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

Optimal. Leaf size=245 \[ -\frac{c \left (\frac{b}{c}+x\right )}{3 b \left (b^2-3 a c\right ) \left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )}+\frac{c \log \left (\sqrt [3]{b} c \sqrt [3]{b^2-3 a c} \left (\frac{b}{c}+x\right )+b^{2/3} \left (b^2-3 a c\right )^{2/3}+c^2 \left (\frac{b}{c}+x\right )^2\right )}{9 b^{5/3} \left (b^2-3 a c\right )^{5/3}}-\frac{2 c \log \left (-\sqrt [3]{b} \sqrt [3]{b^2-3 a c}+b+c x\right )}{9 b^{5/3} \left (b^2-3 a c\right )^{5/3}}+\frac{2 c \tan ^{-1}\left (\frac{\frac{2 (b+c x)}{\sqrt [3]{b^2-3 a c}}+\sqrt [3]{b}}{\sqrt{3} \sqrt [3]{b}}\right )}{3 \sqrt{3} b^{5/3} \left (b^2-3 a c\right )^{5/3}} \]

[Out]

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

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Rubi [A]  time = 0.248543, antiderivative size = 245, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {2067, 199, 200, 31, 634, 617, 204, 628} \[ -\frac{c \left (\frac{b}{c}+x\right )}{3 b \left (b^2-3 a c\right ) \left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )}+\frac{c \log \left (\sqrt [3]{b} c \sqrt [3]{b^2-3 a c} \left (\frac{b}{c}+x\right )+b^{2/3} \left (b^2-3 a c\right )^{2/3}+c^2 \left (\frac{b}{c}+x\right )^2\right )}{9 b^{5/3} \left (b^2-3 a c\right )^{5/3}}-\frac{2 c \log \left (-\sqrt [3]{b} \sqrt [3]{b^2-3 a c}+b+c x\right )}{9 b^{5/3} \left (b^2-3 a c\right )^{5/3}}+\frac{2 c \tan ^{-1}\left (\frac{\frac{2 (b+c x)}{\sqrt [3]{b^2-3 a c}}+\sqrt [3]{b}}{\sqrt{3} \sqrt [3]{b}}\right )}{3 \sqrt{3} b^{5/3} \left (b^2-3 a c\right )^{5/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2067

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]

Rule 199

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] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 200

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 31

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

Rule 634

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 617

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

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]

Rubi steps

\begin{align*} \int \frac{1}{\left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^2} \, dx &=\operatorname{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 )\\ &=-\frac{c \left (\frac{b}{c}+x\right )}{3 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 )}-\frac{(2 c) \operatorname{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 )}{3 b \left (b^2-3 a c\right )}\\ &=-\frac{c \left (\frac{b}{c}+x\right )}{3 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 )}-\frac{\left (2 c^{5/3}\right ) \operatorname{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 )}{9 b^{5/3} \left (b^2-3 a c\right )^{5/3}}-\frac{\left (2 c^{5/3}\right ) \operatorname{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 )}{9 b^{5/3} \left (b^2-3 a c\right )^{5/3}}\\ &=-\frac{c \left (\frac{b}{c}+x\right )}{3 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 )}-\frac{2 c \log \left (\sqrt [3]{b} \left (b^{2/3}-\sqrt [3]{b^2-3 a c}\right )+c x\right )}{9 b^{5/3} \left (b^2-3 a c\right )^{5/3}}+\frac{c \operatorname{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 )}{9 b^{5/3} \left (b^2-3 a c\right )^{5/3}}+\frac{c^{4/3} \operatorname{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 )}{3 b^{4/3} \left (b^2-3 a c\right )^{4/3}}\\ &=-\frac{c \left (\frac{b}{c}+x\right )}{3 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 )}-\frac{2 c \log \left (\sqrt [3]{b} \left (b^{2/3}-\sqrt [3]{b^2-3 a c}\right )+c x\right )}{9 b^{5/3} \left (b^2-3 a c\right )^{5/3}}+\frac{c \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 )}{9 b^{5/3} \left (b^2-3 a c\right )^{5/3}}-\frac{(2 c) \operatorname{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 )}{3 b^{5/3} \left (b^2-3 a c\right )^{5/3}}\\ &=-\frac{c \left (\frac{b}{c}+x\right )}{3 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 )}+\frac{2 c \tan ^{-1}\left (\frac{1+\frac{2 (b+c x)}{\sqrt [3]{b} \sqrt [3]{b^2-3 a c}}}{\sqrt{3}}\right )}{3 \sqrt{3} b^{5/3} \left (b^2-3 a c\right )^{5/3}}-\frac{2 c \log \left (\sqrt [3]{b} \left (b^{2/3}-\sqrt [3]{b^2-3 a c}\right )+c x\right )}{9 b^{5/3} \left (b^2-3 a c\right )^{5/3}}+\frac{c \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 )}{9 b^{5/3} \left (b^2-3 a c\right )^{5/3}}\\ \end{align*}

