\(\int \frac {1}{(3 a b+3 b^2 x+3 b c x^2+c^2 x^3)^3} \, dx\) [14]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [C] (verified)
   Fricas [B] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [F]
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 27, antiderivative size = 305 \[ \int \frac {1}{\left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^3} \, dx=-\frac {c \left (\frac {b}{c}+x\right )}{6 b \left (b^2-3 a c\right ) \left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^2}+\frac {5 c^2 \left (\frac {b}{c}+x\right )}{18 b^2 \left (b^2-3 a c\right )^2 \left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )}-\frac {5 c^2 \arctan \left (\frac {\sqrt [3]{b}+\frac {2 (b+c x)}{\sqrt [3]{b^2-3 a c}}}{\sqrt {3} \sqrt [3]{b}}\right )}{9 \sqrt {3} b^{8/3} \left (b^2-3 a c\right )^{8/3}}+\frac {5 c^2 \log \left (b-\sqrt [3]{b} \sqrt [3]{b^2-3 a c}+c x\right )}{27 b^{8/3} \left (b^2-3 a c\right )^{8/3}}-\frac {5 c^2 \log \left (b^{2/3} \left (b^2-3 a c\right )^{2/3}+\sqrt [3]{b} c \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+c^2 \left (\frac {b}{c}+x\right )^2\right )}{54 b^{8/3} \left (b^2-3 a c\right )^{8/3}} \]

[Out]

-1/6*c*(b/c+x)/b/(-3*a*c+b^2)/(c^2*x^3+3*b*c*x^2+3*b^2*x+3*a*b)^2+5/18*c^2*(b/c+x)/b^2/(-3*a*c+b^2)^2/(c^2*x^3
+3*b*c*x^2+3*b^2*x+3*a*b)+5/27*c^2*ln(b-b^(1/3)*(-3*a*c+b^2)^(1/3)+c*x)/b^(8/3)/(-3*a*c+b^2)^(8/3)-5/54*c^2*ln
(b^(2/3)*(-3*a*c+b^2)^(2/3)+b^(1/3)*c*(-3*a*c+b^2)^(1/3)*(b/c+x)+c^2*(b/c+x)^2)/b^(8/3)/(-3*a*c+b^2)^(8/3)-5/2
7*c^2*arctan(1/3*(b^(1/3)+2*(c*x+b)/(-3*a*c+b^2)^(1/3))/b^(1/3)*3^(1/2))/b^(8/3)/(-3*a*c+b^2)^(8/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2092, 205, 206, 31, 648, 631, 210, 642} \[ \int \frac {1}{\left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^3} \, dx=-\frac {5 c^2 \arctan \left (\frac {\frac {2 (b+c x)}{\sqrt [3]{b^2-3 a c}}+\sqrt [3]{b}}{\sqrt {3} \sqrt [3]{b}}\right )}{9 \sqrt {3} b^{8/3} \left (b^2-3 a c\right )^{8/3}}+\frac {5 c^2 \left (\frac {b}{c}+x\right )}{18 b^2 \left (b^2-3 a c\right )^2 \left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )}-\frac {c \left (\frac {b}{c}+x\right )}{6 b \left (b^2-3 a c\right ) \left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^2}+\frac {5 c^2 \log \left (-\sqrt [3]{b} \sqrt [3]{b^2-3 a c}+b+c x\right )}{27 b^{8/3} \left (b^2-3 a c\right )^{8/3}}-\frac {5 c^2 \log \left (\sqrt [3]{b} c \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+b^{2/3} \left (b^2-3 a c\right )^{2/3}+c^2 \left (\frac {b}{c}+x\right )^2\right )}{54 b^{8/3} \left (b^2-3 a c\right )^{8/3}} \]

[In]

Int[(3*a*b + 3*b^2*x + 3*b*c*x^2 + c^2*x^3)^(-3),x]

[Out]

