Integrand size = 27, antiderivative size = 185 \[ \int \frac {1}{3 a b+3 b^2 x+3 b c x^2+c^2 x^3} \, dx=-\frac {\arctan \left (\frac {\sqrt [3]{b}+\frac {2 (b+c x)}{\sqrt [3]{b^2-3 a c}}}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt {3} b^{2/3} \left (b^2-3 a c\right )^{2/3}}+\frac {\log \left (\sqrt [3]{b} \left (b^{2/3}-\sqrt [3]{b^2-3 a c}\right )+c x\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}-\frac {\log \left (b^{2/3} \left (b^2-3 a c\right )^{2/3}+\sqrt [3]{b} \sqrt [3]{b^2-3 a c} (b+c x)+(b+c x)^2\right )}{6 b^{2/3} \left (b^2-3 a c\right )^{2/3}} \] Output:
-1/3*arctan(1/3*(b^(1/3)+2*(c*x+b)/(-3*a*c+b^2)^(1/3))*3^(1/2)/b^(1/3))*3^ (1/2)/b^(2/3)/(-3*a*c+b^2)^(2/3)+1/3*ln(b^(1/3)*(b^(2/3)-(-3*a*c+b^2)^(1/3 ))+c*x)/b^(2/3)/(-3*a*c+b^2)^(2/3)-1/6*ln(b^(2/3)*(-3*a*c+b^2)^(2/3)+b^(1/ 3)*(-3*a*c+b^2)^(1/3)*(c*x+b)+(c*x+b)^2)/b^(2/3)/(-3*a*c+b^2)^(2/3)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.34 \[ \int \frac {1}{3 a b+3 b^2 x+3 b c x^2+c^2 x^3} \, dx=\frac {1}{3} \text {RootSum}\left [3 a b+3 b^2 \text {$\#$1}+3 b c \text {$\#$1}^2+c^2 \text {$\#$1}^3\&,\frac {\log (x-\text {$\#$1})}{b^2+2 b c \text {$\#$1}+c^2 \text {$\#$1}^2}\&\right ] \] Input:
Integrate[(3*a*b + 3*b^2*x + 3*b*c*x^2 + c^2*x^3)^(-1),x]
Output:
RootSum[3*a*b + 3*b^2*#1 + 3*b*c*#1^2 + c^2*#1^3 & , Log[x - #1]/(b^2 + 2* b*c*#1 + c^2*#1^2) & ]/3
Time = 0.47 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2458, 750, 16, 25, 1142, 27, 1082, 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.
| \(\displaystyle \int \frac {1}{3 a b+3 b^2 x+3 b c x^2+c^2 x^3} \, dx\) |
| \(\Big \downarrow \) 2458 |
| \(\displaystyle \int \frac {1}{b \left (3 a-\frac {b^2}{c}\right )+c^2 \left (\frac {b}{c}+x\right )^3}d\left (\frac {b}{c}+x\right )\) |
| \(\Big \downarrow \) 750 |
| \(\displaystyle \frac {c^{2/3} \int \frac {1}{c^{2/3} \left (\frac {b}{c}+x\right )-\frac {\sqrt [3]{b} \sqrt [3]{b^2-3 a c}}{\sqrt [3]{c}}}d\left (\frac {b}{c}+x\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}+\frac {c^{2/3} \int -\frac {c \left (\frac {b}{c}+x\right )+2 \sqrt [3]{b} \sqrt [3]{b^2-3 a c}}{\sqrt [3]{c} \left (c^{4/3} \left (\frac {b}{c}+x\right )^2+\sqrt [3]{b} \sqrt [3]{c} \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+\frac {b^{2/3} \left (b^2-3 a c\right )^{2/3}}{c^{2/3}}\right )}d\left (\frac {b}{c}+x\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {c^{2/3} \int -\frac {c \left (\frac {b}{c}+x\right )+2 \sqrt [3]{b} \sqrt [3]{b^2-3 a c}}{\sqrt [3]{c} \left (c^{4/3} \left (\frac {b}{c}+x\right )^2+\sqrt [3]{b} \sqrt [3]{c} \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+\frac {b^{2/3} \left (b^2-3 a c\right )^{2/3}}{c^{2/3}}\right )}d\left (\frac {b}{c}+x\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}+\frac {\log \left (\sqrt [3]{b} \sqrt [3]{b^2-3 a c}-c \left (\frac {b}{c}+x\right )\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\log \left (\sqrt [3]{b} \sqrt [3]{b^2-3 a c}-c \left (\frac {b}{c}+x\right )\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}-\frac {c^{2/3} \int \frac {c \left (\frac {b}{c}+x\right )+2 \sqrt [3]{b} \sqrt [3]{b^2-3 a c}}{c^{5/3} \left (\frac {b}{c}+x\right )^2+\sqrt [3]{b} c^{2/3} \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+\frac {b^{2/3} \left (b^2-3 a c\right )^{2/3}}{\sqrt [3]{c}}}d\left (\frac {b}{c}+x\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {\log \left (\sqrt [3]{b} \sqrt [3]{b^2-3 a c}-c \left (\frac {b}{c}+x\right )\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}-\frac {c^{2/3} \left (\frac {3}{2} \sqrt [3]{b} \sqrt [3]{b^2-3 a c} \int \frac {1}{c^{5/3} \left (\frac {b}{c}+x\right )^2+\sqrt [3]{b} c^{2/3} \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+\frac {b^{2/3} \left (b^2-3 a c\right )^{2/3}}{\sqrt [3]{c}}}d\left (\frac {b}{c}+x\right )+\frac {\int \frac {c^{2/3} \left (2 c \left (\frac {b}{c}+x\right )+\sqrt [3]{b} \sqrt [3]{b^2-3 a c}\right )}{c^{5/3} \left (\frac {b}{c}+x\right )^2+\sqrt [3]{b} c^{2/3} \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+\frac {b^{2/3} \left (b^2-3 a c\right )^{2/3}}{\sqrt [3]{c}}}d\left (\frac {b}{c}+x\right )}{2 c^{2/3}}\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\log \left (\sqrt [3]{b} \sqrt [3]{b^2-3 a c}-c \left (\frac {b}{c}+x\right )\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}-\frac {c^{2/3} \left (\frac {3}{2} \sqrt [3]{b} \sqrt [3]{b^2-3 a c} \int \frac {1}{c^{5/3} \left (\frac {b}{c}+x\right )^2+\sqrt [3]{b} c^{2/3} \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+\frac {b^{2/3} \left (b^2-3 a c\right )^{2/3}}{\sqrt [3]{c}}}d\left (\frac {b}{c}+x\right )+\frac {1}{2} \int \frac {2 c \left (\frac {b}{c}+x\right )+\sqrt [3]{b} \sqrt [3]{b^2-3 a c}}{c^{5/3} \left (\frac {b}{c}+x\right )^2+\sqrt [3]{b} c^{2/3} \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+\frac {b^{2/3} \left (b^2-3 a c\right )^{2/3}}{\sqrt [3]{c}}}d\left (\frac {b}{c}+x\right )\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {\log \left (\sqrt [3]{b} \sqrt [3]{b^2-3 a c}-c \left (\frac {b}{c}+x\right )\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}-\frac {c^{2/3} \left (\frac {1}{2} \int \frac {2 c \left (\frac {b}{c}+x\right )+\sqrt [3]{b} \sqrt [3]{b^2-3 a c}}{c^{5/3} \left (\frac {b}{c}+x\right )^2+\sqrt [3]{b} c^{2/3} \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+\frac {b^{2/3} \left (b^2-3 a c\right )^{2/3}}{\sqrt [3]{c}}}d\left (\frac {b}{c}+x\right )-\frac {3 \int \frac {1}{-\left (\frac {2 c \left (\frac {b}{c}+x\right )}{\sqrt [3]{b} \sqrt [3]{b^2-3 a c}}+1\right )^2-3}d\left (\frac {2 c \left (\frac {b}{c}+x\right )}{\sqrt [3]{b} \sqrt [3]{b^2-3 a c}}+1\right )}{c^{2/3}}\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\log \left (\sqrt [3]{b} \sqrt [3]{b^2-3 a c}-c \left (\frac {b}{c}+x\right )\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}-\frac {c^{2/3} \left (\frac {1}{2} \int \frac {2 c \left (\frac {b}{c}+x\right )+\sqrt [3]{b} \sqrt [3]{b^2-3 a c}}{c^{5/3} \left (\frac {b}{c}+x\right )^2+\sqrt [3]{b} c^{2/3} \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+\frac {b^{2/3} \left (b^2-3 a c\right )^{2/3}}{\sqrt [3]{c}}}d\left (\frac {b}{c}+x\right )+\frac {\sqrt {3} \arctan \left (\frac {\frac {2 c \left (\frac {b}{c}+x\right )}{\sqrt [3]{b} \sqrt [3]{b^2-3 a c}}+1}{\sqrt {3}}\right )}{c^{2/3}}\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {\log \left (\sqrt [3]{b} \sqrt [3]{b^2-3 a c}-c \left (\frac {b}{c}+x\right )\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}-\frac {c^{2/3} \left (\frac {\sqrt {3} \arctan \left (\frac {\frac {2 c \left (\frac {b}{c}+x\right )}{\sqrt [3]{b} \sqrt [3]{b^2-3 a c}}+1}{\sqrt {3}}\right )}{c^{2/3}}+\frac {\log \left (\sqrt [3]{b} c \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+b^{2/3} \left (b^2-3 a c\right )^{2/3}+c^2 \left (\frac {b}{c}+x\right )^2\right )}{2 c^{2/3}}\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}\) |
Input:
Int[(3*a*b + 3*b^2*x + 3*b*c*x^2 + c^2*x^3)^(-1),x]
Output:
Log[b^(1/3)*(b^2 - 3*a*c)^(1/3) - c*(b/c + x)]/(3*b^(2/3)*(b^2 - 3*a*c)^(2 /3)) - (c^(2/3)*((Sqrt[3]*ArcTan[(1 + (2*c*(b/c + x))/(b^(1/3)*(b^2 - 3*a* c)^(1/3)))/Sqrt[3]])/c^(2/3) + Log[b^(2/3)*(b^2 - 3*a*c)^(2/3) + b^(1/3)*c *(b^2 - 3*a*c)^(1/3)*(b/c + x) + c^2*(b/c + x)^2]/(2*c^(2/3))))/(3*b^(2/3) *(b^2 - 3*a*c)^(2/3))
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] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{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[(Pn_)^(p_.), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1]/(Exp on[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x - S, x]^p, x], x, x + S] /; BinomialQ[Pn /. x -> x - S, x] || (IntegerQ[Exp on[Pn, x]/2] && TrinomialQ[Pn /. x -> x - S, x])] /; FreeQ[p, x] && PolyQ[P n, x] && GtQ[Expon[Pn, x], 2] && NeQ[Coeff[Pn, x, Expon[Pn, x] - 1], 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.06 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.31
| method | result | size |
| default | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c^{2} \textit {\_Z}^{3}+3 b c \,\textit {\_Z}^{2}+3 b^{2} \textit {\_Z} +3 a b \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2} c^{2}+2 \textit {\_R} b c +b^{2}}\right )}{3}\) | \(57\) |
| risch | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c^{2} \textit {\_Z}^{3}+3 b c \,\textit {\_Z}^{2}+3 b^{2} \textit {\_Z} +3 a b \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2} c^{2}+2 \textit {\_R} b c +b^{2}}\right )}{3}\) | \(57\) |
Input:
int(1/(c^2*x^3+3*b*c*x^2+3*b^2*x+3*a*b),x,method=_RETURNVERBOSE)
Output:
1/3*sum(1/(_R^2*c^2+2*_R*b*c+b^2)*ln(x-_R),_R=RootOf(_Z^3*c^2+3*_Z^2*b*c+3 *_Z*b^2+3*a*b))
Leaf count of result is larger than twice the leaf count of optimal. 383 vs. \(2 (146) = 292\).
