Integrand size = 17, antiderivative size = 50 \[ \int \frac {27-8 x^3}{729-64 x^6} \, dx=-\frac {\arctan \left (\frac {3-4 x}{3 \sqrt {3}}\right )}{18 \sqrt {3}}+\frac {1}{54} \log (3+2 x)-\frac {1}{108} \log \left (9-6 x+4 x^2\right ) \] Output:
-1/54*arctan(1/9*(3-4*x)*3^(1/2))*3^(1/2)+1/54*ln(3+2*x)-1/108*ln(4*x^2-6* x+9)
Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int \frac {27-8 x^3}{729-64 x^6} \, dx=\frac {\arctan \left (\frac {-3+4 x}{3 \sqrt {3}}\right )}{18 \sqrt {3}}+\frac {1}{54} \log (3+2 x)-\frac {1}{108} \log \left (9-6 x+4 x^2\right ) \] Input:
Integrate[(27 - 8*x^3)/(729 - 64*x^6),x]
Output:
ArcTan[(-3 + 4*x)/(3*Sqrt[3])]/(18*Sqrt[3]) + Log[3 + 2*x]/54 - Log[9 - 6* x + 4*x^2]/108
Time = 0.35 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {1386, 750, 16, 27, 1142, 27, 1083, 217, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {27-8 x^3}{729-64 x^6} \, dx\) |
\(\Big \downarrow \) 1386 |
\(\displaystyle \int \frac {1}{8 x^3+27}dx\) |
\(\Big \downarrow \) 750 |
\(\displaystyle \frac {1}{27} \int \frac {2 (3-x)}{4 x^2-6 x+9}dx+\frac {1}{27} \int \frac {1}{2 x+3}dx\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{27} \int \frac {2 (3-x)}{4 x^2-6 x+9}dx+\frac {1}{54} \log (2 x+3)\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{27} \int \frac {3-x}{4 x^2-6 x+9}dx+\frac {1}{54} \log (2 x+3)\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {2}{27} \left (\frac {9}{4} \int \frac {1}{4 x^2-6 x+9}dx-\frac {1}{8} \int -\frac {2 (3-4 x)}{4 x^2-6 x+9}dx\right )+\frac {1}{54} \log (2 x+3)\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{27} \left (\frac {9}{4} \int \frac {1}{4 x^2-6 x+9}dx+\frac {1}{4} \int \frac {3-4 x}{4 x^2-6 x+9}dx\right )+\frac {1}{54} \log (2 x+3)\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {2}{27} \left (\frac {1}{4} \int \frac {3-4 x}{4 x^2-6 x+9}dx-\frac {9}{2} \int \frac {1}{-(8 x-6)^2-108}d(8 x-6)\right )+\frac {1}{54} \log (2 x+3)\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {2}{27} \left (\frac {1}{4} \int \frac {3-4 x}{4 x^2-6 x+9}dx+\frac {1}{4} \sqrt {3} \arctan \left (\frac {8 x-6}{6 \sqrt {3}}\right )\right )+\frac {1}{54} \log (2 x+3)\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {2}{27} \left (\frac {1}{4} \sqrt {3} \arctan \left (\frac {8 x-6}{6 \sqrt {3}}\right )-\frac {1}{8} \log \left (4 x^2-6 x+9\right )\right )+\frac {1}{54} \log (2 x+3)\) |
Input:
Int[(27 - 8*x^3)/(729 - 64*x^6),x]
Output:
Log[3 + 2*x]/54 + (2*((Sqrt[3]*ArcTan[(-6 + 8*x)/(6*Sqrt[3])])/4 - Log[9 - 6*x + 4*x^2]/8))/27
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2) Int[1/ (Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]^2) Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Simp[(-e^2/c)^q Int[u*(d - e*x^n)^p, x], x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && EqQ[p + q, 0 ] && GtQ[d, 0] && LtQ[c, 0] && GtQ[e^2, 0]
Time = 0.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.78
method | result | size |
default | \(-\frac {\ln \left (4 x^{2}-6 x +9\right )}{108}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (8 x -6\right ) \sqrt {3}}{18}\right )}{54}+\frac {\ln \left (2 x +3\right )}{54}\) | \(39\) |
risch | \(\frac {\ln \left (2 x +3\right )}{54}-\frac {\ln \left (4 x^{2}-6 x +9\right )}{108}+\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (2 x -\frac {3}{2}\right ) \sqrt {3}}{9}\right )}{54}\) | \(39\) |