-1/6*(c*(b/c + x))/(b*(b^2 - 3*a*c)*(3*a*b + 3*b^2*x + 3*b*c*x^2 + c^2*x^3)^2) + (5*c^2*(b/c + x))/(18*b^2*(b^
2 - 3*a*c)^2*(3*a*b + 3*b^2*x + 3*b*c*x^2 + c^2*x^3)) - (5*c^2*ArcTan[(b^(1/3) + (2*(b + c*x))/(b^2 - 3*a*c)^(
1/3))/(Sqrt[3]*b^(1/3))])/(9*Sqrt[3]*b^(8/3)*(b^2 - 3*a*c)^(8/3)) + (5*c^2*Log[b - b^(1/3)*(b^2 - 3*a*c)^(1/3)
 + c*x])/(27*b^(8/3)*(b^2 - 3*a*c)^(8/3)) - (5*c^2*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])/(54*b^(8/3)*(b^2 - 3*a*c)^(8/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
 + 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 206

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[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]

Rule 210

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])

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
 FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
 b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 2092

Int[(P3_)^(p_), x_Symbol] :> With[{a = Coeff[P3, x, 0], b = Coeff[P3, x, 1], c = Coeff[P3, x, 2], d = Coeff[P3
, x, 3]}, Subst[Int[Simp[(2*c^3 - 9*b*c*d + 27*a*d^2)/(27*d^2) - (c^2 - 3*b*d)*(x/(3*d)) + d*x^3, x]^p, x], x,
 x + c/(3*d)] /; NeQ[c, 0]] /; FreeQ[p, x] && PolyQ[P3, x, 3]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{\left (b \left (3 a-\frac {b^2}{c}\right )+c^2 x^3\right )^3} \, dx,x,\frac {b}{c}+x\right ) \\ & = -\frac {c \left (\frac {b}{c}+x\right )}{6 b \left (b^2-3 a c\right ) \left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^2}-\frac {(5 c) \text {Subst}\left (\int \frac {1}{\left (b \left (3 a-\frac {b^2}{c}\right )+c^2 x^3\right )^2} \, dx,x,\frac {b}{c}+x\right )}{6 b \left (b^2-3 a c\right )} \\ & = -\frac {c \left (\frac {b}{c}+x\right )}{6 b \left (b^2-3 a c\right ) \left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^2}+\frac {5 c (b+c x)}{18 b^2 \left (b^2-3 a c\right )^2 \left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )}+\frac {\left (5 c^2\right ) \text {Subst}\left (\int \frac {1}{b \left (3 a-\frac {b^2}{c}\right )+c^2 x^3} \, dx,x,\frac {b}{c}+x\right )}{9 b^2 \left (b^2-3 a c\right )^2} \\ & = -\frac {c \left (\frac {b}{c}+x\right )}{6 b \left (b^2-3 a c\right ) \left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^2}+\frac {5 c (b+c x)}{18 b^2 \left (b^2-3 a c\right )^2 \left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )}+\frac {\left (5 c^{8/3}\right ) \text {Subst}\left (\int \frac {1}{-\frac {\sqrt [3]{b} \sqrt [3]{b^2-3 a c}}{\sqrt [3]{c}}+c^{2/3} x} \, dx,x,\frac {b}{c}+x\right )}{27 b^{8/3} \left (b^2-3 a c\right )^{8/3}}+\frac {\left (5 c^{8/3}\right ) \text {Subst}\left (\int \frac {-\frac {2 \sqrt [3]{b} \sqrt [3]{b^2-3 a c}}{\sqrt [3]{c}}-c^{2/3} x}{\frac {b^{2/3} \left (b^2-3 a c\right )^{2/3}}{c^{2/3}}+\sqrt [3]{b} \sqrt [3]{c} \sqrt [3]{b^2-3 a c} x+c^{4/3} x^2} \, dx,x,\frac {b}{c}+x\right )}{27 b^{8/3} \left (b^2-3 a c\right )^{8/3}} \\ & = -\frac {c \left (\frac {b}{c}+x\right )}{6 b \left (b^2-3 a c\right ) \left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^2}+\frac {5 c (b+c x)}{18 