\(\int \frac {1}{(a+\frac {b}{x^3}) x^6} \, dx\) [1976]

Optimal result

Integrand size = 13, antiderivative size = 124 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right ) x^6} \, dx=-\frac {1}{2 b x^2}+\frac {a^{2/3} \arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt {3} b^{5/3}}-\frac {a^{2/3} \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{3 b^{5/3}}+\frac {a^{2/3} \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{6 b^{5/3}} \]

[Out]

-1/2/b/x^2-1/3*a^(2/3)*ln(b^(1/3)+a^(1/3)*x)/b^(5/3)+1/6*a^(2/3)*ln(b^(2/3)-a^(1/3)*b^(1/3)*x+a^(2/3)*x^2)/b^(
5/3)+1/3*a^(2/3)*arctan(1/3*(b^(1/3)-2*a^(1/3)*x)/b^(1/3)*3^(1/2))/b^(5/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 124, 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, 331, 206, 31, 648, 631, 210, 642} \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right ) x^6} \, dx=\frac {a^{2/3} \arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt {3} b^{5/3}}+\frac {a^{2/3} \log \left (a^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3}\right )}{6 b^{5/3}}-\frac {a^{2/3} \log \left (\sqrt [3]{a} x+\sqrt [3]{b}\right )}{3 b^{5/3}}-\frac {1}{2 b x^2} \]

[In]

Int[1/((a + b/x^3)*x^6),x]

[Out]

-1/2*1/(b*x^2) + (a^(2/3)*ArcTan[(b^(1/3) - 2*a^(1/3)*x)/(Sqrt[3]*b^(1/3))])/(Sqrt[3]*b^(5/3)) - (a^(2/3)*Log[
b^(1/3) + a^(1/3)*x])/(3*b^(5/3)) + (a^(2/3)*Log[b^(2/3) - a^(1/3)*b^(1/3)*x + a^(2/3)*x^2])/(6*b^(5/3))

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 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}& = \int \frac {1}{x^3 \left (b+a x^3\right )} \, dx \\ & = -\frac {1}{2 b x^2}-\frac {a \int \frac {1}{b+a x^3} \, dx}{b} \\ & = -\frac {1}{2 b x^2}-\frac {a \int \frac {1}{\sqrt [3]{b}+\sqrt [3]{a} x} \, dx}{3 b^{5/3}}-\frac {a \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}{3 b^{5/3}} \\ & = -\frac {1}{2 b x^2}-\frac {a^{2/3} \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{3 b^{5/3}}+\frac {a^{2/3} \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}{6 b^{5/3}}-\frac {a \int \frac {1}{b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2} \, dx}{2 b^{4/3}} \\ & = -\frac {1}{2 b x^2}-\frac {a^{2/3} \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{3 b^{5/3}}+\frac {a^{2/3} \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{6 b^{5/3}}-\frac {a^{2/3} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{a} x}{\sqrt [3]{b}}\right )}{b^{5/3}} \\ & = -\frac {1}{2 b x^2}+\frac {a^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt {3} b^{5/3}}-\frac {a^{2/3} \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{3 b^{5/3}}+\frac {a^{2/3} \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{6 b^{5/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right ) x^6} \, dx=\frac {-3 b^{2/3}+2 \sqrt {3} a^{2/3} x^2 \arctan \left (\frac {1-\frac {2 \sqrt [3]{a} x}{\sqrt [3]{b}}}{\sqrt {3}}\right )-2 a^{2/3} x^2 \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )+a^{2/3} x^2 \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{6 b^{5/3} x^2} \]

[In]

Integrate[1/((a + b/x^3)*x^6),x]

[Out]

(-3*b^(2/3) + 2*Sqrt[3]*a^(2/3)*x^2*ArcTan[(1 - (2*a^(1/3)*x)/b^(1/3))/Sqrt[3]] - 2*a^(2/3)*x^2*Log[b^(1/3) +
a^(1/3)*x] + a^(2/3)*x^2*Log[b^(2/3) - a^(1/3)*b^(1/3)*x + a^(2/3)*x^2])/(6*b^(5/3)*x^2)

Maple [C] (verified)