Time = 0.07 (sec) , antiderivative size = 383, normalized size of antiderivative = 2.07 \[ \int \frac {1}{3 a b+3 b^2 x+3 b c x^2+c^2 x^3} \, dx=-\frac {6 \, \sqrt {\frac {1}{3}} {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {1}{6}} {\left (b^{3} - 3 \, a b c\right )} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {2}{3}} {\left (c x + b\right )} + {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {1}{3}} {\left (b^{3} - 3 \, a b c\right )}\right )}}{{\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {5}{6}}}\right ) + {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {2}{3}} \log \left (-b^{5} + 3 \, a b^{3} c - {\left (b^{3} c^{2} - 3 \, a b c^{3}\right )} x^{2} - 2 \, {\left (b^{4} c - 3 \, a b^{2} c^{2}\right )} x - {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {2}{3}} {\left (c x + b\right )} - {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {1}{3}} {\left (b^{3} - 3 \, a b c\right )}\right ) - 2 \, {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {2}{3}} \log \left (-b^{4} + 3 \, a b^{2} c - {\left (b^{3} c - 3 \, a b c^{2}\right )} x + {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {2}{3}}\right )}{6 \, {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}} \] Input:
integrate(1/(c^2*x^3+3*b*c*x^2+3*b^2*x+3*a*b),x, algorithm="fricas")
Output:
-1/6*(6*sqrt(1/3)*(b^6 - 6*a*b^4*c + 9*a^2*b^2*c^2)^(1/6)*(b^3 - 3*a*b*c)* arctan(sqrt(1/3)*(2*(b^6 - 6*a*b^4*c + 9*a^2*b^2*c^2)^(2/3)*(c*x + b) + (b ^6 - 6*a*b^4*c + 9*a^2*b^2*c^2)^(1/3)*(b^3 - 3*a*b*c))/(b^6 - 6*a*b^4*c + 9*a^2*b^2*c^2)^(5/6)) + (b^6 - 6*a*b^4*c + 9*a^2*b^2*c^2)^(2/3)*log(-b^5 + 3*a*b^3*c - (b^3*c^2 - 3*a*b*c^3)*x^2 - 2*(b^4*c - 3*a*b^2*c^2)*x - (b^6 - 6*a*b^4*c + 9*a^2*b^2*c^2)^(2/3)*(c*x + b) - (b^6 - 6*a*b^4*c + 9*a^2*b^ 2*c^2)^(1/3)*(b^3 - 3*a*b*c)) - 2*(b^6 - 6*a*b^4*c + 9*a^2*b^2*c^2)^(2/3)* log(-b^4 + 3*a*b^2*c - (b^3*c - 3*a*b*c^2)*x + (b^6 - 6*a*b^4*c + 9*a^2*b^ 2*c^2)^(2/3)))/(b^6 - 6*a*b^4*c + 9*a^2*b^2*c^2)
Time = 0.31 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.29 \[ \int \frac {1}{3 a b+3 b^2 x+3 b c x^2+c^2 x^3} \, dx=\operatorname {RootSum} {\left (t^{3} \cdot \left (243 a^{2} b^{2} c^{2} - 162 a b^{4} c + 27 b^{6}\right ) - 1, \left ( t \mapsto t \log {\left (x + \frac {9 t a b c - 3 t b^{3} + b}{c} \right )} \right )\right )} \] Input:
integrate(1/(c**2*x**3+3*b*c*x**2+3*b**2*x+3*a*b),x)
Output:
RootSum(_t**3*(243*a**2*b**2*c**2 - 162*a*b**4*c + 27*b**6) - 1, Lambda(_t , _t*log(x + (9*_t*a*b*c - 3*_t*b**3 + b)/c)))
\[ \int \frac {1}{3 a b+3 b^2 x+3 b c x^2+c^2 x^3} \, dx=\int { \frac {1}{c^{2} x^{3} + 3 \, b c x^{2} + 3 \, b^{2} x + 3 \, a b} \,d x } \] Input:
integrate(1/(c^2*x^3+3*b*c*x^2+3*b^2*x+3*a*b),x, algorithm="maxima")
Output:
integrate(1/(c^2*x^3 + 3*b*c*x^2 + 3*b^2*x + 3*a*b), x)
