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

Optimal. Leaf size=300 \[ -\frac{\sqrt [4]{d} \left (\sqrt{d} x^2+2\right ) \sqrt{\frac{d x^4+4}{\left (\sqrt{d} x^2+2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x}{\sqrt{2}}\right ),\frac{1}{2}\right )}{2 \sqrt{2} \sqrt{d x^4+4} \left (2 b-a \sqrt{d}\right )}+\frac{\sqrt{b} \tan ^{-1}\left (\frac{x \sqrt{a^2 d+4 b^2}}{\sqrt{a} \sqrt{b} \sqrt{d x^4+4}}\right )}{2 \sqrt{a} \sqrt{a^2 d+4 b^2}}+\frac{\left (\sqrt{d} x^2+2\right ) \sqrt{\frac{d x^4+4}{\left (\sqrt{d} x^2+2\right )^2}} \left (a \sqrt{d}+2 b\right ) \Pi \left (-\frac{\left (2 b-a \sqrt{d}\right )^2}{8 a b \sqrt{d}};2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x}{\sqrt{2}}\right )|\frac{1}{2}\right )}{4 \sqrt{2} a \sqrt [4]{d} \sqrt{d x^4+4} \left (2 b-a \sqrt{d}\right )} \]

[Out]

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

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Rubi [A]  time = 0.223249, antiderivative size = 300, 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} \[ \frac{\sqrt{b} \tan ^{-1}\left (\frac{x \sqrt{a^2 d+4 b^2}}{\sqrt{a} \sqrt{b} \sqrt{d x^4+4}}\right )}{2 \sqrt{a} \sqrt{a^2 d+4 b^2}}-\frac{\sqrt [4]{d} \left (\sqrt{d} x^2+2\right ) \sqrt{\frac{d x^4+4}{\left (\sqrt{d} x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x}{\sqrt{2}}\right )|\frac{1}{2}\right )}{2 \sqrt{2} \sqrt{d x^4+4} \left (2 b-a \sqrt{d}\right )}+\frac{\left (\sqrt{d} x^2+2\right ) \sqrt{\frac{d x^4+4}{\left (\sqrt{d} x^2+2\right )^2}} \left (a \sqrt{d}+2 b\right ) \Pi \left (-\frac{\left (2 b-a \sqrt{d}\right )^2}{8 a b \sqrt{d}};2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x}{\sqrt{2}}\right )|\frac{1}{2}\right )}{4 \sqrt{2} a \sqrt [4]{d} \sqrt{d x^4+4} \left (2 b-a \sqrt{d}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*ArcTan[(Sqrt[4*b^2 + a^2*d]*x)/(Sqrt[a]*Sqrt[b]*Sqrt[4 + d*x^4])])/(2*Sqrt[a]*Sqrt[4*b^2 + a^2*d]) -
(d^(1/4)*(2 + Sqrt[d]*x^2)*Sqrt[(4 + d*x^4)/(2 + Sqrt[d]*x^2)^2]*EllipticF[2*ArcTan[(d^(1/4)*x)/Sqrt[2]], 1/2]
)/(2*Sqrt[2]*(2*b - a*Sqrt[d])*Sqrt[4 + d*x^4]) + ((2*b + a*Sqrt[d])*(2 + Sqrt[d]*x^2)*Sqrt[(4 + d*x^4)/(2 + S
qrt[d]*x^2)^2]*EllipticPi[-(2*b - a*Sqrt[d])^2/(8*a*b*Sqrt[d]), 2*ArcTan[(d^(1/4)*x)/Sqrt[2]], 1/2])/(4*Sqrt[2
]*a*(2*b - a*Sqrt[d])*d^(1/4)*Sqrt[4 + d*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+d x^4}} \, dx &=\frac{(2 b) \int \frac{1+\frac{\sqrt{d} x^2}{2}}{\left (a+b x^2\right ) \sqrt{4+d x^4}} \, dx}{2 b-a \sqrt{d}}-\frac{\sqrt{d} \int \frac{1}{\sqrt{4+d x^4}} \, dx}{2 b-a \sqrt{d}}\\ &=\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{4 b^2+a^2 d} x}{\sqrt{a} \sqrt{b} \sqrt{4+d x^4}}\right )}{2 \sqrt{a} \sqrt{4 b^2+a^2 d}}-\frac{\sqrt [4]{d} \left (2+\sqrt{d} x^2\right ) \sqrt{\frac{4+d x^4}{\left (2+\sqrt{d} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x}{\sqrt{2}}\right )|\frac{1}{2}\right )}{2 \sqrt{2} \left (2 b-a \sqrt{d}\right ) \sqrt{4+d x^4}}+\frac{\left (2 b+a \sqrt{d}\right ) \left (2+\sqrt{d} x^2\right ) \sqrt{\frac{4+d x^4}{\left (2+\sqrt{d} x^2\right )^2}} \Pi \left (-\frac{\left (2 b-a \sqrt{d}\right )^2}{8 a b \sqrt{d}};2 \tan ^{-1}\left (\frac{\sqrt [4]{d} x}{\sqrt{2}}\right )|\frac{1}{2}\right )}{4 \sqrt{2} a \left (2 b-a \sqrt{d}\right ) \sqrt [4]{d} \sqrt{4+d x^4}}\\ \end{align*}

Mathematica [C]  time = 0.111654, size = 65, normalized size = 0.22 \[ -\frac{i \Pi \left (-\frac{2 i b}{a \sqrt{d}};\left .i \sinh ^{-1}\left (\frac{\sqrt{i \sqrt{d}} x}{\sqrt{2}}\right )\right |-1\right )}{\sqrt{2} a \sqrt{i \sqrt{d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*EllipticPi[((-2*I)*b)/(a*Sqrt[d]), I*ArcSinh[(Sqrt[I*Sqrt[d]]*x)/Sqrt[2]], -1])/(Sqrt[2]*a*Sqrt[I*Sqrt[d
]])

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Maple [C]  time = 0.023, size = 86, normalized size = 0.3 \begin{align*}{\frac{1}{a}\sqrt{1-{\frac{i}{2}}\sqrt{d}{x}^{2}}\sqrt{1+{\frac{i}{2}}\sqrt{d}{x}^{2}}{\it EllipticPi} \left ( \sqrt{{\frac{i}{2}}\sqrt{d}}x,{\frac{2\,ib}{a}{\frac{1}{\sqrt{d}}}},{\sqrt{-{\frac{i}{2}}\sqrt{d}}{\frac{1}{\sqrt{{\frac{i}{2}}\sqrt{d}}}}} \right ){\frac{1}{\sqrt{{\frac{i}{2}}\sqrt{d}}}}{\frac{1}{\sqrt{d{x}^{4}+4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{d 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)/(d*x^4+4)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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{d x^{4} + 4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{d 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)/(d*x^4+4)^(1/2),x, algorithm="giac")

[Out]

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