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

Optimal. Leaf size=210 \[ \frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^4}}-\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{b^{3/4} \sqrt {a+b x^4}}+\frac {x \sqrt {a+b x^4}}{\sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 210, 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 = {305, 220, 1196} \[ \frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^4}}-\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{b^{3/4} \sqrt {a+b x^4}}+\frac {x \sqrt {a+b x^4}}{\sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[a + b*x^4],x]

[Out]

(x*Sqrt[a + b*x^4])/(Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) - (a^(1/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqr
t[a] + Sqrt[b]*x^2)^2]*EllipticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(b^(3/4)*Sqrt[a + b*x^4]) + (a^(1/4)*(Sq
rt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2]
)/(2*b^(3/4)*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 305

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

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 51, normalized size = 0.24 \[ \frac {x^3 \sqrt {\frac {b x^4}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {b x^4}{a}\right )}{3 \sqrt {a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[a + b*x^4],x]

[Out]

(x^3*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, -((b*x^4)/a)])/(3*Sqrt[a + b*x^4])

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{2}}{\sqrt {b x^{4} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2/sqrt(b*x^4 + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/sqrt(b*x^4 + a), x)

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maple [C]  time = 0.01, size = 97, normalized size = 0.46 \[ \frac {i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )\right ) \sqrt {a}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/sqrt(b*x^4 + a), x)

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

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 = 1.49, size = 37, normalized size = 0.18 \[ \frac {x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(7/4))

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