\(\int \frac {1}{x^3 (a+b x^6)^2} \, dx\) [1342]

Optimal result

Integrand size = 13, antiderivative size = 152 \[ \int \frac {1}{x^3 \left (a+b x^6\right )^2} \, dx=-\frac {2}{3 a^2 x^2}+\frac {1}{6 a x^2 \left (a+b x^6\right )}+\frac {2 \sqrt [3]{b} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x^2}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{7/3}}+\frac {2 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x^2\right )}{9 a^{7/3}}-\frac {\sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x^2+b^{2/3} x^4\right )}{9 a^{7/3}} \]

[Out]

-2/3/a^2/x^2+1/6/a/x^2/(b*x^6+a)+2/9*b^(1/3)*ln(a^(1/3)+b^(1/3)*x^2)/a^(7/3)-1/9*b^(1/3)*ln(a^(2/3)-a^(1/3)*b^
(1/3)*x^2+b^(2/3)*x^4)/a^(7/3)+2/9*b^(1/3)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x^2)/a^(1/3)*3^(1/2))/a^(7/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {281, 296, 331, 298, 31, 648, 631, 210, 642} \[ \int \frac {1}{x^3 \left (a+b x^6\right )^2} \, dx=\frac {2 \sqrt [3]{b} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x^2}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{7/3}}-\frac {\sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x^2+b^{2/3} x^4\right )}{9 a^{7/3}}+\frac {2 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x^2\right )}{9 a^{7/3}}-\frac {2}{3 a^2 x^2}+\frac {1}{6 a x^2 \left (a+b x^6\right )} \]

[In]

Int[1/(x^3*(a + b*x^6)^2),x]

[Out]

-2/(3*a^2*x^2) + 1/(6*a*x^2*(a + b*x^6)) + (2*b^(1/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*x^2)/(Sqrt[3]*a^(1/3))])/(3*
Sqrt[3]*a^(7/3)) + (2*b^(1/3)*Log[a^(1/3) + b^(1/3)*x^2])/(9*a^(7/3)) - (b^(1/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)
*x^2 + b^(2/3)*x^4])/(9*a^(7/3))

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 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && 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 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x^2 \left (a+b x^3\right )^2} \, dx,x,x^2\right ) \\ & = \frac {1}{6 a x^2 \left (a+b x^6\right )}+\frac {2 \text {Subst}\left (\int \frac {1}{x^2 \left (a+b x^3\right )} \, dx,x,x^2\right )}{3 a} \\ & = -\frac {2}{3 a^2 x^2}+\frac {1}{6 a x^2 \left (a+b x^6\right )}-\frac {(2 b) \text {Subst}\left (\int \frac {x}{a+b x^3} \, dx,x,x^2\right )}{3 a^2} \\ & = -\frac {2}{3 a^2 x^2}+\frac {1}{6 a x^2 \left (a+b x^6\right )}+\frac {\left (2 b^{2/3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,x^2\right )}{9 a^{7/3}}-\frac {\left (2 b^{2/3}\right ) \text {Subst}\left (\int \frac {\sqrt [3]{a}+\sqrt [3]{b} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,x^2\right )}{9 a^{7/3}} \\ & = -\frac {2}{3 a^2 x^2}+\frac {1}{6 a x^2 \left (a+b x^6\right )}+\frac {2 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x^2\right )}{9 a^{7/3}}-\frac {\sqrt [3]{b} \text {Subst}\left (\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,x,x^2\right )}{9 a^{7/3}}-\frac {b^{2/3} \text {Subst}\left (\int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,x^2\right )}{3 a^2} \\ & = -\frac {2}{3 a^2 x^2}+\frac {1}{6 a x^2 \left (a+b x^6\right )}+\frac {2 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x^2\right )}{9 a^{7/3}}-\frac {\sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x^2+b^{2/3} x^4\right )}{9 a^{7/3}}-\frac {\left (2 \sqrt [3]{b}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x^2}{\sqrt [3]{a}}\right )}{3 a^{7/3}} \\ & = -\frac {2}{3 a^2 x^2}+\frac {1}{6 a x^2 \left (a+b x^6\right )}+\frac {2 \sqrt [3]{b} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x^2}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{7/3}}+\frac {2 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x^2\right )}{9 a^{7/3}}-\frac {\sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x^2+b^{2/3} x^4\right )}{9 a^{7/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.37 \[ \int \frac {1}{x^3 \left (a+b x^6\right )^2} \, dx=\frac {-\frac {9 \sqrt [3]{a}}{x^2}-\frac {3 \sqrt [3]{a} b x^4}{a+b x^6}+4 \sqrt {3} \sqrt [3]{b} \arctan \left (\sqrt {3}-\frac {2 \sqrt [6]{b} x}{\sqrt [6]{a}}\right )+4 \sqrt {3} \sqrt [3]{b} \arctan \left (\sqrt {3}+\frac {2 \sqrt [6]{b} x}{\sqrt [6]{a}}\right )+4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x^2\right )-2 \sqrt [3]{b} \log \left (\sqrt [3]{a}-\sqrt {3} \sqrt [6]{a} \sqrt [6]{b} x+\sqrt [3]{b} x^2\right )-2 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt {3} \sqrt [6]{a} \sqrt [6]{b} x+\sqrt [3]{b} x^2\right )}{18 a^{7/3}} \]

