Integrand size = 23, antiderivative size = 373 \[ \int \frac {A+B x+C x^2}{2-6 x+3 x^3} \, dx=\frac {1}{9} C \log \left (2-6 x+3 x^3\right )+\frac {\left (3 A+2 \left (C+\sqrt {6} B \cos \left (\frac {1}{6} \left (\pi +2 \arcsin \left (\frac {\sqrt {\frac {3}{2}}}{2}\right )\right )\right )\right )\right ) \log \left (3 x-2 \sqrt {6} \cos \left (\frac {1}{6} \left (\pi +2 \arcsin \left (\frac {\sqrt {\frac {3}{2}}}{2}\right )\right )\right )\right ) \sec \left (\frac {1}{3} \arcsin \left (\frac {\sqrt {\frac {3}{2}}}{2}\right )\right )}{24 \sqrt {3} \left (\cos \left (\frac {1}{6} \left (\pi +2 \arcsin \left (\frac {\sqrt {\frac {3}{2}}}{2}\right )\right )\right )-\sin \left (\frac {1}{3} \arcsin \left (\frac {\sqrt {\frac {3}{2}}}{2}\right )\right )\right )}+\frac {\log \left (3 x-2 \sqrt {6} \sin \left (\frac {1}{3} \arcsin \left (\frac {\sqrt {\frac {3}{2}}}{2}\right )\right )\right ) \left (3 A+2 \left (C+\sqrt {6} B \sin \left (\frac {1}{3} \arcsin \left (\frac {\sqrt {\frac {3}{2}}}{2}\right )\right )\right )\right )}{18 \left (1-2 \cos \left (\frac {2}{3} \arcsin \left (\frac {\sqrt {\frac {3}{2}}}{2}\right )\right )\right )}+\frac {\log \left (3 x+2 \sqrt {6} \sin \left (\frac {1}{3} \left (\pi +\arcsin \left (\frac {\sqrt {\frac {3}{2}}}{2}\right )\right )\right )\right ) \sec \left (\frac {1}{3} \arcsin \left (\frac {\sqrt {\frac {3}{2}}}{2}\right )\right ) \left (3 A+2 C-2 \sqrt {6} B \sin \left (\frac {1}{3} \left (\pi +\arcsin \left (\frac {\sqrt {\frac {3}{2}}}{2}\right )\right )\right )\right )}{24 \sqrt {3} \left (\sin \left (\frac {1}{3} \arcsin \left (\frac {\sqrt {\frac {3}{2}}}{2}\right )\right )+\sin \left (\frac {1}{3} \left (\pi +\arcsin \left (\frac {\sqrt {\frac {3}{2}}}{2}\right )\right )\right )\right )} \] Output:
1/9*C*ln(3*x^3-6*x+2)+1/72*(3*A+2*C+2*6^(1/2)*B*cos(1/6*Pi+1/3*arcsin(1/4* 6^(1/2))))*ln(3*x-2*6^(1/2)*cos(1/6*Pi+1/3*arcsin(1/4*6^(1/2))))*sec(1/3*a rcsin(1/4*6^(1/2)))*3^(1/2)/(cos(1/6*Pi+1/3*arcsin(1/4*6^(1/2)))-sin(1/3*a rcsin(1/4*6^(1/2))))+ln(3*x-2*6^(1/2)*sin(1/3*arcsin(1/4*6^(1/2))))*(3*A+2 *C+2*6^(1/2)*B*sin(1/3*arcsin(1/4*6^(1/2))))/(18-36*cos(2/3*arcsin(1/4*6^( 1/2))))+1/72*ln(3*x+2*6^(1/2)*sin(1/3*Pi+1/3*arcsin(1/4*6^(1/2))))*sec(1/3 *arcsin(1/4*6^(1/2)))*(3*A+2*C-2*6^(1/2)*B*sin(1/3*Pi+1/3*arcsin(1/4*6^(1/ 2))))*3^(1/2)/(sin(1/3*arcsin(1/4*6^(1/2)))+sin(1/3*Pi+1/3*arcsin(1/4*6^(1 /2))))
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.17 \[ \int \frac {A+B x+C x^2}{2-6 x+3 x^3} \, dx=\frac {1}{3} \text {RootSum}\left [2-6 \text {$\#$1}+3 \text {$\#$1}^3\&,\frac {A \log (x-\text {$\#$1})+B \log (x-\text {$\#$1}) \text {$\#$1}+C \log (x-\text {$\#$1}) \text {$\#$1}^2}{-2+3 \text {$\#$1}^2}\&\right ] \] Input:
Integrate[(A + B*x + C*x^2)/(2 - 6*x + 3*x^3),x]
Output:
RootSum[2 - 6*#1 + 3*#1^3 & , (A*Log[x - #1] + B*Log[x - #1]*#1 + C*Log[x - #1]*#1^2)/(-2 + 3*#1^2) & ]/3
Result contains complex when optimal does not.
