\(\int \frac {-5 \log (x)+(1+x) \log ^2(x)+(-20-4 x \log ^2(x)) \log ^3(\frac {-5+(1+x) \log (x)}{\log (x)})}{-5 x \log (x)+(x+x^2) \log ^2(x)} \, dx\) [9512]

Optimal result

Integrand size = 58, antiderivative size = 28 \[ \int \frac {-5 \log (x)+(1+x) \log ^2(x)+\left (-20-4 x \log ^2(x)\right ) \log ^3\left (\frac {-5+(1+x) \log (x)}{\log (x)}\right )}{-5 x \log (x)+\left (x+x^2\right ) \log ^2(x)} \, dx=2 \left (-3+\frac {1}{2 e}\right )+\log (x)-\log ^4\left (1+x-\frac {5}{\log (x)}\right ) \]

[Out]

exp(-1)-6+ln(x)-ln(x-5/ln(x)+1)^4

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {6873, 6874, 6818} \[ \int \frac {-5 \log (x)+(1+x) \log ^2(x)+\left (-20-4 x \log ^2(x)\right ) \log ^3\left (\frac {-5+(1+x) \log (x)}{\log (x)}\right )}{-5 x \log (x)+\left (x+x^2\right ) \log ^2(x)} \, dx=\log (x)-\log ^4\left (x-\frac {5}{\log (x)}+1\right ) \]

[In]

Int[(-5*Log[x] + (1 + x)*Log[x]^2 + (-20 - 4*x*Log[x]^2)*Log[(-5 + (1 + x)*Log[x])/Log[x]]^3)/(-5*x*Log[x] + (
x + x^2)*Log[x]^2),x]

[Out]

Log[x] - Log[1 + x - 5/Log[x]]^4

Rule 6818

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]

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 {5 \log (x)-(1+x) \log ^2(x)-\left (-20-4 x \log ^2(x)\right ) \log ^3\left (\frac {-5+(1+x) \log (x)}{\log (x)}\right )}{x \log (x) (5-\log (x)-x \log (x))} \, dx \\ & = \int \left (\frac {1}{x}-\frac {4 \left (5+x \log ^2(x)\right ) \log ^3\left (1+x-\frac {5}{\log (x)}\right )}{x \log (x) (-5+\log (x)+x \log (x))}\right ) \, dx \\ & = \log (x)-4 \int \frac {\left (5+x \log ^2(x)\right ) \log ^3\left (1+x-\frac {5}{\log (x)}\right )}{x \log (x) (-5+\log (x)+x \log (x))} \, dx \\ & = \log (x)-\log ^4\left (1+x-\frac {5}{\log (x)}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {-5 \log (x)+(1+x) \log ^2(x)+\left (-20-4 x \log ^2(x)\right ) \log ^3\left (\frac {-5+(1+x) \log (x)}{\log (x)}\right )}{-5 x \log (x)+\left (x+x^2\right ) \log ^2(x)} \, dx=\log (x)-\log ^4\left (1+x-\frac {5}{\log (x)}\right ) \]

[In]

Integrate[(-5*Log[x] + (1 + x)*Log[x]^2 + (-20 - 4*x*Log[x]^2)*Log[(-5 + (1 + x)*Log[x])/Log[x]]^3)/(-5*x*Log[
x] + (x + x^2)*Log[x]^2),x]

[Out]

Log[x] - Log[1 + x - 5/Log[x]]^4

Maple [A] (verified)

Time = 3.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79

method	result	size
default	\(\ln \left (x \right )-\ln \left (\frac {x \ln \left (x \right )+\ln \left (x \right )-5}{\ln \left (x \right )}\right )^{4}\)	\(22\)

[In]

int(((-4*x*ln(x)^2-20)*ln((ln(x)*(1+x)-5)/ln(x))^3+(1+x)*ln(x)^2-5*ln(x))/((x^2+x)*ln(x)^2-5*x*ln(x)),x,method
=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {-5 \log (x)+(1+x) \log ^2(x)+\left (-20-4 x \log ^2(x)\right ) \log ^3\left (\frac {-5+(1+x) \log (x)}{\log (x)}\right )}{-5 x \log (x)+\left (x+x^2\right ) \log ^2(x)} \, dx=-\log \left (\frac {{\left (x + 1\right )} \log \left (x\right ) - 5}{\log \left (x\right )}\right )^{4} + \log \left (x\right ) \]

[In]

integrate(((-4*x*log(x)^2-20)*log((log(x)*(1+x)-5)/log(x))^3+(1+x)*log(x)^2-5*log(x))/((x^2+x)*log(x)^2-5*x*lo
g(x)),x, algorithm="fricas")

[Out]

-log(((x + 1)*log(x) - 5)/log(x))^4 + log(x)