[In]

Integrate[1/(x^3*(a + b*x^6)^2),x]

[Out]

((-9*a^(1/3))/x^2 - (3*a^(1/3)*b*x^4)/(a + b*x^6) + 4*Sqrt[3]*b^(1/3)*ArcTan[Sqrt[3] - (2*b^(1/6)*x)/a^(1/6)]
+ 4*Sqrt[3]*b^(1/3)*ArcTan[Sqrt[3] + (2*b^(1/6)*x)/a^(1/6)] + 4*b^(1/3)*Log[a^(1/3) + b^(1/3)*x^2] - 2*b^(1/3)
*Log[a^(1/3) - Sqrt[3]*a^(1/6)*b^(1/6)*x + b^(1/3)*x^2] - 2*b^(1/3)*Log[a^(1/3) + Sqrt[3]*a^(1/6)*b^(1/6)*x +
b^(1/3)*x^2])/(18*a^(7/3))

Maple [C] (verified)

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

Time = 4.41 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.49

method	result	size
risch	\(\frac {-\frac {2 b \,x^{6}}{3 a^{2}}-\frac {1}{2 a}}{x^{2} \left (b \,x^{6}+a \right )}+\frac {2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{7} \textit {\_Z}^{3}-b \right )}{\sum }\textit {\_R} \ln \left (\left (-7 a^{7} \textit {\_R}^{3}+6 b \right ) x^{2}-a^{5} \textit {\_R}^{2}\right )\right )}{9}\)	\(75\)
default	\(-\frac {b \left (\frac {x^{4}}{3 b \,x^{6}+3 a}-\frac {4 \ln \left (x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {2 \ln \left (x^{4}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x^{2}+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {4 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x^{2}}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{2 a^{2}}-\frac {1}{2 a^{2} x^{2}}\)	\(126\)

[In]

int(1/x^3/(b*x^6+a)^2,x,method=_RETURNVERBOSE)

[Out]

(-2/3*b/a^2*x^6-1/2/a)/x^2/(b*x^6+a)+2/9*sum(_R*ln((-7*_R^3*a^7+6*b)*x^2-a^5*_R^2),_R=RootOf(_Z^3*a^7-b))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x^3 \left (a+b x^6\right )^2} \, dx=-\frac {12 \, b x^{6} + 4 \, \sqrt {3} {\left (b x^{8} + a x^{2}\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \arctan \left (\frac {2}{3} \, \sqrt {3} x^{2} \left (\frac {b}{a}\right )^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) + 2 \, {\left (b x^{8} + a x^{2}\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b x^{4} - a x^{2} \left (\frac {b}{a}\right )^{\frac {2}{3}} + a \left (\frac {b}{a}\right )^{\frac {1}{3}}\right ) - 4 \, {\left (b x^{8} + a x^{2}\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b x^{2} + a \left (\frac {b}{a}\right )^{\frac {2}{3}}\right ) + 9 \, a}{18 \, {\left (a^{2} b x^{8} + a^{3} x^{2}\right )}} \]

[In]

integrate(1/x^3/(b*x^6+a)^2,x, algorithm="fricas")

[Out]

-1/18*(12*b*x^6 + 4*sqrt(3)*(b*x^8 + a*x^2)*(b/a)^(1/3)*arctan(2/3*sqrt(3)*x^2*(b/a)^(1/3) - 1/3*sqrt(3)) + 2*
(b*x^8 + a*x^2)*(b/a)^(1/3)*log(b*x^4 - a*x^2*(b/a)^(2/3) + a*(b/a)^(1/3)) - 4*(b*x^8 + a*x^2)*(b/a)^(1/3)*log
(b*x^2 + a*(b/a)^(2/3)) + 9*a)/(a^2*b*x^8 + a^3*x^2)

