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

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

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 347, normalized size of antiderivative = 1.00, 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 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 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 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*}

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Mathematica [C]  time = 0.15, 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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fricas [F]  time = 10.93, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c x^{4} - a}}{c e x^{6} + c d x^{4} + a e x^{2} + a d}, x\right ) \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 110, normalized size = 0.32 \[ \frac {\sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \EllipticPi \left (\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}\, x , -\frac {i \sqrt {a}\, e}{\sqrt {c}\, d}, \frac {\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}\right )}{\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}-a}\, d} \]

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)*(I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)*(-I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(-c*x^4-a)^(1/
2)*EllipticPi((-I/a^(1/2)*c^(1/2))^(1/2)*x,-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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-c x^{4} - a} {\left (e x^{2} + d\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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