3.2086 \(\int \frac {x^2}{\sqrt {a+\frac {b}{x^4}}} \, dx\)

Optimal. Leaf size=110 \[ \frac {b^{3/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{6 a^{5/4} \sqrt {a+\frac {b}{x^4}}}+\frac {x^3 \sqrt {a+\frac {b}{x^4}}}{3 a} \]

[Out]

1/3*x^3*(a+b/x^4)^(1/2)/a+1/6*b^(3/4)*(cos(2*arccot(a^(1/4)*x/b^(1/4)))^2)^(1/2)/cos(2*arccot(a^(1/4)*x/b^(1/4
)))*EllipticF(sin(2*arccot(a^(1/4)*x/b^(1/4))),1/2*2^(1/2))*(a^(1/2)+b^(1/2)/x^2)*((a+b/x^4)/(a^(1/2)+b^(1/2)/
x^2)^2)^(1/2)/a^(5/4)/(a+b/x^4)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {335, 325, 220} \[ \frac {b^{3/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{6 a^{5/4} \sqrt {a+\frac {b}{x^4}}}+\frac {x^3 \sqrt {a+\frac {b}{x^4}}}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[a + b/x^4],x]

[Out]

(Sqrt[a + b/x^4]*x^3)/(3*a) + (b^(3/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*Ell
ipticF[2*ArcCot[(a^(1/4)*x)/b^(1/4)], 1/2])/(6*a^(5/4)*Sqrt[a + b/x^4])

Rule 220

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

Rule 325

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

Rule 335

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

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 64, normalized size = 0.58 \[ \frac {-b \sqrt {\frac {a x^4}{b}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {a x^4}{b}\right )+a x^4+b}{3 a x \sqrt {a+\frac {b}{x^4}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[a + b/x^4],x]

[Out]

(b + a*x^4 - b*Sqrt[1 + (a*x^4)/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((a*x^4)/b)])/(3*a*Sqrt[a + b/x^4]*x)

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{6} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{a x^{4} + b}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^6*sqrt((a*x^4 + b)/x^4)/(a*x^4 + b), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/sqrt(a + b/x^4), x)

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maple [C]  time = 0.01, size = 124, normalized size = 1.13 \[ \frac {\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a \,x^{5}+\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b x -\sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, b \EllipticF \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )}{3 \sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((I*a^(1/2)/b^(1/2))^(1/2)*a*x^5-b*(-(I*a^(1/2)*x^2-b^(1/2))/b^(1/2))^(1/2)*((I*a^(1/2)*x^2+b^(1/2))/b^(1/
2))^(1/2)*EllipticF((I*a^(1/2)/b^(1/2))^(1/2)*x,I)+(I*a^(1/2)/b^(1/2))^(1/2)*b*x)/((a*x^4+b)/x^4)^(1/2)/x^2/a/
(I*a^(1/2)/b^(1/2))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/sqrt(a + b/x^4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2}{\sqrt {a+\frac {b}{x^4}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b/x^4)^(1/2),x)

[Out]

int(x^2/(a + b/x^4)^(1/2), x)

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sympy [C]  time = 1.30, size = 42, normalized size = 0.38 \[ - \frac {x^{3} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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