3.171 \(\int \frac{1}{(a+b x^2) \sqrt{4+5 x^4}} \, dx\)

Optimal. Leaf size=310 \[ \frac{\sqrt [4]{5} \left (\sqrt{5} x^2+2\right ) \sqrt{\frac{5 x^4+4}{\left (\sqrt{5} x^2+2\right )^2}} \left (\sqrt{5} a+2 b\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{5} x}{\sqrt{2}}\right ),\frac{1}{2}\right )}{2 \sqrt{2} \sqrt{5 x^4+4} \left (5 a^2-4 b^2\right )}+\frac{\sqrt{b} \tan ^{-1}\left (\frac{x \sqrt{5 a^2+4 b^2}}{\sqrt{a} \sqrt{b} \sqrt{5 x^4+4}}\right )}{2 \sqrt{a} \sqrt{5 a^2+4 b^2}}-\frac{\left (\sqrt{5} x^2+2\right ) \sqrt{\frac{5 x^4+4}{\left (\sqrt{5} x^2+2\right )^2}} \left (\sqrt{5} a+2 b\right )^2 \Pi \left (-\frac{\left (\sqrt{5} a-2 b\right )^2}{8 \sqrt{5} a b};2 \tan ^{-1}\left (\frac{\sqrt [4]{5} x}{\sqrt{2}}\right )|\frac{1}{2}\right )}{4 \sqrt{2} \sqrt [4]{5} a \sqrt{5 x^4+4} \left (5 a^2-4 b^2\right )} \]

[Out]

(Sqrt[b]*ArcTan[(Sqrt[5*a^2 + 4*b^2]*x)/(Sqrt[a]*Sqrt[b]*Sqrt[4 + 5*x^4])])/(2*Sqrt[a]*Sqrt[5*a^2 + 4*b^2]) +
(5^(1/4)*(Sqrt[5]*a + 2*b)*(2 + Sqrt[5]*x^2)*Sqrt[(4 + 5*x^4)/(2 + Sqrt[5]*x^2)^2]*EllipticF[2*ArcTan[(5^(1/4)
*x)/Sqrt[2]], 1/2])/(2*Sqrt[2]*(5*a^2 - 4*b^2)*Sqrt[4 + 5*x^4]) - ((Sqrt[5]*a + 2*b)^2*(2 + Sqrt[5]*x^2)*Sqrt[
(4 + 5*x^4)/(2 + Sqrt[5]*x^2)^2]*EllipticPi[-(Sqrt[5]*a - 2*b)^2/(8*Sqrt[5]*a*b), 2*ArcTan[(5^(1/4)*x)/Sqrt[2]
], 1/2])/(4*Sqrt[2]*5^(1/4)*a*(5*a^2 - 4*b^2)*Sqrt[4 + 5*x^4])

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Rubi [A]  time = 0.272687, antiderivative size = 310, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {1217, 220, 1707} \[ \frac{\sqrt{b} \tan ^{-1}\left (\frac{x \sqrt{5 a^2+4 b^2}}{\sqrt{a} \sqrt{b} \sqrt{5 x^4+4}}\right )}{2 \sqrt{a} \sqrt{5 a^2+4 b^2}}+\frac{\sqrt [4]{5} \left (\sqrt{5} x^2+2\right ) \sqrt{\frac{5 x^4+4}{\left (\sqrt{5} x^2+2\right )^2}} \left (\sqrt{5} a+2 b\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{5} x}{\sqrt{2}}\right )|\frac{1}{2}\right )}{2 \sqrt{2} \sqrt{5 x^4+4} \left (5 a^2-4 b^2\right )}-\frac{\left (\sqrt{5} x^2+2\right ) \sqrt{\frac{5 x^4+4}{\left (\sqrt{5} x^2+2\right )^2}} \left (\sqrt{5} a+2 b\right )^2 \Pi \left (-\frac{\left (\sqrt{5} a-2 b\right )^2}{8 \sqrt{5} a b};2 \tan ^{-1}\left (\frac{\sqrt [4]{5} x}{\sqrt{2}}\right )|\frac{1}{2}\right )}{4 \sqrt{2} \sqrt [4]{5} a \sqrt{5 x^4+4} \left (5 a^2-4 b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*ArcTan[(Sqrt[5*a^2 + 4*b^2]*x)/(Sqrt[a]*Sqrt[b]*Sqrt[4 + 5*x^4])])/(2*Sqrt[a]*Sqrt[5*a^2 + 4*b^2]) +
(5^(1/4)*(Sqrt[5]*a + 2*b)*(2 + Sqrt[5]*x^2)*Sqrt[(4 + 5*x^4)/(2 + Sqrt[5]*x^2)^2]*EllipticF[2*ArcTan[(5^(1/4)
*x)/Sqrt[2]], 1/2])/(2*Sqrt[2]*(5*a^2 - 4*b^2)*Sqrt[4 + 5*x^4]) - ((Sqrt[5]*a + 2*b)^2*(2 + Sqrt[5]*x^2)*Sqrt[
(4 + 5*x^4)/(2 + Sqrt[5]*x^2)^2]*EllipticPi[-(Sqrt[5]*a - 2*b)^2/(8*Sqrt[5]*a*b), 2*ArcTan[(5^(1/4)*x)/Sqrt[2]
], 1/2])/(4*Sqrt[2]*5^(1/4)*a*(5*a^2 - 4*b^2)*Sqrt[4 + 5*x^4])