b^2 \left (b^2-3 a c\right )^2 \left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )}+\frac {5 c^2 \log \left (\sqrt [3]{b} \left (b^{2/3}-\sqrt [3]{b^2-3 a c}\right )+c x\right )}{27 b^{8/3} \left (b^2-3 a c\right )^{8/3}}-\frac {\left (5 c^2\right ) \text {Subst}\left (\int \frac {\sqrt [3]{b} \sqrt [3]{c} \sqrt [3]{b^2-3 a c}+2 c^{4/3} x}{\frac {b^{2/3} \left (b^2-3 a c\right )^{2/3}}{c^{2/3}}+\sqrt [3]{b} \sqrt [3]{c} \sqrt [3]{b^2-3 a c} x+c^{4/3} x^2} \, dx,x,\frac {b}{c}+x\right )}{54 b^{8/3} \left (b^2-3 a c\right )^{8/3}}-\frac {\left (5 c^{7/3}\right ) \text {Subst}\left (\int \frac {1}{\frac {b^{2/3} \left (b^2-3 a c\right )^{2/3}}{c^{2/3}}+\sqrt [3]{b} \sqrt [3]{c} \sqrt [3]{b^2-3 a c} x+c^{4/3} x^2} \, dx,x,\frac {b}{c}+x\right )}{18 b^{7/3} \left (b^2-3 a c\right )^{7/3}} \\ & = -\frac {c \left (\frac {b}{c}+x\right )}{6 b \left (b^2-3 a c\right ) \left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^2}+\frac {5 c (b+c x)}{18 b^2 \left (b^2-3 a c\right )^2 \left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )}+\frac {5 c^2 \log \left (\sqrt [3]{b} \left (b^{2/3}-\sqrt [3]{b^2-3 a c}\right )+c x\right )}{27 b^{8/3} \left (b^2-3 a c\right )^{8/3}}-\frac {5 c^2 \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 )}{54 b^{8/3} \left (b^2-3 a c\right )^{8/3}}+\frac {\left (5 c^2\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 c \left (\frac {b}{c}+x\right )}{\sqrt [3]{b} \sqrt [3]{b^2-3 a c}}\right )}{9 b^{8/3} \left (b^2-3 a c\right )^{8/3}} \\ & = -\frac {c \left (\frac {b}{c}+x\right )}{6 b \left (b^2-3 a c\right ) \left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^2}+\frac {5 c (b+c x)}{18 b^2 \left (b^2-3 a c\right )^2 \left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )}-\frac {5 c^2 \tan ^{-1}\left (\frac {1+\frac {2 (b+c x)}{\sqrt [3]{b} \sqrt [3]{b^2-3 a c}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{8/3} \left (b^2-3 a c\right )^{8/3}}+\frac {5 c^2 \log \left (\sqrt [3]{b} \left (b^{2/3}-\sqrt [3]{b^2-3 a c}\right )+c x\right )}{27 b^{8/3} \left (b^2-3 a c\right )^{8/3}}-\frac {5 c^2 \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 )}{54 b^{8/3} \left (b^2-3 a c\right )^{8/3}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.06 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.49 \[ \int \frac {1}{\left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^3} \, dx=\frac {-\frac {3 (b+c x) \left (3 b^3-15 b^2 c x-5 c^3 x^3-3 b c \left (8 a+5 c x^2\right )\right )}{\left (3 a b+x \left (3 b^2+3 b c x+c^2 x^2\right )\right )^2}+10 c^2 \text {RootSum}\left [3 a b+3 b^2 \text {$\#$1}+3 b c \text {$\#$1}^2+c^2 \text {$\#$1}^3\&,\frac {\log (x-\text {$\#$1})}{b^2+2 b c \text {$\#$1}+c^2 \text {$\#$1}^2}\&\right ]}{54 \left (b^3-3 a b c\right )^2} \]

[In]

Integrate[(3*a*b + 3*b^2*x + 3*b*c*x^2 + c^2*x^3)^(-3),x]

[Out]

((-3*(b + c*x)*(3*b^3 - 15*b^2*c*x - 5*c^3*x^3 - 3*b*c*(8*a + 5*c*x^2)))/(3*a*b + x*(3*b^2 + 3*b*c*x + c^2*x^2
))^2 + 10*c^2*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) & ]
)/(54*(b^3 - 3*a*b*c)^2)

