\(\int \frac {1}{(a+b x^3)^2} \, dx\) [67]

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

Optimal result

Integrand size = 9, antiderivative size = 112 \[ \int \frac {1}{\left (a+b x^3\right )^2} \, dx=\frac {x}{3 a \left (a+b x^3\right )}+\frac {2 \sqrt [3]{\frac {a}{b}} \left (\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{2 \sqrt [3]{\frac {a}{b}}-x}\right )+\frac {1}{2} \log \left (\frac {\left (\sqrt [3]{\frac {a}{b}}+x\right )^2}{\left (\frac {a}{b}\right )^{2/3}-\sqrt [3]{\frac {a}{b}} x+x^2}\right )\right )}{9 a^2} \]

[Out]

1/3*x/a/(b*x^3+a)+2/9/a^2*(a/b)^(1/3)*(1/2*ln((x+(a/b)^(1/3))^2/(x^2-(a/b)^(1/3)*x+(a/b)^(2/3)))+3^(1/2)*arcta
n(3^(1/2)*x/(2*(a/b)^(1/3)-x)))

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.20, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.778, Rules used = {205, 206, 31, 648, 631, 210, 642} \[ \int \frac {1}{\left (a+b x^3\right )^2} \, dx=-\frac {2 \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} \sqrt [3]{b}}-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{9 a^{5/3} \sqrt [3]{b}}+\frac {2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{5/3} \sqrt [3]{b}}+\frac {x}{3 a \left (a+b x^3\right )} \]

[In]

Int[(a + b*x^3)^(-2),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.05 \[ \int \frac {1}{\left (a+b x^3\right )^2} \, dx=\frac {\frac {3 a^{2/3} x}{a+b x^3}-\frac {2 \sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}+\frac {2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b}}-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{\sqrt [3]{b}}}{9 a^{5/3}} \]

[In]

Integrate[(a + b*x^3)^(-2),x]

[Out]

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

Maple [C] (verified)

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

Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.41

method result size
risch \(\frac {x}{3 a \left (b \,x^{3}+a \right )}+\frac {2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}\right )}{9 a b}\) \(46\)
default \(\frac {x}{3 a \left (b \,x^{3}+a \right )}+\frac {\frac {2 \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}}{a}\) \(112\)

[In]

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

[Out]

1/3*x/a/(b*x^3+a)+2/9/a/b*sum(1/_R^2*ln(x-_R),_R=RootOf(_Z^3*b+a))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 389, normalized size of antiderivative = 3.47 \[ \int \frac {1}{\left (a+b x^3\right )^2} \, dx=\left [\frac {3 \, a^{2} b x + 3 \, \sqrt {\frac {1}{3}} {\left (a b^{2} x^{3} + a^{2} b\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \log \left (\frac {2 \, a b x^{3} - 3 \, \left (a^{2} b\right )^{\frac {1}{3}} a x - a^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, a b x^{2} + \left (a^{2} b\right )^{\frac {2}{3}} x - \left (a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}}}{b x^{3} + a}\right ) - {\left (b x^{3} + a\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{2} - \left (a^{2} b\right )^{\frac {2}{3}} x + \left (a^{2} b\right )^{\frac {1}{3}} a\right ) + 2 \, {\left (b x^{3} + a\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x + \left (a^{2} b\right )^{\frac {2}{3}}\right )}{9 \, {\left (a^{3} b^{2} x^{3} + a^{4} b\right )}}, \frac {3 \, a^{2} b x + 6 \, \sqrt {\frac {1}{3}} {\left (a b^{2} x^{3} + a^{2} b\right )} \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left (a^{2} b\right )^{\frac {2}{3}} x - \left (a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}}}{a^{2}}\right ) - {\left (b x^{3} + a\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{2} - \left (a^{2} b\right )^{\frac {2}{3}} x + \left (a^{2} b\right )^{\frac {1}{3}} a\right ) + 2 \, {\left (b x^{3} + a\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x + \left (a^{2} b\right )^{\frac {2}{3}}\right )}{9 \, {\left (a^{3} b^{2} x^{3} + a^{4} b\right )}}\right ] \]

[In]

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

[Out]

