3.1988 \(\int \frac {1}{(a+\frac {b}{x^3})^2 x^5} \, dx\)

Optimal. Leaf size=136 \[ \frac {\log \left (a^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3}\right )}{18 a^{2/3} b^{4/3}}-\frac {\log \left (\sqrt [3]{a} x+\sqrt [3]{b}\right )}{9 a^{2/3} b^{4/3}}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{3 \sqrt {3} a^{2/3} b^{4/3}}+\frac {x^2}{3 b \left (a x^3+b\right )} \]

[Out]

1/3*x^2/b/(a*x^3+b)-1/9*ln(b^(1/3)+a^(1/3)*x)/a^(2/3)/b^(4/3)+1/18*ln(b^(2/3)-a^(1/3)*b^(1/3)*x+a^(2/3)*x^2)/a
^(2/3)/b^(4/3)-1/9*arctan(1/3*(b^(1/3)-2*a^(1/3)*x)/b^(1/3)*3^(1/2))/a^(2/3)/b^(4/3)*3^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 136, normalized size of antiderivative = 1.00, 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 = {263, 290, 292, 31, 634, 617, 204, 628} \[ \frac {\log \left (a^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3}\right )}{18 a^{2/3} b^{4/3}}-\frac {\log \left (\sqrt [3]{a} x+\sqrt [3]{b}\right )}{9 a^{2/3} b^{4/3}}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{3 \sqrt {3} a^{2/3} b^{4/3}}+\frac {x^2}{3 b \left (a x^3+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^3)^2*x^5),x]

[Out]

x^2/(3*b*(b + a*x^3)) - ArcTan[(b^(1/3) - 2*a^(1/3)*x)/(Sqrt[3]*b^(1/3))]/(3*Sqrt[3]*a^(2/3)*b^(4/3)) - Log[b^
(1/3) + a^(1/3)*x]/(9*a^(2/3)*b^(4/3)) + Log[b^(2/3) - a^(1/3)*b^(1/3)*x + a^(2/3)*x^2]/(18*a^(2/3)*b^(4/3))

Rule 31

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

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 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 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 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 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 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]

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 119, normalized size = 0.88 \[ \frac {\frac {\log \left (a^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3}\right )}{a^{2/3}}-\frac {2 \log \left (\sqrt [3]{a} x+\sqrt [3]{b}\right )}{a^{2/3}}-\frac {2 \sqrt {3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{a} x}{\sqrt [3]{b}}}{\sqrt {3}}\right )}{a^{2/3}}+\frac {6 \sqrt [3]{b} x^2}{a x^3+b}}{18 b^{4/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^3)^2*x^5),x]

[Out]

((6*b^(1/3)*x^2)/(b + a*x^3) - (2*Sqrt[3]*ArcTan[(1 - (2*a^(1/3)*x)/b^(1/3))/Sqrt[3]])/a^(2/3) - (2*Log[b^(1/3
) + a^(1/3)*x])/a^(2/3) + Log[b^(2/3) - a^(1/3)*b^(1/3)*x + a^(2/3)*x^2]/a^(2/3))/(18*b^(4/3))