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.15 (sec) , antiderivative size = 277, normalized size of antiderivative = 0.91

method result size
risch \(\frac {\frac {5 c^{4} x^{4}}{18 \left (9 a^{2} c^{2}-6 a \,b^{2} c +b^{4}\right ) b^{2}}+\frac {10 c^{3} x^{3}}{9 b \left (9 a^{2} c^{2}-6 a \,b^{2} c +b^{4}\right )}+\frac {5 c^{2} x^{2}}{3 \left (9 a^{2} c^{2}-6 a \,b^{2} c +b^{4}\right )}+\frac {2 \left (2 a c +b^{2}\right ) c x}{3 b \left (9 a^{2} c^{2}-6 a \,b^{2} c +b^{4}\right )}+\frac {9 \left (8 a c -b^{2}\right )}{486 a^{2} c^{2}-324 a \,b^{2} c +54 b^{4}}}{\left (c^{2} x^{3}+3 b c \,x^{2}+3 b^{2} x +3 a b \right )^{2}}+\frac {5 c^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c^{2} \textit {\_Z}^{3}+3 b \,\textit {\_Z}^{2} c +3 b^{2} \textit {\_Z} +3 a b \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\left (9 a^{2} c^{2}-6 a \,b^{2} c +b^{4}\right ) \left (\textit {\_R}^{2} c^{2}+2 \textit {\_R} b c +b^{2}\right )}\right )}{27 b^{2}}\) \(277\)
default \(\frac {\frac {5 c^{4} x^{4}}{18 \left (9 a^{2} c^{2}-6 a \,b^{2} c +b^{4}\right ) b^{2}}+\frac {10 c^{3} x^{3}}{9 b \left (9 a^{2} c^{2}-6 a \,b^{2} c +b^{4}\right )}+\frac {5 c^{2} x^{2}}{3 \left (9 a^{2} c^{2}-6 a \,b^{2} c +b^{4}\right )}+\frac {2 \left (2 a c +b^{2}\right ) c x}{3 b \left (9 a^{2} c^{2}-6 a \,b^{2} c +b^{4}\right )}+\frac {9 \left (8 a c -b^{2}\right )}{486 a^{2} c^{2}-324 a \,b^{2} c +54 b^{4}}}{\left (c^{2} x^{3}+3 b c \,x^{2}+3 b^{2} x +3 a b \right )^{2}}+\frac {5 c^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c^{2} \textit {\_Z}^{3}+3 b \,\textit {\_Z}^{2} c +3 b^{2} \textit {\_Z} +3 a b \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2} c^{2}+2 \textit {\_R} b c +b^{2}}\right )}{3 \left (81 a^{2} c^{2}-54 a \,b^{2} c +9 b^{4}\right ) b^{2}}\) \(279\)

[In]

int(1/(c^2*x^3+3*b*c*x^2+3*b^2*x+3*a*b)^3,x,method=_RETURNVERBOSE)

[Out]

9*(5/162*c^4/(9*a^2*c^2-6*a*b^2*c+b^4)/b^2*x^4+10/81/b*c^3/(9*a^2*c^2-6*a*b^2*c+b^4)*x^3+5/27*c^2/(9*a^2*c^2-6
*a*b^2*c+b^4)*x^2+2/27*(2*a*c+b^2)*c/b/(9*a^2*c^2-6*a*b^2*c+b^4)*x+1/54*(8*a*c-b^2)/(9*a^2*c^2-6*a*b^2*c+b^4))
/(c^2*x^3+3*b*c*x^2+3*b^2*x+3*a*b)^2+5/27*c^2/b^2*sum(1/(9*a^2*c^2-6*a*b^2*c+b^4)/(_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))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1268 vs. \(2 (262) = 524\).

Time = 0.28 (sec) , antiderivative size = 1268, normalized size of antiderivative = 4.16 \[ \int \frac {1}{\left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^3} \, dx=\text {Too large to display} \]

[In]

integrate(1/(c^2*x^3+3*b*c*x^2+3*b^2*x+3*a*b)^3,x, algorithm="fricas")

[Out]

