\(\int \frac {1}{(a+\frac {b}{x^3})^2 x^3} \, dx\) [1986]

Optimal result

Integrand size = 13, antiderivative size = 134 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^3} \, dx=-\frac {x}{3 a \left (b+a x^3\right )}-\frac {\arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{3 \sqrt {3} a^{4/3} b^{2/3}}+\frac {\log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{9 a^{4/3} b^{2/3}}-\frac {\log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{18 a^{4/3} b^{2/3}} \]

[Out]

-1/3*x/a/(a*x^3+b)+1/9*ln(b^(1/3)+a^(1/3)*x)/a^(4/3)/b^(2/3)-1/18*ln(b^(2/3)-a^(1/3)*b^(1/3)*x+a^(2/3)*x^2)/a^
(4/3)/b^(2/3)-1/9*arctan(1/3*(b^(1/3)-2*a^(1/3)*x)/b^(1/3)*3^(1/2))/a^(4/3)/b^(2/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {269, 294, 206, 31, 648, 631, 210, 642} \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^3} \, dx=-\frac {\arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{3 \sqrt {3} a^{4/3} b^{2/3}}-\frac {\log \left (a^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3}\right )}{18 a^{4/3} b^{2/3}}+\frac {\log \left (\sqrt [3]{a} x+\sqrt [3]{b}\right )}{9 a^{4/3} b^{2/3}}-\frac {x}{3 a \left (a x^3+b\right )} \]

[In]

Int[1/((a + b/x^3)^2*x^3),x]

[Out]

-1/3*x/(a*(b + a*x^3)) - ArcTan[(b^(1/3) - 2*a^(1/3)*x)/(Sqrt[3]*b^(1/3))]/(3*Sqrt[3]*a^(4/3)*b^(2/3)) + Log[b
^(1/3) + a^(1/3)*x]/(9*a^(4/3)*b^(2/3)) - Log[b^(2/3) - a^(1/3)*b^(1/3)*x + a^(2/3)*x^2]/(18*a^(4/3)*b^(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 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 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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}& = \int \frac {x^3}{\left (b+a x^3\right )^2} \, dx \\ & = -\frac {x}{3 a \left (b+a x^3\right )}+\frac {\int \frac {1}{b+a x^3} \, dx}{3 a} \\ & = -\frac {x}{3 a \left (b+a x^3\right )}+\frac {\int \frac {1}{\sqrt [3]{b}+\sqrt [3]{a} x} \, dx}{9 a b^{2/3}}+\frac {\int \frac {2 \sqrt [3]{b}-\sqrt [3]{a} x}{b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2} \, dx}{9 a b^{2/3}} \\ & = -\frac {x}{3 a \left (b+a x^3\right )}+\frac {\log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{9 a^{4/3} b^{2/3}}-\frac {\int \frac {-\sqrt [3]{a} \sqrt [3]{b}+2 a^{2/3} x}{b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2} \, dx}{18 a^{4/3} b^{2/3}}+\frac {\int \frac {1}{b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2} \, dx}{6 a \sqrt [3]{b}} \\ & = -\frac {x}{3 a \left (b+a x^3\right )}+\frac {\log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{9 a^{4/3} b^{2/3}}-\frac {\log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{18 a^{4/3} b^{2/3}}+\frac {\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{a} x}{\sqrt [3]{b}}\right )}{3 a^{4/3} b^{2/3}} \\ & = -\frac {x}{3 a \left (b+a x^3\right )}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{3 \sqrt {3} a^{4/3} b^{2/3}}+\frac {\log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{9 a^{4/3} b^{2/3}}-\frac {\log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{18 a^{4/3} b^{2/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^3} \, dx=\frac {-\frac {6 \sqrt [3]{a} x}{b+a x^3}-\frac {2 \sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{a} x}{\sqrt [3]{b}}}{\sqrt {3}}\right )}{b^{2/3}}+\frac {2 \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{b^{2/3}}-\frac {\log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{b^{2/3}}}{18 a^{4/3}} \]

[In]

Integrate[1/((a + b/x^3)^2*x^3),x]

[Out]

((-6*a^(1/3)*x)/(b + a*x^3) - (2*Sqrt[3]*ArcTan[(1 - (2*a^(1/3)*x)/b^(1/3))/Sqrt[3]])/b^(2/3) + (2*Log[b^(1/3)
+ a^(1/3)*x])/b^(2/3) - Log[b^(2/3) - a^(1/3)*b^(1/3)*x + a^(2/3)*x^2]/b^(2/3))/(18*a^(4/3))

