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

Optimal. Leaf size=412 \[ \frac{\sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a-b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{4} \left (\frac{b}{\sqrt{a} \sqrt{c}}+2\right )\right )}{2 \sqrt [4]{a} \sqrt{-a+b x^2-c x^4} \left (\sqrt{c} d-\sqrt{a} e\right )}-\frac{a^{3/4} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a-b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\frac{\sqrt{c} d}{\sqrt{a}}+e\right )^2 \Pi \left (-\frac{\left (\sqrt{c} d-\sqrt{a} e\right )^2}{4 \sqrt{a} \sqrt{c} d e};2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (\frac{b}{\sqrt{a} \sqrt{c}}+2\right )\right )}{4 \sqrt [4]{c} d \sqrt{-a+b x^2-c x^4} \left (c d^2-a e^2\right )}+\frac{\sqrt{e} \tan ^{-1}\left (\frac{x \sqrt{-e (a e+b d)-c d^2}}{\sqrt{d} \sqrt{e} \sqrt{-a+b x^2-c x^4}}\right )}{2 \sqrt{d} \sqrt{-e (a e+b d)-c d^2}} \]

[Out]

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

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Rubi [A]  time = 0.362714, antiderivative size = 412, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {1216, 1103, 1706} \[ -\frac{a^{3/4} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a-b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\frac{\sqrt{c} d}{\sqrt{a}}+e\right )^2 \Pi \left (-\frac{\left (\sqrt{c} d-\sqrt{a} e\right )^2}{4 \sqrt{a} \sqrt{c} d e};2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (\frac{b}{\sqrt{a} \sqrt{c}}+2\right )\right )}{4 \sqrt [4]{c} d \sqrt{-a+b x^2-c x^4} \left (c d^2-a e^2\right )}+\frac{\sqrt{e} \tan ^{-1}\left (\frac{x \sqrt{-e (a e+b d)-c d^2}}{\sqrt{d} \sqrt{e} \sqrt{-a+b x^2-c x^4}}\right )}{2 \sqrt{d} \sqrt{-e (a e+b d)-c d^2}}+\frac{\sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a-b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (\frac{b}{\sqrt{a} \sqrt{c}}+2\right )\right )}{2 \sqrt [4]{a} \sqrt{-a+b x^2-c x^4} \left (\sqrt{c} d-\sqrt{a} e\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1216

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

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1706

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[
{q = Rt[B/A, 2]}, -Simp[((B*d - A*e)*ArcTan[(Rt[-b + (c*d)/e + (a*e)/d, 2]*x)/Sqrt[a + b*x^2 + c*x^4]])/(2*d*e
*Rt[-b + (c*d)/e + (a*e)/d, 2]), x] + Simp[((B*d + A*e)*(A + B*x^2)*Sqrt[(A^2*(a + b*x^2 + 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 - (b*A)/(4*a*B)])/(4*d*e*A*q*Sqrt[
a + b*x^2 + c*x^4]), x]] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + 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+b x^2-c x^4}} \, dx &=\frac{\sqrt{c} \int \frac{1}{\sqrt{-a+b x^2-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+b x^2-c x^4}} \, dx}{\sqrt{c} d-\sqrt{a} e}\\ &=\frac{\sqrt{e} \tan ^{-1}\left (\frac{\sqrt{-c d^2-e (b d+a e)} x}{\sqrt{d} \sqrt{e} \sqrt{-a+b x^2-c x^4}}\right )}{2 \sqrt{d} \sqrt{-c d^2-e (b d+a e)}}+\frac{\sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a-b x^2+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}{4} \left (2+\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{2 \sqrt [4]{a} \left (\sqrt{c} d-\sqrt{a} e\right ) \sqrt{-a+b x^2-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-b x^2+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}{4} \left (2+\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{4 \sqrt [4]{c} d \left (\sqrt{c} d-\sqrt{a} e\right ) \sqrt{-a+b x^2-c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.229861, size = 207, normalized size = 0.5 \[ -\frac{i \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}-b}+1} \sqrt{1-\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}} \Pi \left (-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{2 c d};i \sinh ^{-1}\left (\sqrt{2} \sqrt{-\frac{c}{b+\sqrt{b^2-4 a c}}} x\right )|-\frac{b+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}-b}\right )}{\sqrt{2} d \sqrt{-\frac{c}{\sqrt{b^2-4 a c}+b}} \sqrt{-a+b x^2-c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.023, size = 199, normalized size = 0.5 \begin{align*}{\frac{1}{d}\sqrt{1+{\frac{{x}^{2}}{2\,a}\sqrt{-4\,ac+{b}^{2}}}-{\frac{b{x}^{2}}{2\,a}}}\sqrt{1-{\frac{b{x}^{2}}{2\,a}}-{\frac{{x}^{2}}{2\,a}\sqrt{-4\,ac+{b}^{2}}}}{\it EllipticPi} \left ( \sqrt{-{\frac{1}{2\,a} \left ( \sqrt{-4\,ac+{b}^{2}}-b \right ) }}x,2\,{\frac{ae}{ \left ( \sqrt{-4\,ac+{b}^{2}}-b \right ) d}},{\frac{\sqrt{2}}{2}\sqrt{{\frac{1}{a} \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) }}{\frac{1}{\sqrt{-{\frac{1}{2\,a} \left ( \sqrt{-4\,ac+{b}^{2}}-b \right ) }}}}} \right ){\frac{1}{\sqrt{-{\frac{1}{2\,a}\sqrt{-4\,ac+{b}^{2}}}+{\frac{b}{2\,a}}}}}{\frac{1}{\sqrt{-c{x}^{4}+b{x}^{2}-a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate(1/(sqrt(-c*x^4 + b*x^2 - a)*(e*x^2 + d)), 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/(e*x^2+d)/(-c*x^4+b*x^2-a)^(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 (d + e x^{2}\right ) \sqrt{- a + b x^{2} - c x^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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