-1/54*(9*b^10 - 126*a*b^8*c + 513*a^2*b^6*c^2 - 648*a^3*b^4*c^3 - 15*(b^6*c^4 - 6*a*b^4*c^5 + 9*a^2*b^2*c^6)*x
^4 - 60*(b^7*c^3 - 6*a*b^5*c^4 + 9*a^2*b^3*c^5)*x^3 - 90*(b^8*c^2 - 6*a*b^6*c^3 + 9*a^2*b^4*c^4)*x^2 + 10*sqrt
(3)*(9*a^2*b^5*c^2 - 27*a^3*b^3*c^3 + (b^3*c^6 - 3*a*b*c^7)*x^6 + 6*(b^4*c^5 - 3*a*b^2*c^6)*x^5 + 15*(b^5*c^4
- 3*a*b^3*c^5)*x^4 + 6*(3*b^6*c^3 - 8*a*b^4*c^4 - 3*a^2*b^2*c^5)*x^3 + 9*(b^7*c^2 - a*b^5*c^3 - 6*a^2*b^3*c^4)
*x^2 + 18*(a*b^6*c^2 - 3*a^2*b^4*c^3)*x)*(b^6 - 6*a*b^4*c + 9*a^2*b^2*c^2)^(1/6)*arctan(1/3*(2*sqrt(3)*(b^6 -
6*a*b^4*c + 9*a^2*b^2*c^2)^(2/3)*(c*x + b) + sqrt(3)*(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)) + 5*(c^6*x^6 + 6*b*c^5*x^5 + 15*b^2*c^4*x^4 + 18*a*b^3*c^2*x + 9*a^2*
b^2*c^2 + 6*(3*b^3*c^3 + a*b*c^4)*x^3 + 9*(b^4*c^2 + 2*a*b^2*c^3)*x^2)*(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)) - 10*(c^6*x^6 + 6*b*c^5*x^5 + 1
5*b^2*c^4*x^4 + 18*a*b^3*c^2*x + 9*a^2*b^2*c^2 + 6*(3*b^3*c^3 + a*b*c^4)*x^3 + 9*(b^4*c^2 + 2*a*b^2*c^3)*x^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)) - 36*(b^9*c - 4*a*b^7*c^2 - 3*a^2*b^5*c^3 + 18*a^3*b^3*c^4)*x)/(9*a^2*b^14 - 108*a^3*b^12*c
+ 486*a^4*b^10*c^2 - 972*a^5*b^8*c^3 + 729*a^6*b^6*c^4 + (b^12*c^4 - 12*a*b^10*c^5 + 54*a^2*b^8*c^6 - 108*a^3*
b^6*c^7 + 81*a^4*b^4*c^8)*x^6 + 6*(b^13*c^3 - 12*a*b^11*c^4 + 54*a^2*b^9*c^5 - 108*a^3*b^7*c^6 + 81*a^4*b^5*c^
7)*x^5 + 15*(b^14*c^2 - 12*a*b^12*c^3 + 54*a^2*b^10*c^4 - 108*a^3*b^8*c^5 + 81*a^4*b^6*c^6)*x^4 + 6*(3*b^15*c
- 35*a*b^13*c^2 + 150*a^2*b^11*c^3 - 270*a^3*b^9*c^4 + 135*a^4*b^7*c^5 + 81*a^5*b^5*c^6)*x^3 + 9*(b^16 - 10*a*
b^14*c + 30*a^2*b^12*c^2 - 135*a^4*b^8*c^4 + 162*a^5*b^6*c^5)*x^2 + 18*(a*b^15 - 12*a^2*b^13*c + 54*a^3*b^11*c
^2 - 108*a^4*b^9*c^3 + 81*a^5*b^7*c^4)*x)

Sympy [A] (verification not implemented)

