3.169 \(\int \frac{1}{(d+e x^2) \sqrt{-a-c x^4}} \, dx\)

Optimal. Leaf size=347 \[ \frac{\sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{2 \sqrt [4]{a} \sqrt{-a-c x^4} \left (\sqrt{c} d-\sqrt{a} e\right )}-\frac{a^{3/4} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\frac{\sqrt{c} d}{\sqrt{a}}+e\right )^2 \Pi \left (-\frac{\left (\sqrt{c} d-\sqrt{a} e\right )^2}{4 \sqrt{a} \sqrt{c} d e};2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 \sqrt [4]{c} d \sqrt{-a-c x^4} \left (c d^2-a e^2\right )}+\frac{\sqrt{e} \tan ^{-1}\left (\frac{x \sqrt{-a e^2-c d^2}}{\sqrt{d} \sqrt{e} \sqrt{-a-c x^4}}\right )}{2 \sqrt{d} \sqrt{-a e^2-c d^2}} \]

[Out]

(Sqrt[e]*ArcTan[(Sqrt[-(c*d^2) - a*e^2]*x)/(Sqrt[d]*Sqrt[e]*Sqrt[-a - c*x^4])])/(2*Sqrt[d]*Sqrt[-(c*d^2) - a*e
^2]) + (c^(1/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4
)*x)/a^(1/4)], 1/2])/(2*a^(1/4)*(Sqrt[c]*d - Sqrt[a]*e)*Sqrt[-a - c*x^4]) - (a^(3/4)*((Sqrt[c]*d)/Sqrt[a] + e)
^2*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticPi[-(Sqrt[c]*d - Sqrt[a]*e)^2/(
4*Sqrt[a]*Sqrt[c]*d*e), 2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(4*c^(1/4)*d*(c*d^2 - a*e^2)*Sqrt[-a - c*x^4])

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Rubi [A]  time = 0.295035, antiderivative size = 347, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1217, 220, 1707} \[ -\frac{a^{3/4} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\frac{\sqrt{c} d}{\sqrt{a}}+e\right )^2 \Pi \left (-\frac{\left (\sqrt{c} d-\sqrt{a} e\right )^2}{4 \sqrt{a} \sqrt{c} d e};2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 \sqrt [4]{c} d \sqrt{-a-c x^4} \left (c d^2-a e^2\right )}+\frac{\sqrt{e} \tan ^{-1}\left (\frac{x \sqrt{-a e^2-c d^2}}{\sqrt{d} \sqrt{e} \sqrt{-a-c x^4}}\right )}{2 \sqrt{d} \sqrt{-a e^2-c d^2}}+\frac{\sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} \sqrt{-a-c x^4} \left (\sqrt{c} d-\sqrt{a} e\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[e]*ArcTan[(Sqrt[-(c*d^2) - a*e^2]*x)/(Sqrt[d]*Sqrt[e]*Sqrt[-a - c*x^4])])/(2*Sqrt[d]*Sqrt[-(c*d^2) - a*e
^2]) + (c^(1/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4
)*x)/a^(1/4)], 1/2])/(2*a^(1/4)*(Sqrt[c]*d - Sqrt[a]*e)*Sqrt[-a - c*x^4]) - (a^(3/4)*((Sqrt[c]*d)/Sqrt[a] + e)
^2*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticPi[-(Sqrt[c]*d - Sqrt[a]*e)^2/(
4*Sqrt[a]*Sqrt[c]*d*e), 2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(4*c^(1/4)*d*(c*d^2 - a*e^2)*Sqrt[-a - c*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 (d+e x^2\right ) \sqrt{-a-c x^4}} \, dx &=\frac{\sqrt{c} \int \frac{1}{\sqrt{-a-c x^4}} \, dx}{\sqrt{c} d-\sqrt{a} e}-\frac{\left (\sqrt{a} e\right ) \int \frac{1+\frac{\sqrt{c} x^2}{\sqrt{a}}}{\left (d+e x^2\right ) \sqrt{-a-c x^4}} \, dx}{\sqrt{c} d-\sqrt{a} e}\\ &=\frac{\sqrt{e} \tan ^{-1}\left (\frac{\sqrt{-c d^2-a e^2} x}{\sqrt{d} \sqrt{e} \sqrt{-a-c x^4}}\right )}{2 \sqrt{d} \sqrt{-c d^2-a e^2}}+\frac{\sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} \left (\sqrt{c} d-\sqrt{a} e\right ) \sqrt{-a-c x^4}}-\frac{\sqrt [4]{a} \left (\frac{\sqrt{c} d}{\sqrt{a}}+e\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \Pi \left (-\frac{\left (\sqrt{c} d-\sqrt{a} e\right )^2}{4 \sqrt{a} \sqrt{c} d e};2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 \sqrt [4]{c} d \left (\sqrt{c} d-\sqrt{a} e\right ) \sqrt{-a-c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.147194, size = 98, normalized size = 0.28 \[ -\frac{i \sqrt{\frac{c x^4}{a}+1} \Pi \left (-\frac{i \sqrt{a} e}{\sqrt{c} d};\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{c}}{\sqrt{a}}} x\right )\right |-1\right )}{d \sqrt{\frac{i \sqrt{c}}{\sqrt{a}}} \sqrt{-a-c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[1 + (c*x^4)/a]*EllipticPi[((-I)*Sqrt[a]*e)/(Sqrt[c]*d), I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[a]]*x], -1]
)/(Sqrt[(I*Sqrt[c])/Sqrt[a]]*d*Sqrt[-a - c*x^4])

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Maple [C]  time = 0.178, size = 110, normalized size = 0.3 \begin{align*}{\frac{1}{d}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\it EllipticPi} \left ( x\sqrt{{-i\sqrt{c}{\frac{1}{\sqrt{a}}}}},{\frac{-ie}{d}\sqrt{a}{\frac{1}{\sqrt{c}}}},{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\frac{1}{\sqrt{{-i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}} \right ){\frac{1}{\sqrt{{-i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{-c{x}^{4}-a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x^2+d)/(-c*x^4-a)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x^2+d)/(-c*x^4-a)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(-c*x^4 - a)*(e*x^2 + d)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-c*x^4 - a)/(c*e*x^6 + c*d*x^4 + a*e*x^2 + a*d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x**2+d)/(-c*x**4-a)**(1/2),x)

[Out]

Integral(1/(sqrt(-a - c*x**4)*(d + e*x**2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x^2+d)/(-c*x^4-a)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(-c*x^4 - a)*(e*x^2 + d)), x)