3.1982 \(\int \frac{x}{(a+\frac{b}{x^3})^2} \, dx\)

Optimal. Leaf size=146 \[ -\frac{5 b^{2/3} \log \left (a^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3}\right )}{18 a^{8/3}}+\frac{5 b^{2/3} \log \left (\sqrt [3]{a} x+\sqrt [3]{b}\right )}{9 a^{8/3}}+\frac{5 b^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt{3} \sqrt [3]{b}}\right )}{3 \sqrt{3} a^{8/3}}+\frac{5 x^2}{6 a^2}-\frac{x^5}{3 a \left (a x^3+b\right )} \]

[Out]

(5*x^2)/(6*a^2) - x^5/(3*a*(b + a*x^3)) + (5*b^(2/3)*ArcTan[(b^(1/3) - 2*a^(1/3)*x)/(Sqrt[3]*b^(1/3))])/(3*Sqr
t[3]*a^(8/3)) + (5*b^(2/3)*Log[b^(1/3) + a^(1/3)*x])/(9*a^(8/3)) - (5*b^(2/3)*Log[b^(2/3) - a^(1/3)*b^(1/3)*x
+ a^(2/3)*x^2])/(18*a^(8/3))

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Rubi [A]  time = 0.0758377, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.818, Rules used = {263, 288, 321, 292, 31, 634, 617, 204, 628} \[ -\frac{5 b^{2/3} \log \left (a^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3}\right )}{18 a^{8/3}}+\frac{5 b^{2/3} \log \left (\sqrt [3]{a} x+\sqrt [3]{b}\right )}{9 a^{8/3}}+\frac{5 b^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt{3} \sqrt [3]{b}}\right )}{3 \sqrt{3} a^{8/3}}+\frac{5 x^2}{6 a^2}-\frac{x^5}{3 a \left (a x^3+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b/x^3)^2,x]

[Out]

(5*x^2)/(6*a^2) - x^5/(3*a*(b + a*x^3)) + (5*b^(2/3)*ArcTan[(b^(1/3) - 2*a^(1/3)*x)/(Sqrt[3]*b^(1/3))])/(3*Sqr
t[3]*a^(8/3)) + (5*b^(2/3)*Log[b^(1/3) + a^(1/3)*x])/(9*a^(8/3)) - (5*b^(2/3)*Log[b^(2/3) - a^(1/3)*b^(1/3)*x
+ a^(2/3)*x^2])/(18*a^(8/3))

Rule 263

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 288

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 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 292

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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 634

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]

Rule 617

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

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]

Rubi steps

\begin{align*} \int \frac{x}{\left (a+\frac{b}{x^3}\right )^2} \, dx &=\int \frac{x^7}{\left (b+a x^3\right )^2} \, dx\\ &=-\frac{x^5}{3 a \left (b+a x^3\right )}+\frac{5 \int \frac{x^4}{b+a x^3} \, dx}{3 a}\\ &=\frac{5 x^2}{6 a^2}-\frac{x^5}{3 a \left (b+a x^3\right )}-\frac{(5 b) \int \frac{x}{b+a x^3} \, dx}{3 a^2}\\ &=\frac{5 x^2}{6 a^2}-\frac{x^5}{3 a \left (b+a x^3\right )}+\frac{\left (5 b^{2/3}\right ) \int \frac{1}{\sqrt [3]{b}+\sqrt [3]{a} x} \, dx}{9 a^{7/3}}-\frac{\left (5 b^{2/3}\right ) \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^{7/3}}\\ &=\frac{5 x^2}{6 a^2}-\frac{x^5}{3 a \left (b+a x^3\right )}+\frac{5 b^{2/3} \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{9 a^{8/3}}-\frac{\left (5 b^{2/3}\right ) \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^{8/3}}-\frac{(5 b) \int \frac{1}{b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2} \, dx}{6 a^{7/3}}\\ &=\frac{5 x^2}{6 a^2}-\frac{x^5}{3 a \left (b+a x^3\right )}+\frac{5 b^{2/3} \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{9 a^{8/3}}-\frac{5 b^{2/3} \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{18 a^{8/3}}-\frac{\left (5 b^{2/3}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{a} x}{\sqrt [3]{b}}\right )}{3 a^{8/3}}\\ &=\frac{5 x^2}{6 a^2}-\frac{x^5}{3 a \left (b+a x^3\right )}+\frac{5 b^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt{3} \sqrt [3]{b}}\right )}{3 \sqrt{3} a^{8/3}}+\frac{5 b^{2/3} \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{9 a^{8/3}}-\frac{5 b^{2/3} \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{18 a^{8/3}}\\ \end{align*}

Mathematica [A]  time = 0.078319, size = 131, normalized size = 0.9 \[ \frac{-5 b^{2/3} \log \left (a^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3}\right )+\frac{6 a^{2/3} b x^2}{a x^3+b}+9 a^{2/3} x^2+10 b^{2/3} \log \left (\sqrt [3]{a} x+\sqrt [3]{b}\right )+10 \sqrt{3} b^{2/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{a} x}{\sqrt [3]{b}}}{\sqrt{3}}\right )}{18 a^{8/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b/x^3)^2,x]

[Out]

