\(\int \frac {1+2 x^3+3 x^3 \log (x)}{x^4+(x+x^4) \log (x)+x \log ^2(x)} \, dx\) [9175]

Optimal result

Integrand size = 35, antiderivative size = 17 \[ \int \frac {1+2 x^3+3 x^3 \log (x)}{x^4+\left (x+x^4\right ) \log (x)+x \log ^2(x)} \, dx=-\log (3 (1+\log (x)))+\log \left (x^3+\log (x)\right ) \]

[Out]

ln(ln(x)+x^3)-ln(3*ln(x)+3)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6873, 6874, 2339, 29, 6816} \[ \int \frac {1+2 x^3+3 x^3 \log (x)}{x^4+\left (x+x^4\right ) \log (x)+x \log ^2(x)} \, dx=\log \left (x^3+\log (x)\right )-\log (\log (x)+1) \]

[In]

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

[Out]

-Log[1 + Log[x]] + Log[x^3 + Log[x]]

Rule 29

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

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1+2 x^3+3 x^3 \log (x)}{x^4+\left (x+x^4\right ) \log (x)+x \log ^2(x)} \, dx=-\log (1+\log (x))+\log \left (x^3+\log (x)\right ) \]

[In]

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

[Out]

-Log[1 + Log[x]] + Log[x^3 + Log[x]]

Maple [A] (verified)

Time = 1.05 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
default	\(-\ln \left (\ln \left (x \right )+1\right )+\ln \left (\ln \left (x \right )+x^{3}\right )\)	\(16\)
norman	\(-\ln \left (\ln \left (x \right )+1\right )+\ln \left (\ln \left (x \right )+x^{3}\right )\)	\(16\)
risch	\(-\ln \left (\ln \left (x \right )+1\right )+\ln \left (\ln \left (x \right )+x^{3}\right )\)	\(16\)
parallelrisch	\(-\ln \left (\ln \left (x \right )+1\right )+\ln \left (\ln \left (x \right )+x^{3}\right )\)	\(16\)

[In]

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

[Out]

-ln(ln(x)+1)+ln(ln(x)+x^3)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1+2 x^3+3 x^3 \log (x)}{x^4+\left (x+x^4\right ) \log (x)+x \log ^2(x)} \, dx=\log \left (x^{3} + \log \left (x\right )\right ) - \log \left (\log \left (x\right ) + 1\right ) \]

[In]

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

[Out]

log(x^3 + log(x)) - log(log(x) + 1)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {1+2 x^3+3 x^3 \log (x)}{x^4+\left (x+x^4\right ) \log (x)+x \log ^2(x)} \, dx=\log {\left (x^{3} + \log {\left (x \right )} \right )} - \log {\left (\log {\left (x \right )} + 1 \right )} \]

[In]

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

[Out]

log(x**3 + log(x)) - log(log(x) + 1)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1+2 x^3+3 x^3 \log (x)}{x^4+\left (x+x^4\right ) \log (x)+x \log ^2(x)} \, dx=\log \left (x^{3} + \log \left (x\right )\right ) - \log \left (\log \left (x\right ) + 1\right ) \]

[In]

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

[Out]

log(x^3 + log(x)) - log(log(x) + 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1+2 x^3+3 x^3 \log (x)}{x^4+\left (x+x^4\right ) \log (x)+x \log ^2(x)} \, dx=\log \left (x^{3} + \log \left (x\right )\right ) - \log \left (\log \left (x\right ) + 1\right ) \]

[In]

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

[Out]

log(x^3 + log(x)) - log(log(x) + 1)

Mupad [B] (verification not implemented)

Time = 12.60 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1+2 x^3+3 x^3 \log (x)}{x^4+\left (x+x^4\right ) \log (x)+x \log ^2(x)} \, dx=\ln \left (\ln \left (x\right )+x^3\right )-\ln \left (\ln \left (x\right )+1\right ) \]

[In]

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

[Out]

log(log(x) + x^3) - log(log(x) + 1)