3.789 \(\int \sqrt [4]{a+b x^2} \, dx\)

Optimal. Leaf size=75 \[ \frac {2 a^{3/2} \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{3 \sqrt {b} \left (a+b x^2\right )^{3/4}}+\frac {2}{3} x \sqrt [4]{a+b x^2} \]

[Out]

2/3*x*(b*x^2+a)^(1/4)+2/3*a^(3/2)*(1+b*x^2/a)^(3/4)*(cos(1/2*arctan(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arcta
n(x*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))/(b*x^2+a)^(3/4)/b^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {195, 233, 231} \[ \frac {2 a^{3/2} \left (\frac {b x^2}{a}+1\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 \sqrt {b} \left (a+b x^2\right )^{3/4}}+\frac {2}{3} x \sqrt [4]{a+b x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(1/4),x]

[Out]

(2*x*(a + b*x^2)^(1/4))/3 + (2*a^(3/2)*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(3*S
qrt[b]*(a + b*x^2)^(3/4))

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 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(3/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 233

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(3/4)/(a + b*x^2)^(3/4), Int[1/(1 + (b*x^2
)/a)^(3/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 46, normalized size = 0.61 \[ \frac {x \sqrt [4]{a+b x^2} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{2};-\frac {b x^2}{a}\right )}{\sqrt [4]{\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(1/4),x]

[Out]

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

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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x^{2} + a\right )}^{\frac {1}{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(1/4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{2} + a\right )}^{\frac {1}{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/4),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(1/4), x)

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maple [F]  time = 0.31, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{2}+a \right )^{\frac {1}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(1/4),x)

[Out]

int((b*x^2+a)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{2} + a\right )}^{\frac {1}{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/4),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(1/4), x)

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mupad [B]  time = 4.66, size = 37, normalized size = 0.49 \[ \frac {x\,{\left (b\,x^2+a\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{1/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/4),x)

[Out]

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

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sympy [C]  time = 0.81, size = 26, normalized size = 0.35 \[ \sqrt [4]{a} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(1/4),x)

[Out]

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

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