(9*a^(2/3)*x^2 + (6*a^(2/3)*b*x^2)/(b + a*x^3) + 10*Sqrt[3]*b^(2/3)*ArcTan[(1 - (2*a^(1/3)*x)/b^(1/3))/Sqrt[3]
] + 10*b^(2/3)*Log[b^(1/3) + a^(1/3)*x] - 5*b^(2/3)*Log[b^(2/3) - a^(1/3)*b^(1/3)*x + a^(2/3)*x^2])/(18*a^(8/3
))

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Maple [A]  time = 0.007, size = 120, normalized size = 0.8 \begin{align*}{\frac{{x}^{2}}{2\,{a}^{2}}}+{\frac{b{x}^{2}}{3\,{a}^{2} \left ( a{x}^{3}+b \right ) }}+{\frac{5\,b}{9\,{a}^{3}}\ln \left ( x+\sqrt [3]{{\frac{b}{a}}} \right ){\frac{1}{\sqrt [3]{{\frac{b}{a}}}}}}-{\frac{5\,b}{18\,{a}^{3}}\ln \left ({x}^{2}-\sqrt [3]{{\frac{b}{a}}}x+ \left ({\frac{b}{a}} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{\frac{b}{a}}}}}}-{\frac{5\,b\sqrt{3}}{9\,{a}^{3}}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{x{\frac{1}{\sqrt [3]{{\frac{b}{a}}}}}}-1 \right ) } \right ){\frac{1}{\sqrt [3]{{\frac{b}{a}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a+b/x^3)^2,x)

[Out]

1/2*x^2/a^2+1/3*b/a^2*x^2/(a*x^3+b)+5/9*b/a^3/(b/a)^(1/3)*ln(x+(b/a)^(1/3))-5/18*b/a^3/(b/a)^(1/3)*ln(x^2-(b/a
)^(1/3)*x+(b/a)^(2/3))-5/9*b/a^3*3^(1/2)/(b/a)^(1/3)*arctan(1/3*3^(1/2)*(2/(b/a)^(1/3)*x-1))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.4309, size = 393, normalized size = 2.69 \begin{align*} \frac{9 \, a x^{5} + 15 \, b x^{2} - 10 \, \sqrt{3}{\left (a x^{3} + b\right )} \left (\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}} \arctan \left (\frac{2 \, \sqrt{3} a x \left (\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}} - \sqrt{3} b}{3 \, b}\right ) - 5 \,{\left (a x^{3} + b\right )} \left (\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}} \log \left (b x^{2} - a x \left (\frac{b^{2}}{a^{2}}\right )^{\frac{2}{3}} + b \left (\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}}\right ) + 10 \,{\left (a x^{3} + b\right )} \left (\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}} \log \left (b x + a \left (\frac{b^{2}}{a^{2}}\right )^{\frac{2}{3}}\right )}{18 \,{\left (a^{3} x^{3} + a^{2} b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(9*a*x^5 + 15*b*x^2 - 10*sqrt(3)*(a*x^3 + b)*(b^2/a^2)^(1/3)*arctan(1/3*(2*sqrt(3)*a*x*(b^2/a^2)^(1/3) -
sqrt(3)*b)/b) - 5*(a*x^3 + b)*(b^2/a^2)^(1/3)*log(b*x^2 - a*x*(b^2/a^2)^(2/3) + b*(b^2/a^2)^(1/3)) + 10*(a*x^3
+ b)*(b^2/a^2)^(1/3)*log(b*x + a*(b^2/a^2)^(2/3)))/(a^3*x^3 + a^2*b)

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Sympy [A]  time = 0.484293, size = 58, normalized size = 0.4 \begin{align*} \frac{b x^{2}}{3 a^{3} x^{3} + 3 a^{2} b} + \operatorname{RootSum}{\left (729 t^{3} a^{8} - 125 b^{2}, \left ( t \mapsto t \log{\left (\frac{81 t^{2} a^{5}}{25 b} + x \right )} \right )\right )} + \frac{x^{2}}{2 a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x**2/(3*a**3*x**3 + 3*a**2*b) + RootSum(729*_t**3*a**8 - 125*b**2, Lambda(_t, _t*log(81*_t**2*a**5/(25*b) +
x))) + x**2/(2*a**2)

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Giac [A]  time = 1.16221, size = 178, normalized size = 1.22 \begin{align*} \frac{x^{2}}{2 \, a^{2}} + \frac{b x^{2}}{3 \,{\left (a x^{3} + b\right )} a^{2}} + \frac{5 \, \left (-\frac{b}{a}\right )^{\frac{2}{3}} \log \left ({\left | x - \left (-\frac{b}{a}\right )^{\frac{1}{3}} \right |}\right )}{9 \, a^{2}} + \frac{5 \, \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^{4}} - \frac{5 \, \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 )}{18 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2/a^2 + 1/3*b*x^2/((a*x^3 + b)*a^2) + 5/9*(-b/a)^(2/3)*log(abs(x - (-b/a)^(1/3)))/a^2 + 5/9*sqrt(3)*(-a^
2*b)^(2/3)*arctan(1/3*sqrt(3)*(2*x + (-b/a)^(1/3))/(-b/a)^(1/3))/a^4 - 5/18*(-a^2*b)^(2/3)*log(x^2 + x*(-b/a)^
(1/3) + (-b/a)^(2/3))/a^4