Time = 1.39 (sec) , antiderivative size = 474, normalized size of antiderivative = 1.55 \[ \int \frac {1}{\left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^3} \, dx=\frac {24 a b^{2} c - 3 b^{4} + 30 b^{2} c^{2} x^{2} + 20 b c^{3} x^{3} + 5 c^{4} x^{4} + x \left (24 a b c^{2} + 12 b^{3} c\right )}{1458 a^{4} b^{4} c^{2} - 972 a^{3} b^{6} c + 162 a^{2} b^{8} + x^{6} \cdot \left (162 a^{2} b^{2} c^{6} - 108 a b^{4} c^{5} + 18 b^{6} c^{4}\right ) + x^{5} \cdot \left (972 a^{2} b^{3} c^{5} - 648 a b^{5} c^{4} + 108 b^{7} c^{3}\right ) + x^{4} \cdot \left (2430 a^{2} b^{4} c^{4} - 1620 a b^{6} c^{3} + 270 b^{8} c^{2}\right ) + x^{3} \cdot \left (972 a^{3} b^{3} c^{4} + 2268 a^{2} b^{5} c^{3} - 1836 a b^{7} c^{2} + 324 b^{9} c\right ) + x^{2} \cdot \left (2916 a^{3} b^{4} c^{3} - 486 a^{2} b^{6} c^{2} - 648 a b^{8} c + 162 b^{10}\right ) + x \left (2916 a^{3} b^{5} c^{2} - 1944 a^{2} b^{7} c + 324 a b^{9}\right )} + \operatorname {RootSum} {\left (t^{3} \cdot \left (129140163 a^{8} b^{8} c^{8} - 344373768 a^{7} b^{10} c^{7} + 401769396 a^{6} b^{12} c^{6} - 267846264 a^{5} b^{14} c^{5} + 111602610 a^{4} b^{16} c^{4} - 29760696 a^{3} b^{18} c^{3} + 4960116 a^{2} b^{20} c^{2} - 472392 a b^{22} c + 19683 b^{24}\right ) - 125 c^{6}, \left ( t \mapsto t \log {\left (x + \frac {729 t a^{3} b^{3} c^{3} - 729 t a^{2} b^{5} c^{2} + 243 t a b^{7} c - 27 t b^{9} + 5 b c^{2}}{5 c^{3}} \right )} \right )\right )} \]

[In]

integrate(1/(c**2*x**3+3*b*c*x**2+3*b**2*x+3*a*b)**3,x)

[Out]

(24*a*b**2*c - 3*b**4 + 30*b**2*c**2*x**2 + 20*b*c**3*x**3 + 5*c**4*x**4 + x*(24*a*b*c**2 + 12*b**3*c))/(1458*
a**4*b**4*c**2 - 972*a**3*b**6*c + 162*a**2*b**8 + x**6*(162*a**2*b**2*c**6 - 108*a*b**4*c**5 + 18*b**6*c**4)
+ x**5*(972*a**2*b**3*c**5 - 648*a*b**5*c**4 + 108*b**7*c**3) + x**4*(2430*a**2*b**4*c**4 - 1620*a*b**6*c**3 +
 270*b**8*c**2) + x**3*(972*a**3*b**3*c**4 + 2268*a**2*b**5*c**3 - 1836*a*b**7*c**2 + 324*b**9*c) + x**2*(2916
*a**3*b**4*c**3 - 486*a**2*b**6*c**2 - 648*a*b**8*c + 162*b**10) + x*(2916*a**3*b**5*c**2 - 1944*a**2*b**7*c +
 324*a*b**9)) + RootSum(_t**3*(129140163*a**8*b**8*c**8 - 344373768*a**7*b**10*c**7 + 401769396*a**6*b**12*c**
6 - 267846264*a**5*b**14*c**5 + 111602610*a**4*b**16*c**4 - 29760696*a**3*b**18*c**3 + 4960116*a**2*b**20*c**2
 - 472392*a*b**22*c + 19683*b**24) - 125*c**6, Lambda(_t, _t*log(x + (729*_t*a**3*b**3*c**3 - 729*_t*a**2*b**5
*c**2 + 243*_t*a*b**7*c - 27*_t*b**9 + 5*b*c**2)/(5*c**3))))

Maxima [F]

\[ \int \frac {1}{\left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^3} \, dx=\int { \frac {1}{{\left (c^{2} x^{3} + 3 \, b c x^{2} + 3 \, b^{2} x + 3 \, a b\right )}^{3}} \,d x } \]

[In]

integrate(1/(c^2*x^3+3*b*c*x^2+3*b^2*x+3*a*b)^3,x, algorithm="maxima")

[Out]