________________________________________________________________________________________

fricas [A]  time = 0.90, size = 402, normalized size = 2.96 \[ \left [\frac {6 \, a^{2} b x^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (a^{2} b x^{3} + a b^{2}\right )} \sqrt {\frac {\left (-a^{2} b\right )^{\frac {1}{3}}}{b}} \log \left (\frac {2 \, a^{2} x^{3} - a b + 3 \, \sqrt {\frac {1}{3}} {\left (a b x + 2 \, \left (-a^{2} b\right )^{\frac {2}{3}} x^{2} + \left (-a^{2} b\right )^{\frac {1}{3}} b\right )} \sqrt {\frac {\left (-a^{2} b\right )^{\frac {1}{3}}}{b}} - 3 \, \left (-a^{2} b\right )^{\frac {2}{3}} x}{a x^{3} + b}\right ) + {\left (a x^{3} + b\right )} \left (-a^{2} b\right )^{\frac {2}{3}} \log \left (a^{2} x^{2} + \left (-a^{2} b\right )^{\frac {1}{3}} a x + \left (-a^{2} b\right )^{\frac {2}{3}}\right ) - 2 \, {\left (a x^{3} + b\right )} \left (-a^{2} b\right )^{\frac {2}{3}} \log \left (a x - \left (-a^{2} b\right )^{\frac {1}{3}}\right )}{18 \, {\left (a^{3} b^{2} x^{3} + a^{2} b^{3}\right )}}, \frac {6 \, a^{2} b x^{2} + 6 \, \sqrt {\frac {1}{3}} {\left (a^{2} b x^{3} + a b^{2}\right )} \sqrt {-\frac {\left (-a^{2} b\right )^{\frac {1}{3}}}{b}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, a x + \left (-a^{2} b\right )^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-a^{2} b\right )^{\frac {1}{3}}}{b}}}{a}\right ) + {\left (a x^{3} + b\right )} \left (-a^{2} b\right )^{\frac {2}{3}} \log \left (a^{2} x^{2} + \left (-a^{2} b\right )^{\frac {1}{3}} a x + \left (-a^{2} b\right )^{\frac {2}{3}}\right ) - 2 \, {\left (a x^{3} + b\right )} \left (-a^{2} b\right )^{\frac {2}{3}} \log \left (a x - \left (-a^{2} b\right )^{\frac {1}{3}}\right )}{18 \, {\left (a^{3} b^{2} x^{3} + a^{2} b^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^3)^2/x^5,x, algorithm="fricas")

[Out]

[1/18*(6*a^2*b*x^2 + 3*sqrt(1/3)*(a^2*b*x^3 + a*b^2)*sqrt((-a^2*b)^(1/3)/b)*log((2*a^2*x^3 - a*b + 3*sqrt(1/3)
*(a*b*x + 2*(-a^2*b)^(2/3)*x^2 + (-a^2*b)^(1/3)*b)*sqrt((-a^2*b)^(1/3)/b) - 3*(-a^2*b)^(2/3)*x)/(a*x^3 + b)) +
(a*x^3 + b)*(-a^2*b)^(2/3)*log(a^2*x^2 + (-a^2*b)^(1/3)*a*x + (-a^2*b)^(2/3)) - 2*(a*x^3 + b)*(-a^2*b)^(2/3)*
log(a*x - (-a^2*b)^(1/3)))/(a^3*b^2*x^3 + a^2*b^3), 1/18*(6*a^2*b*x^2 + 6*sqrt(1/3)*(a^2*b*x^3 + a*b^2)*sqrt(-
(-a^2*b)^(1/3)/b)*arctan(sqrt(1/3)*(2*a*x + (-a^2*b)^(1/3))*sqrt(-(-a^2*b)^(1/3)/b)/a) + (a*x^3 + b)*(-a^2*b)^
(2/3)*log(a^2*x^2 + (-a^2*b)^(1/3)*a*x + (-a^2*b)^(2/3)) - 2*(a*x^3 + b)*(-a^2*b)^(2/3)*log(a*x - (-a^2*b)^(1/
3)))/(a^3*b^2*x^3 + a^2*b^3)]

