\(\int \frac {1}{(a+\frac {b}{x^3})^2 x^2} \, dx\) [1985]

Optimal result

Integrand size = 13, antiderivative size = 136 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^2} \, dx=-\frac {x^2}{3 a \left (b+a x^3\right )}-\frac {2 \arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{3 \sqrt {3} a^{5/3} \sqrt [3]{b}}-\frac {2 \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{9 a^{5/3} \sqrt [3]{b}}+\frac {\log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{9 a^{5/3} \sqrt [3]{b}} \]

[Out]

-1/3*x^2/a/(a*x^3+b)-2/9*ln(b^(1/3)+a^(1/3)*x)/a^(5/3)/b^(1/3)+1/9*ln(b^(2/3)-a^(1/3)*b^(1/3)*x+a^(2/3)*x^2)/a
^(5/3)/b^(1/3)-2/9*arctan(1/3*(b^(1/3)-2*a^(1/3)*x)/b^(1/3)*3^(1/2))/a^(5/3)/b^(1/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 136, 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, 298, 31, 648, 631, 210, 642} \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^2} \, dx=-\frac {2 \arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{3 \sqrt {3} a^{5/3} \sqrt [3]{b}}+\frac {\log \left (a^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3}\right )}{9 a^{5/3} \sqrt [3]{b}}-\frac {2 \log \left (\sqrt [3]{a} x+\sqrt [3]{b}\right )}{9 a^{5/3} \sqrt [3]{b}}-\frac {x^2}{3 a \left (a x^3+b\right )} \]

[In]

Int[1/((a + b/x^3)^2*x^2),x]

[Out]

-1/3*x^2/(a*(b + a*x^3)) - (2*ArcTan[(b^(1/3) - 2*a^(1/3)*x)/(Sqrt[3]*b^(1/3))])/(3*Sqrt[3]*a^(5/3)*b^(1/3)) -
(2*Log[b^(1/3) + a^(1/3)*x])/(9*a^(5/3)*b^(1/3)) + Log[b^(2/3) - a^(1/3)*b^(1/3)*x + a^(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 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 298

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Dist[-(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(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 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^4}{\left (b+a x^3\right )^2} \, dx \\ & = -\frac {x^2}{3 a \left (b+a x^3\right )}+\frac {2 \int \frac {x}{b+a x^3} \, dx}{3 a} \\ & = -\frac {x^2}{3 a \left (b+a x^3\right )}-\frac {2 \int \frac {1}{\sqrt [3]{b}+\sqrt [3]{a} x} \, dx}{9 a^{4/3} \sqrt [3]{b}}+\frac {2 \int \frac {\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^{4/3} \sqrt [3]{b}} \\ & = -\frac {x^2}{3 a \left (b+a x^3\right )}-\frac {2 \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{9 a^{5/3} \sqrt [3]{b}}+\frac {\int \frac {1}{b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2} \, dx}{3 a^{4/3}}+\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}{9 a^{5/3} \sqrt [3]{b}} \\ & = -\frac {x^2}{3 a \left (b+a x^3\right )}-\frac {2 \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{9 a^{5/3} \sqrt [3]{b}}+\frac {\log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{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]{a} x}{\sqrt [3]{b}}\right )}{3 a^{5/3} \sqrt [3]{b}} \\ & = -\frac {x^2}{3 a \left (b+a x^3\right )}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{3 \sqrt {3} a^{5/3} \sqrt [3]{b}}-\frac {2 \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{9 a^{5/3} \sqrt [3]{b}}+\frac {\log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{9 a^{5/3} \sqrt [3]{b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^2} \, dx=\frac {-\frac {3 a^{2/3} x^2}{b+a x^3}-\frac {2 \sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{a} x}{\sqrt [3]{b}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}-\frac {2 \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{\sqrt [3]{b}}+\frac {\log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{\sqrt [3]{b}}}{9 a^{5/3}} \]

[In]

Integrate[1/((a + b/x^3)^2*x^2),x]

[Out]

((-3*a^(2/3)*x^2)/(b + a*x^3) - (2*Sqrt[3]*ArcTan[(1 - (2*a^(1/3)*x)/b^(1/3))/Sqrt[3]])/b^(1/3) - (2*Log[b^(1/
3) + a^(1/3)*x])/b^(1/3) + Log[b^(2/3) - a^(1/3)*b^(1/3)*x + a^(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 = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.33