meijerg | \(-\frac {x \left (\ln \left (1-\frac {2 \left (x^{6}\right )^{\frac {1}{6}}}{3}\right )-\ln \left (1+\frac {2 \left (x^{6}\right )^{\frac {1}{6}}}{3}\right )+\frac {\ln \left (1-\frac {2 \left (x^{6}\right )^{\frac {1}{6}}}{3}+\frac {4 \left (x^{6}\right )^{\frac {1}{3}}}{9}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}{3-\left (x^{6}\right )^{\frac {1}{6}}}\right )-\frac {\ln \left (1+\frac {2 \left (x^{6}\right )^{\frac {1}{6}}}{3}+\frac {4 \left (x^{6}\right )^{\frac {1}{3}}}{9}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}{3+\left (x^{6}\right )^{\frac {1}{6}}}\right )\right )}{108 \left (x^{6}\right )^{\frac {1}{6}}}+\frac {x^{4} \left (\ln \left (1-\frac {4 \left (x^{6}\right )^{\frac {1}{3}}}{9}\right )-\frac {\ln \left (1+\frac {4 \left (x^{6}\right )^{\frac {1}{3}}}{9}+\frac {16 \left (x^{6}\right )^{\frac {2}{3}}}{81}\right )}{2}+\sqrt {3}\, \arctan \left (\frac {2 \sqrt {3}\, \left (x^{6}\right )^{\frac {1}{3}}}{9 \left (1+\frac {2 \left (x^{6}\right )^{\frac {1}{3}}}{9}\right )}\right )\right )}{108 \left (x^{6}\right )^{\frac {2}{3}}}\) | \(191\) |
Input:
int((-8*x^3+27)/(-64*x^6+729),x,method=_RETURNVERBOSE)
Output:
-1/108*ln(4*x^2-6*x+9)+1/54*3^(1/2)*arctan(1/18*(8*x-6)*3^(1/2))+1/54*ln(2 *x+3)
Time = 0.08 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.76 \[ \int \frac {27-8 x^3}{729-64 x^6} \, dx=\frac {1}{54} \, \sqrt {3} \arctan \left (\frac {1}{9} \, \sqrt {3} {\left (4 \, x - 3\right )}\right ) - \frac {1}{108} \, \log \left (4 \, x^{2} - 6 \, x + 9\right ) + \frac {1}{54} \, \log \left (2 \, x + 3\right ) \] Input:
integrate((-8*x^3+27)/(-64*x^6+729),x, algorithm="fricas")
Output:
1/54*sqrt(3)*arctan(1/9*sqrt(3)*(4*x - 3)) - 1/108*log(4*x^2 - 6*x + 9) + 1/54*log(2*x + 3)
Time = 0.10 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.96 \[ \int \frac {27-8 x^3}{729-64 x^6} \, dx=\frac {\log {\left (x + \frac {3}{2} \right )}}{54} - \frac {\log {\left (x^{2} - \frac {3 x}{2} + \frac {9}{4} \right )}}{108} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {4 \sqrt {3} x}{9} - \frac {\sqrt {3}}{3} \right )}}{54} \] Input:
integrate((-8*x**3+27)/(-64*x**6+729),x)
Output:
log(x + 3/2)/54 - log(x**2 - 3*x/2 + 9/4)/108 + sqrt(3)*atan(4*sqrt(3)*x/9 - sqrt(3)/3)/54
Time = 0.11 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.76 \[ \int \frac {27-8 x^3}{729-64 x^6} \, dx=\frac {1}{54} \, \sqrt {3} \arctan \left (\frac {1}{9} \, \sqrt {3} {\left (4 \, x - 3\right )}\right ) - \frac {1}{108} \, \log \left (4 \, x^{2} - 6 \, x + 9\right ) + \frac {1}{54} \, \log \left (2 \, x + 3\right ) \] Input:
integrate((-8*x^3+27)/(-64*x^6+729),x, algorithm="maxima")
Output:
1/54*sqrt(3)*arctan(1/9*sqrt(3)*(4*x - 3)) - 1/108*log(4*x^2 - 6*x + 9) + 1/54*log(2*x + 3)
Time = 0.12 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.70 \[ \int \frac {27-8 x^3}{729-64 x^6} \, dx=\frac {1}{54} \, \sqrt {3} \arctan \left (\frac {1}{9} \, \sqrt {3} {\left (4 \, x - 3\right )}\right ) - \frac {1}{108} \, \log \left (x^{2} - \frac {3}{2} \, x + \frac {9}{4}\right ) + \frac {1}{54} \, \log \left ({\left | x + \frac {3}{2} \right |}\right ) \] Input:
integrate((-8*x^3+27)/(-64*x^6+729),x, algorithm="giac")
Output:
1/54*sqrt(3)*arctan(1/9*sqrt(3)*(4*x - 3)) - 1/108*log(x^2 - 3/2*x + 9/4) + 1/54*log(abs(x + 3/2))
Time = 5.92 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.92 \[ \int \frac {27-8 x^3}{729-64 x^6} \, dx=\frac {\ln \left (x+\frac {3}{2}\right )}{54}-\ln \left (x-\frac {3}{4}-\frac {\sqrt {3}\,3{}\mathrm {i}}{4}\right )\,\left (\frac {1}{108}+\frac {\sqrt {3}\,1{}\mathrm {i}}{108}\right )+\ln \left (x-\frac {3}{4}+\frac {\sqrt {3}\,3{}\mathrm {i}}{4}\right )\,\left (-\frac {1}{108}+\frac {\sqrt {3}\,1{}\mathrm {i}}{108}\right ) \] Input:
int((8*x^3 - 27)/(64*x^6 - 729),x)
Output:
log(x + 3/2)/54 - log(x - (3^(1/2)*3i)/4 - 3/4)*((3^(1/2)*1i)/108 + 1/108) + log(x + (3^(1/2)*3i)/4 - 3/4)*((3^(1/2)*1i)/108 - 1/108)
Time = 0.15 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.76 \[ \int \frac {27-8 x^3}{729-64 x^6} \, dx=\frac {\sqrt {3}\, \mathit {atan} \left (\frac {4 x -3}{3 \sqrt {3}}\right )}{54}-\frac {\mathrm {log}\left (4 x^{2}-6 x +9\right )}{108}+\frac {\mathrm {log}\left (2 x +3\right )}{54} \] Input:
int((-8*x^3+27)/(-64*x^6+729),x)
Output:
(2*sqrt(3)*atan((4*x - 3)/(3*sqrt(3))) - log(4*x**2 - 6*x + 9) + 2*log(2*x + 3))/108