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]

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

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Rubi [A]  time = 0.31, antiderivative size = 188, normalized size of antiderivative = 1.00, 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 31

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

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 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 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 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]

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 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]

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*}

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Mathematica [C]  time = 0.02, size = 63, normalized size = 0.34 \[ \frac {1}{3} \text {RootSum}\left [\text {$\#$1}^3 c^2+3 \text {$\#$1}^2 b c+3 \text {$\#$1} b^2+3 a b\& ,\frac {\log (x-\text {$\#$1})}{\text {$\#$1}^2 c^2+2 \text {$\#$1} b c+b^2}\& \right ] \]

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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fricas [B]  time = 0.86, size = 387, normalized size = 2.06 \[ -\frac {2 \, \sqrt {3} {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {1}{6}} {\left (b^{3} - 3 \, a b c\right )} \arctan \left (\frac {2 \, \sqrt {3} {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {2}{3}} {\left (c x + b\right )} + \sqrt {3} {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {1}{3}} {\left (b^{3} - 3 \, a b c\right )}}{3 \, {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {5}{6}}}\right ) + {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {2}{3}} \log \left (-b^{5} + 3 \, a b^{3} c - {\left (b^{3} c^{2} - 3 \, a b c^{3}\right )} x^{2} - 2 \, {\left (b^{4} c - 3 \, a b^{2} c^{2}\right )} x - {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {2}{3}} {\left (c x + b\right )} - {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {1}{3}} {\left (b^{3} - 3 \, a b c\right )}\right ) - 2 \, {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {2}{3}} \log \left (-b^{4} + 3 \, a b^{2} c - {\left (b^{3} c - 3 \, a b c^{2}\right )} x + {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {2}{3}}\right )}{6 \, {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}} \]

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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giac [A]  time = 0.36, size = 212, normalized size = 1.13 \[ \frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} c x + \sqrt {3} b - \sqrt {3} {\left (-b^{3} + 3 \, a b c\right )}^{\frac {1}{3}}}{c x + b + {\left (-b^{3} + 3 \, a b c\right )}^{\frac {1}{3}}}\right )}{3 \, {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {1}{3}}} - \frac {\log \left (4 \, {\left (\sqrt {3} c x + \sqrt {3} b - \sqrt {3} {\left (-b^{3} + 3 \, a b c\right )}^{\frac {1}{3}}\right )}^{2} + 4 \, {\left (c x + b + {\left (-b^{3} + 3 \, a b c\right )}^{\frac {1}{3}}\right )}^{2}\right )}{6 \, {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {1}{3}}} + \frac {\log \left ({\left | c x + b + {\left (-b^{3} + 3 \, a b c\right )}^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{6} - 6 \, a b^{4} c + 9 \, a^{2} b^{2} c^{2}\right )}^{\frac {1}{3}}} \]

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(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)

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maple [C]  time = 0.00, size = 57, normalized size = 0.30 \[ \frac {\ln \left (-\RootOf \left (c^{2} \textit {\_Z}^{3}+3 b c \,\textit {\_Z}^{2}+3 b^{2} \textit {\_Z} +3 a b \right )+x \right )}{3 \RootOf \left (c^{2} \textit {\_Z}^{3}+3 b c \,\textit {\_Z}^{2}+3 b^{2} \textit {\_Z} +3 a b \right )^{2} c^{2}+6 \RootOf \left (c^{2} \textit {\_Z}^{3}+3 b c \,\textit {\_Z}^{2}+3 b^{2} \textit {\_Z} +3 a b \right ) b c +3 b^{2}} \]

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(-_R+x),_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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{c^{2} x^{3} + 3 \, b c x^{2} + 3 \, b^{2} x + 3 \, a b}\,{d x} \]

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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mupad [B]  time = 0.49, size = 174, normalized size = 0.93 \[ \frac {\ln \left (b+b^{1/3}\,{\left (3\,a\,c-b^2\right )}^{1/3}+c\,x\right )}{3\,b^{2/3}\,{\left (3\,a\,c-b^2\right )}^{2/3}}+\frac {\ln \left (3\,b\,c^3+3\,c^4\,x+\frac {3\,b^{1/3}\,c^3\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (3\,a\,c-b^2\right )}^{1/3}}{2}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,b^{2/3}\,{\left (3\,a\,c-b^2\right )}^{2/3}}-\frac {\ln \left (3\,b\,c^3+3\,c^4\,x-\frac {3\,b^{1/3}\,c^3\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (3\,a\,c-b^2\right )}^{1/3}}{2}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,b^{2/3}\,{\left (3\,a\,c-b^2\right )}^{2/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))

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sympy [A]  time = 0.41, size = 53, normalized size = 0.28 \[ \operatorname {RootSum} {\left (t^{3} \left (243 a^{2} b^{2} c^{2} - 162 a b^{4} c + 27 b^{6}\right ) - 1, \left (t \mapsto t \log {\left (x + \frac {9 t a b c - 3 t b^{3} + b}{c} \right )} \right )\right )} \]

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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