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

Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.44

method	result	size
risch	\(-\frac {1}{2 x^{2} b}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b^{5} \textit {\_Z}^{3}+a^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} b^{5}-3 a^{2}\right ) x -a \,b^{2} \textit {\_R} \right )\right )}{3}\)	\(54\)
default	\(-\frac {\left (\frac {\ln \left (x +\left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {b}{a}\right )^{\frac {1}{3}} x +\left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{6 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {b}{a}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}\right ) a}{b}-\frac {1}{2 x^{2} b}\)	\(106\)

[In]

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

[Out]

-1/2/x^2/b+1/3*sum(_R*ln((-4*_R^3*b^5-3*a^2)*x-a*b^2*_R),_R=RootOf(_Z^3*b^5+a^2))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.15 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right ) x^6} \, dx=\frac {2 \, \sqrt {3} x^{2} \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}} - \sqrt {3} a}{3 \, a}\right ) - x^{2} \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a^{2} x^{2} + a b x \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} + b^{2} \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}}\right ) + 2 \, x^{2} \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x - b \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}}\right ) - 3}{6 \, b x^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.26 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right ) x^6} \, dx=\operatorname {RootSum} {\left (27 t^{3} b^{5} + a^{2}, \left ( t \mapsto t \log {\left (- \frac {3 t b^{2}}{a} + x \right )} \right )\right )} - \frac {1}{2 b x^{2}} \]

[In]

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

[Out]

RootSum(27*_t**3*b**5 + a**2, Lambda(_t, _t*log(-3*_t*b**2/a + x))) - 1/(2*b*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right ) x^6} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{3 \, b \left (\frac {b}{a}\right )^{\frac {2}{3}}} + \frac {\log \left (x^{2} - x \left (\frac {b}{a}\right )^{\frac {1}{3}} + \left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{6 \, b \left (\frac {b}{a}\right )^{\frac {2}{3}}} - \frac {\log \left (x + \left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{3 \, b \left (\frac {b}{a}\right )^{\frac {2}{3}}} - \frac {1}{2 \, b x^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right ) x^6} \, dx=\frac {a \left (-\frac {b}{a}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {b}{a}\right )^{\frac {1}{3}} \right |}\right )}{3 \, b^{2}} - \frac {\sqrt {3} \left (-a^{2} b\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {b}{a}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{3 \, b^{2}} - \frac {\left (-a^{2} b\right )^{\frac {1}{3}} \log \left (x^{2} + x \left (-\frac {b}{a}\right )^{\frac {1}{3}} + \left (-\frac {b}{a}\right )^{\frac {2}{3}}\right )}{6 \, b^{2}} - \frac {1}{2 \, b x^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.80 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\left (a+\frac {b}{x^3}\right ) x^6} \, dx=\frac {a^{2/3}\,\ln \left ({\left (-b\right )}^{7/3}-a^{1/3}\,b^2\,x\right )}{3\,{\left (-b\right )}^{5/3}}-\frac {1}{2\,b\,x^2}-\frac {a^{2/3}\,\ln \left (3\,a^3\,b^2\,x+3\,a^{8/3}\,{\left (-b\right )}^{7/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,{\left (-b\right )}^{5/3}}+\frac {a^{2/3}\,\ln \left (3\,a^3\,b^2\,x-9\,a^{8/3}\,{\left (-b\right )}^{7/3}\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{{\left (-b\right )}^{5/3}} \]

[In]

int(1/(x^6*(a + b/x^3)),x)

[Out]

(a^(2/3)*log((-b)^(7/3) - a^(1/3)*b^2*x))/(3*(-b)^(5/3)) - 1/(2*b*x^2) - (a^(2/3)*log(3*a^3*b^2*x + 3*a^(8/3)*
(-b)^(7/3)*((3^(1/2)*1i)/2 + 1/2))*((3^(1/2)*1i)/2 + 1/2))/(3*(-b)^(5/3)) + (a^(2/3)*log(3*a^3*b^2*x - 9*a^(8/
3)*(-b)^(7/3)*((3^(1/2)*1i)/6 - 1/6))*((3^(1/2)*1i)/6 - 1/6))/(-b)^(5/3)