Time = 11.39 (sec) , antiderivative size = 930, normalized size of antiderivative = 2.49, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2525, 27, 2485, 25, 1200, 2009}
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.
| \(\displaystyle \int \frac {A+B x+C x^2}{3 x^3-6 x+2} \, dx\) |
| \(\Big \downarrow \) 2525 |
| \(\displaystyle \frac {1}{9} \int \frac {3 (3 A+2 C+3 B x)}{3 x^3-6 x+2}dx+\frac {1}{9} C \log \left (3 x^3-6 x+2\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \int \frac {3 A+2 C+3 B x}{3 x^3-6 x+2}dx+\frac {1}{9} C \log \left (3 x^3-6 x+2\right )\) |
| \(\Big \downarrow \) 2485 |
| \(\displaystyle \frac {1}{9} C \log \left (3 x^3-6 x+2\right )+3 \int -\frac {3 A+2 C+3 B x}{\left (3 x+\sqrt [3]{9-3 i \sqrt {15}}+\frac {2\ 3^{2/3}}{\sqrt [3]{3-i \sqrt {15}}}\right ) \left (-9 x^2+3 \left (\frac {2\ 3^{2/3}}{\sqrt [3]{3-i \sqrt {15}}}+\sqrt [3]{9-3 i \sqrt {15}}\right ) x-\left (9-3 i \sqrt {15}\right )^{2/3}-\frac {12 \sqrt [3]{3}}{\left (3-i \sqrt {15}\right )^{2/3}}+6\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{9} C \log \left (3 x^3-6 x+2\right )-3 \int \frac {3 A+2 C+3 B x}{\left (3 x+\sqrt [3]{9-3 i \sqrt {15}}+\frac {2\ 3^{2/3}}{\sqrt [3]{3-i \sqrt {15}}}\right ) \left (-9 x^2+3 \left (\frac {2\ 3^{2/3}}{\sqrt [3]{3-i \sqrt {15}}}+\sqrt [3]{9-3 i \sqrt {15}}\right ) x-\left (9-3 i \sqrt {15}\right )^{2/3}-\frac {12 \sqrt [3]{3}}{\left (3-i \sqrt {15}\right )^{2/3}}+6\right )}dx\) |
| \(\Big \downarrow \) 1200 |
| \(\displaystyle \frac {1}{9} C \log \left (3 x^3-6 x+2\right )-3 \int \left (\frac {\left (3-i \sqrt {15}\right )^{2/3} \left (-3 \sqrt [3]{3-i \sqrt {15}} A+\sqrt [3]{3} \left (2 \sqrt [3]{3}+\left (3-i \sqrt {15}\right )^{2/3}\right ) B-2 \sqrt [3]{3-i \sqrt {15}} C\right )}{9 \left (4 \sqrt [3]{3}+\left (3^{2/3}-i \sqrt [6]{3} \sqrt {5}\right ) \sqrt [3]{3-i \sqrt {15}}+2 \left (3-i \sqrt {15}\right )^{2/3}\right ) \left (3 \sqrt [3]{3-i \sqrt {15}} x+\sqrt [3]{3} \left (3-i \sqrt {15}\right )^{2/3}+2\ 3^{2/3}\right )}+\frac {\left (3-i \sqrt {15}\right )^{2/3} \left (-6 \sqrt [3]{3} \left (3 i+\sqrt {15}+2 i \sqrt [3]{9-3 i \sqrt {15}}\right ) A-3 \left (4 i \sqrt [3]{3}+\sqrt [6]{3} \left (i \sqrt {3}+\sqrt {5}\right ) \sqrt [3]{3-i \sqrt {15}}-2 i \left (3-i \sqrt {15}\right )^{2/3}\right ) B-4 \sqrt [3]{3} \left (3 i+\sqrt {15}+2 i \sqrt [3]{9-3 i \sqrt {15}}\right ) C+3 \left (3 i \left (3-i \sqrt {15}\right )^{2/3} A-\sqrt [3]{3} \left (3 i+\sqrt {15}+2 i \sqrt [3]{9-3 i \sqrt {15}}\right ) B+2 i \left (3-i \sqrt {15}\right )^{2/3} C\right ) x\right )}{27 \left (4 \sqrt [3]{3}+\left (3^{2/3}-i \sqrt [6]{3} \sqrt {5}\right ) \sqrt [3]{3-i \sqrt {15}}+2 \left (3-i \sqrt {15}\right )^{2/3}\right ) \left (3 i \left (3-i \sqrt {15}\right )^{2/3} x^2-3^{2/3} \left (i 3^{2/3}+\sqrt [6]{3} \sqrt {5}+2 i \sqrt [3]{3-i \sqrt {15}}\right ) x+i \left (4 \sqrt [3]{3}+\left (3^{2/3}-i \sqrt [6]{3} \sqrt {5}\right ) \sqrt [3]{3-i \sqrt {15}}-2 \left (3-i \sqrt {15}\right )^{2/3}\right )\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{9} C \log \left (3 x^3-6 x+2\right )-3 \left (\frac {\left (3 \left (6 i \sqrt {3}+6 \sqrt {5}+3 i \sqrt [6]{3} \left (3-i \sqrt {15}\right )^{2/3}+\sqrt {5} \left (9-3 i \sqrt {15}\right )^{2/3}\right ) A-6 \left (i \sqrt {3}-3 \sqrt {5}-2 i \sqrt [6]{3} \left (3-i \sqrt {15}\right )^{2/3}\right ) B+2 \left (6 i \sqrt {3}+6 \sqrt {5}+3 i \sqrt [6]{3} \left (3-i \sqrt {15}\right )^{2/3}+\sqrt {5} \left (9-3 i \sqrt {15}\right )^{2/3}\right ) C\right ) \arctan \left (\frac {\sqrt [3]{3} \left (3 i+\sqrt {15}+2 i \sqrt [3]{9-3 i \sqrt {15}}\right )-6 i \left (3-i \sqrt {15}\right )^{2/3} x}{3 \sqrt {2 \left (3^{2/3}+3 i \sqrt [6]{3} \sqrt {5}-2 \sqrt [3]{3} \left (3-i \sqrt {15}\right )^{2/3}+2 \left (3-i \sqrt {15}\right )^{4/3}\right )}}\right )}{9\ 3^{5/6} \left (4 \sqrt [3]{3}+\left (3^{2/3}-i \sqrt [6]{3} \sqrt {5}\right ) \sqrt [3]{3-i \sqrt {15}}+2 \left (3-i \sqrt {15}\right )^{2/3}\right ) \sqrt {2 \left (3^{2/3}+3 i \sqrt [6]{3} \sqrt {5}-2 \sqrt [3]{3} \left (3-i \sqrt {15}\right )^{2/3}+2 \left (3-i \sqrt {15}\right )^{4/3}\right )}}-\frac {\sqrt [3]{3-i \sqrt {15}} \left (3 \sqrt [3]{3-i \sqrt {15}} A-\sqrt [3]{3} \left (2 \sqrt [3]{3}+\left (3-i \sqrt {15}\right )^{2/3}\right ) B+2 \sqrt [3]{3-i \sqrt {15}} C\right ) \log \left (3^{2/3} \sqrt [3]{3-i \sqrt {15}} x+\left (3-i \sqrt {15}\right )^{2/3}+2 \sqrt [3]{3}\right )}{27 \left (4 \sqrt [3]{3}+\left (3^{2/3}-i \sqrt [6]{3} \sqrt {5}\right ) \sqrt [3]{3-i \sqrt {15}}+2 \left (3-i \sqrt {15}\right )^{2/3}\right )}+\frac {\left (3 \left (3-i \sqrt {15}\right )^{2/3} A-\sqrt [3]{3} \left (3-i \sqrt {15}+2 \sqrt [3]{9-3 i \sqrt {15}}\right ) B+2 \left (3-i \sqrt {15}\right )^{2/3} C\right ) \log \left (3 i \left (3-i \sqrt {15}\right )^{2/3} x^2-3^{2/3} \left (i 3^{2/3}+\sqrt [6]{3} \sqrt {5}+2 i \sqrt [3]{3-i \sqrt {15}}\right ) x+i \left (4 \sqrt [3]{3}+\left (3^{2/3}-i \sqrt [6]{3} \sqrt {5}\right ) \sqrt [3]{3-i \sqrt {15}}-2 \left (3-i \sqrt {15}\right )^{2/3}\right )\right )}{54 \left (4 \sqrt [3]{3}+\left (3^{2/3}-i \sqrt [6]{3} \sqrt {5}\right ) \sqrt [3]{3-i \sqrt {15}}+2 \left (3-i \sqrt {15}\right )^{2/3}\right )}\right )\) |
Input:
Int[(A + B*x + C*x^2)/(2 - 6*x + 3*x^3),x]
Output:
-3*(((3*((6*I)*Sqrt[3] + 6*Sqrt[5] + (3*I)*3^(1/6)*(3 - I*Sqrt[15])^(2/3) + Sqrt[5]*(9 - (3*I)*Sqrt[15])^(2/3))*A - 6*(I*Sqrt[3] - 3*Sqrt[5] - (2*I) *3^(1/6)*(3 - I*Sqrt[15])^(2/3))*B + 2*((6*I)*Sqrt[3] + 6*Sqrt[5] + (3*I)* 3^(1/6)*(3 - I*Sqrt[15])^(2/3) + Sqrt[5]*(9 - (3*I)*Sqrt[15])^(2/3))*C)*Ar cTan[(3^(1/3)*(3*I + Sqrt[15] + (2*I)*(9 - (3*I)*Sqrt[15])^(1/3)) - (6*I)* (3 - I*Sqrt[15])^(2/3)*x)/(3*Sqrt[2*(3^(2/3) + (3*I)*3^(1/6)*Sqrt[5] - 2*3 ^(1/3)*(3 - I*Sqrt[15])^(2/3) + 2*(3 - I*Sqrt[15])^(4/3))])])/(9*3^(5/6)*( 4*3^(1/3) + (3^(2/3) - I*3^(1/6)*Sqrt[5])*(3 - I*Sqrt[15])^(1/3) + 2*(3 - I*Sqrt[15])^(2/3))*Sqrt[2*(3^(2/3) + (3*I)*3^(1/6)*Sqrt[5] - 2*3^(1/3)*(3 - I*Sqrt[15])^(2/3) + 2*(3 - I*Sqrt[15])^(4/3))]) - ((3 - I*Sqrt[15])^(1/3 )*(3*(3 - I*Sqrt[15])^(1/3)*A - 3^(1/3)*(2*3^(1/3) + (3 - I*Sqrt[15])^(2/3 ))*B + 2*(3 - I*Sqrt[15])^(1/3)*C)*Log[2*3^(1/3) + (3 - I*Sqrt[15])^(2/3) + 3^(2/3)*(3 - I*Sqrt[15])^(1/3)*x])/(27*(4*3^(1/3) + (3^(2/3) - I*3^(1/6) *Sqrt[5])*(3 - I*Sqrt[15])^(1/3) + 2*(3 - I*Sqrt[15])^(2/3))) + ((3*(3 - I *Sqrt[15])^(2/3)*A - 