\(\int \frac {5}{32-64 x+32 x^2+(32-32 x) \log (16)+8 \log ^2(16)} \, dx\) [6665]

Optimal result

Integrand size = 28, antiderivative size = 13 \[ \int \frac {5}{32-64 x+32 x^2+(32-32 x) \log (16)+8 \log ^2(16)} \, dx=\frac {5}{16 (2-2 x+\log (16))} \]

[Out]

5/8/(4-4*x+8*ln(2))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 2006, 27, 32} \[ \int \frac {5}{32-64 x+32 x^2+(32-32 x) \log (16)+8 \log ^2(16)} \, dx=\frac {5}{16 (-2 x+2+\log (16))} \]

[In]

Int[5/(32 - 64*x + 32*x^2 + (32 - 32*x)*Log[16] + 8*Log[16]^2),x]

[Out]

5/(16*(2 - 2*x + Log[16]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2006

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && QuadraticQ[u, x] &&  !QuadraticMatch
Q[u, x]

Rubi steps \begin{align*} \text {integral}& = 5 \int \frac {1}{32-64 x+32 x^2+(32-32 x) \log (16)+8 \log ^2(16)} \, dx \\ & = 5 \int \frac {1}{32 x^2-32 x (2+\log (16))+8 (2+\log (16))^2} \, dx \\ & = 5 \int \frac {1}{8 (-2+2 x-\log (16))^2} \, dx \\ & = \frac {5}{8} \int \frac {1}{(-2+2 x-\log (16))^2} \, dx \\ & = \frac {5}{16 (2-2 x+\log (16))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {5}{32-64 x+32 x^2+(32-32 x) \log (16)+8 \log ^2(16)} \, dx=\frac {5}{8 (4-4 x+\log (256))} \]

[In]

Integrate[5/(32 - 64*x + 32*x^2 + (32 - 32*x)*Log[16] + 8*Log[16]^2),x]

[Out]

5/(8*(4 - 4*x + Log[256]))

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92

method	result	size
default	\(-\frac {5}{32 \left (-2 \ln \left (2\right )+x -1\right )}\)	\(12\)
risch	\(\frac {5}{64 \left (\ln \left (2\right )-\frac {x}{2}+\frac {1}{2}\right )}\)	\(12\)
gosper	\(\frac {5}{32 \left (2 \ln \left (2\right )-x +1\right )}\)	\(14\)
norman	\(\frac {5}{32 \left (2 \ln \left (2\right )-x +1\right )}\)	\(14\)
parallelrisch	\(\frac {5}{32 \left (2 \ln \left (2\right )-x +1\right )}\)	\(14\)
meijerg	\(-\frac {5 x}{32 \left (-1-2 \ln \left (2\right )\right ) \left (1+2 \ln \left (2\right )\right ) \left (1-\frac {x}{1+2 \ln \left (2\right )}\right )}\)	\(35\)

[In]

int(5/(128*ln(2)^2+4*(-32*x+32)*ln(2)+32*x^2-64*x+32),x,method=_RETURNVERBOSE)

[Out]

-5/32/(-2*ln(2)+x-1)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {5}{32-64 x+32 x^2+(32-32 x) \log (16)+8 \log ^2(16)} \, dx=-\frac {5}{32 \, {\left (x - 2 \, \log \left (2\right ) - 1\right )}} \]

[In]

integrate(5/(128*log(2)^2+4*(-32*x+32)*log(2)+32*x^2-64*x+32),x, algorithm="fricas")

[Out]

-5/32/(x - 2*log(2) - 1)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {5}{32-64 x+32 x^2+(32-32 x) \log (16)+8 \log ^2(16)} \, dx=- \frac {5}{32 x - 64 \log {\left (2 \right )} - 32} \]

[In]

integrate(5/(128*ln(2)**2+4*(-32*x+32)*ln(2)+32*x**2-64*x+32),x)

[Out]

-5/(32*x - 64*log(2) - 32)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {5}{32-64 x+32 x^2+(32-32 x) \log (16)+8 \log ^2(16)} \, dx=-\frac {5}{32 \, {\left (x - 2 \, \log \left (2\right ) - 1\right )}} \]

[In]

integrate(5/(128*log(2)^2+4*(-32*x+32)*log(2)+32*x^2-64*x+32),x, algorithm="maxima")

[Out]

-5/32/(x - 2*log(2) - 1)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {5}{32-64 x+32 x^2+(32-32 x) \log (16)+8 \log ^2(16)} \, dx=-\frac {5}{32 \, {\left (x - 2 \, \log \left (2\right ) - 1\right )}} \]

[In]

integrate(5/(128*log(2)^2+4*(-32*x+32)*log(2)+32*x^2-64*x+32),x, algorithm="giac")

[Out]

-5/32/(x - 2*log(2) - 1)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {5}{32-64 x+32 x^2+(32-32 x) \log (16)+8 \log ^2(16)} \, dx=\frac {5}{32\,\left (\ln \left (4\right )-x+1\right )} \]

[In]

int(5/(128*log(2)^2 - 4*log(2)*(32*x - 32) - 64*x + 32*x^2 + 32),x)

[Out]

5/(32*(log(4) - x + 1))