\(\int \frac {3+2 x^2+(1-4 x) \log (4)+2 \log ^2(4)}{x^2-2 x \log (4)+\log ^2(4)} \, dx\) [3363]

Optimal result

Integrand size = 37, antiderivative size = 18 \[ \int \frac {3+2 x^2+(1-4 x) \log (4)+2 \log ^2(4)}{x^2-2 x \log (4)+\log ^2(4)} \, dx=2 x+\frac {-3-x}{x-\log (4)} \]

[Out]

2*x+(-3-x)/(x-2*ln(2))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {27, 1864} \[ \int \frac {3+2 x^2+(1-4 x) \log (4)+2 \log ^2(4)}{x^2-2 x \log (4)+\log ^2(4)} \, dx=2 x-\frac {3+\log (4)}{x-\log (4)} \]

[In]

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

[Out]

2*x - (3 + Log[4])/(x - Log[4])

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1864

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int \frac {3+2 x^2+(1-4 x) \log (4)+2 \log ^2(4)}{x^2-2 x \log (4)+\log ^2(4)} \, dx=\frac {-3-\log (4)}{x-\log (4)}+2 (x-\log (4)) \]

[In]

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

[Out]

(-3 - Log[4])/(x - Log[4]) + 2*(x - Log[4])

Maple [A] (verified)

Time = 2.48 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17

method	result	size
default	\(2 x -\frac {2 \ln \left (2\right )+3}{x -2 \ln \left (2\right )}\)	\(21\)
risch	\(2 x +\frac {\ln \left (2\right )}{\ln \left (2\right )-\frac {x}{2}}+\frac {3}{2 \left (\ln \left (2\right )-\frac {x}{2}\right )}\)	\(26\)
gosper	\(\frac {-2 x^{2}+3+8 \ln \left (2\right )^{2}+2 \ln \left (2\right )}{2 \ln \left (2\right )-x}\)	\(29\)
norman	\(\frac {-2 x^{2}+3+8 \ln \left (2\right )^{2}+2 \ln \left (2\right )}{2 \ln \left (2\right )-x}\)	\(29\)
parallelrisch	\(\frac {-2 x^{2}+3+8 \ln \left (2\right )^{2}+2 \ln \left (2\right )}{2 \ln \left (2\right )-x}\)	\(29\)
meijerg	\(\frac {2 x}{1-\frac {x}{2 \ln \left (2\right )}}+\frac {3 x}{4 \ln \left (2\right )^{2} \left (1-\frac {x}{2 \ln \left (2\right )}\right )}+\frac {x}{2 \ln \left (2\right ) \left (1-\frac {x}{2 \ln \left (2\right )}\right )}-8 \ln \left (2\right ) \left (\frac {x}{2 \ln \left (2\right ) \left (1-\frac {x}{2 \ln \left (2\right )}\right )}+\ln \left (1-\frac {x}{2 \ln \left (2\right )}\right )\right )-4 \ln \left (2\right ) \left (-\frac {x \left (-\frac {3 x}{2 \ln \left (2\right )}+6\right )}{6 \ln \left (2\right ) \left (1-\frac {x}{2 \ln \left (2\right )}\right )}-2 \ln \left (1-\frac {x}{2 \ln \left (2\right )}\right )\right )\)	\(129\)

[In]

int((8*ln(2)^2+2*(-4*x+1)*ln(2)+2*x^2+3)/(4*ln(2)^2-4*x*ln(2)+x^2),x,method=_RETURNVERBOSE)

[Out]

2*x-(2*ln(2)+3)/(x-2*ln(2))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \frac {3+2 x^2+(1-4 x) \log (4)+2 \log ^2(4)}{x^2-2 x \log (4)+\log ^2(4)} \, dx=\frac {2 \, x^{2} - 2 \, {\left (2 \, x + 1\right )} \log \left (2\right ) - 3}{x - 2 \, \log \left (2\right )} \]

[In]

integrate((8*log(2)^2+2*(-4*x+1)*log(2)+2*x^2+3)/(4*log(2)^2-4*x*log(2)+x^2),x, algorithm="fricas")

[Out]

(2*x^2 - 2*(2*x + 1)*log(2) - 3)/(x - 2*log(2))

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {3+2 x^2+(1-4 x) \log (4)+2 \log ^2(4)}{x^2-2 x \log (4)+\log ^2(4)} \, dx=2 x + \frac {-3 - 2 \log {\left (2 \right )}}{x - 2 \log {\left (2 \right )}} \]

[In]

integrate((8*ln(2)**2+2*(-4*x+1)*ln(2)+2*x**2+3)/(4*ln(2)**2-4*x*ln(2)+x**2),x)

[Out]

2*x + (-3 - 2*log(2))/(x - 2*log(2))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {3+2 x^2+(1-4 x) \log (4)+2 \log ^2(4)}{x^2-2 x \log (4)+\log ^2(4)} \, dx=2 \, x - \frac {2 \, \log \left (2\right ) + 3}{x - 2 \, \log \left (2\right )} \]

[In]

integrate((8*log(2)^2+2*(-4*x+1)*log(2)+2*x^2+3)/(4*log(2)^2-4*x*log(2)+x^2),x, algorithm="maxima")

[Out]

2*x - (2*log(2) + 3)/(x - 2*log(2))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {3+2 x^2+(1-4 x) \log (4)+2 \log ^2(4)}{x^2-2 x \log (4)+\log ^2(4)} \, dx=2 \, x - \frac {2 \, \log \left (2\right ) + 3}{x - 2 \, \log \left (2\right )} \]

[In]

integrate((8*log(2)^2+2*(-4*x+1)*log(2)+2*x^2+3)/(4*log(2)^2-4*x*log(2)+x^2),x, algorithm="giac")

[Out]

2*x - (2*log(2) + 3)/(x - 2*log(2))

Mupad [F(-1)]

Timed out. \[ \int \frac {3+2 x^2+(1-4 x) \log (4)+2 \log ^2(4)}{x^2-2 x \log (4)+\log ^2(4)} \, dx=\int \frac {8\,{\ln \left (2\right )}^2-2\,\ln \left (2\right )\,\left (4\,x-1\right )+2\,x^2+3}{x^2-4\,\ln \left (2\right )\,x+4\,{\ln \left (2\right )}^2} \,d x \]

[In]

int((8*log(2)^2 - 2*log(2)*(4*x - 1) + 2*x^2 + 3)/(4*log(2)^2 - 4*x*log(2) + x^2),x)

[Out]

int((8*log(2)^2 - 2*log(2)*(4*x - 1) + 2*x^2 + 3)/(4*log(2)^2 - 4*x*log(2) + x^2), x)