\(\int -\frac {16 (i \pi +\log (\log (16)))}{-3+9 x-9 x^2+3 x^3} \, dx\) [9720]

Optimal result

Integrand size = 28, antiderivative size = 20 \[ \int -\frac {16 (i \pi +\log (\log (16)))}{-3+9 x-9 x^2+3 x^3} \, dx=\frac {8 (i \pi +\log (\log (16)))}{3 (1-x)^2} \]

[Out]

8/3*ln(-4*ln(2))/(1-x)^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {12, 2083, 32} \[ \int -\frac {16 (i \pi +\log (\log (16)))}{-3+9 x-9 x^2+3 x^3} \, dx=\frac {8 (\log (\log (16))+i \pi )}{3 (1-x)^2} \]

[In]

Int[(-16*(I*Pi + Log[Log[16]]))/(-3 + 9*x - 9*x^2 + 3*x^3),x]

[Out]

(8*(I*Pi + Log[Log[16]]))/(3*(1 - x)^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 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 2083

Int[(P_)^(p_), x_Symbol] :> With[{u = Factor[P]}, Int[ExpandIntegrand[u^p, x], x] /;  !SumQ[NonfreeFactors[u,
x]]] /; PolyQ[P, x] && ILtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = -\left ((16 (i \pi +\log (\log (16)))) \int \frac {1}{-3+9 x-9 x^2+3 x^3} \, dx\right ) \\ & = -\left ((16 (i \pi +\log (\log (16)))) \int \frac {1}{3 (-1+x)^3} \, dx\right ) \\ & = -\left (\frac {1}{3} (16 (i \pi +\log (\log (16)))) \int \frac {1}{(-1+x)^3} \, dx\right ) \\ & = \frac {8 (i \pi +\log (\log (16)))}{3 (1-x)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int -\frac {16 (i \pi +\log (\log (16)))}{-3+9 x-9 x^2+3 x^3} \, dx=\frac {8 i (\pi -i \log (\log (16)))}{3 (-1+x)^2} \]

[In]

Integrate[(-16*(I*Pi + Log[Log[16]]))/(-3 + 9*x - 9*x^2 + 3*x^3),x]

[Out]

(((8*I)/3)*(Pi - I*Log[Log[16]]))/(-1 + x)^2

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65

method	result	size
default	\(\frac {8 \ln \left (-4 \ln \left (2\right )\right )}{3 \left (-1+x \right )^{2}}\)	\(13\)
gosper	\(\frac {8 \ln \left (-4 \ln \left (2\right )\right )}{3 \left (x^{2}-2 x +1\right )}\)	\(18\)
parallelrisch	\(\frac {8 \ln \left (-4 \ln \left (2\right )\right )}{3 \left (x^{2}-2 x +1\right )}\)	\(18\)
norman	\(\frac {\frac {16 \ln \left (2\right )}{3}+\frac {8 \ln \left (\ln \left (2\right )\right )}{3}+\frac {8 i \pi }{3}}{\left (-1+x \right )^{2}}\)	\(21\)
risch	\(\frac {16 \ln \left (2\right )}{3 \left (x^{2}-2 x +1\right )}+\frac {8 \ln \left (\ln \left (2\right )\right )}{3 \left (x^{2}-2 x +1\right )}+\frac {8 i \pi }{3 \left (x^{2}-2 x +1\right )}\)	\(45\)

[In]

int(-16*ln(-4*ln(2))/(3*x^3-9*x^2+9*x-3),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int -\frac {16 (i \pi +\log (\log (16)))}{-3+9 x-9 x^2+3 x^3} \, dx=\frac {8 \, \log \left (-4 \, \log \left (2\right )\right )}{3 \, {\left (x^{2} - 2 \, x + 1\right )}} \]

[In]

integrate(-16*log(-4*log(2))/(3*x^3-9*x^2+9*x-3),x, algorithm="fricas")

[Out]

8/3*log(-4*log(2))/(x^2 - 2*x + 1)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45 \[ \int -\frac {16 (i \pi +\log (\log (16)))}{-3+9 x-9 x^2+3 x^3} \, dx=- \frac {- 32 \log {\left (2 \right )} - 16 \log {\left (\log {\left (2 \right )} \right )} - 16 i \pi }{6 x^{2} - 12 x + 6} \]

[In]

integrate(-16*ln(-4*ln(2))/(3*x**3-9*x**2+9*x-3),x)

[Out]

-(-32*log(2) - 16*log(log(2)) - 16*I*pi)/(6*x**2 - 12*x + 6)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int -\frac {16 (i \pi +\log (\log (16)))}{-3+9 x-9 x^2+3 x^3} \, dx=\frac {8 \, \log \left (-4 \, \log \left (2\right )\right )}{3 \, {\left (x^{2} - 2 \, x + 1\right )}} \]

[In]

integrate(-16*log(-4*log(2))/(3*x^3-9*x^2+9*x-3),x, algorithm="maxima")

[Out]

8/3*log(-4*log(2))/(x^2 - 2*x + 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int -\frac {16 (i \pi +\log (\log (16)))}{-3+9 x-9 x^2+3 x^3} \, dx=\frac {8 \, \log \left (-4 \, \log \left (2\right )\right )}{3 \, {\left (x - 1\right )}^{2}} \]

[In]

integrate(-16*log(-4*log(2))/(3*x^3-9*x^2+9*x-3),x, algorithm="giac")

[Out]

8/3*log(-4*log(2))/(x - 1)^2

Mupad [B] (verification not implemented)

Time = 15.65 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int -\frac {16 (i \pi +\log (\log (16)))}{-3+9 x-9 x^2+3 x^3} \, dx=\frac {8\,\ln \left (-\ln \left (16\right )\right )}{3\,\left (x^2-2\,x+1\right )} \]

[In]

int(-(16*log(-4*log(2)))/(9*x - 9*x^2 + 3*x^3 - 3),x)

[Out]

(8*log(-log(16)))/(3*(x^2 - 2*x + 1))