Maple [C] (verified)

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

Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.32

method	result	size
risch	\(-\frac {x}{3 a \left (a \,x^{3}+b \right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{3}+b \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}}{9 a^{2}}\)	\(43\)
default	\(-\frac {x}{3 a \left (a \,x^{3}+b \right )}+\frac {\frac {\ln \left (x +\left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {b}{a}\right )^{\frac {1}{3}} x +\left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{6 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {b}{a}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}}{3 a}\)	\(112\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 391, normalized size of antiderivative = 2.92 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^3} \, dx=\left [-\frac {6 \, a b^{2} x - 3 \, \sqrt {\frac {1}{3}} {\left (a^{2} b x^{3} + a b^{2}\right )} \sqrt {-\frac {\left (a b^{2}\right )^{\frac {1}{3}}}{a}} \log \left (\frac {2 \, a b x^{3} - 3 \, \left (a b^{2}\right )^{\frac {1}{3}} b x - b^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, a b x^{2} + \left (a b^{2}\right )^{\frac {2}{3}} x - \left (a b^{2}\right )^{\frac {1}{3}} b\right )} \sqrt {-\frac {\left (a b^{2}\right )^{\frac {1}{3}}}{a}}}{a x^{3} + b}\right ) + {\left (a x^{3} + b\right )} \left (a b^{2}\right )^{\frac {2}{3}} \log \left (a b x^{2} - \left (a b^{2}\right )^{\frac {2}{3}} x + \left (a b^{2}\right )^{\frac {1}{3}} b\right ) - 2 \, {\left (a x^{3} + b\right )} \left (a b^{2}\right )^{\frac {2}{3}} \log \left (a b x + \left (a b^{2}\right )^{\frac {2}{3}}\right )}{18 \, {\left (a^{3} b^{2} x^{3} + a^{2} b^{3}\right )}}, -\frac {6 \, a b^{2} x - 6 \, \sqrt {\frac {1}{3}} {\left (a^{2} b x^{3} + a b^{2}\right )} \sqrt {\frac {\left (a b^{2}\right )^{\frac {1}{3}}}{a}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left (a b^{2}\right )^{\frac {2}{3}} x - \left (a b^{2}\right )^{\frac {1}{3}} b\right )} \sqrt {\frac {\left (a b^{2}\right )^{\frac {1}{3}}}{a}}}{b^{2}}\right ) + {\left (a x^{3} + b\right )} \left (a b^{2}\right )^{\frac {2}{3}} \log \left (a b x^{2} - \left (a b^{2}\right )^{\frac {2}{3}} x + \left (a b^{2}\right )^{\frac {1}{3}} b\right ) - 2 \, {\left (a x^{3} + b\right )} \left (a b^{2}\right )^{\frac {2}{3}} \log \left (a b x + \left (a b^{2}\right )^{\frac {2}{3}}\right )}{18 \, {\left (a^{3} b^{2} x^{3} + a^{2} b^{3}\right )}}\right ] \]

[In]

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

[Out]

[-1/18*(6*a*b^2*x - 3*sqrt(1/3)*(a^2*b*x^3 + a*b^2)*sqrt(-(a*b^2)^(1/3)/a)*log((2*a*b*x^3 - 3*(a*b^2)^(1/3)*b*
x - b^2 + 3*sqrt(1/3)*(2*a*b*x^2 + (a*b^2)^(2/3)*x - (a*b^2)^(1/3)*b)*sqrt(-(a*b^2)^(1/3)/a))/(a*x^3 + b)) + (
a*x^3 + b)*(a*b^2)^(2/3)*log(a*b*x^2 - (a*b^2)^(2/3)*x + (a*b^2)^(1/3)*b) - 2*(a*x^3 + b)*(a*b^2)^(2/3)*log(a*
b*x + (a*b^2)^(2/3)))/(a^3*b^2*x^3 + a^2*b^3), -1/18*(6*a*b^2*x - 6*sqrt(1/3)*(a^2*b*x^3 + a*b^2)*sqrt((a*b^2)
^(1/3)/a)*arctan(sqrt(1/3)*(2*(a*b^2)^(2/3)*x - (a*b^2)^(1/3)*b)*sqrt((a*b^2)^(1/3)/a)/b^2) + (a*x^3 + b)*(a*b
^2)^(2/3)*log(a*b*x^2 - (a*b^2)^(2/3)*x + (a*b^2)^(1/3)*b) - 2*(a*x^3 + b)*(a*b^2)^(2/3)*log(a*b*x + (a*b^2)^(
2/3)))/(a^3*b^2*x^3 + a^2*b^3)]