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.39 \[ \int \frac {1}{x^3 \left (a+b x^6\right )^2} \, dx=\frac {- 3 a - 4 b x^{6}}{6 a^{3} x^{2} + 6 a^{2} b x^{8}} + \operatorname {RootSum} {\left (729 t^{3} a^{7} - 8 b, \left ( t \mapsto t \log {\left (\frac {81 t^{2} a^{5}}{4 b} + x^{2} \right )} \right )\right )} \]

[In]

integrate(1/x**3/(b*x**6+a)**2,x)

[Out]

(-3*a - 4*b*x**6)/(6*a**3*x**2 + 6*a**2*b*x**8) + RootSum(729*_t**3*a**7 - 8*b, Lambda(_t, _t*log(81*_t**2*a**
5/(4*b) + x**2)))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^3 \left (a+b x^6\right )^2} \, dx=-\frac {4 \, b x^{6} + 3 \, a}{6 \, {\left (a^{2} b x^{8} + a^{3} x^{2}\right )}} - \frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, x^{2} - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {\log \left (x^{4} - x^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 \, a^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {2 \, \log \left (x^{2} + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, a^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} \]

[In]

integrate(1/x^3/(b*x^6+a)^2,x, algorithm="maxima")

[Out]

-1/6*(4*b*x^6 + 3*a)/(a^2*b*x^8 + a^3*x^2) - 2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - (a/b)^(1/3))/(a/b)^(1/3))
/(a^2*(a/b)^(1/3)) - 1/9*log(x^4 - x^2*(a/b)^(1/3) + (a/b)^(2/3))/(a^2*(a/b)^(1/3)) + 2/9*log(x^2 + (a/b)^(1/3
))/(a^2*(a/b)^(1/3))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x^3 \left (a+b x^6\right )^2} \, dx=\frac {2 \, b \left (-\frac {a}{b}\right )^{\frac {2}{3}} \log \left ({\left | x^{2} - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{3}} + \frac {2 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x^{2} + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{3} b} - \frac {4 \, b x^{6} + 3 \, a}{6 \, {\left (b x^{8} + a x^{2}\right )} a^{2}} - \frac {\left (-a b^{2}\right )^{\frac {2}{3}} \log \left (x^{4} + x^{2} \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 \, a^{3} b} \]

[In]

integrate(1/x^3/(b*x^6+a)^2,x, algorithm="giac")

[Out]

2/9*b*(-a/b)^(2/3)*log(abs(x^2 - (-a/b)^(1/3)))/a^3 + 2/9*sqrt(3)*(-a*b^2)^(2/3)*arctan(1/3*sqrt(3)*(2*x^2 + (
-a/b)^(1/3))/(-a/b)^(1/3))/(a^3*b) - 1/6*(4*b*x^6 + 3*a)/((b*x^8 + a*x^2)*a^2) - 1/9*(-a*b^2)^(2/3)*log(x^4 +
x^2*(-a/b)^(1/3) + (-a/b)^(2/3))/(a^3*b)

Mupad [B] (verification not implemented)

Time = 5.67 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^3 \left (a+b x^6\right )^2} \, dx=\frac {2\,b^{1/3}\,\ln \left (a^{1/3}+b^{1/3}\,x^2\right )}{9\,a^{7/3}}-\frac {\frac {1}{2\,a}+\frac {2\,b\,x^6}{3\,a^2}}{b\,x^8+a\,x^2}-\frac {2\,b^{1/3}\,\ln \left (6912\,a^7\,b^6-6912\,a^{20/3}\,b^{19/3}\,x^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{7/3}}+\frac {b^{1/3}\,\ln \left (6912\,a^7\,b^6+31104\,a^{20/3}\,b^{19/3}\,x^2\,\left (-\frac {1}{9}+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )\right )\,\left (-\frac {1}{9}+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )}{a^{7/3}} \]

[In]

int(1/(x^3*(a + b*x^6)^2),x)

[Out]

(2*b^(1/3)*log(a^(1/3) + b^(1/3)*x^2))/(9*a^(7/3)) - (1/(2*a) + (2*b*x^6)/(3*a^2))/(a*x^2 + b*x^8) - (2*b^(1/3
)*log(6912*a^7*b^6 - 6912*a^(20/3)*b^(19/3)*x^2*((3^(1/2)*1i)/2 + 1/2))*((3^(1/2)*1i)/2 + 1/2))/(9*a^(7/3)) +
(b^(1/3)*log(6912*a^7*b^6 + 31104*a^(20/3)*b^(19/3)*x^2*((3^(1/2)*1i)/9 - 1/9))*((3^(1/2)*1i)/9 - 1/9))/a^(7/3
)