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

Optimal. Leaf size=24 \[ \log \left (\log \left (\frac {4}{-4+x+5 \left (1+x^2\right )-\log (2)-\log (x)}\right )\right ) \]

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Rubi [A]  time = 0.14, antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 2, number of rules used = 2, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6, 6684} \begin {gather*} \log \left (\log \left (\frac {4}{5 x^2+x-\log (x)+1-\log (2)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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 6684

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.28, size = 20, normalized size = 0.83 \begin {gather*} \log \left (\log \left (\frac {4}{1+x+5 x^2-\log (2 x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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])]]

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fricas [A]  time = 0.95, size = 22, normalized size = 0.92 \begin {gather*} \log \left (\log \left (\frac {4}{5 \, x^{2} + x - \log \relax (2) - \log \relax (x) + 1}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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giac [A]  time = 0.36, size = 23, normalized size = 0.96 \begin {gather*} \log \left (-2 \, \log \relax (2) + \log \left (5 \, x^{2} + x - \log \relax (2) - \log \relax (x) + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [C]  time = 0.08, size = 80, normalized size = 3.33

method	result	size
risch	\(\ln \left (\ln \left (\ln \relax (x )+\ln \relax (2)-5 x^{2}-x -1\right )-\frac {i \left (-2 \pi \mathrm {csgn}\left (\frac {i}{\ln \relax (x )+\ln \relax (2)-5 x^{2}-x -1}\right )^{2}+2 \pi \mathrm {csgn}\left (\frac {i}{\ln \relax (x )+\ln \relax (2)-5 x^{2}-x -1}\right )^{3}-4 i \ln \relax (2)+2 \pi \right )}{2}\right )\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[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(ln(x)+ln(2)-5*x^2-x-1)-1/2*I*(-2*Pi*csgn(I/(ln(x)+ln(2)-5*x^2-x-1))^2+2*Pi*csgn(I/(ln(x)+ln(2)-5*x^2-x-1
))^3-4*I*ln(2)+2*Pi))

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maxima [A]  time = 0.64, size = 23, normalized size = 0.96 \begin {gather*} \log \left (-2 \, \log \relax (2) + \log \left (5 \, x^{2} + x - \log \relax (2) - \log \relax (x) + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 16.65, size = 20, normalized size = 0.83 \begin {gather*} \ln \left (\ln \left (\frac {4}{x-\ln \left (2\,x\right )+5\,x^2+1}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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sympy [A]  time = 0.56, size = 20, normalized size = 0.83 \begin {gather*} \log {\left (\log {\left (- \frac {4}{- 5 x^{2} - x + \log {\relax (x )} - 1 + \log {\relax (2 )}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))))

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