Rule 1217

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

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 1707

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

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b x^2\right ) \sqrt{4+5 x^4}} \, dx &=-\frac{\left (2 b \left (\sqrt{5} a+2 b\right )\right ) \int \frac{1+\frac{\sqrt{5} x^2}{2}}{\left (a+b x^2\right ) \sqrt{4+5 x^4}} \, dx}{5 a^2-4 b^2}+\frac{\left (5 a+2 \sqrt{5} b\right ) \int \frac{1}{\sqrt{4+5 x^4}} \, dx}{5 a^2-4 b^2}\\ &=\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{5 a^2+4 b^2} x}{\sqrt{a} \sqrt{b} \sqrt{4+5 x^4}}\right )}{2 \sqrt{a} \sqrt{5 a^2+4 b^2}}+\frac{\sqrt [4]{5} \left (\sqrt{5} a+2 b\right ) \left (2+\sqrt{5} x^2\right ) \sqrt{\frac{4+5 x^4}{\left (2+\sqrt{5} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{5} x}{\sqrt{2}}\right )|\frac{1}{2}\right )}{2 \sqrt{2} \left (5 a^2-4 b^2\right ) \sqrt{4+5 x^4}}-\frac{\left (\sqrt{5} a+2 b\right )^2 \left (2+\sqrt{5} x^2\right ) \sqrt{\frac{4+5 x^4}{\left (2+\sqrt{5} x^2\right )^2}} \Pi \left (-\frac{\left (\sqrt{5} a-2 b\right )^2}{8 \sqrt{5} a b};2 \tan ^{-1}\left (\frac{\sqrt [4]{5} x}{\sqrt{2}}\right )|\frac{1}{2}\right )}{4 \sqrt{2} \sqrt [4]{5} a \left (5 a^2-4 b^2\right ) \sqrt{4+5 x^4}}\\ \end{align*}

Mathematica [C]  time = 0.103303, size = 50, normalized size = 0.16 \[ -\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \Pi \left (-\frac{2 i b}{\sqrt{5} a};\left .i \sinh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{5} x\right )\right |-1\right )}{\sqrt [4]{5} a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/2 - I/2)*EllipticPi[((-2*I)*b)/(Sqrt[5]*a), I*ArcSinh[(1/2 + I/2)*5^(1/4)*x], -1])/(5^(1/4)*a)

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Maple [C]  time = 0.382, size = 86, normalized size = 0.3 \begin{align*}{\frac{1}{a\sqrt{{\frac{i}{2}}\sqrt{5}}}\sqrt{1-{\frac{i}{2}}{x}^{2}\sqrt{5}}\sqrt{1+{\frac{i}{2}}{x}^{2}\sqrt{5}}{\it EllipticPi} \left ( \sqrt{{\frac{i}{2}}\sqrt{5}}x,{\frac{{\frac{2\,i}{5}}\sqrt{5}b}{a}},{\frac{\sqrt{-{\frac{i}{2}}\sqrt{5}}}{\sqrt{{\frac{i}{2}}\sqrt{5}}}} \right ){\frac{1}{\sqrt{5\,{x}^{4}+4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2+a)/(5*x^4+4)^(1/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{5 \, x^{4} + 4}{\left (b x^{2} + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)/(5*x^4+4)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(5*x^4 + 4)*(b*x^2 + a)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{5 \, x^{4} + 4}}{5 \, b x^{6} + 5 \, a x^{4} + 4 \, b x^{2} + 4 \, a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)/(5*x^4+4)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(5*x^4 + 4)/(5*b*x^6 + 5*a*x^4 + 4*b*x^2 + 4*a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b x^{2}\right ) \sqrt{5 x^{4} + 4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**2+a)/(5*x**4+4)**(1/2),x)

[Out]

Integral(1/((a + b*x**2)*sqrt(5*x**4 + 4)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{5 \, x^{4} + 4}{\left (b x^{2} + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)/(5*x^4+4)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(5*x^4 + 4)*(b*x^2 + a)), x)