3.1.76 \(\int \frac {1}{x^3 (a+b x^3)^2} \, dx\) [76]

Optimal. Leaf size=126 \[ \frac {-\frac {1}{2 a x^2}-\frac {5 b x}{6 a^2}}{a+b x^3}-\frac {5 \sqrt [3]{\frac {a}{b}} b \left (\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{2 \sqrt [3]{\frac {a}{b}}-x}\right )+\frac {1}{2} \log \left (\frac {\left (\sqrt [3]{\frac {a}{b}}+x\right )^2}{\left (\frac {a}{b}\right )^{2/3}-\sqrt [3]{\frac {a}{b}} x+x^2}\right )\right )}{9 a^3} \]

[Out]

-(1/2/a/x^2+5/6*b*x/a^2)/(b*x^3+a)-5/9*b/a^3*(a/b)^(1/3)*(1/2*ln((x+(a/b)^(1/3))^2/(x^2-(a/b)^(1/3)*x+(a/b)^(2
/3)))+3^(1/2)*arctan(3^(1/2)*x/(2*(a/b)^(1/3)-x)))

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Rubi [A]
time = 0.05, antiderivative size = 146, normalized size of antiderivative = 1.16, 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 = {296, 331, 206, 31, 648, 631, 210, 642} \begin {gather*} \frac {5 b^{2/3} \arctan \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 (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}{6 a^2 x^2}+\frac {1}{3 a x^2 \left (a+b x^3\right )} \end {gather*}

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 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 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 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 {gather*} \begin {aligned} \text {Integral} &=\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 ) \text {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} \arctan \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 {aligned} \end {gather*}

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Mathematica [A]
time = 0.05, size = 129, normalized size = 1.02 \begin {gather*} \frac {-\frac {9 a^{2/3}}{x^2}-\frac {6 a^{2/3} b x}{a+b x^3}+10 \sqrt {3} b^{2/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )-10 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+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 {gather*}

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 = 0.03, size = 118, normalized size = 0.94

method	result	size
risch	\(\frac {-\frac {5 b \,x^{3}}{6 a^{2}}-\frac {1}{2 a}}{x^{2} \left (b \,x^{3}+a \right )}+\frac {5 \left (\munderset {\textit {\_R} =\mathit {RootOf}\left (a^{8} \textit {\_Z}^{3}+b^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \textit {\_R}^{3} a^{8}-3 b^{2}\right ) x -a^{3} b \textit {\_R} \right )\right )}{9}\)	\(74\)
default	\(-\frac {b \left (\frac {x}{3 b \,x^{3}+3 a}+\frac {5 \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {5 \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {5 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{a^{2}}-\frac {1}{2 a^{2} x^{2}}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.41, size = 128, normalized size = 1.02 \begin {gather*} -\frac {5 \, b x^{3} + 3 \, a}{6 \, {\left (a^{2} b x^{5} + a^{3} x^{2}\right )}} - \frac {5 \, \sqrt {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^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {5 \, \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, a^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {5 \, \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, a^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(5*b*x^3 + 3*a)/(a^2*b*x^5 + a^3*x^2) - 5/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(
a^2*(a/b)^(2/3)) + 5/18*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a^2*(a/b)^(2/3)) - 5/9*log(x + (a/b)^(1/3))/(a
^2*(a/b)^(2/3))

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Fricas [A]
time = 0.60, size = 187, normalized size = 1.48 \begin {gather*} -\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 {gather*}

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 = 0.17, size = 58, normalized size = 0.46 \begin {gather*} \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 {gather*}

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 = 0.38, size = 131, normalized size = 1.04 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 0.46, size = 146, normalized size = 1.16 \begin {gather*} \frac {5\,{\left (-1\right )}^{1/3}\,b^{2/3}\,\ln \left ({\left (-1\right )}^{1/3}\,a^{1/3}-b^{1/3}\,x\right )}{9\,a^{8/3}}-\frac {\frac {1}{2\,a}+\frac {5\,b\,x^3}{6\,a^2}}{b\,x^5+a\,x^2}-\frac {5\,{\left (-1\right )}^{1/3}\,b^{2/3}\,\ln \left ({\left (-1\right )}^{1/3}\,a^{1/3}+2\,b^{1/3}\,x+{\left (-1\right )}^{5/6}\,\sqrt {3}\,a^{1/3}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{8/3}}+\frac {5\,{\left (-1\right )}^{1/3}\,b^{2/3}\,\ln \left ({\left (-1\right )}^{1/3}\,a^{1/3}+2\,b^{1/3}\,x-{\left (-1\right )}^{5/6}\,\sqrt {3}\,a^{1/3}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{8/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*x^3)^2),x)

[Out]

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

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