3^(1/3)*(3 - I*Sqrt[15] + 2*(9 - (3*I)*Sqrt[15])^(1/3 ))*B + 2*(3 - I*Sqrt[15])^(2/3)*C)*Log[I*(4*3^(1/3) + (3^(2/3) - I*3^(1/6) *Sqrt[5])*(3 - I*Sqrt[15])^(1/3) - 2*(3 - I*Sqrt[15])^(2/3)) - 3^(2/3)*(I* 3^(2/3) + 3^(1/6)*Sqrt[5] + (2*I)*(3 - I*Sqrt[15])^(1/3))*x + (3*I)*(3 - I *Sqrt[15])^(2/3)*x^2])/(54*(4*3^(1/3) + (3^(2/3) - I*3^(1/6)*Sqrt[5])*(3 - I*Sqrt[15])^(1/3) + 2*(3 - I*Sqrt[15])^(2/3)))) + (C*Log[2 - 6*x + 3*x...
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In tegersQ[n]
Int[((e_.) + (f_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (d_.)*(x_)^3)^(p_), x_S ymbol] :> With[{r = Rt[-9*a*d^2 + Sqrt[3]*d*Sqrt[4*b^3*d + 27*a^2*d^2], 3]} , Simp[1/d^(2*p) Int[(e + f*x)^m*Simp[18^(1/3)*b*(d/(3*r)) - r/18^(1/3) + d*x, x]^p*Simp[b*(d/3) + 12^(1/3)*b^2*(d^2/(3*r^2)) + r^2/(3*12^(1/3)) - d *(2^(1/3)*b*(d/(3^(1/3)*r)) - r/18^(1/3))*x + d^2*x^2, x]^p, x], x]] /; Fre eQ[{a, b, d, e, f, m}, x] && NeQ[4*b^3 + 27*a^2*d, 0] && ILtQ[p, 0]
Int[(Pm_)/(Qn_), x_Symbol] :> With[{m = Expon[Pm, x], n = Expon[Qn, x]}, Si mp[Coeff[Pm, x, m]*(Log[Qn]/(n*Coeff[Qn, x, n])), x] + Simp[1/(n*Coeff[Qn, x, n]) Int[ExpandToSum[n*Coeff[Qn, x, n]*Pm - Coeff[Pm, x, m]*D[Qn, x], x ]/Qn, x], x] /; EqQ[m, n - 1]] /; PolyQ[Pm, x] && PolyQ[Qn, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.06 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.12
| method | result | size |
| default | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (3 \textit {\_Z}^{3}-6 \textit {\_Z} +2\right )}{\sum }\frac {\left (C \,\textit {\_R}^{2}+B \textit {\_R} +A \right ) \ln \left (x -\textit {\_R} \right )}{3 \textit {\_R}^{2}-2}\right )}{3}\) | \(43\) |
| risch | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (3 \textit {\_Z}^{3}-6 \textit {\_Z} +2\right )}{\sum }\frac {\left (C \,\textit {\_R}^{2}+B \textit {\_R} +A \right ) \ln \left (x -\textit {\_R} \right )}{3 \textit {\_R}^{2}-2}\right )}{3}\) | \(43\) |
Input:
int((C*x^2+B*x+A)/(3*x^3-6*x+2),x,method=_RETURNVERBOSE)
Output:
1/3*sum((C*_R^2+B*_R+A)/(3*_R^2-2)*ln(x-_R),_R=RootOf(3*_Z^3-6*_Z+2))
Result contains complex when optimal does not.
