3.1014 \(\int \frac {\sqrt [6]{a+b x^2}}{x^2} \, dx\)

Optimal. Leaf size=266 \[ \frac {\sqrt {2-\sqrt {3}} \sqrt [6]{a+b x^2} \left (1-\sqrt [3]{\frac {a}{a+b x^2}}\right ) \sqrt {\frac {\left (\frac {a}{a+b x^2}\right )^{2/3}+\sqrt [3]{\frac {a}{a+b x^2}}+1}{\left (-\sqrt [3]{\frac {a}{a+b x^2}}-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\sin ^{-1}\left (\frac {-\sqrt [3]{\frac {a}{a+b x^2}}+\sqrt {3}+1}{-\sqrt [3]{\frac {a}{a+b x^2}}-\sqrt {3}+1}\right ),4 \sqrt {3}-7\right )}{\sqrt [4]{3} x \sqrt [3]{\frac {a}{a+b x^2}} \sqrt {-\frac {1-\sqrt [3]{\frac {a}{a+b x^2}}}{\left (-\sqrt [3]{\frac {a}{a+b x^2}}-\sqrt {3}+1\right )^2}}}-\frac {\sqrt [6]{a+b x^2}}{x} \]

[Out]

-(b*x^2+a)^(1/6)/x+1/3*(b*x^2+a)^(1/6)*(1-(a/(b*x^2+a))^(1/3))*EllipticF((1-(a/(b*x^2+a))^(1/3)+3^(1/2))/(1-(a
/(b*x^2+a))^(1/3)-3^(1/2)),2*I-I*3^(1/2))*(1/2*6^(1/2)-1/2*2^(1/2))*((1+(a/(b*x^2+a))^(1/3)+(a/(b*x^2+a))^(2/3
))/(1-(a/(b*x^2+a))^(1/3)-3^(1/2))^2)^(1/2)*3^(3/4)/x/(a/(b*x^2+a))^(1/3)/((-1+(a/(b*x^2+a))^(1/3))/(1-(a/(b*x
^2+a))^(1/3)-3^(1/2))^2)^(1/2)

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Rubi [A]  time = 0.21, antiderivative size = 266, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {277, 241, 236, 219} \[ \frac {\sqrt {2-\sqrt {3}} \sqrt [6]{a+b x^2} \left (1-\sqrt [3]{\frac {a}{a+b x^2}}\right ) \sqrt {\frac {\left (\frac {a}{a+b x^2}\right )^{2/3}+\sqrt [3]{\frac {a}{a+b x^2}}+1}{\left (-\sqrt [3]{\frac {a}{a+b x^2}}-\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {-\sqrt [3]{\frac {a}{b x^2+a}}+\sqrt {3}+1}{-\sqrt [3]{\frac {a}{b x^2+a}}-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{\sqrt [4]{3} x \sqrt [3]{\frac {a}{a+b x^2}} \sqrt {-\frac {1-\sqrt [3]{\frac {a}{a+b x^2}}}{\left (-\sqrt [3]{\frac {a}{a+b x^2}}-\sqrt {3}+1\right )^2}}}-\frac {\sqrt [6]{a+b x^2}}{x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(1/6)/x^2,x]

[Out]

-((a + b*x^2)^(1/6)/x) + (Sqrt[2 - Sqrt[3]]*(a + b*x^2)^(1/6)*(1 - (a/(a + b*x^2))^(1/3))*Sqrt[(1 + (a/(a + b*
x^2))^(1/3) + (a/(a + b*x^2))^(2/3))/(1 - Sqrt[3] - (a/(a + b*x^2))^(1/3))^2]*EllipticF[ArcSin[(1 + Sqrt[3] -
(a/(a + b*x^2))^(1/3))/(1 - Sqrt[3] - (a/(a + b*x^2))^(1/3))], -7 + 4*Sqrt[3]])/(3^(1/4)*x*(a/(a + b*x^2))^(1/
3)*Sqrt[-((1 - (a/(a + b*x^2))^(1/3))/(1 - Sqrt[3] - (a/(a + b*x^2))^(1/3))^2)])

Rule 219

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 - Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 + Sqrt[3
])*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[-((s*(s + r*x))/((1 - S
qrt[3])*s + r*x)^2)]), x]] /; FreeQ[{a, b}, x] && NegQ[a]