Mathematica [C]  time = 0.054183, size = 112, normalized size = 0.46 \[ -\frac{2 c \text{RootSum}\left [3 \text{$\#$1}^2 b c+\text{$\#$1}^3 c^2+3 \text{$\#$1} b^2+3 a b\& ,\frac{\log (x-\text{$\#$1})}{\text{$\#$1}^2 c^2+2 \text{$\#$1} b c+b^2}\& \right ]+\frac{3 (b+c x)}{3 a b+x \left (3 b^2+3 b c x+c^2 x^2\right )}}{9 \left (b^3-3 a b c\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*(b + c*x))/(3*a*b + x*(3*b^2 + 3*b*c*x + c^2*x^2)) + 2*c*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) & ])/(9*(b^3 - 3*a*b*c))

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Maple [C]  time = 0.007, size = 136, normalized size = 0.6 \begin{align*}{\frac{1}{{c}^{2}{x}^{3}+3\,bc{x}^{2}+3\,{b}^{2}x+3\,ab} \left ({\frac{cx}{3\,b \left ( 3\,ac-{b}^{2} \right ) }}+{\frac{1}{9\,ac-3\,{b}^{2}}} \right ) }+{\frac{2\,c}{9\,b \left ( 3\,ac-{b}^{2} \right ) }\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{3}{c}^{2}+3\,{{\it \_Z}}^{2}bc+3\,{\it \_Z}\,{b}^{2}+3\,ab \right ) }{\frac{\ln \left ( x-{\it \_R} \right ) }{{{\it \_R}}^{2}{c}^{2}+2\,{\it \_R}\,bc+{b}^{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/3*c/b/(3*a*c-b^2)*x+1/3/(3*a*c-b^2))/(c^2*x^3+3*b*c*x^2+3*b^2*x+3*a*b)+2/9*c/b/(3*a*c-b^2)*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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{\frac{1}{3} \,{\left (\frac{2 \, \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 )}{{\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac{1}{3}}} - \frac{\log \left ({\left (\sqrt{3} c x + \sqrt{3} b - \sqrt{3}{\left (-b^{3} + 3 \, a b c\right )}^{\frac{1}{3}}\right )}^{2} +{\left (c x + b +{\left (-b^{3} + 3 \, a b c\right )}^{\frac{1}{3}}\right )}^{2}\right )}{{\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac{1}{3}}} + \frac{2 \, \log \left ({\left | c x + b +{\left (-b^{3} + 3 \, a b c\right )}^{\frac{1}{3}} \right |}\right )}{{\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac{1}{3}}}\right )} c}{3 \,{\left (b^{3} - 3 \, a b c\right )}} - \frac{c x + b}{3 \,{\left (3 \, a b^{4} - 9 \, a^{2} b^{2} c +{\left (b^{3} c^{2} - 3 \, a b c^{3}\right )} x^{3} + 3 \,{\left (b^{4} c - 3 \, a b^{2} c^{2}\right )} x^{2} + 3 \,{\left (b^{5} - 3 \, a b^{3} c\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*c*integrate(1/(c^2*x^3 + 3*b*c*x^2 + 3*b^2*x + 3*a*b), x)/(b^3 - 3*a*b*c) - 1/3*(c*x + b)/(3*a*b^4 - 9*a^
2*b^2*c + (b^3*c^2 - 3*a*b*c^3)*x^3 + 3*(b^4*c - 3*a*b^2*c^2)*x^2 + 3*(b^5 - 3*a*b^3*c)*x)

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Fricas [B]  time = 1.44711, size = 1536, normalized size = 6.27 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*(3*b^7 - 18*a*b^5*c + 27*a^2*b^3*c^2 - 2*sqrt(3)*(b^6 - 6*a*b^4*c + 9*a^2*b^2*c^2)^(1/6)*(3*a*b^4*c - 9*a
^2*b^2*c^2 + (b^3*c^3 - 3*a*b*c^4)*x^3 + 3*(b^4*c^2 - 3*a*b^2*c^3)*x^2 + 3*(b^5*c - 3*a*b^3*c^2)*x)*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)) - (b^6 - 6*a*b^4*c + 9*a^2*b^2*c^2)^(2/3)*(c^3*x^3
+ 3*b*c^2*x^2 + 3*b^2*c*x + 3*a*b*c)*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)*(c^3*x^3 + 3*b*c^2*x^2 + 3*b^2*c*x + 3*a*b*c)*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)) + 3*(b^6*c - 6*a*b^4*c^2 + 9*a^2*b^2
*c^3)*x)/(3*a*b^10 - 27*a^2*b^8*c + 81*a^3*b^6*c^2 - 81*a^4*b^4*c^3 + (b^9*c^2 - 9*a*b^7*c^3 + 27*a^2*b^5*c^4
- 27*a^3*b^3*c^5)*x^3 + 3*(b^10*c - 9*a*b^8*c^2 + 27*a^2*b^6*c^3 - 27*a^3*b^4*c^4)*x^2 + 3*(b^11 - 9*a*b^9*c +
27*a^2*b^7*c^2 - 27*a^3*b^5*c^3)*x)