5/9*c^2*integrate(1/(c^2*x^3 + 3*b*c*x^2 + 3*b^2*x + 3*a*b), x)/(b^6 - 6*a*b^4*c + 9*a^2*b^2*c^2) + 1/18*(5*c^
4*x^4 + 20*b*c^3*x^3 + 30*b^2*c^2*x^2 - 3*b^4 + 24*a*b^2*c + 12*(b^3*c + 2*a*b*c^2)*x)/(9*a^2*b^8 - 54*a^3*b^6
*c + 81*a^4*b^4*c^2 + (b^6*c^4 - 6*a*b^4*c^5 + 9*a^2*b^2*c^6)*x^6 + 6*(b^7*c^3 - 6*a*b^5*c^4 + 9*a^2*b^3*c^5)*
x^5 + 15*(b^8*c^2 - 6*a*b^6*c^3 + 9*a^2*b^4*c^4)*x^4 + 6*(3*b^9*c - 17*a*b^7*c^2 + 21*a^2*b^5*c^3 + 9*a^3*b^3*
c^4)*x^3 + 9*(b^10 - 4*a*b^8*c - 3*a^2*b^6*c^2 + 18*a^3*b^4*c^3)*x^2 + 18*(a*b^9 - 6*a^2*b^7*c + 9*a^3*b^5*c^2
)*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 366, normalized size of antiderivative = 1.20 \[ \int \frac {1}{\left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^3} \, dx=\frac {5 \, {\left (2 \, \sqrt {3} \left (\frac {c^{6}}{b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} c x + \sqrt {3} b - \sqrt {3} {\left (-b^{3} + 3 \, a b c\right )}^{\frac {1}{3}}}{c x + b + {\left (-b^{3} + 3 \, a b c\right )}^{\frac {1}{3}}}\right ) - \left (\frac {c^{6}}{b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}}\right )^{\frac {1}{3}} \log \left (4 \, {\left (\sqrt {3} c x + \sqrt {3} b - \sqrt {3} {\left (-b^{3} + 3 \, a b c\right )}^{\frac {1}{3}}\right )}^{2} + 4 \, {\left (c x + b + {\left (-b^{3} + 3 \, a b c\right )}^{\frac {1}{3}}\right )}^{2}\right ) + 2 \, \left (\frac {c^{6}}{b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}}\right )^{\frac {1}{3}} \log \left ({\left | c x + b + {\left (-b^{3} + 3 \, a b c\right )}^{\frac {1}{3}} \right |}\right )\right )}}{54 \, {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}} + \frac {5 \, c^{4} x^{4} + 20 \, b c^{3} x^{3} + 30 \, b^{2} c^{2} x^{2} + 12 \, b^{3} c x + 24 \, a b c^{2} x - 3 \, b^{4} + 24 \, a b^{2} c}{18 \, {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )} {\left (c^{2} x^{3} + 3 \, b c x^{2} + 3 \, b^{2} x + 3 \, a b\right )}^{2}} \]

[In]

integrate(1/(c^2*x^3+3*b*c*x^2+3*b^2*x+3*a*b)^3,x, algorithm="giac")

[Out]

5/54*(2*sqrt(3)*(c^6/(b^6 - 6*a*b^4*c + 9*a^2*b^2*c^2))^(1/3)*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))) - (c^6/(b^6 - 6*a*b^4*c + 9*a^2*b^2*c^2))^(1/3)*log(4*(s
qrt(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) + 2*(c^6/
(b^6 - 6*a*b^4*c + 9*a^2*b^2*c^2))^(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) + 1/18*(5*c^4*x^4 + 20*b*c^3*x^3 + 30*b^2*c^2*x^2 + 12*b^3*c*x + 24*a*b*c^2*x - 3*b^4 + 24*a*b^2*c)/(
(b^6 - 6*a*b^4*c + 9*a^2*b^2*c^2)*(c^2*x^3 + 3*b*c*x^2 + 3*b^2*x + 3*a*b)^2)

Mupad [B] (verification not implemented)

