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

Optimal. Leaf size=299 \[ -\frac {a^{3/4} \sqrt [4]{c} e \sqrt {1-\frac {c x^4}{a}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{2 d \sqrt {a-c x^4} \left (c d^2-a e^2\right )}+\frac {\sqrt [4]{a} \sqrt {1-\frac {c x^4}{a}} \left (3 c d^2-a e^2\right ) \Pi \left (-\frac {\sqrt {a} e}{\sqrt {c} d};\left .\sin ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{2 \sqrt [4]{c} d^2 \sqrt {a-c x^4} \left (c d^2-a e^2\right )}-\frac {e^2 x \sqrt {a-c x^4}}{2 d \left (d+e x^2\right ) \left (c d^2-a e^2\right )}-\frac {\sqrt [4]{a} \sqrt [4]{c} \sqrt {1-\frac {c x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{2 d \sqrt {a-c x^4} \left (\sqrt {a} e+\sqrt {c} d\right )} \]

[Out]

-1/2*e^2*x*(-c*x^4+a)^(1/2)/d/(-a*e^2+c*d^2)/(e*x^2+d)-1/2*a^(3/4)*c^(1/4)*e*EllipticE(c^(1/4)*x/a^(1/4),I)*(1
-c*x^4/a)^(1/2)/d/(-a*e^2+c*d^2)/(-c*x^4+a)^(1/2)+1/2*a^(1/4)*(-a*e^2+3*c*d^2)*EllipticPi(c^(1/4)*x/a^(1/4),-e
*a^(1/2)/d/c^(1/2),I)*(1-c*x^4/a)^(1/2)/c^(1/4)/d^2/(-a*e^2+c*d^2)/(-c*x^4+a)^(1/2)-1/2*a^(1/4)*c^(1/4)*Ellipt
icF(c^(1/4)*x/a^(1/4),I)*(1-c*x^4/a)^(1/2)/d/(e*a^(1/2)+d*c^(1/2))/(-c*x^4+a)^(1/2)

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Rubi [A]  time = 0.36, antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {1224, 1717, 1201, 224, 221, 1200, 1199, 424, 1219, 1218} \[ -\frac {a^{3/4} \sqrt [4]{c} e \sqrt {1-\frac {c x^4}{a}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{2 d \sqrt {a-c x^4} \left (c d^2-a e^2\right )}-\frac {e^2 x \sqrt {a-c x^4}}{2 d \left (d+e x^2\right ) \left (c d^2-a e^2\right )}+\frac {\sqrt [4]{a} \sqrt {1-\frac {c x^4}{a}} \left (3 c d^2-a e^2\right ) \Pi \left (-\frac {\sqrt {a} e}{\sqrt {c} d};\left .\sin ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{2 \sqrt [4]{c} d^2 \sqrt {a-c x^4} \left (c d^2-a e^2\right )}-\frac {\sqrt [4]{a} \sqrt [4]{c} \sqrt {1-\frac {c x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{2 d \sqrt {a-c x^4} \left (\sqrt {a} e+\sqrt {c} d\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + (b*x^4)/a]/Sqrt[a + b*x^4], Int[1/Sqrt[1 + (b*x^4)
/a], x], x] /; FreeQ[{a, b}, x] && NegQ[b/a] &&  !GtQ[a, 0]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 1199

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + (e*x^2)/d]/Sqrt
[1 - (e*x^2)/d], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

Rule 1200

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + (c*x^4)/a]/Sqrt[a + c*x^4], In
t[(d + e*x^2)/Sqrt[1 + (c*x^4)/a], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] &&
!GtQ[a, 0]

Rule 1201

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

Rule 1218

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(c/a), 4]}, Simp[(1*Ellipt
icPi[-(e/(d*q^2)), ArcSin[q*x], -1])/(d*Sqrt[a]*q), x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]

Rule 1219

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[Sqrt[1 + (c*x^4)/a]/Sqrt[a + c*x^4]
, Int[1/((d + e*x^2)*Sqrt[1 + (c*x^4)/a]), x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] &&  !GtQ[a, 0]

