3.21.8 \(\int \frac {(-b+a^3 x^3) \sqrt [3]{b+a^3 x^3}}{x^5} \, dx\)

Optimal. Leaf size=142 \[ \frac {\left (b-3 a^3 x^3\right ) \sqrt [3]{a^3 x^3+b}}{4 x^4}-\frac {1}{3} a^4 \log \left (\sqrt [3]{a^3 x^3+b}-a x\right )-\frac {a^4 \tan ^{-1}\left (\frac {\sqrt {3} a x}{2 \sqrt [3]{a^3 x^3+b}+a x}\right )}{\sqrt {3}}+\frac {1}{6} a^4 \log \left (a x \sqrt [3]{a^3 x^3+b}+\left (a^3 x^3+b\right )^{2/3}+a^2 x^2\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 151, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {451, 277, 331, 292, 31, 634, 617, 204, 628} \begin {gather*} -\frac {a^3 \sqrt [3]{a^3 x^3+b}}{x}+\frac {\left (a^3 x^3+b\right )^{4/3}}{4 x^4}-\frac {1}{3} a^4 \log \left (1-\frac {a x}{\sqrt [3]{a^3 x^3+b}}\right )-\frac {a^4 \tan ^{-1}\left (\frac {\frac {2 a x}{\sqrt [3]{a^3 x^3+b}}+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{6} a^4 \log \left (\frac {a x}{\sqrt [3]{a^3 x^3+b}}+\frac {a^2 x^2}{\left (a^3 x^3+b\right )^{2/3}}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-b + a^3*x^3)*(b + a^3*x^3)^(1/3))/x^5,x]

[Out]

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

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 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 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 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 451

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a,
b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1]))

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

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Mathematica [C]  time = 0.06, size = 74, normalized size = 0.52 \begin {gather*} \frac {\sqrt [3]{a^3 x^3+b} \left (-\frac {4 a^3 x^3 \, _2F_1\left (-\frac {1}{3},-\frac {1}{3};\frac {2}{3};-\frac {a^3 x^3}{b}\right )}{\sqrt [3]{\frac {a^3 x^3}{b}+1}}+a^3 x^3+b\right )}{4 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-b + a^3*x^3)*(b + a^3*x^3)^(1/3))/x^5,x]

[Out]

((b + a^3*x^3)^(1/3)*(b + a^3*x^3 - (4*a^3*x^3*Hypergeometric2F1[-1/3, -1/3, 2/3, -((a^3*x^3)/b)])/(1 + (a^3*x
^3)/b)^(1/3)))/(4*x^4)

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IntegrateAlgebraic [A]  time = 0.26, size = 142, normalized size = 1.00 \begin {gather*} \frac {\left (b-3 a^3 x^3\right ) \sqrt [3]{b+a^3 x^3}}{4 x^4}-\frac {a^4 \tan ^{-1}\left (\frac {\sqrt {3} a x}{a x+2 \sqrt [3]{b+a^3 x^3}}\right )}{\sqrt {3}}-\frac {1}{3} a^4 \log \left (-a x+\sqrt [3]{b+a^3 x^3}\right )+\frac {1}{6} a^4 \log \left (a^2 x^2+a x \sqrt [3]{b+a^3 x^3}+\left (b+a^3 x^3\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-b + a^3*x^3)*(b + a^3*x^3)^(1/3))/x^5,x]

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (a^{3} x^{3} + b\right )}^{\frac {1}{3}} {\left (a^{3} x^{3} - b\right )}}{x^{5}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (a^{3} x^{3}-b \right ) \left (a^{3} x^{3}+b \right )^{\frac {1}{3}}}{x^{5}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 133, normalized size = 0.94 \begin {gather*} \frac {1}{6} \, {\left (2 \, \sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (a + \frac {2 \, {\left (a^{3} x^{3} + b\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, a}\right ) + a \log \left (a^{2} + \frac {{\left (a^{3} x^{3} + b\right )}^{\frac {1}{3}} a}{x} + \frac {{\left (a^{3} x^{3} + b\right )}^{\frac {2}{3}}}{x^{2}}\right ) - 2 \, a \log \left (-a + \frac {{\left (a^{3} x^{3} + b\right )}^{\frac {1}{3}}}{x}\right ) - \frac {6 \, {\left (a^{3} x^{3} + b\right )}^{\frac {1}{3}}}{x}\right )} a^{3} + \frac {{\left (a^{3} x^{3} + b\right )}^{\frac {4}{3}}}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*sqrt(3)*a*arctan(1/3*sqrt(3)*(a + 2*(a^3*x^3 + b)^(1/3)/x)/a) + a*log(a^2 + (a^3*x^3 + b)^(1/3)*a/x + (
a^3*x^3 + b)^(2/3)/x^2) - 2*a*log(-a + (a^3*x^3 + b)^(1/3)/x) - 6*(a^3*x^3 + b)^(1/3)/x)*a^3 + 1/4*(a^3*x^3 +
b)^(4/3)/x^4

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mupad [B]  time = 1.63, size = 66, normalized size = 0.46 \begin {gather*} \frac {{\left (a^3\,x^3+b\right )}^{4/3}}{4\,x^4}-\frac {a^3\,{\left (a^3\,x^3+b\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},-\frac {1}{3};\ \frac {2}{3};\ -\frac {a^3\,x^3}{b}\right )}{x\,{\left (\frac {a^3\,x^3}{b}+1\right )}^{1/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b + a^3*x^3)^(4/3)/(4*x^4) - (a^3*(b + a^3*x^3)^(1/3)*hypergeom([-1/3, -1/3], 2/3, -(a^3*x^3)/b))/(x*((a^3*x^
3)/b + 1)^(1/3))

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sympy [C]  time = 2.63, size = 114, normalized size = 0.80 \begin {gather*} - \frac {a^{4} \sqrt [3]{1 + \frac {b}{a^{3} x^{3}}} \Gamma \left (- \frac {4}{3}\right )}{3 \Gamma \left (- \frac {1}{3}\right )} + \frac {a^{3} \sqrt [3]{b} \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, - \frac {1}{3} \\ \frac {2}{3} \end {matrix}\middle | {\frac {a^{3} x^{3} e^{i \pi }}{b}} \right )}}{3 x \Gamma \left (\frac {2}{3}\right )} - \frac {a b \sqrt [3]{1 + \frac {b}{a^{3} x^{3}}} \Gamma \left (- \frac {4}{3}\right )}{3 x^{3} \Gamma \left (- \frac {1}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**4*(1 + b/(a**3*x**3))**(1/3)*gamma(-4/3)/(3*gamma(-1/3)) + a**3*b**(1/3)*gamma(-1/3)*hyper((-1/3, -1/3), (
2/3,), a**3*x**3*exp_polar(I*pi)/b)/(3*x*gamma(2/3)) - a*b*(1 + b/(a**3*x**3))**(1/3)*gamma(-4/3)/(3*x**3*gamm
a(-1/3))

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