x \right )\right )+4 \ln \left (\ln \left (1+x \right )-\ln \left (3 x \right )\right )\)	\(31\)
default	\(4 \ln \left (x \right )+4 \ln \left (1+x \right )-4 \ln \left (\ln \left (1+x \right )\right )+4 \ln \left (\ln \left (3\right )-\ln \left (1+x \right )+\ln \left (x \right )\right )\)	\(33\)
parallelrisch	\(-4 \ln \left (\ln \left (1+x \right )\right )+4 \ln \left (\ln \left (3 x \right )-\ln \left (1+x \right )\right )+4 \ln \left (3 x \right )+4 \ln \left (1+x \right )\)	\(35\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {(4+4 x) \log (1+x)+(-4-8 x) \log ^2(1+x)+\log (3 x) (-4 x+(4+8 x) \log (1+x))}{\left (x+x^2\right ) \log (3 x) \log (1+x)+\left (-x-x^2\right ) \log ^2(1+x)} \, dx=4 \, \log \left (x^{2} + x\right ) + 4 \, \log \left (\log \left (3 \, x\right ) - \log \left (x + 1\right )\right ) - 4 \, \log \left (\log \left (x + 1\right )\right ) \]

[In]

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

[Out]

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {(4+4 x) \log (1+x)+(-4-8 x) \log ^2(1+x)+\log (3 x) (-4 x+(4+8 x) \log (1+x))}{\left (x+x^2\right ) \log (3 x) \log (1+x)+\left (-x-x^2\right ) \log ^2(1+x)} \, dx=\text {Exception raised: PolynomialError} \]

[In]

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

[Out]

Exception raised: PolynomialError >> 1/(x**2 + x) contains an element of the set of generators.

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {(4+4 x) \log (1+x)+(-4-8 x) \log ^2(1+x)+\log (3 x) (-4 x+(4+8 x) \log (1+x))}{\left (x+x^2\right ) \log (3 x) \log (1+x)+\left (-x-x^2\right ) \log ^2(1+x)} \, dx=4 \, \log \left (x + 1\right ) + 4 \, \log \left (x\right ) + 4 \, \log \left (-\log \left (3\right ) + \log \left (x + 1\right ) - \log \left (x\right )\right ) - 4 \, \log \left (\log \left (x + 1\right )\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {(4+4 x) \log (1+x)+(-4-8 x) \log ^2(1+x)+\log (3 x) (-4 x+(4+8 x) \log (1+x))}{\left (x+x^2\right ) \log (3 x) \log (1+x)+\left (-x-x^2\right ) \log ^2(1+x)} \, dx=4 \, \log \left (x + 1\right ) + 4 \, \log \left (x\right ) + 4 \, \log \left (\log \left (3 \, x\right ) - \log \left (x + 1\right )\right ) - 4 \, \log \left (\log \left (x + 1\right )\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.48 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {(4+4 x) \log (1+x)+(-4-8 x) \log ^2(1+x)+\log (3 x) (-4 x+(4+8 x) \log (1+x))}{\left (x+x^2\right ) \log (3 x) \log (1+x)+\left (-x-x^2\right ) \log ^2(1+x)} \, dx=4\,\ln \left (x\,\left (x+1\right )\right )-4\,\ln \left (\ln \left (x+1\right )\right )+4\,\ln \left (\ln \left (3\,x\right )-\ln \left (x+1\right )\right ) \]

[In]

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

[Out]

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