3.51.6 \(\int \frac {-228+20 \log (2)}{11520-2880 x+180 x^2+(-1440+180 x) \log (2)+45 \log ^2(2)} \, dx\)

Optimal. Leaf size=23 \[ \frac {2 \left (-3+x+\frac {4+x}{9}\right )}{5 (-16+2 x+\log (2))} \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 21, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {12, 1981, 27, 32} \begin {gather*} -\frac {2 (57-\log (32))}{45 (-2 x+16-\log (2))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-228 + 20*Log[2])/(11520 - 2880*x + 180*x^2 + (-1440 + 180*x)*Log[2] + 45*Log[2]^2),x]

[Out]

(-2*(57 - Log[32]))/(45*(16 - 2*x - Log[2]))

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 1981

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left ((4 (57-\log (32))) \int \frac {1}{11520-2880 x+180 x^2+(-1440+180 x) \log (2)+45 \log ^2(2)} \, dx\right )\\ &=-\left ((4 (57-\log (32))) \int \frac {1}{180 x^2-180 x (16-\log (2))+45 (16-\log (2))^2} \, dx\right )\\ &=-\left ((4 (57-\log (32))) \int \frac {1}{45 (-16+2 x+\log (2))^2} \, dx\right )\\ &=-\left (\frac {1}{45} (4 (57-\log (32))) \int \frac {1}{(-16+2 x+\log (2))^2} \, dx\right )\\ &=-\frac {2 (57-\log (32))}{45 (16-2 x-\log (2))}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 17, normalized size = 0.74 \begin {gather*} -\frac {2 (-57+\log (32))}{45 (-16+2 x+\log (2))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-228 + 20*Log[2])/(11520 - 2880*x + 180*x^2 + (-1440 + 180*x)*Log[2] + 45*Log[2]^2),x]

[Out]

(-2*(-57 + Log[32]))/(45*(-16 + 2*x + Log[2]))

________________________________________________________________________________________

fricas [A]  time = 0.85, size = 17, normalized size = 0.74 \begin {gather*} -\frac {2 \, {\left (5 \, \log \relax (2) - 57\right )}}{45 \, {\left (2 \, x + \log \relax (2) - 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*log(2)-228)/(45*log(2)^2+(180*x-1440)*log(2)+180*x^2-2880*x+11520),x, algorithm="fricas")

[Out]

-2/45*(5*log(2) - 57)/(2*x + log(2) - 16)

________________________________________________________________________________________

giac [A]  time = 0.15, size = 17, normalized size = 0.74 \begin {gather*} -\frac {2 \, {\left (5 \, \log \relax (2) - 57\right )}}{45 \, {\left (2 \, x + \log \relax (2) - 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*log(2)-228)/(45*log(2)^2+(180*x-1440)*log(2)+180*x^2-2880*x+11520),x, algorithm="giac")

[Out]

-2/45*(5*log(2) - 57)/(2*x + log(2) - 16)

________________________________________________________________________________________

maple [A]  time = 0.22, size = 17, normalized size = 0.74

method	result	size
norman	\(\frac {\frac {38}{15}-\frac {2 \ln \relax (2)}{9}}{\ln \relax (2)+2 x -16}\)	\(17\)
gosper	\(-\frac {2 \left (5 \ln \relax (2)-57\right )}{45 \left (\ln \relax (2)+2 x -16\right )}\)	\(18\)
default	\(-\frac {20 \ln \relax (2)-228}{90 \left (\ln \relax (2)+2 x -16\right )}\)	\(18\)
risch	\(\frac {38}{15 \left (\ln \relax (2)+2 x -16\right )}-\frac {2 \ln \relax (2)}{9 \left (\ln \relax (2)+2 x -16\right )}\)	\(26\)
meijerg	\(\frac {38 x}{15 \left (\frac {\ln \relax (2)}{2}-8\right ) \left (-\ln \relax (2)+16\right ) \left (1-\frac {2 x}{-\ln \relax (2)+16}\right )}-\frac {2 \ln \relax (2) x}{9 \left (\frac {\ln \relax (2)}{2}-8\right ) \left (-\ln \relax (2)+16\right ) \left (1-\frac {2 x}{-\ln \relax (2)+16}\right )}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*ln(2)-228)/(45*ln(2)^2+(180*x-1440)*ln(2)+180*x^2-2880*x+11520),x,method=_RETURNVERBOSE)

[Out]

(38/15-2/9*ln(2))/(ln(2)+2*x-16)

________________________________________________________________________________________

maxima [A]  time = 0.35, size = 17, normalized size = 0.74 \begin {gather*} -\frac {2 \, {\left (5 \, \log \relax (2) - 57\right )}}{45 \, {\left (2 \, x + \log \relax (2) - 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*log(2)-228)/(45*log(2)^2+(180*x-1440)*log(2)+180*x^2-2880*x+11520),x, algorithm="maxima")

[Out]

-2/45*(5*log(2) - 57)/(2*x + log(2) - 16)

________________________________________________________________________________________

mupad [B]  time = 3.56, size = 17, normalized size = 0.74 \begin {gather*} -\frac {\frac {2\,\ln \relax (2)}{9}-\frac {38}{15}}{2\,x+\ln \relax (2)-16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*log(2) - 228)/(log(2)*(180*x - 1440) - 2880*x + 45*log(2)^2 + 180*x^2 + 11520),x)

[Out]

-((2*log(2))/9 - 38/15)/(2*x + log(2) - 16)

________________________________________________________________________________________

sympy [A]  time = 0.16, size = 17, normalized size = 0.74 \begin {gather*} - \frac {-228 + 20 \log {\relax (2 )}}{180 x - 1440 + 90 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*ln(2)-228)/(45*ln(2)**2+(180*x-1440)*ln(2)+180*x**2-2880*x+11520),x)

[Out]

-(-228 + 20*log(2))/(180*x - 1440 + 90*log(2))

________________________________________________________________________________________