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.29 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^3} \, dx=- \frac {x}{3 a^{2} x^{3} + 3 a b} + \operatorname {RootSum} {\left (729 t^{3} a^{4} b^{2} - 1, \left ( t \mapsto t \log {\left (9 t a b + x \right )} \right )\right )} \]

[In]

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

[Out]

-x/(3*a**2*x**3 + 3*a*b) + RootSum(729*_t**3*a**4*b**2 - 1, Lambda(_t, _t*log(9*_t*a*b + x)))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^3} \, dx=-\frac {x}{3 \, {\left (a^{2} x^{3} + a b\right )}} + \frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{9 \, a^{2} \left (\frac {b}{a}\right )^{\frac {2}{3}}} - \frac {\log \left (x^{2} - x \left (\frac {b}{a}\right )^{\frac {1}{3}} + \left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{18 \, a^{2} \left (\frac {b}{a}\right )^{\frac {2}{3}}} + \frac {\log \left (x + \left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{9 \, a^{2} \left (\frac {b}{a}\right )^{\frac {2}{3}}} \]

[In]

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

[Out]

-1/3*x/(a^2*x^3 + a*b) + 1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - (b/a)^(1/3))/(b/a)^(1/3))/(a^2*(b/a)^(2/3)) - 1
/18*log(x^2 - x*(b/a)^(1/3) + (b/a)^(2/3))/(a^2*(b/a)^(2/3)) + 1/9*log(x + (b/a)^(1/3))/(a^2*(b/a)^(2/3))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.97 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^3} \, dx=-\frac {\left (-\frac {b}{a}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {b}{a}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a b} - \frac {x}{3 \, {\left (a x^{3} + b\right )} a} + \frac {\sqrt {3} \left (-a^{2} b\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {b}{a}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{9 \, a^{2} b} + \frac {\left (-a^{2} b\right )^{\frac {1}{3}} \log \left (x^{2} + x \left (-\frac {b}{a}\right )^{\frac {1}{3}} + \left (-\frac {b}{a}\right )^{\frac {2}{3}}\right )}{18 \, a^{2} b} \]

[In]

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

[Out]

-1/9*(-b/a)^(1/3)*log(abs(x - (-b/a)^(1/3)))/(a*b) - 1/3*x/((a*x^3 + b)*a) + 1/9*sqrt(3)*(-a^2*b)^(1/3)*arctan
(1/3*sqrt(3)*(2*x + (-b/a)^(1/3))/(-b/a)^(1/3))/(a^2*b) + 1/18*(-a^2*b)^(1/3)*log(x^2 + x*(-b/a)^(1/3) + (-b/a
)^(2/3))/(a^2*b)

Mupad [B] (verification not implemented)

Time = 5.99 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^3} \, dx=\frac {\ln \left (a^{1/3}\,x+b^{1/3}\right )}{9\,a^{4/3}\,b^{2/3}}-\frac {x}{3\,a\,\left (a\,x^3+b\right )}+\frac {\ln \left (a\,x+\frac {a^{2/3}\,b^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{18\,a^{4/3}\,b^{2/3}}-\frac {\ln \left (a\,x-\frac {a^{2/3}\,b^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{18\,a^{4/3}\,b^{2/3}} \]

[In]

int(1/(x^3*(a + b/x^3)^2),x)

[Out]

log(a^(1/3)*x + b^(1/3))/(9*a^(4/3)*b^(2/3)) - x/(3*a*(b + a*x^3)) + (log(a*x + (a^(2/3)*b^(1/3)*(3^(1/2)*1i -
1))/2)*(3^(1/2)*1i - 1))/(18*a^(4/3)*b^(2/3)) - (log(a*x - (a^(2/3)*b^(1/3)*(3^(1/2)*1i + 1))/2)*(3^(1/2)*1i
+ 1))/(18*a^(4/3)*b^(2/3))