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

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

[Out]

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

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Rubi [A]  time = 0.0775369, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.615, Rules used = {290, 325, 200, 31, 634, 617, 204, 628} \[ \frac{5 b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{8/3}}-\frac{5 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{8/3}}+\frac{5 b^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{8/3}}-\frac{5}{6 a^2 x^2}+\frac{1}{3 a x^2 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x^3)^2),x]

[Out]

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

Rule 290

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] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

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 200

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

Mathematica [A]  time = 0.0873572, size = 129, normalized size = 0.88 \[ \frac{5 b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )-\frac{6 a^{2/3} b x}{a+b x^3}-\frac{9 a^{2/3}}{x^2}-10 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+10 \sqrt{3} b^{2/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{18 a^{8/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*x^3)^2),x]

[Out]

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

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Maple [A]  time = 0.008, size = 115, normalized size = 0.8 \begin{align*} -{\frac{bx}{3\,{a}^{2} \left ( b{x}^{3}+a \right ) }}-{\frac{5}{9\,{a}^{2}}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}+{\frac{5}{18\,{a}^{2}}\ln \left ({x}^{2}-\sqrt [3]{{\frac{a}{b}}}x+ \left ({\frac{a}{b}} \right ) ^{{\frac{2}{3}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}-{\frac{5\,\sqrt{3}}{9\,{a}^{2}}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{x{\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-1 \right ) } \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}-{\frac{1}{2\,{a}^{2}{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.50234, size = 428, normalized size = 2.93 \begin{align*} -\frac{15 \, b x^{3} - 10 \, \sqrt{3}{\left (b x^{5} + a x^{2}\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{2}{3}} - \sqrt{3} b}{3 \, b}\right ) + 5 \,{\left (b x^{5} + a x^{2}\right )} \left (-\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}} \log \left (b^{2} x^{2} + a b x \left (-\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}} + a^{2} \left (-\frac{b^{2}}{a^{2}}\right )^{\frac{2}{3}}\right ) - 10 \,{\left (b x^{5} + a x^{2}\right )} \left (-\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}} \log \left (b x - a \left (-\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}}\right ) + 9 \, a}{18 \,{\left (a^{2} b x^{5} + a^{3} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.818298, size = 56, normalized size = 0.38 \begin{align*} - \frac{3 a + 5 b x^{3}}{6 a^{3} x^{2} + 6 a^{2} b x^{5}} + \operatorname{RootSum}{\left (729 t^{3} a^{8} + 125 b^{2}, \left ( t \mapsto t \log{\left (- \frac{9 t a^{3}}{5 b} + x \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*a + 5*b*x**3)/(6*a**3*x**2 + 6*a**2*b*x**5) + RootSum(729*_t**3*a**8 + 125*b**2, Lambda(_t, _t*log(-9*_t*a
**3/(5*b) + x)))

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Giac [A]  time = 1.15906, size = 177, normalized size = 1.21 \begin{align*} \frac{5 \, b \left (-\frac{a}{b}\right )^{\frac{1}{3}} \log \left ({\left | x - \left (-\frac{a}{b}\right )^{\frac{1}{3}} \right |}\right )}{9 \, a^{3}} - \frac{b x}{3 \,{\left (b x^{3} + a\right )} a^{2}} - \frac{5 \, \sqrt{3} \left (-a b^{2}\right )^{\frac{1}{3}} \arctan \left (\frac{\sqrt{3}{\left (2 \, x + \left (-\frac{a}{b}\right )^{\frac{1}{3}}\right )}}{3 \, \left (-\frac{a}{b}\right )^{\frac{1}{3}}}\right )}{9 \, a^{3}} - \frac{5 \, \left (-a b^{2}\right )^{\frac{1}{3}} \log \left (x^{2} + x \left (-\frac{a}{b}\right )^{\frac{1}{3}} + \left (-\frac{a}{b}\right )^{\frac{2}{3}}\right )}{18 \, a^{3}} - \frac{1}{2 \, a^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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