\(\int \frac {2-80 x-4 x^2-5 x^3+5 x \log (x)}{-80 x-5 x^3+5 x \log (x)} \, dx\) [7916]

Optimal result

Integrand size = 37, antiderivative size = 17 \[ \int \frac {2-80 x-4 x^2-5 x^3+5 x \log (x)}{-80 x-5 x^3+5 x \log (x)} \, dx=-4+x+\frac {2}{5} \log \left (-16-x^2+\log (x)\right ) \]

[Out]

2/5*ln(ln(x)-16-x^2)-4+x

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {6873, 12, 6874, 6816} \[ \int \frac {2-80 x-4 x^2-5 x^3+5 x \log (x)}{-80 x-5 x^3+5 x \log (x)} \, dx=\frac {2}{5} \log \left (x^2-\log (x)+16\right )+x \]

[In]

Int[(2 - 80*x - 4*x^2 - 5*x^3 + 5*x*Log[x])/(-80*x - 5*x^3 + 5*x*Log[x]),x]

[Out]

x + (2*Log[16 + x^2 - Log[x]])/5

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 {-2+80 x+4 x^2+5 x^3-5 x \log (x)}{5 x \left (16+x^2-\log (x)\right )} \, dx \\ & = \frac {1}{5} \int \frac {-2+80 x+4 x^2+5 x^3-5 x \log (x)}{x \left (16+x^2-\log (x)\right )} \, dx \\ & = \frac {1}{5} \int \left (5+\frac {2 \left (-1+2 x^2\right )}{x \left (16+x^2-\log (x)\right )}\right ) \, dx \\ & = x+\frac {2}{5} \int \frac {-1+2 x^2}{x \left (16+x^2-\log (x)\right )} \, dx \\ & = x+\frac {2}{5} \log \left (16+x^2-\log (x)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18 \[ \int \frac {2-80 x-4 x^2-5 x^3+5 x \log (x)}{-80 x-5 x^3+5 x \log (x)} \, dx=\frac {1}{5} \left (5 x+2 \log \left (16+x^2-\log (x)\right )\right ) \]

[In]

Integrate[(2 - 80*x - 4*x^2 - 5*x^3 + 5*x*Log[x])/(-80*x - 5*x^3 + 5*x*Log[x]),x]

[Out]

(5*x + 2*Log[16 + x^2 - Log[x]])/5

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88

method	result	size
default	\(x +\frac {2 \ln \left (\ln \left (x \right )-16-x^{2}\right )}{5}\)	\(15\)
norman	\(x +\frac {2 \ln \left (x^{2}-\ln \left (x \right )+16\right )}{5}\)	\(15\)
risch	\(x +\frac {2 \ln \left (\ln \left (x \right )-16-x^{2}\right )}{5}\)	\(15\)
parallelrisch	\(x +\frac {2 \ln \left (x^{2}-\ln \left (x \right )+16\right )}{5}\)	\(15\)

[In]

int((5*x*ln(x)-5*x^3-4*x^2-80*x+2)/(5*x*ln(x)-5*x^3-80*x),x,method=_RETURNVERBOSE)

[Out]

x+2/5*ln(ln(x)-16-x^2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {2-80 x-4 x^2-5 x^3+5 x \log (x)}{-80 x-5 x^3+5 x \log (x)} \, dx=x + \frac {2}{5} \, \log \left (-x^{2} + \log \left (x\right ) - 16\right ) \]

[In]

integrate((5*x*log(x)-5*x^3-4*x^2-80*x+2)/(5*x*log(x)-5*x^3-80*x),x, algorithm="fricas")

[Out]

x + 2/5*log(-x^2 + log(x) - 16)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {2-80 x-4 x^2-5 x^3+5 x \log (x)}{-80 x-5 x^3+5 x \log (x)} \, dx=x + \frac {2 \log {\left (- x^{2} + \log {\left (x \right )} - 16 \right )}}{5} \]

[In]

integrate((5*x*ln(x)-5*x**3-4*x**2-80*x+2)/(5*x*ln(x)-5*x**3-80*x),x)

[Out]

x + 2*log(-x**2 + log(x) - 16)/5

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {2-80 x-4 x^2-5 x^3+5 x \log (x)}{-80 x-5 x^3+5 x \log (x)} \, dx=x + \frac {2}{5} \, \log \left (-x^{2} + \log \left (x\right ) - 16\right ) \]

[In]

integrate((5*x*log(x)-5*x^3-4*x^2-80*x+2)/(5*x*log(x)-5*x^3-80*x),x, algorithm="maxima")

[Out]

x + 2/5*log(-x^2 + log(x) - 16)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {2-80 x-4 x^2-5 x^3+5 x \log (x)}{-80 x-5 x^3+5 x \log (x)} \, dx=x + \frac {2}{5} \, \log \left (-x^{2} + \log \left (x\right ) - 16\right ) \]

[In]

integrate((5*x*log(x)-5*x^3-4*x^2-80*x+2)/(5*x*log(x)-5*x^3-80*x),x, algorithm="giac")

[Out]

x + 2/5*log(-x^2 + log(x) - 16)

Mupad [B] (verification not implemented)

Time = 12.95 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {2-80 x-4 x^2-5 x^3+5 x \log (x)}{-80 x-5 x^3+5 x \log (x)} \, dx=x+\frac {2\,\ln \left (x^2-\ln \left (x\right )+16\right )}{5} \]

[In]

int((80*x - 5*x*log(x) + 4*x^2 + 5*x^3 - 2)/(80*x - 5*x*log(x) + 5*x^3),x)

[Out]

x + (2*log(x^2 - log(x) + 16))/5