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

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

[Out]

1/2*arctan(x*(-a*e^2-b*d*e-c*d^2)^(1/2)/d^(1/2)/e^(1/2)/(-c*x^4+b*x^2-a)^(1/2))*e^(1/2)/d^(1/2)/(-a*e^2-b*d*e-
c*d^2)^(1/2)+1/2*c^(1/4)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticF
(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*(2+b/a^(1/2)/c^(1/2))^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4-b*x^2+a)/(a^(
1/2)+x^2*c^(1/2))^2)^(1/2)/a^(1/4)/(-e*a^(1/2)+d*c^(1/2))/(-c*x^4+b*x^2-a)^(1/2)-1/4*a^(3/4)*(cos(2*arctan(c^(
1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticPi(sin(2*arctan(c^(1/4)*x/a^(1/4))),-1/4*(-
e*a^(1/2)+d*c^(1/2))^2/d/e/a^(1/2)/c^(1/2),1/2*(2+b/a^(1/2)/c^(1/2))^(1/2))*(a^(1/2)+x^2*c^(1/2))*(e+d*c^(1/2)
/a^(1/2))^2*((c*x^4-b*x^2+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/c^(1/4)/d/(-a*e^2+c*d^2)/(-c*x^4+b*x^2-a)^(1/2)

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Rubi [A]  time = 0.36, antiderivative size = 412, normalized size of antiderivative = 1.00, 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 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 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 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*}

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Mathematica [C]  time = 0.22, size = 207, normalized size = 0.50 \[ -\frac {i \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}-b}+1} \sqrt {1-\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}} \Pi \left (-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c d};i \sinh ^{-1}\left (\sqrt {2} \sqrt {-\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|-\frac {b+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}-b}\right )}{\sqrt {2} d \sqrt {-\frac {c}{\sqrt {b^2-4 a c}+b}} \sqrt {-a+b x^2-c x^4}} \]

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[-1/2
*((b + Sqrt[b^2 - 4*a*c])*e)/(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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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-c x^{4} + b x^{2} - a} {\left (e x^{2} + d\right )}}\,{d x} \]

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)

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maple [A]  time = 0.03, size = 199, normalized size = 0.48 \[ \frac {\sqrt {-\frac {b \,x^{2}}{2 a}+\frac {\sqrt {-4 a c +b^{2}}\, x^{2}}{2 a}+1}\, \sqrt {-\frac {b \,x^{2}}{2 a}-\frac {\sqrt {-4 a c +b^{2}}\, x^{2}}{2 a}+1}\, \EllipticPi \left (\sqrt {-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}}\, x , \frac {2 a e}{\left (-b +\sqrt {-4 a c +b^{2}}\right ) d}, \frac {\sqrt {2}\, \sqrt {\frac {b +\sqrt {-4 a c +b^{2}}}{a}}}{2 \sqrt {-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}}}\right )}{\sqrt {\frac {b}{2 a}-\frac {\sqrt {-4 a c +b^{2}}}{2 a}}\, \sqrt {-c \,x^{4}+b \,x^{2}-a}\, d} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-c x^{4} + b x^{2} - a} {\left (e x^{2} + d\right )}}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (d + e x^{2}\right ) \sqrt {- a + b x^{2} - c x^{4}}}\, dx \]

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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