Time = 1.41 (sec) , antiderivative size = 7031, normalized size of antiderivative = 18.85 \[ \int \frac {A+B x+C x^2}{2-6 x+3 x^3} \, dx=\text {Too large to display} \] Input:
integrate((C*x^2+B*x+A)/(3*x^3-6*x+2),x, algorithm="fricas")
Output:
Too large to include
Time = 9.21 (sec) , antiderivative size = 277, normalized size of antiderivative = 0.74 \[ \int \frac {A+B x+C x^2}{2-6 x+3 x^3} \, dx=\operatorname {RootSum} {\left (4860 t^{3} - 1620 t^{2} C + t \left (- 162 A^{2} - 162 A B - 216 A C - 108 B^{2} - 108 B C + 108 C^{2}\right ) + 9 A^{3} + 36 A^{2} C - 18 A B^{2} + 18 A B C + 36 A C^{2} - 6 B^{3} + 12 B C^{2} + 4 C^{3}, \left ( t \mapsto t \log {\left (x + \frac {- 3240 t^{2} A - 1620 t^{2} B - 2160 t^{2} C - 270 t A^{2} + 360 t A C + 180 t B^{2} + 360 t B C + 360 t C^{2} + 72 A^{3} + 108 A^{2} B + 174 A^{2} C + 84 A B^{2} + 144 A B C + 96 A C^{2} + 24 B^{3} + 36 B^{2} C + 28 B C^{2} + 8 C^{3}}{27 A^{3} + 108 A^{2} B + 54 A^{2} C + 54 A B^{2} + 144 A B C + 36 A C^{2} - 6 B^{3} + 36 B^{2} C + 48 B C^{2} + 8 C^{3}} \right )} \right )\right )} \] Input:
integrate((C*x**2+B*x+A)/(3*x**3-6*x+2),x)
Output:
RootSum(4860*_t**3 - 1620*_t**2*C + _t*(-162*A**2 - 162*A*B - 216*A*C - 10 8*B**2 - 108*B*C + 108*C**2) + 9*A**3 + 36*A**2*C - 18*A*B**2 + 18*A*B*C + 36*A*C**2 - 6*B**3 + 12*B*C**2 + 4*C**3, Lambda(_t, _t*log(x + (-3240*_t* *2*A - 1620*_t**2*B - 2160*_t**2*C - 270*_t*A**2 + 360*_t*A*C + 180*_t*B** 2 + 360*_t*B*C + 360*_t*C**2 + 72*A**3 + 108*A**2*B + 174*A**2*C + 84*A*B* *2 + 144*A*B*C + 96*A*C**2 + 24*B**3 + 36*B**2*C + 28*B*C**2 + 8*C**3)/(27 *A**3 + 108*A**2*B + 54*A**2*C + 54*A*B**2 + 144*A*B*C + 36*A*C**2 - 6*B** 3 + 36*B**2*C + 48*B*C**2 + 8*C**3))))
\[ \int \frac {A+B x+C x^2}{2-6 x+3 x^3} \, dx=\int { \frac {C x^{2} + B x + A}{3 \, x^{3} - 6 \, x + 2} \,d x } \] Input:
integrate((C*x^2+B*x+A)/(3*x^3-6*x+2),x, algorithm="maxima")
Output:
integrate((C*x^2 + B*x + A)/(3*x^3 - 6*x + 2), x)