[1/9*(3*a^2*b*x + 3*sqrt(1/3)*(a*b^2*x^3 + a^2*b)*sqrt(-(a^2*b)^(1/3)/b)*log((2*a*b*x^3 - 3*(a^2*b)^(1/3)*a*x
- a^2 + 3*sqrt(1/3)*(2*a*b*x^2 + (a^2*b)^(2/3)*x - (a^2*b)^(1/3)*a)*sqrt(-(a^2*b)^(1/3)/b))/(b*x^3 + a)) - (b*
x^3 + a)*(a^2*b)^(2/3)*log(a*b*x^2 - (a^2*b)^(2/3)*x + (a^2*b)^(1/3)*a) + 2*(b*x^3 + a)*(a^2*b)^(2/3)*log(a*b*
x + (a^2*b)^(2/3)))/(a^3*b^2*x^3 + a^4*b), 1/9*(3*a^2*b*x + 6*sqrt(1/3)*(a*b^2*x^3 + a^2*b)*sqrt((a^2*b)^(1/3)
/b)*arctan(sqrt(1/3)*(2*(a^2*b)^(2/3)*x - (a^2*b)^(1/3)*a)*sqrt((a^2*b)^(1/3)/b)/a^2) - (b*x^3 + a)*(a^2*b)^(2
/3)*log(a*b*x^2 - (a^2*b)^(2/3)*x + (a^2*b)^(1/3)*a) + 2*(b*x^3 + a)*(a^2*b)^(2/3)*log(a*b*x + (a^2*b)^(2/3)))
/(a^3*b^2*x^3 + a^4*b)]

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.35 \[ \int \frac {1}{\left (a+b x^3\right )^2} \, dx=\frac {x}{3 a^{2} + 3 a b x^{3}} + \operatorname {RootSum} {\left (729 t^{3} a^{5} b - 8, \left ( t \mapsto t \log {\left (\frac {9 t a^{2}}{2} + x \right )} \right )\right )} \]

[In]

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

[Out]

x/(3*a**2 + 3*a*b*x**3) + RootSum(729*_t**3*a**5*b - 8, Lambda(_t, _t*log(9*_t*a**2/2 + x)))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.09 \[ \int \frac {1}{\left (a+b x^3\right )^2} \, dx=\frac {x}{3 \, {\left (a b x^{3} + a^{2}\right )}} + \frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a b \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {\log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 \, a b \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {2 \, \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, a b \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]

[In]

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

[Out]

1/3*x/(a*b*x^3 + a^2) + 2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(a*b*(a/b)^(2/3)) - 1/
9*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a*b*(a/b)^(2/3)) + 2/9*log(x + (a/b)^(1/3))/(a*b*(a/b)^(2/3))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.13 \[ \int \frac {1}{\left (a+b x^3\right )^2} \, dx=-\frac {2 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{2}} + \frac {x}{3 \, {\left (b x^{3} + a\right )} a} + \frac {2 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{2} b} + \frac {\left (-a b^{2}\right )^{\frac {1}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 \, a^{2} b} \]

[In]

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

[Out]

-2/9*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/a^2 + 1/3*x/((b*x^3 + a)*a) + 2/9*sqrt(3)*(-a*b^2)^(1/3)*arctan(1
/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/(a^2*b) + 1/9*(-a*b^2)^(1/3)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(
2/3))/(a^2*b)

Mupad [B] (verification not implemented)

Time = 16.11 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.14 \[ \int \frac {1}{\left (a+b x^3\right )^2} \, dx=\frac {x}{3\,a\,\left (b\,x^3+a\right )}+\frac {2\,\ln \left (\frac {2\,b^{5/3}}{a^{2/3}}+\frac {2\,b^2\,x}{a}\right )}{9\,a^{5/3}\,b^{1/3}}+\frac {\ln \left (\frac {2\,b^2\,x}{a}+\frac {b^{5/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{a^{2/3}}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{9\,a^{5/3}\,b^{1/3}}-\frac {\ln \left (\frac {2\,b^2\,x}{a}-\frac {b^{5/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{a^{2/3}}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{9\,a^{5/3}\,b^{1/3}} \]

[In]

int(1/(a + b*x^3)^2,x)

[Out]

x/(3*a*(a + b*x^3)) + (2*log((2*b^(5/3))/a^(2/3) + (2*b^2*x)/a))/(9*a^(5/3)*b^(1/3)) + (log((2*b^2*x)/a + (b^(
5/3)*(3^(1/2)*1i - 1))/a^(2/3))*(3^(1/2)*1i - 1))/(9*a^(5/3)*b^(1/3)) - (log((2*b^2*x)/a - (b^(5/3)*(3^(1/2)*1
i + 1))/a^(2/3))*(3^(1/2)*1i + 1))/(9*a^(5/3)*b^(1/3))