________________________________________________________________________________________

giac [A]  time = 0.18, size = 129, normalized size = 0.95 \[ \frac {x^{2}}{3 \, {\left (a x^{3} + b\right )} b} - \frac {\left (-\frac {b}{a}\right )^{\frac {2}{3}} \log \left ({\left | x - \left (-\frac {b}{a}\right )^{\frac {1}{3}} \right |}\right )}{9 \, b^{2}} - \frac {\sqrt {3} \left (-a^{2} b\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {b}{a}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{9 \, a^{2} b^{2}} + \frac {\left (-a^{2} b\right )^{\frac {2}{3}} \log \left (x^{2} + x \left (-\frac {b}{a}\right )^{\frac {1}{3}} + \left (-\frac {b}{a}\right )^{\frac {2}{3}}\right )}{18 \, a^{2} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^3)^2/x^5,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.00, size = 117, normalized size = 0.86 \[ \frac {x^{2}}{3 \left (a \,x^{3}+b \right ) b}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {b}{a}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 \left (\frac {b}{a}\right )^{\frac {1}{3}} a b}-\frac {\ln \left (x +\left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{9 \left (\frac {b}{a}\right )^{\frac {1}{3}} a b}+\frac {\ln \left (x^{2}-\left (\frac {b}{a}\right )^{\frac {1}{3}} x +\left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{18 \left (\frac {b}{a}\right )^{\frac {1}{3}} a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x^3)^2/x^5,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.86, size = 124, normalized size = 0.91 \[ \frac {x^{2}}{3 \, {\left (a b x^{3} + b^{2}\right )}} + \frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{9 \, a b \left (\frac {b}{a}\right )^{\frac {1}{3}}} + \frac {\log \left (x^{2} - x \left (\frac {b}{a}\right )^{\frac {1}{3}} + \left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{18 \, a b \left (\frac {b}{a}\right )^{\frac {1}{3}}} - \frac {\log \left (x + \left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{9 \, a b \left (\frac {b}{a}\right )^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^3)^2/x^5,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.32, size = 138, normalized size = 1.01 \[ \frac {x^2}{3\,b\,\left (a\,x^3+b\right )}+\frac {{\left (-1\right )}^{1/3}\,\ln \left (\frac {{\left (-1\right )}^{2/3}\,a^{2/3}}{9\,b^{5/3}}+\frac {a\,x}{9\,b^2}\right )}{9\,a^{2/3}\,b^{4/3}}-\frac {{\left (-1\right )}^{1/3}\,\ln \left ({\left (-1\right )}^{2/3}\,b^{1/3}-2\,a^{1/3}\,x+{\left (-1\right )}^{1/6}\,\sqrt {3}\,b^{1/3}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{2/3}\,b^{4/3}}+\frac {{\left (-1\right )}^{1/3}\,\ln \left (2\,a^{1/3}\,x-{\left (-1\right )}^{2/3}\,b^{1/3}+{\left (-1\right )}^{1/6}\,\sqrt {3}\,b^{1/3}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{2/3}\,b^{4/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(a + b/x^3)^2),x)

[Out]

x^2/(3*b*(b + a*x^3)) + ((-1)^(1/3)*log(((-1)^(2/3)*a^(2/3))/(9*b^(5/3)) + (a*x)/(9*b^2)))/(9*a^(2/3)*b^(4/3))
- ((-1)^(1/3)*log((-1)^(2/3)*b^(1/3) - 2*a^(1/3)*x + (-1)^(1/6)*3^(1/2)*b^(1/3))*((3^(1/2)*1i)/2 + 1/2))/(9*a
^(2/3)*b^(4/3)) + ((-1)^(1/3)*log(2*a^(1/3)*x - (-1)^(2/3)*b^(1/3) + (-1)^(1/6)*3^(1/2)*b^(1/3))*((3^(1/2)*1i)
/2 - 1/2))/(9*a^(2/3)*b^(4/3))

________________________________________________________________________________________

sympy [A]  time = 0.50, size = 44, normalized size = 0.32 \[ \frac {x^{2}}{3 a b x^{3} + 3 b^{2}} + \operatorname {RootSum} {\left (729 t^{3} a^{2} b^{4} + 1, \left (t \mapsto t \log {\left (81 t^{2} a b^{3} + x \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**3)**2/x**5,x)

[Out]

x**2/(3*a*b*x**3 + 3*b**2) + RootSum(729*_t**3*a**2*b**4 + 1, Lambda(_t, _t*log(81*_t**2*a*b**3 + x)))

________________________________________________________________________________________