Rule 236

Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Dist[(3*Sqrt[b*x^2])/(2*b*x), Subst[Int[1/Sqrt[-a + x^3], x], x
, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b}, x]

Rule 241

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a/(a + b*x^n))^(p + 1/n)*(a + b*x^n)^(p + 1/n), Subst[In
t[1/(1 - b*x^n)^(p + 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)] && LtQ[Denominator[p + 1/n], Denominator[p]]

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]

Rubi steps

\begin {align*} \int \frac {\sqrt [6]{a+b x^2}}{x^2} \, dx &=-\frac {\sqrt [6]{a+b x^2}}{x}+\frac {1}{3} b \int \frac {1}{\left (a+b x^2\right )^{5/6}} \, dx\\ &=-\frac {\sqrt [6]{a+b x^2}}{x}+\frac {b \operatorname {Subst}\left (\int \frac {1}{\left (1-b x^2\right )^{2/3}} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{3 \sqrt [3]{\frac {a}{a+b x^2}} \sqrt [3]{a+b x^2}}\\ &=-\frac {\sqrt [6]{a+b x^2}}{x}-\frac {\left (\sqrt {-\frac {b x^2}{a+b x^2}} \sqrt [6]{a+b x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x^3}} \, dx,x,\sqrt [3]{\frac {a}{a+b x^2}}\right )}{2 x \sqrt [3]{\frac {a}{a+b x^2}}}\\ &=-\frac {\sqrt [6]{a+b x^2}}{x}+\frac {\sqrt {2-\sqrt {3}} \sqrt {-\frac {b x^2}{a+b x^2}} \sqrt [6]{a+b x^2} \left (1-\sqrt [3]{\frac {a}{a+b x^2}}\right ) \sqrt {\frac {1+\sqrt [3]{\frac {a}{a+b x^2}}+\left (\frac {a}{a+b x^2}\right )^{2/3}}{\left (1-\sqrt {3}-\sqrt [3]{\frac {a}{a+b x^2}}\right )^2}} F\left (\sin ^{-1}\left (\frac {1+\sqrt {3}-\sqrt [3]{\frac {a}{a+b x^2}}}{1-\sqrt {3}-\sqrt [3]{\frac {a}{a+b x^2}}}\right )|-7+4 \sqrt {3}\right )}{\sqrt [4]{3} x \sqrt [3]{\frac {a}{a+b x^2}} \sqrt {-\frac {1-\sqrt [3]{\frac {a}{a+b x^2}}}{\left (1-\sqrt {3}-\sqrt [3]{\frac {a}{a+b x^2}}\right )^2}} \sqrt {-1+\frac {a}{a+b x^2}}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 49, normalized size = 0.18 \[ -\frac {\sqrt [6]{a+b x^2} \, _2F_1\left (-\frac {1}{2},-\frac {1}{6};\frac {1}{2};-\frac {b x^2}{a}\right )}{x \sqrt [6]{\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(1/6)/x^2,x]

[Out]

-(((a + b*x^2)^(1/6)*Hypergeometric2F1[-1/2, -1/6, 1/2, -((b*x^2)/a)])/(x*(1 + (b*x^2)/a)^(1/6)))

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{6}}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/6)/x^2,x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(1/6)/x^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {1}{6}}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/6)/x^2,x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(1/6)/x^2, x)

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maple [F]  time = 0.31, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{6}}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(1/6)/x^2,x)

[Out]

int((b*x^2+a)^(1/6)/x^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {1}{6}}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/6)/x^2,x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(1/6)/x^2, x)

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mupad [B]  time = 5.08, size = 40, normalized size = 0.15 \[ -\frac {3\,{\left (b\,x^2+a\right )}^{1/6}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{6},\frac {1}{3};\ \frac {4}{3};\ -\frac {a}{b\,x^2}\right )}{2\,x\,{\left (\frac {a}{b\,x^2}+1\right )}^{1/6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/6)/x^2,x)

[Out]

-(3*(a + b*x^2)^(1/6)*hypergeom([-1/6, 1/3], 4/3, -a/(b*x^2)))/(2*x*(a/(b*x^2) + 1)^(1/6))

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sympy [A]  time = 0.98, size = 29, normalized size = 0.11 \[ - \frac {\sqrt [6]{a} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{6} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(1/6)/x**2,x)

[Out]

-a**(1/6)*hyper((-1/2, -1/6), (1/2,), b*x**2*exp_polar(I*pi)/a)/x

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