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

Optimal. Leaf size=21 \[ -x+\left (\log (\log (5))+\log \left (\log \left (-2 x+5 x^2\right )\right )\right )^2 \]

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Rubi [A]  time = 0.61, antiderivative size = 19, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {1593, 6688, 6742, 6686} \begin {gather*} \log ^2(\log (5) \log (-((2-5 x) x)))-x \end {gather*}

Antiderivative was successfully verified.

[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 1593

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 6686

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 6688

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

Rule 6742

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

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 18, normalized size = 0.86 \begin {gather*} -x+\log ^2(\log (5) \log (x (-2+5 x))) \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.46, size = 33, normalized size = 1.57 \begin {gather*} 2 \, \log \left (\log \relax (5)\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.24, size = 33, normalized size = 1.57 \begin {gather*} 2 \, \log \left (\log \relax (5)\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.11, size = 34, normalized size = 1.62

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

Verification of antiderivative is not currently implemented for this CAS.

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

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maxima [A]  time = 0.51, size = 31, normalized size = 1.48 \begin {gather*} \log \left (\log \left (5 \, x - 2\right ) + \log \relax (x)\right )^{2} + 2 \, \log \left (\log \left (5 \, x - 2\right ) + \log \relax (x)\right ) \log \left (\log \relax (5)\right ) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 4.59, size = 33, normalized size = 1.57 \begin {gather*} {\ln \left (\ln \left (5\,x^2-2\,x\right )\right )}^2+2\,\ln \left (\ln \relax (5)\right )\,\ln \left (\ln \left (5\,x^2-2\,x\right )\right )-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.45, size = 32, normalized size = 1.52 \begin {gather*} - x + \log {\left (\log {\left (5 x^{2} - 2 x \right )} \right )}^{2} + 2 \log {\left (\log {\relax (5 )} \right )} \log {\left (\log {\left (5 x^{2} - 2 x \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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