Time = 0.14 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.15 \[ \int \frac {1}{3 a b+3 b^2 x+3 b c x^2+c^2 x^3} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} c x + \sqrt {3} b - \sqrt {3} {\left (-b^{3} + 3 \, a b c\right )}^{\frac {1}{3}}}{c x + b + {\left (-b^{3} + 3 \, a b c\right )}^{\frac {1}{3}}}\right )}{3 \, {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {1}{3}}} - \frac {\log \left (4 \, {\left (\sqrt {3} c x + \sqrt {3} b - \sqrt {3} {\left (-b^{3} + 3 \, a b c\right )}^{\frac {1}{3}}\right )}^{2} + 4 \, {\left (c x + b + {\left (-b^{3} + 3 \, a b c\right )}^{\frac {1}{3}}\right )}^{2}\right )}{6 \, {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {1}{3}}} + \frac {\log \left ({\left | c x + b + {\left (-b^{3} + 3 \, a b c\right )}^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {1}{3}}} \] Input:
integrate(1/(c^2*x^3+3*b*c*x^2+3*b^2*x+3*a*b),x, algorithm="giac")
Output:
1/3*sqrt(3)*arctan((sqrt(3)*c*x + sqrt(3)*b - sqrt(3)*(-b^3 + 3*a*b*c)^(1/ 3))/(c*x + b + (-b^3 + 3*a*b*c)^(1/3)))/(b^6 - 6*a*b^4*c + 9*a^2*b^2*c^2)^ (1/3) - 1/6*log(4*(sqrt(3)*c*x + sqrt(3)*b - sqrt(3)*(-b^3 + 3*a*b*c)^(1/3 ))^2 + 4*(c*x + b + (-b^3 + 3*a*b*c)^(1/3))^2)/(b^6 - 6*a*b^4*c + 9*a^2*b^ 2*c^2)^(1/3) + 1/3*log(abs(c*x + b + (-b^3 + 3*a*b*c)^(1/3)))/(b^6 - 6*a*b ^4*c + 9*a^2*b^2*c^2)^(1/3)
Time = 12.84 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.94 \[ \int \frac {1}{3 a b+3 b^2 x+3 b c x^2+c^2 x^3} \, dx=\frac {\ln \left (b+b^{1/3}\,{\left (3\,a\,c-b^2\right )}^{1/3}+c\,x\right )}{3\,b^{2/3}\,{\left (3\,a\,c-b^2\right )}^{2/3}}+\frac {\ln \left (3\,b\,c^3+3\,c^4\,x+\frac {3\,b^{1/3}\,c^3\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (3\,a\,c-b^2\right )}^{1/3}}{2}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,b^{2/3}\,{\left (3\,a\,c-b^2\right )}^{2/3}}-\frac {\ln \left (3\,b\,c^3+3\,c^4\,x-\frac {3\,b^{1/3}\,c^3\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (3\,a\,c-b^2\right )}^{1/3}}{2}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,b^{2/3}\,{\left (3\,a\,c-b^2\right )}^{2/3}} \] Input:
int(1/(3*a*b + 3*b^2*x + c^2*x^3 + 3*b*c*x^2),x)
Output:
log(b + b^(1/3)*(3*a*c - b^2)^(1/3) + c*x)/(3*b^(2/3)*(3*a*c - b^2)^(2/3)) + (log(3*b*c^3 + 3*c^4*x + (3*b^(1/3)*c^3*(3^(1/2)*1i - 1)*(3*a*c - b^2)^ (1/3))/2)*(3^(1/2)*1i - 1))/(6*b^(2/3)*(3*a*c - b^2)^(2/3)) - (log(3*b*c^3 + 3*c^4*x - (3*b^(1/3)*c^3*(3^(1/2)*1i + 1)*(3*a*c - b^2)^(1/3))/2)*(3^(1 /2)*1i + 1))/(6*b^(2/3)*(3*a*c - b^2)^(2/3))
Time = 0.17 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.88 \[ \int \frac {1}{3 a b+3 b^2 x+3 b c x^2+c^2 x^3} \, dx=\frac {-2 \sqrt {3}\, \mathit {atan} \left (\frac {b^{\frac {1}{3}} \left (3 a c -b^{2}\right )^{\frac {1}{3}}-2 b -2 c x}{b^{\frac {1}{3}} \left (3 a c -b^{2}\right )^{\frac {1}{3}} \sqrt {3}}\right )-\mathrm {log}\left (b^{\frac {2}{3}} \left (3 a c -b^{2}\right )^{\frac {2}{3}}-b^{\frac {4}{3}} \left (3 a c -b^{2}\right )^{\frac {1}{3}}-b^{\frac {1}{3}} \left (3 a c -b^{2}\right )^{\frac {1}{3}} c x +b^{2}+2 b c x +c^{2} x^{2}\right )+2 \,\mathrm {log}\left (b^{\frac {1}{3}} \left (3 a c -b^{2}\right )^{\frac {1}{3}}+b +c x \right )}{6 b^{\frac {2}{3}} \left (3 a c -b^{2}\right )^{\frac {2}{3}}} \] Input:
int(1/(c^2*x^3+3*b*c*x^2+3*b^2*x+3*a*b),x)
Output:
(b**(1/3)*(3*a*c - b**2)**(1/3)*( - 2*sqrt(3)*atan((b**(1/3)*(3*a*c - b**2 )**(1/3) - 2*b - 2*c*x)/(b**(1/3)*(3*a*c - b**2)**(1/3)*sqrt(3))) - log(b* *(2/3)*(3*a*c - b**2)**(2/3) - b**(1/3)*(3*a*c - b**2)**(1/3)*b - b**(1/3) *(3*a*c - b**2)**(1/3)*c*x + b**2 + 2*b*c*x + c**2*x**2) + 2*log(b**(1/3)* (3*a*c - b**2)**(1/3) + b + c*x)))/(6*b*(3*a*c - b**2))