\(\int \frac {(2 x-5 x^2) \log (-2 x+5 x^2)+(-4+20 x) \log (\log (5))+(-4+20 x) \log (\log (-2 x+5 x^2))}{(-2 x+5 x^2) \log (-2 x+5 x^2)} \, dx\) [6695]

Optimal result

Integrand size = 71, antiderivative size = 21 \[ \int \frac {\left (2 x-5 x^2\right ) \log \left (-2 x+5 x^2\right )+(-4+20 x) \log (\log (5))+(-4+20 x) \log \left (\log \left (-2 x+5 x^2\right )\right )}{\left (-2 x+5 x^2\right ) \log \left (-2 x+5 x^2\right )} \, dx=-x+\left (\log (\log (5))+\log \left (\log \left (-2 x+5 x^2\right )\right )\right )^2 \]

[Out]

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {1607, 6820, 6874, 6818} \[ \int \frac {\left (2 x-5 x^2\right ) \log \left (-2 x+5 x^2\right )+(-4+20 x) \log (\log (5))+(-4+20 x) \log \left (\log \left (-2 x+5 x^2\right )\right )}{\left (-2 x+5 x^2\right ) \log \left (-2 x+5 x^2\right )} \, dx=\log ^2(\log (5) \log (-((2-5 x) x)))-x \]

[In]

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

[Out]

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

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {\left (2 x-5 x^2\right ) \log \left (-2 x+5 x^2\right )+(-4+20 x) \log (\log (5))+(-4+20 x) \log \left (\log \left (-2 x+5 x^2\right )\right )}{\left (-2 x+5 x^2\right ) \log \left (-2 x+5 x^2\right )} \, dx=-x+\log ^2(\log (5) \log (x (-2+5 x))) \]

[In]

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

[Out]

-x + Log[Log[5]*Log[x*(-2 + 5*x)]]^2

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62

method	result	size
default	\(-x +2 \ln \left (\ln \left (5 x^{2}-2 x \right )\right ) \ln \left (\ln \left (5\right )\right )+{\ln \left (\ln \left (5 x^{2}-2 x \right )\right )}^{2}\)	\(34\)
norman	\(-x +2 \ln \left (\ln \left (5 x^{2}-2 x \right )\right ) \ln \left (\ln \left (5\right )\right )+{\ln \left (\ln \left (5 x^{2}-2 x \right )\right )}^{2}\)	\(34\)
parts	\(-x +2 \ln \left (\ln \left (5 x^{2}-2 x \right )\right ) \ln \left (\ln \left (5\right )\right )+{\ln \left (\ln \left (5 x^{2}-2 x \right )\right )}^{2}\)	\(34\)

[In]

int(((20*x-4)*ln(ln(5*x^2-2*x))+(20*x-4)*ln(ln(5))+(-5*x^2+2*x)*ln(5*x^2-2*x))/(5*x^2-2*x)/ln(5*x^2-2*x),x,met
hod=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57 \[ \int \frac {\left (2 x-5 x^2\right ) \log \left (-2 x+5 x^2\right )+(-4+20 x) \log (\log (5))+(-4+20 x) \log \left (\log \left (-2 x+5 x^2\right )\right )}{\left (-2 x+5 x^2\right ) \log \left (-2 x+5 x^2\right )} \, dx=2 \, \log \left (\log \left (5\right )\right ) \log \left (\log \left (5 \, x^{2} - 2 \, x\right )\right ) + \log \left (\log \left (5 \, x^{2} - 2 \, x\right )\right )^{2} - x \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52 \[ \int \frac {\left (2 x-5 x^2\right ) \log \left (-2 x+5 x^2\right )+(-4+20 x) \log (\log (5))+(-4+20 x) \log \left (\log \left (-2 x+5 x^2\right )\right )}{\left (-2 x+5 x^2\right ) \log \left (-2 x+5 x^2\right )} \, dx=- x + \log {\left (\log {\left (5 x^{2} - 2 x \right )} \right )}^{2} + 2 \log {\left (\log {\left (5 \right )} \right )} \log {\left (\log {\left (5 x^{2} - 2 x \right )} \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int \frac {\left (2 x-5 x^2\right ) \log \left (-2 x+5 x^2\right )+(-4+20 x) \log (\log (5))+(-4+20 x) \log \left (\log \left (-2 x+5 x^2\right )\right )}{\left (-2 x+5 x^2\right ) \log \left (-2 x+5 x^2\right )} \, dx=\log \left (\log \left (5 \, x - 2\right ) + \log \left (x\right )\right )^{2} + 2 \, \log \left (\log \left (5 \, x - 2\right ) + \log \left (x\right )\right ) \log \left (\log \left (5\right )\right ) - x \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57 \[ \int \frac {\left (2 x-5 x^2\right ) \log \left (-2 x+5 x^2\right )+(-4+20 x) \log (\log (5))+(-4+20 x) \log \left (\log \left (-2 x+5 x^2\right )\right )}{\left (-2 x+5 x^2\right ) \log \left (-2 x+5 x^2\right )} \, dx=2 \, \log \left (\log \left (5\right )\right ) \log \left (\log \left (5 \, x^{2} - 2 \, x\right )\right ) + \log \left (\log \left (5 \, x^{2} - 2 \, x\right )\right )^{2} - x \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57 \[ \int \frac {\left (2 x-5 x^2\right ) \log \left (-2 x+5 x^2\right )+(-4+20 x) \log (\log (5))+(-4+20 x) \log \left (\log \left (-2 x+5 x^2\right )\right )}{\left (-2 x+5 x^2\right ) \log \left (-2 x+5 x^2\right )} \, dx={\ln \left (\ln \left (5\,x^2-2\,x\right )\right )}^2+2\,\ln \left (\ln \left (5\right )\right )\,\ln \left (\ln \left (5\,x^2-2\,x\right )\right )-x \]

[In]

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

[Out]

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