Exception generated. \[ \int \frac {A+B x+C x^2}{2-6 x+3 x^3} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((C*x^2+B*x+A)/(3*x^3-6*x+2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Time = 12.43 (sec) , antiderivative size = 322, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x+C x^2}{2-6 x+3 x^3} \, dx=\sum _{k=1}^3\ln \left (-x\,\left (-3\,B^2+6\,C^2+3\,A\,C\right )+2\,C^2+\mathrm {root}\left (z^3-\frac {C\,z^2}{3}-\frac {z\,\left (162\,A^2+108\,B^2-108\,C^2+162\,A\,B+216\,A\,C+108\,B\,C\right )}{4860}+\frac {A\,B\,C}{270}+\frac {B\,C^2}{405}+\frac {A^2\,C}{135}+\frac {A\,C^2}{135}-\frac {A\,B^2}{270}+\frac {A^3}{540}+\frac {C^3}{1215}-\frac {B^3}{810},z,k\right )\,\left (18\,B-36\,C+x\,\left (27\,A+90\,C\right )-\mathrm {root}\left (z^3-\frac {C\,z^2}{3}-\frac {z\,\left (162\,A^2+108\,B^2-108\,C^2+162\,A\,B+216\,A\,C+108\,B\,C\right )}{4860}+\frac {A\,B\,C}{270}+\frac {B\,C^2}{405}+\frac {A^2\,C}{135}+\frac {A\,C^2}{135}-\frac {A\,B^2}{270}+\frac {A^3}{540}+\frac {C^3}{1215}-\frac {B^3}{810},z,k\right )\,\left (324\,x-162\right )\right )+3\,A\,B\right )\,\mathrm {root}\left (z^3-\frac {C\,z^2}{3}-\frac {z\,\left (162\,A^2+108\,B^2-108\,C^2+162\,A\,B+216\,A\,C+108\,B\,C\right )}{4860}+\frac {A\,B\,C}{270}+\frac {B\,C^2}{405}+\frac {A^2\,C}{135}+\frac {A\,C^2}{135}-\frac {A\,B^2}{270}+\frac {A^3}{540}+\frac {C^3}{1215}-\frac {B^3}{810},z,k\right ) \] Input:
int((A + B*x + C*x^2)/(3*x^3 - 6*x + 2),x)
Output:
symsum(log(2*C^2 - x*(6*C^2 - 3*B^2 + 3*A*C) + root(z^3 - (C*z^2)/3 - (z*( 162*A^2 + 108*B^2 - 108*C^2 + 162*A*B + 216*A*C + 108*B*C))/4860 + (A*B*C) /270 + (B*C^2)/405 + (A^2*C)/135 + (A*C^2)/135 - (A*B^2)/270 + A^3/540 + C ^3/1215 - B^3/810, z, k)*(18*B - 36*C + x*(27*A + 90*C) - root(z^3 - (C*z^ 2)/3 - (z*(162*A^2 + 108*B^2 - 108*C^2 + 162*A*B + 216*A*C + 108*B*C))/486 0 + (A*B*C)/270 + (B*C^2)/405 + (A^2*C)/135 + (A*C^2)/135 - (A*B^2)/270 + A^3/540 + C^3/1215 - B^3/810, z, k)*(324*x - 162)) + 3*A*B)*root(z^3 - (C* z^2)/3 - (z*(162*A^2 + 108*B^2 - 108*C^2 + 162*A*B + 216*A*C + 108*B*C))/4 860 + (A*B*C)/270 + (B*C^2)/405 + (A^2*C)/135 + (A*C^2)/135 - (A*B^2)/270 + A^3/540 + C^3/1215 - B^3/810, z, k), k, 1, 3)
\[ \int \frac {A+B x+C x^2}{2-6 x+3 x^3} \, dx=\left (\int \frac {x^{2}}{3 x^{3}-6 x +2}d x \right ) c +\left (\int \frac {x}{3 x^{3}-6 x +2}d x \right ) b +\left (\int \frac {1}{3 x^{3}-6 x +2}d x \right ) a \] Input:
int((C*x^2+B*x+A)/(3*x^3-6*x+2),x)
Output:
int(x**2/(3*x**3 - 6*x + 2),x)*c + int(x/(3*x**3 - 6*x + 2),x)*b + int(1/( 3*x**3 - 6*x + 2),x)*a