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {-5 \log (x)+(1+x) \log ^2(x)+\left (-20-4 x \log ^2(x)\right ) \log ^3\left (\frac {-5+(1+x) \log (x)}{\log (x)}\right )}{-5 x \log (x)+\left (x+x^2\right ) \log ^2(x)} \, dx=\log {\left (x \right )} - \log {\left (\frac {\left (x + 1\right ) \log {\left (x \right )} - 5}{\log {\left (x \right )}} \right )}^{4} \]

[In]

integrate(((-4*x*ln(x)**2-20)*ln((ln(x)*(1+x)-5)/ln(x))**3+(1+x)*ln(x)**2-5*ln(x))/((x**2+x)*ln(x)**2-5*x*ln(x
)),x)

[Out]

log(x) - log(((x + 1)*log(x) - 5)/log(x))**4

Maxima [F]

\[ \int \frac {-5 \log (x)+(1+x) \log ^2(x)+\left (-20-4 x \log ^2(x)\right ) \log ^3\left (\frac {-5+(1+x) \log (x)}{\log (x)}\right )}{-5 x \log (x)+\left (x+x^2\right ) \log ^2(x)} \, dx=\int { -\frac {4 \, {\left (x \log \left (x\right )^{2} + 5\right )} \log \left (\frac {{\left (x + 1\right )} \log \left (x\right ) - 5}{\log \left (x\right )}\right )^{3} - {\left (x + 1\right )} \log \left (x\right )^{2} + 5 \, \log \left (x\right )}{{\left (x^{2} + x\right )} \log \left (x\right )^{2} - 5 \, x \log \left (x\right )} \,d x } \]

[In]

integrate(((-4*x*log(x)^2-20)*log((log(x)*(1+x)-5)/log(x))^3+(1+x)*log(x)^2-5*log(x))/((x^2+x)*log(x)^2-5*x*lo
g(x)),x, algorithm="maxima")

[Out]

-integrate((4*(x*log(x)^2 + 5)*log(((x + 1)*log(x) - 5)/log(x))^3 - (x + 1)*log(x)^2 + 5*log(x))/((x^2 + x)*lo
g(x)^2 - 5*x*log(x)), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (20) = 40\).

Time = 0.43 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.57 \[ \int \frac {-5 \log (x)+(1+x) \log ^2(x)+\left (-20-4 x \log ^2(x)\right ) \log ^3\left (\frac {-5+(1+x) \log (x)}{\log (x)}\right )}{-5 x \log (x)+\left (x+x^2\right ) \log ^2(x)} \, dx=-4 \, {\left (\log \left (x \log \left (x\right ) + \log \left (x\right ) - 5\right ) - \log \left (\log \left (x\right )\right )\right )} \log \left (x \log \left (x\right ) + \log \left (x\right ) - 5\right )^{3} - 6 \, \log \left (x \log \left (x\right ) + \log \left (x\right ) - 5\right )^{2} \log \left (\log \left (x\right )\right )^{2} + 4 \, \log \left (x \log \left (x\right ) + \log \left (x\right ) - 5\right ) \log \left (\log \left (x\right )\right )^{3} - \log \left (\log \left (x\right )\right )^{4} + \log \left (x\right ) \]

[In]

integrate(((-4*x*log(x)^2-20)*log((log(x)*(1+x)-5)/log(x))^3+(1+x)*log(x)^2-5*log(x))/((x^2+x)*log(x)^2-5*x*lo
g(x)),x, algorithm="giac")

[Out]

-4*(log(x*log(x) + log(x) - 5) - log(log(x)))*log(x*log(x) + log(x) - 5)^3 - 6*log(x*log(x) + log(x) - 5)^2*lo
g(log(x))^2 + 4*log(x*log(x) + log(x) - 5)*log(log(x))^3 - log(log(x))^4 + log(x)

Mupad [B] (verification not implemented)

Time = 15.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {-5 \log (x)+(1+x) \log ^2(x)+\left (-20-4 x \log ^2(x)\right ) \log ^3\left (\frac {-5+(1+x) \log (x)}{\log (x)}\right )}{-5 x \log (x)+\left (x+x^2\right ) \log ^2(x)} \, dx=\ln \left (x\right )-{\ln \left (\frac {\ln \left (x\right )\,\left (x+1\right )-5}{\ln \left (x\right )}\right )}^4 \]

[In]

int(-(5*log(x) - log(x)^2*(x + 1) + log((log(x)*(x + 1) - 5)/log(x))^3*(4*x*log(x)^2 + 20))/(log(x)^2*(x + x^2
) - 5*x*log(x)),x)

[Out]

log(x) - log((log(x)*(x + 1) - 5)/log(x))^4