\(\int \frac {125}{1-10 x+25 x^2+(8-40 x) \log (16)+16 \log ^2(16)} \, dx\) [3943]

Optimal result

Integrand size = 28, antiderivative size = 20 \[ \int \frac {125}{1-10 x+25 x^2+(8-40 x) \log (16)+16 \log ^2(16)} \, dx=\frac {25 x}{x-x (x+4 (x-\log (16)))} \]

[Out]

25/(x-x*(5*x-16*ln(2)))*x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 2006, 27, 32} \[ \int \frac {125}{1-10 x+25 x^2+(8-40 x) \log (16)+16 \log ^2(16)} \, dx=\frac {25}{-5 x+1+4 \log (16)} \]

[In]

Int[125/(1 - 10*x + 25*x^2 + (8 - 40*x)*Log[16] + 16*Log[16]^2),x]

[Out]

25/(1 - 5*x + 4*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}& = 125 \int \frac {1}{1-10 x+25 x^2+(8-40 x) \log (16)+16 \log ^2(16)} \, dx \\ & = 125 \int \frac {1}{25 x^2-10 x (1+4 \log (16))+(1+4 \log (16))^2} \, dx \\ & = 125 \int \frac {1}{(-1+5 x-4 \log (16))^2} \, dx \\ & = \frac {25}{1-5 x+4 \log (16)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65 \[ \int \frac {125}{1-10 x+25 x^2+(8-40 x) \log (16)+16 \log ^2(16)} \, dx=\frac {25}{1-5 x+4 \log (16)} \]

[In]

Integrate[125/(1 - 10*x + 25*x^2 + (8 - 40*x)*Log[16] + 16*Log[16]^2),x]

[Out]

25/(1 - 5*x + 4*Log[16])

Maple [A] (verified)

Time = 3.43 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60

method	result	size
risch	\(\frac {25}{16 \left (\ln \left (2\right )-\frac {5 x}{16}+\frac {1}{16}\right )}\)	\(12\)
gosper	\(\frac {25}{16 \ln \left (2\right )-5 x +1}\)	\(14\)
default	\(-\frac {25}{-16 \ln \left (2\right )+5 x -1}\)	\(14\)
norman	\(\frac {25}{16 \ln \left (2\right )-5 x +1}\)	\(14\)
parallelrisch	\(\frac {25}{16 \ln \left (2\right )-5 x +1}\)	\(14\)
meijerg	\(-\frac {25 x}{\left (-\frac {16 \ln \left (2\right )}{5}-\frac {1}{5}\right ) \left (16 \ln \left (2\right )+1\right ) \left (1-\frac {5 x}{16 \ln \left (2\right )+1}\right )}\)	\(35\)

[In]

int(125/(256*ln(2)^2+4*(-40*x+8)*ln(2)+25*x^2-10*x+1),x,method=_RETURNVERBOSE)

[Out]

25/16/(ln(2)-5/16*x+1/16)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65 \[ \int \frac {125}{1-10 x+25 x^2+(8-40 x) \log (16)+16 \log ^2(16)} \, dx=-\frac {25}{5 \, x - 16 \, \log \left (2\right ) - 1} \]

[In]

integrate(125/(256*log(2)^2+4*(-40*x+8)*log(2)+25*x^2-10*x+1),x, algorithm="fricas")

[Out]

-25/(5*x - 16*log(2) - 1)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int \frac {125}{1-10 x+25 x^2+(8-40 x) \log (16)+16 \log ^2(16)} \, dx=- \frac {125}{25 x - 80 \log {\left (2 \right )} - 5} \]

[In]

integrate(125/(256*ln(2)**2+4*(-40*x+8)*ln(2)+25*x**2-10*x+1),x)

[Out]

-125/(25*x - 80*log(2) - 5)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65 \[ \int \frac {125}{1-10 x+25 x^2+(8-40 x) \log (16)+16 \log ^2(16)} \, dx=-\frac {25}{5 \, x - 16 \, \log \left (2\right ) - 1} \]

[In]

integrate(125/(256*log(2)^2+4*(-40*x+8)*log(2)+25*x^2-10*x+1),x, algorithm="maxima")

[Out]

-25/(5*x - 16*log(2) - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65 \[ \int \frac {125}{1-10 x+25 x^2+(8-40 x) \log (16)+16 \log ^2(16)} \, dx=-\frac {25}{5 \, x - 16 \, \log \left (2\right ) - 1} \]

[In]

integrate(125/(256*log(2)^2+4*(-40*x+8)*log(2)+25*x^2-10*x+1),x, algorithm="giac")

[Out]

-25/(5*x - 16*log(2) - 1)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65 \[ \int \frac {125}{1-10 x+25 x^2+(8-40 x) \log (16)+16 \log ^2(16)} \, dx=\frac {25}{16\,\ln \left (2\right )-5\,x+1} \]

[In]

int(125/(256*log(2)^2 - 4*log(2)*(40*x - 8) - 10*x + 25*x^2 + 1),x)

[Out]

25/(16*log(2) - 5*x + 1)