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

Optimal. Leaf size=146 \[ -\frac{2 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{9 a^{7/3}}+\frac{4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{7/3}}+\frac{4 \sqrt [3]{b} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{7/3}}-\frac{4}{3 a^2 x}+\frac{1}{3 a x \left (a+b x^3\right )} \]

[Out]

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

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Rubi [A]  time = 0.0774414, 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, 292, 31, 634, 617, 204, 628} \[ -\frac{2 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{9 a^{7/3}}+\frac{4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{7/3}}+\frac{4 \sqrt [3]{b} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{7/3}}-\frac{4}{3 a^2 x}+\frac{1}{3 a x \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x^3)^2),x]

[Out]

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

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

Mathematica [A]  time = 0.0901279, size = 131, normalized size = 0.9 \[ \frac{-2 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )-\frac{3 \sqrt [3]{a} b x^2}{a+b x^3}+4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+4 \sqrt{3} \sqrt [3]{b} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt{3}}\right )-\frac{9 \sqrt [3]{a}}{x}}{9 a^{7/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*x^3)^2),x]

[Out]

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

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Maple [A]  time = 0.009, size = 117, normalized size = 0.8 \begin{align*} -{\frac{b{x}^{2}}{3\,{a}^{2} \left ( b{x}^{3}+a \right ) }}+{\frac{4}{9\,{a}^{2}}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{2}{9\,{a}^{2}}\ln \left ({x}^{2}-\sqrt [3]{{\frac{a}{b}}}x+ \left ({\frac{a}{b}} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{4\,\sqrt{3}}{9\,{a}^{2}}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{x{\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-1 \right ) } \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{1}{{a}^{2}x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^2/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.44544, size = 350, normalized size = 2.4 \begin{align*} -\frac{12 \, b x^{3} + 4 \, \sqrt{3}{\left (b x^{4} + a x\right )} \left (\frac{b}{a}\right )^{\frac{1}{3}} \arctan \left (\frac{2}{3} \, \sqrt{3} x \left (\frac{b}{a}\right )^{\frac{1}{3}} - \frac{1}{3} \, \sqrt{3}\right ) + 2 \,{\left (b x^{4} + a x\right )} \left (\frac{b}{a}\right )^{\frac{1}{3}} \log \left (b x^{2} - a x \left (\frac{b}{a}\right )^{\frac{2}{3}} + a \left (\frac{b}{a}\right )^{\frac{1}{3}}\right ) - 4 \,{\left (b x^{4} + a x\right )} \left (\frac{b}{a}\right )^{\frac{1}{3}} \log \left (b x + a \left (\frac{b}{a}\right )^{\frac{2}{3}}\right ) + 9 \, a}{9 \,{\left (a^{2} b x^{4} + a^{3} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.73716, size = 54, normalized size = 0.37 \begin{align*} - \frac{3 a + 4 b x^{3}}{3 a^{3} x + 3 a^{2} b x^{4}} + \operatorname{RootSum}{\left (729 t^{3} a^{7} - 64 b, \left ( t \mapsto t \log{\left (\frac{81 t^{2} a^{5}}{16 b} + x \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*a + 4*b*x**3)/(3*a**3*x + 3*a**2*b*x**4) + RootSum(729*_t**3*a**7 - 64*b, Lambda(_t, _t*log(81*_t**2*a**5/
(16*b) + x)))

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Giac [A]  time = 1.14025, size = 188, normalized size = 1.29 \begin{align*} \frac{4 \, b \left (-\frac{a}{b}\right )^{\frac{2}{3}} \log \left ({\left | x - \left (-\frac{a}{b}\right )^{\frac{1}{3}} \right |}\right )}{9 \, a^{3}} + \frac{4 \, \sqrt{3} \left (-a b^{2}\right )^{\frac{2}{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} b} - \frac{4 \, b x^{3} + 3 \, a}{3 \,{\left (b x^{4} + a x\right )} a^{2}} - \frac{2 \, \left (-a b^{2}\right )^{\frac{2}{3}} \log \left (x^{2} + x \left (-\frac{a}{b}\right )^{\frac{1}{3}} + \left (-\frac{a}{b}\right )^{\frac{2}{3}}\right )}{9 \, a^{3} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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