Time = 10.19 (sec) , antiderivative size = 483, normalized size of antiderivative = 1.58 \[ \int \frac {1}{\left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^3} \, dx=\frac {\frac {8\,a\,c-b^2}{6\,\left (9\,a^2\,c^2-6\,a\,b^2\,c+b^4\right )}+\frac {5\,c^2\,x^2}{3\,\left (9\,a^2\,c^2-6\,a\,b^2\,c+b^4\right )}+\frac {10\,c^3\,x^3}{9\,b\,\left (9\,a^2\,c^2-6\,a\,b^2\,c+b^4\right )}+\frac {5\,c^4\,x^4}{18\,b^2\,\left (9\,a^2\,c^2-6\,a\,b^2\,c+b^4\right )}+\frac {2\,c\,x\,\left (b^2+2\,a\,c\right )}{3\,b\,\left (9\,a^2\,c^2-6\,a\,b^2\,c+b^4\right )}}{x^2\,\left (9\,b^4+18\,a\,c\,b^2\right )+9\,a^2\,b^2+c^4\,x^6+x^3\,\left (18\,b^3\,c+6\,a\,b\,c^2\right )+6\,b\,c^3\,x^5+15\,b^2\,c^2\,x^4+18\,a\,b^3\,x}+\frac {5\,c^2\,\ln \left (b\,{\left (3\,a\,c-b^2\right )}^{8/3}-b^{19/3}+c\,x\,{\left (3\,a\,c-b^2\right )}^{8/3}+27\,a^3\,b^{1/3}\,c^3-27\,a^2\,b^{7/3}\,c^2+9\,a\,b^{13/3}\,c\right )}{27\,b^{8/3}\,{\left (3\,a\,c-b^2\right )}^{8/3}}-\frac {5\,c^2\,\ln \left (2\,b-b^{1/3}\,{\left (3\,a\,c-b^2\right )}^{1/3}+2\,c\,x-\sqrt {3}\,b^{1/3}\,{\left (3\,a\,c-b^2\right )}^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{27\,b^{8/3}\,{\left (3\,a\,c-b^2\right )}^{8/3}}+\frac {5\,c^2\,\ln \left (2\,b-b^{1/3}\,{\left (3\,a\,c-b^2\right )}^{1/3}+2\,c\,x+\sqrt {3}\,b^{1/3}\,{\left (3\,a\,c-b^2\right )}^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{27\,b^{8/3}\,{\left (3\,a\,c-b^2\right )}^{8/3}} \]

[In]

int(1/(3*a*b + 3*b^2*x + c^2*x^3 + 3*b*c*x^2)^3,x)

[Out]

((8*a*c - b^2)/(6*(b^4 + 9*a^2*c^2 - 6*a*b^2*c)) + (5*c^2*x^2)/(3*(b^4 + 9*a^2*c^2 - 6*a*b^2*c)) + (10*c^3*x^3
)/(9*b*(b^4 + 9*a^2*c^2 - 6*a*b^2*c)) + (5*c^4*x^4)/(18*b^2*(b^4 + 9*a^2*c^2 - 6*a*b^2*c)) + (2*c*x*(2*a*c + b
^2))/(3*b*(b^4 + 9*a^2*c^2 - 6*a*b^2*c)))/(x^2*(9*b^4 + 18*a*b^2*c) + 9*a^2*b^2 + c^4*x^6 + x^3*(18*b^3*c + 6*
a*b*c^2) + 6*b*c^3*x^5 + 15*b^2*c^2*x^4 + 18*a*b^3*x) + (5*c^2*log(b*(3*a*c - b^2)^(8/3) - b^(19/3) + c*x*(3*a
*c - b^2)^(8/3) + 27*a^3*b^(1/3)*c^3 - 27*a^2*b^(7/3)*c^2 + 9*a*b^(13/3)*c))/(27*b^(8/3)*(3*a*c - b^2)^(8/3))
- (5*c^2*log(2*b - b^(1/3)*(3*a*c - b^2)^(1/3) + 2*c*x - 3^(1/2)*b^(1/3)*(3*a*c - b^2)^(1/3)*1i)*((3^(1/2)*1i)
/2 + 1/2))/(27*b^(8/3)*(3*a*c - b^2)^(8/3)) + (5*c^2*log(2*b - b^(1/3)*(3*a*c - b^2)^(1/3) + 2*c*x + 3^(1/2)*b
^(1/3)*(3*a*c - b^2)^(1/3)*1i)*((3^(1/2)*1i)/2 - 1/2))/(27*b^(8/3)*(3*a*c - b^2)^(8/3))