\(\int \frac {-1+x+10 x^2}{(-x-x^2-5 x^3+x \log (2)+x \log (x)) \log (-\frac {4}{-1-x-5 x^2+\log (2)+\log (x)})} \, dx\) [1414]

Optimal result

Integrand size = 54, antiderivative size = 24 \[ \int \frac {-1+x+10 x^2}{\left (-x-x^2-5 x^3+x \log (2)+x \log (x)\right ) \log \left (-\frac {4}{-1-x-5 x^2+\log (2)+\log (x)}\right )} \, dx=\log \left (\log \left (\frac {4}{-4+x+5 \left (1+x^2\right )-\log (2)-\log (x)}\right )\right ) \]

[Out]

ln(ln(4/(x+1-ln(x)+5*x^2-ln(2))))

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6, 6816} \[ \int \frac {-1+x+10 x^2}{\left (-x-x^2-5 x^3+x \log (2)+x \log (x)\right ) \log \left (-\frac {4}{-1-x-5 x^2+\log (2)+\log (x)}\right )} \, dx=\log \left (\log \left (\frac {4}{5 x^2+x-\log (x)+1-\log (2)}\right )\right ) \]

[In]

Int[(-1 + x + 10*x^2)/((-x - x^2 - 5*x^3 + x*Log[2] + x*Log[x])*Log[-4/(-1 - x - 5*x^2 + Log[2] + Log[x])]),x]

[Out]

Log[Log[4/(1 + x + 5*x^2 - Log[2] - Log[x])]]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 6816

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-1+x+10 x^2}{\left (-x^2-5 x^3+x (-1+\log (2))+x \log (x)\right ) \log \left (-\frac {4}{-1-x-5 x^2+\log (2)+\log (x)}\right )} \, dx \\ & = \log \left (\log \left (\frac {4}{1+x+5 x^2-\log (2)-\log (x)}\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-1+x+10 x^2}{\left (-x-x^2-5 x^3+x \log (2)+x \log (x)\right ) \log \left (-\frac {4}{-1-x-5 x^2+\log (2)+\log (x)}\right )} \, dx=\log \left (\log \left (\frac {4}{1+x+5 x^2-\log (2 x)}\right )\right ) \]

[In]

Integrate[(-1 + x + 10*x^2)/((-x - x^2 - 5*x^3 + x*Log[2] + x*Log[x])*Log[-4/(-1 - x - 5*x^2 + Log[2] + Log[x]
)]),x]

[Out]

Log[Log[4/(1 + x + 5*x^2 - Log[2*x])]]

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88

method	result	size
parallelrisch	\(\ln \left (\ln \left (-\frac {4}{\ln \left (x \right )+\ln \left (2\right )-5 x^{2}-x -1}\right )\right )\)	\(21\)
default	\(\ln \left (2 \ln \left (2\right )+\ln \left (-\frac {1}{\ln \left (x \right )+\ln \left (2\right )-5 x^{2}-x -1}\right )\right )\)	\(26\)
risch	\(\ln \left (\ln \left (\ln \left (x \right )+\ln \left (2\right )-5 x^{2}-x -1\right )-\frac {i \left (-2 \pi \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )+\ln \left (2\right )-5 x^{2}-x -1}\right )^{2}+2 \pi \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )+\ln \left (2\right )-5 x^{2}-x -1}\right )^{3}-4 i \ln \left (2\right )+2 \pi \right )}{2}\right )\)	\(80\)

[In]

int((10*x^2+x-1)/(x*ln(x)+x*ln(2)-5*x^3-x^2-x)/ln(-4/(ln(x)+ln(2)-5*x^2-x-1)),x,method=_RETURNVERBOSE)

[Out]

ln(ln(-4/(ln(x)+ln(2)-5*x^2-x-1)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-1+x+10 x^2}{\left (-x-x^2-5 x^3+x \log (2)+x \log (x)\right ) \log \left (-\frac {4}{-1-x-5 x^2+\log (2)+\log (x)}\right )} \, dx=\log \left (\log \left (\frac {4}{5 \, x^{2} + x - \log \left (2\right ) - \log \left (x\right ) + 1}\right )\right ) \]

[In]

integrate((10*x^2+x-1)/(x*log(x)+x*log(2)-5*x^3-x^2-x)/log(-4/(log(x)+log(2)-5*x^2-x-1)),x, algorithm="fricas"
)

[Out]

log(log(4/(5*x^2 + x - log(2) - log(x) + 1)))

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-1+x+10 x^2}{\left (-x-x^2-5 x^3+x \log (2)+x \log (x)\right ) \log \left (-\frac {4}{-1-x-5 x^2+\log (2)+\log (x)}\right )} \, dx=\log {\left (\log {\left (- \frac {4}{- 5 x^{2} - x + \log {\left (x \right )} - 1 + \log {\left (2 \right )}} \right )} \right )} \]

[In]

integrate((10*x**2+x-1)/(x*ln(x)+x*ln(2)-5*x**3-x**2-x)/ln(-4/(ln(x)+ln(2)-5*x**2-x-1)),x)

[Out]

log(log(-4/(-5*x**2 - x + log(x) - 1 + log(2))))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {-1+x+10 x^2}{\left (-x-x^2-5 x^3+x \log (2)+x \log (x)\right ) \log \left (-\frac {4}{-1-x-5 x^2+\log (2)+\log (x)}\right )} \, dx=\log \left (-2 \, \log \left (2\right ) + \log \left (5 \, x^{2} + x - \log \left (2\right ) - \log \left (x\right ) + 1\right )\right ) \]

[In]

integrate((10*x^2+x-1)/(x*log(x)+x*log(2)-5*x^3-x^2-x)/log(-4/(log(x)+log(2)-5*x^2-x-1)),x, algorithm="maxima"
)

[Out]

log(-2*log(2) + log(5*x^2 + x - log(2) - log(x) + 1))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {-1+x+10 x^2}{\left (-x-x^2-5 x^3+x \log (2)+x \log (x)\right ) \log \left (-\frac {4}{-1-x-5 x^2+\log (2)+\log (x)}\right )} \, dx=\log \left (-2 \, \log \left (2\right ) + \log \left (5 \, x^{2} + x - \log \left (2\right ) - \log \left (x\right ) + 1\right )\right ) \]

[In]

integrate((10*x^2+x-1)/(x*log(x)+x*log(2)-5*x^3-x^2-x)/log(-4/(log(x)+log(2)-5*x^2-x-1)),x, algorithm="giac")

[Out]

log(-2*log(2) + log(5*x^2 + x - log(2) - log(x) + 1))

Mupad [B] (verification not implemented)

Time = 26.52 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-1+x+10 x^2}{\left (-x-x^2-5 x^3+x \log (2)+x \log (x)\right ) \log \left (-\frac {4}{-1-x-5 x^2+\log (2)+\log (x)}\right )} \, dx=\ln \left (\ln \left (\frac {4}{x-\ln \left (2\,x\right )+5\,x^2+1}\right )\right ) \]

[In]

int(-(x + 10*x^2 - 1)/(log(4/(x - log(2) - log(x) + 5*x^2 + 1))*(x - x*log(2) - x*log(x) + x^2 + 5*x^3)),x)

[Out]

log(log(4/(x - log(2*x) + 5*x^2 + 1)))