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

Optimal. Leaf size=188 \[ -\frac{\log \left (\sqrt [3]{b} c \sqrt [3]{b^2-3 a c} \left (\frac{b}{c}+x\right )+b^{2/3} \left (b^2-3 a c\right )^{2/3}+c^2 \left (\frac{b}{c}+x\right )^2\right )}{6 b^{2/3} \left (b^2-3 a c\right )^{2/3}}+\frac{\log \left (-\sqrt [3]{b} \sqrt [3]{b^2-3 a c}+b+c x\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}-\frac{\tan ^{-1}\left (\frac{\frac{2 (b+c x)}{\sqrt [3]{b^2-3 a c}}+\sqrt [3]{b}}{\sqrt{3} \sqrt [3]{b}}\right )}{\sqrt{3} b^{2/3} \left (b^2-3 a c\right )^{2/3}} \]

[Out]

-(ArcTan[(b^(1/3) + (2*(b + c*x))/(b^2 - 3*a*c)^(1/3))/(Sqrt[3]*b^(1/3))]/(Sqrt[3]*b^(2/3)*(b^2 - 3*a*c)^(2/3)
)) + Log[b - b^(1/3)*(b^2 - 3*a*c)^(1/3) + c*x]/(3*b^(2/3)*(b^2 - 3*a*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]/(6*b^(2/3)*(b^2 - 3*a*c)^(2/3))

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Rubi [A]  time = 0.311988, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {2067, 200, 31, 634, 617, 204, 628} \[ -\frac{\log \left (\sqrt [3]{b} c \sqrt [3]{b^2-3 a c} \left (\frac{b}{c}+x\right )+b^{2/3} \left (b^2-3 a c\right )^{2/3}+c^2 \left (\frac{b}{c}+x\right )^2\right )}{6 b^{2/3} \left (b^2-3 a c\right )^{2/3}}+\frac{\log \left (-\sqrt [3]{b} \sqrt [3]{b^2-3 a c}+b+c x\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}-\frac{\tan ^{-1}\left (\frac{\frac{2 (b+c x)}{\sqrt [3]{b^2-3 a c}}+\sqrt [3]{b}}{\sqrt{3} \sqrt [3]{b}}\right )}{\sqrt{3} b^{2/3} \left (b^2-3 a c\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[(3*a*b + 3*b^2*x + 3*b*c*x^2 + c^2*x^3)^(-1),x]

[Out]

-(ArcTan[(b^(1/3) + (2*(b + c*x))/(b^2 - 3*a*c)^(1/3))/(Sqrt[3]*b^(1/3))]/(Sqrt[3]*b^(2/3)*(b^2 - 3*a*c)^(2/3)
)) + Log[b - b^(1/3)*(b^2 - 3*a*c)^(1/3) + c*x]/(3*b^(2/3)*(b^2 - 3*a*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]/(6*b^(2/3)*(b^2 - 3*a*c)^(2/3))

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 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}{3 a b+3 b^2 x+3 b c x^2+c^2 x^3} \, dx &=\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 )\\ &=\frac{c^{2/3} \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 )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}+\frac{c^{2/3} \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 )}{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{\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 )}{6 b^{2/3} \left (b^2-3 a c\right )^{2/3}}-\frac{\sqrt [3]{c} \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 )}{2 \sqrt [3]{b} \sqrt [3]{b^2-3 a c}}\\ &=\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}}+\frac{\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 )}{b^{2/3} \left (b^2-3 a c\right )^{2/3}}\\ &=-\frac{\tan ^{-1}\left (\frac{1+\frac{2 (b+c x)}{\sqrt [3]{b} \sqrt [3]{b^2-3 a c}}}{\sqrt{3}}\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}}\\ \end{align*}

Mathematica [C]  time = 0.016193, size = 63, normalized size = 0.34 \[ \frac{1}{3} \text{RootSum}\left [3 \text{$\#$1}^2 b c+\text{$\#$1}^3 c^2+3 \text{$\#$1} b^2+3 a b\& ,\frac{\log (x-\text{$\#$1})}{\text{$\#$1}^2 c^2+2 \text{$\#$1} b c+b^2}\& \right ] \]

Antiderivative was successfully verified.

[In]

Integrate[(3*a*b + 3*b^2*x + 3*b*c*x^2 + c^2*x^3)^(-1),x]

[Out]

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

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Maple [C]  time = 0.003, size = 57, normalized size = 0.3 \begin{align*}{\frac{1}{3}\sum _{{\it \_R}={\it RootOf} \left ({c}^{2}{{\it \_Z}}^{3}+3\,bc{{\it \_Z}}^{2}+3\,{b}^{2}{\it \_Z}+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),x)

[Out]

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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{c^{2} x^{3} + 3 \, b c x^{2} + 3 \, b^{2} x + 3 \, a b}\,{d x} \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),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.41423, size = 878, normalized size = 4.67 \begin{align*} -\frac{2 \, \sqrt{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{2 \, \sqrt{3}{\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac{2}{3}}{\left (c x + b\right )} + \sqrt{3}{\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 )}}{3 \,{\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 )}} \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),x, algorithm="fricas")

[Out]

-1/6*(2*sqrt(3)*(b^6 - 6*a*b^4*c + 9*a^2*b^2*c^2)^(1/6)*(b^3 - 3*a*b*c)*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)*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)

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Sympy [A]  time = 0.375449, size = 53, normalized size = 0.28 \begin{align*} \operatorname{RootSum}{\left (t^{3} \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 )} \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),x)

[Out]

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)))

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Giac [A]  time = 1.11204, size = 281, normalized size = 1.49 \begin{align*} \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 ({\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 )}{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}}} \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),x, algorithm="giac")

[Out]

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((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)/(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)