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

Optimal. Leaf size=20 \[ \frac {8 (i \pi +\log (\log (16)))}{3 (1-x)^2} \]

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Rubi [A]  time = 0.01, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {12, 2058, 32} \begin {gather*} \frac {8 (\log (\log (16))+i \pi )}{3 (1-x)^2} \end {gather*}

Antiderivative was successfully verified.

[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 2058

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 20, normalized size = 1.00 \begin {gather*} \frac {8 i (\pi -i \log (\log (16)))}{3 (-1+x)^2} \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.87, size = 17, normalized size = 0.85 \begin {gather*} \frac {8 \, \log \left (-4 \, \log \relax (2)\right )}{3 \, {\left (x^{2} - 2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.23, size = 12, normalized size = 0.60 \begin {gather*} \frac {8 \, \log \left (-4 \, \log \relax (2)\right )}{3 \, {\left (x - 1\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.04, size = 13, normalized size = 0.65

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

Verification of antiderivative is not currently implemented for this CAS.

[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))/(x-1)^2

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maxima [A]  time = 0.35, size = 17, normalized size = 0.85 \begin {gather*} \frac {8 \, \log \left (-4 \, \log \relax (2)\right )}{3 \, {\left (x^{2} - 2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 5.85, size = 19, normalized size = 0.95 \begin {gather*} \frac {8\,\ln \left (-\ln \left (16\right )\right )}{3\,\left (x^2-2\,x+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.11, size = 29, normalized size = 1.45 \begin {gather*} - \frac {- 32 \log {\relax (2 )} - 16 \log {\left (\log {\relax (2 )} \right )} - 16 i \pi }{6 x^{2} - 12 x + 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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