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Sympy [A]  time = 1.49114, size = 192, normalized size = 0.78 \begin{align*} \frac{b + c x}{27 a^{2} b^{2} c - 9 a b^{4} + x^{3} \left (9 a b c^{3} - 3 b^{3} c^{2}\right ) + x^{2} \left (27 a b^{2} c^{2} - 9 b^{4} c\right ) + x \left (27 a b^{3} c - 9 b^{5}\right )} + \operatorname{RootSum}{\left (t^{3} \left (177147 a^{5} b^{5} c^{5} - 295245 a^{4} b^{7} c^{4} + 196830 a^{3} b^{9} c^{3} - 65610 a^{2} b^{11} c^{2} + 10935 a b^{13} c - 729 b^{15}\right ) - 8 c^{3}, \left ( t \mapsto t \log{\left (x + \frac{81 t a^{2} b^{2} c^{2} - 54 t a b^{4} c + 9 t b^{6} + 2 b c}{2 c^{2}} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b + c*x)/(27*a**2*b**2*c - 9*a*b**4 + x**3*(9*a*b*c**3 - 3*b**3*c**2) + x**2*(27*a*b**2*c**2 - 9*b**4*c) + x*
(27*a*b**3*c - 9*b**5)) + RootSum(_t**3*(177147*a**5*b**5*c**5 - 295245*a**4*b**7*c**4 + 196830*a**3*b**9*c**3
- 65610*a**2*b**11*c**2 + 10935*a*b**13*c - 729*b**15) - 8*c**3, Lambda(_t, _t*log(x + (81*_t*a**2*b**2*c**2
- 54*_t*a*b**4*c + 9*_t*b**6 + 2*b*c)/(2*c**2))))

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Giac [B]  time = 1.1683, size = 551, normalized size = 2.25 \begin{align*} \frac{2}{9} \, \sqrt{3} \left (-\frac{c^{3}}{b^{15} - 15 \, a b^{13} c + 90 \, a^{2} b^{11} c^{2} - 270 \, a^{3} b^{9} c^{3} + 405 \, a^{4} b^{7} c^{4} - 243 \, a^{5} b^{5} c^{5}}\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 ) - \frac{1}{9} \, \left (-\frac{c^{3}}{b^{15} - 15 \, a b^{13} c + 90 \, a^{2} b^{11} c^{2} - 270 \, a^{3} b^{9} c^{3} + 405 \, a^{4} b^{7} c^{4} - 243 \, a^{5} b^{5} c^{5}}\right )^{\frac{1}{3}} \log \left ({\left (\sqrt{3} c x + \sqrt{3} b - \sqrt{3}{\left (-b^{3} + 3 \, a b c\right )}^{\frac{1}{3}}\right )}^{2} +{\left (c x + b +{\left (-b^{3} + 3 \, a b c\right )}^{\frac{1}{3}}\right )}^{2}\right ) + \frac{2}{9} \, \left (-\frac{c^{3}}{b^{15} - 15 \, a b^{13} c + 90 \, a^{2} b^{11} c^{2} - 270 \, a^{3} b^{9} c^{3} + 405 \, a^{4} b^{7} c^{4} - 243 \, a^{5} b^{5} c^{5}}\right )^{\frac{1}{3}} \log \left ({\left | 3 \, b^{4} - 9 \, a b^{2} c + 3 \,{\left (b^{3} c - 3 \, a b c^{2}\right )} x + 3 \,{\left (b^{3} - 3 \, a b c\right )}{\left (-b^{3} + 3 \, a b c\right )}^{\frac{1}{3}} \right |}\right ) - \frac{c x + b}{3 \,{\left (c^{2} x^{3} + 3 \, b c x^{2} + 3 \, b^{2} x + 3 \, a b\right )}{\left (b^{3} - 3 \, a b c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*sqrt(3)*(-c^3/(b^15 - 15*a*b^13*c + 90*a^2*b^11*c^2 - 270*a^3*b^9*c^3 + 405*a^4*b^7*c^4 - 243*a^5*b^5*c^5)
)^(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)))
- 1/9*(-c^3/(b^15 - 15*a*b^13*c + 90*a^2*b^11*c^2 - 270*a^3*b^9*c^3 + 405*a^4*b^7*c^4 - 243*a^5*b^5*c^5))^(1/3
)*log((sqrt(3)*c*x + sqrt(3)*b - sqrt(3)*(-b^3 + 3*a*b*c)^(1/3))^2 + (c*x + b + (-b^3 + 3*a*b*c)^(1/3))^2) + 2
/9*(-c^3/(b^15 - 15*a*b^13*c + 90*a^2*b^11*c^2 - 270*a^3*b^9*c^3 + 405*a^4*b^7*c^4 - 243*a^5*b^5*c^5))^(1/3)*l
og(abs(3*b^4 - 9*a*b^2*c + 3*(b^3*c - 3*a*b*c^2)*x + 3*(b^3 - 3*a*b*c)*(-b^3 + 3*a*b*c)^(1/3))) - 1/3*(c*x + b
)/((c^2*x^3 + 3*b*c*x^2 + 3*b^2*x + 3*a*b)*(b^3 - 3*a*b*c))