method	result	size
risch	\(-\frac {x^{2}}{3 a \left (a \,x^{3}+b \right )}+\frac {2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{3}+b \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{9 a^{2}}\)	\(45\)
default	\(-\frac {x^{2}}{3 a \left (a \,x^{3}+b \right )}+\frac {-\frac {2 \ln \left (x +\left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{9 a \left (\frac {b}{a}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {b}{a}\right )^{\frac {1}{3}} x +\left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{9 a \left (\frac {b}{a}\right )^{\frac {1}{3}}}+\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {b}{a}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 a \left (\frac {b}{a}\right )^{\frac {1}{3}}}}{a}\)	\(114\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.94 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^2} \, dx=\left [-\frac {3 \, a^{2} b x^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (a^{2} b x^{3} + a b^{2}\right )} \sqrt {\frac {\left (-a^{2} b\right )^{\frac {1}{3}}}{b}} \log \left (\frac {2 \, a^{2} x^{3} - a b + 3 \, \sqrt {\frac {1}{3}} {\left (a b x + 2 \, \left (-a^{2} b\right )^{\frac {2}{3}} x^{2} + \left (-a^{2} b\right )^{\frac {1}{3}} b\right )} \sqrt {\frac {\left (-a^{2} b\right )^{\frac {1}{3}}}{b}} - 3 \, \left (-a^{2} b\right )^{\frac {2}{3}} x}{a x^{3} + b}\right ) - {\left (a x^{3} + b\right )} \left (-a^{2} b\right )^{\frac {2}{3}} \log \left (a^{2} x^{2} + \left (-a^{2} b\right )^{\frac {1}{3}} a x + \left (-a^{2} b\right )^{\frac {2}{3}}\right ) + 2 \, {\left (a x^{3} + b\right )} \left (-a^{2} b\right )^{\frac {2}{3}} \log \left (a x - \left (-a^{2} b\right )^{\frac {1}{3}}\right )}{9 \, {\left (a^{4} b x^{3} + a^{3} b^{2}\right )}}, -\frac {3 \, a^{2} b x^{2} - 6 \, \sqrt {\frac {1}{3}} {\left (a^{2} b x^{3} + a b^{2}\right )} \sqrt {-\frac {\left (-a^{2} b\right )^{\frac {1}{3}}}{b}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, a x + \left (-a^{2} b\right )^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-a^{2} b\right )^{\frac {1}{3}}}{b}}}{a}\right ) - {\left (a x^{3} + b\right )} \left (-a^{2} b\right )^{\frac {2}{3}} \log \left (a^{2} x^{2} + \left (-a^{2} b\right )^{\frac {1}{3}} a x + \left (-a^{2} b\right )^{\frac {2}{3}}\right ) + 2 \, {\left (a x^{3} + b\right )} \left (-a^{2} b\right )^{\frac {2}{3}} \log \left (a x - \left (-a^{2} b\right )^{\frac {1}{3}}\right )}{9 \, {\left (a^{4} b x^{3} + a^{3} b^{2}\right )}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.32 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^2} \, dx=- \frac {x^{2}}{3 a^{2} x^{3} + 3 a b} + \operatorname {RootSum} {\left (729 t^{3} a^{5} b + 8, \left ( t \mapsto t \log {\left (\frac {81 t^{2} a^{3} b}{4} + x \right )} \right )\right )} \]

[In]

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

[Out]

-x**2/(3*a**2*x**3 + 3*a*b) + RootSum(729*_t**3*a**5*b + 8, Lambda(_t, _t*log(81*_t**2*a**3*b/4 + x)))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^2} \, dx=-\frac {x^{2}}{3 \, {\left (a^{2} x^{3} + a b\right )}} + \frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{9 \, a^{2} \left (\frac {b}{a}\right )^{\frac {1}{3}}} + \frac {\log \left (x^{2} - x \left (\frac {b}{a}\right )^{\frac {1}{3}} + \left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{9 \, a^{2} \left (\frac {b}{a}\right )^{\frac {1}{3}}} - \frac {2 \, \log \left (x + \left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{9 \, a^{2} \left (\frac {b}{a}\right )^{\frac {1}{3}}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.97 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^2} \, dx=-\frac {x^{2}}{3 \, {\left (a x^{3} + b\right )} a} - \frac {2 \, \left (-\frac {b}{a}\right )^{\frac {2}{3}} \log \left ({\left | x - \left (-\frac {b}{a}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a b} - \frac {2 \, \sqrt {3} \left (-a^{2} b\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {b}{a}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{9 \, a^{3} b} + \frac {\left (-a^{2} b\right )^{\frac {2}{3}} \log \left (x^{2} + x \left (-\frac {b}{a}\right )^{\frac {1}{3}} + \left (-\frac {b}{a}\right )^{\frac {2}{3}}\right )}{9 \, a^{3} b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.10 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.01 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right )^2 x^2} \, dx=\frac {2\,\ln \left (\frac {4\,b^{1/3}}{9\,{\left (-a\right )}^{4/3}}+\frac {4\,x}{9\,a}\right )}{9\,{\left (-a\right )}^{5/3}\,b^{1/3}}-\frac {x^2}{3\,a\,\left (a\,x^3+b\right )}+\frac {\ln \left (\frac {4\,x}{9\,a}+\frac {b^{1/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{9\,{\left (-a\right )}^{4/3}}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{9\,{\left (-a\right )}^{5/3}\,b^{1/3}}-\frac {\ln \left (\frac {4\,x}{9\,a}+\frac {b^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{9\,{\left (-a\right )}^{4/3}}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{9\,{\left (-a\right )}^{5/3}\,b^{1/3}} \]

[In]

int(1/(x^2*(a + b/x^3)^2),x)

[Out]

(2*log((4*b^(1/3))/(9*(-a)^(4/3)) + (4*x)/(9*a)))/(9*(-a)^(5/3)*b^(1/3)) - x^2/(3*a*(b + a*x^3)) + (log((4*x)/
(9*a) + (b^(1/3)*(3^(1/2)*1i - 1)^2)/(9*(-a)^(4/3)))*(3^(1/2)*1i - 1))/(9*(-a)^(5/3)*b^(1/3)) - (log((4*x)/(9*
a) + (b^(1/3)*(3^(1/2)*1i + 1)^2)/(9*(-a)^(4/3)))*(3^(1/2)*1i + 1))/(9*(-a)^(5/3)*b^(1/3))