3.1013 \(\int \sqrt [6]{a+b x^2} \, dx\)

Optimal. Leaf size=273 \[ \frac {3^{3/4} \sqrt {2-\sqrt {3}} a \sqrt [6]{a+b x^2} \left (1-\sqrt [3]{\frac {a}{a+b x^2}}\right ) \sqrt {\frac {\left (\frac {a}{a+b x^2}\right )^{2/3}+\sqrt [3]{\frac {a}{a+b x^2}}+1}{\left (-\sqrt [3]{\frac {a}{a+b x^2}}-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\sin ^{-1}\left (\frac {-\sqrt [3]{\frac {a}{a+b x^2}}+\sqrt {3}+1}{-\sqrt [3]{\frac {a}{a+b x^2}}-\sqrt {3}+1}\right ),4 \sqrt {3}-7\right )}{4 b x \sqrt [3]{\frac {a}{a+b x^2}} \sqrt {-\frac {1-\sqrt [3]{\frac {a}{a+b x^2}}}{\left (-\sqrt [3]{\frac {a}{a+b x^2}}-\sqrt {3}+1\right )^2}}}+\frac {3}{4} x \sqrt [6]{a+b x^2} \]

[Out]

3/4*x*(b*x^2+a)^(1/6)+1/4*3^(3/4)*a*(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)/b/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 = 273, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {195, 241, 236, 219} \[ \frac {3}{4} x \sqrt [6]{a+b x^2}+\frac {3^{3/4} \sqrt {2-\sqrt {3}} a \sqrt [6]{a+b x^2} \left (1-\sqrt [3]{\frac {a}{a+b x^2}}\right ) \sqrt {\frac {\left (\frac {a}{a+b x^2}\right )^{2/3}+\sqrt [3]{\frac {a}{a+b x^2}}+1}{\left (-\sqrt [3]{\frac {a}{a+b x^2}}-\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {-\sqrt [3]{\frac {a}{b x^2+a}}+\sqrt {3}+1}{-\sqrt [3]{\frac {a}{b x^2+a}}-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{4 b x \sqrt [3]{\frac {a}{a+b x^2}} \sqrt {-\frac {1-\sqrt [3]{\frac {a}{a+b x^2}}}{\left (-\sqrt [3]{\frac {a}{a+b x^2}}-\sqrt {3}+1\right )^2}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(1/6),x]

[Out]

(3*x*(a + b*x^2)^(1/6))/4 + (3^(3/4)*Sqrt[2 - Sqrt[3]]*a*(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]])/(4*b*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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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]]

Rubi steps

\begin {align*} \int \sqrt [6]{a+b x^2} \, dx &=\frac {3}{4} x \sqrt [6]{a+b x^2}+\frac {1}{4} a \int \frac {1}{\left (a+b x^2\right )^{5/6}} \, dx\\ &=\frac {3}{4} x \sqrt [6]{a+b x^2}+\frac {a \operatorname {Subst}\left (\int \frac {1}{\left (1-b x^2\right )^{2/3}} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{4 \sqrt [3]{\frac {a}{a+b x^2}} \sqrt [3]{a+b x^2}}\\ &=\frac {3}{4} x \sqrt [6]{a+b x^2}-\frac {\left (3 a \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 )}{8 b x \sqrt [3]{\frac {a}{a+b x^2}}}\\ &=\frac {3}{4} x \sqrt [6]{a+b x^2}+\frac {3^{3/4} \sqrt {2-\sqrt {3}} a \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 )}{4 b 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 = 46, normalized size = 0.17 \[ \frac {x \sqrt [6]{a+b x^2} \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {3}{2};-\frac {b x^2}{a}\right )}{\sqrt [6]{\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(1/6),x]

[Out]

(x*(a + b*x^2)^(1/6)*Hypergeometric2F1[-1/6, 1/2, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^(1/6)

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x^{2} + a\right )}^{\frac {1}{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^(1/6), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^(1/6), x)

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maple [F]  time = 0.33, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{2}+a \right )^{\frac {1}{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^(1/6), x)

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mupad [B]  time = 4.88, size = 37, normalized size = 0.14 \[ \frac {x\,{\left (b\,x^2+a\right )}^{1/6}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{6},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{1/6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/6),x)

[Out]

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

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sympy [A]  time = 0.89, size = 26, normalized size = 0.10 \[ \sqrt [6]{a} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{6}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(1/6)*x*hyper((-1/6, 1/2), (3/2,), b*x**2*exp_polar(I*pi)/a)

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