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

Optimal. Leaf size=25 \[ x^2+\log \left ((1+x-\log (2 x))^2\right )+5 \log \left (\log ^2(4 x)\right ) \]

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Rubi [A]  time = 0.50, antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 4, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {6742, 6684, 2302, 29} \begin {gather*} x^2+2 \log (x-\log (2 x)+1)+10 \log (\log (4 x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x^2 + 2*Log[1 + x - Log[2*x]] + 10*Log[Log[4*x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6684

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

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 \left (\frac {2 \left (-1+x+x^2+x^3-x^2 \log (2 x)\right )}{x (1+x-\log (2 x))}+\frac {10}{x \log (4 x)}\right ) \, dx\\ &=2 \int \frac {-1+x+x^2+x^3-x^2 \log (2 x)}{x (1+x-\log (2 x))} \, dx+10 \int \frac {1}{x \log (4 x)} \, dx\\ &=2 \int \left (x+\frac {-1+x}{x (1+x-\log (2 x))}\right ) \, dx+10 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (4 x)\right )\\ &=x^2+10 \log (\log (4 x))+2 \int \frac {-1+x}{x (1+x-\log (2 x))} \, dx\\ &=x^2+2 \log (1+x-\log (2 x))+10 \log (\log (4 x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 27, normalized size = 1.08 \begin {gather*} 2 \left (\frac {x^2}{2}+\log (1+x-\log (2 x))+5 \log (\log (4 x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.70, size = 26, normalized size = 1.04 \begin {gather*} x^{2} + 2 \, \log \left (-x + \log \left (2 \, x\right ) - 1\right ) + 10 \, \log \left (\log \relax (2) + \log \left (2 \, x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2*log(2*x)-2*x^3-2*x^2-2*x+2)*log(4*x)+10*log(2*x)-10*x-10)/(x*log(2*x)-x^2-x)/log(4*x),x, alg
orithm="fricas")

[Out]

x^2 + 2*log(-x + log(2*x) - 1) + 10*log(log(2) + log(2*x))

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giac [A]  time = 0.50, size = 26, normalized size = 1.04 \begin {gather*} x^{2} + 2 \, \log \left (-x + \log \relax (2) + \log \relax (x) - 1\right ) + 10 \, \log \left (2 \, \log \relax (2) + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2*log(2*x)-2*x^3-2*x^2-2*x+2)*log(4*x)+10*log(2*x)-10*x-10)/(x*log(2*x)-x^2-x)/log(4*x),x, alg
orithm="giac")

[Out]

x^2 + 2*log(-x + log(2) + log(x) - 1) + 10*log(2*log(2) + log(x))

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maple [C]  time = 0.12, size = 36, normalized size = 1.44

method	result	size
risch	\(x^{2}+2 \ln \left (\ln \relax (x )-\frac {i \left (2 i \ln \relax (2)-2 i x -2 i\right )}{2}\right )+10 \ln \left (\ln \relax (x )+2 \ln \relax (2)\right )\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2*ln(2*x)-2*x^3-2*x^2-2*x+2)*ln(4*x)+10*ln(2*x)-10*x-10)/(x*ln(2*x)-x^2-x)/ln(4*x),x,method=_RETURNV
ERBOSE)

[Out]

x^2+2*ln(ln(x)-1/2*I*(2*I*ln(2)-2*I*x-2*I))+10*ln(ln(x)+2*ln(2))

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maxima [A]  time = 0.94, size = 26, normalized size = 1.04 \begin {gather*} x^{2} + 2 \, \log \left (-x + \log \relax (2) + \log \relax (x) - 1\right ) + 10 \, \log \left (2 \, \log \relax (2) + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2*log(2*x)-2*x^3-2*x^2-2*x+2)*log(4*x)+10*log(2*x)-10*x-10)/(x*log(2*x)-x^2-x)/log(4*x),x, alg
orithm="maxima")

[Out]

x^2 + 2*log(-x + log(2) + log(x) - 1) + 10*log(2*log(2) + log(x))

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mupad [B]  time = 0.64, size = 53, normalized size = 2.12 \begin {gather*} 12\,\ln \relax (x)-10\,\ln \left (x-1\right )+2\,\ln \left (\frac {8\,x-\ln \left (256\right )-8\,\ln \relax (x)+8}{x}\right )+10\,\ln \left (\frac {\left (16\,\ln \relax (2)+8\,\ln \relax (x)\right )\,\left (x-1\right )}{x}\right )+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12*log(x) - 10*log(x - 1) + 2*log((8*x - log(256) - 8*log(x) + 8)/x) + 10*log(((16*log(2) + 8*log(x))*(x - 1))
/x) + x^2

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sympy [A]  time = 0.62, size = 26, normalized size = 1.04 \begin {gather*} x^{2} + 10 \log {\left (\log {\left (2 x \right )} + \log {\relax (2 )} \right )} + 2 \log {\left (- x + \log {\left (2 x \right )} - 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2 + 10*log(log(2*x) + log(2)) + 2*log(-x + log(2*x) - 1)

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