Rule 1224

Int[((d_) + (e_.)*(x_)^2)^(q_)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> -Simp[(e^2*x*(d + e*x^2)^(q + 1)*Sqrt[a
+ c*x^4])/(2*d*(q + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*d*(q + 1)*(c*d^2 + a*e^2)), Int[((d + e*x^2)^(q + 1)*
Simp[a*e^2*(2*q + 3) + 2*c*d^2*(q + 1) - 2*e*c*d*(q + 1)*x^2 + c*e^2*(2*q + 5)*x^4, x])/Sqrt[a + c*x^4], x], x
] /; FreeQ[{a, c, d, e}, x] && ILtQ[q, -1]

Rule 1717

Int[(P4x_)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{A = Coeff[P4x, x, 0], B = Coe
ff[P4x, x, 2], C = Coeff[P4x, x, 4]}, -Dist[(e^2)^(-1), Int[(C*d - B*e - C*e*x^2)/Sqrt[a + c*x^4], x], x] + Di
st[(C*d^2 - B*d*e + A*e^2)/e^2, Int[1/((d + e*x^2)*Sqrt[a + c*x^4]), x], x]] /; FreeQ[{a, c, d, e}, x] && Poly
Q[P4x, x^2, 2] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0]

Rubi steps

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

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Mathematica [C]  time = 0.96, size = 508, normalized size = 1.70 \[ \frac {-3 i c d^3 \sqrt {1-\frac {c x^4}{a}} \Pi \left (-\frac {\sqrt {a} e}{\sqrt {c} d};\left .i \sinh ^{-1}\left (\sqrt {-\frac {\sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )-3 i c d^2 e x^2 \sqrt {1-\frac {c x^4}{a}} \Pi \left (-\frac {\sqrt {a} e}{\sqrt {c} d};\left .i \sinh ^{-1}\left (\sqrt {-\frac {\sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )+i a e^3 x^2 \sqrt {1-\frac {c x^4}{a}} \Pi \left (-\frac {\sqrt {a} e}{\sqrt {c} d};\left .i \sinh ^{-1}\left (\sqrt {-\frac {\sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )+c d e^2 x^5 \sqrt {-\frac {\sqrt {c}}{\sqrt {a}}}+i a d e^2 \sqrt {1-\frac {c x^4}{a}} \Pi \left (-\frac {\sqrt {a} e}{\sqrt {c} d};\left .i \sinh ^{-1}\left (\sqrt {-\frac {\sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )-a d e^2 x \sqrt {-\frac {\sqrt {c}}{\sqrt {a}}}-i \sqrt {c} d \sqrt {1-\frac {c x^4}{a}} \left (d+e x^2\right ) \left (\sqrt {a} e-\sqrt {c} d\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt {-\frac {\sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )+i \sqrt {a} \sqrt {c} d e \sqrt {1-\frac {c x^4}{a}} \left (d+e x^2\right ) E\left (\left .i \sinh ^{-1}\left (\sqrt {-\frac {\sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )}{2 d^2 \sqrt {-\frac {\sqrt {c}}{\sqrt {a}}} \sqrt {a-c x^4} \left (d+e x^2\right ) \left (c d^2-a e^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a*Sqrt[-(Sqrt[c]/Sqrt[a])]*d*e^2*x) + Sqrt[-(Sqrt[c]/Sqrt[a])]*c*d*e^2*x^5 + I*Sqrt[a]*Sqrt[c]*d*e*(d + e*x
^2)*Sqrt[1 - (c*x^4)/a]*EllipticE[I*ArcSinh[Sqrt[-(Sqrt[c]/Sqrt[a])]*x], -1] - I*Sqrt[c]*d*(-(Sqrt[c]*d) + Sqr
t[a]*e)*(d + e*x^2)*Sqrt[1 - (c*x^4)/a]*EllipticF[I*ArcSinh[Sqrt[-(Sqrt[c]/Sqrt[a])]*x], -1] - (3*I)*c*d^3*Sqr
t[1 - (c*x^4)/a]*EllipticPi[-((Sqrt[a]*e)/(Sqrt[c]*d)), I*ArcSinh[Sqrt[-(Sqrt[c]/Sqrt[a])]*x], -1] + I*a*d*e^2
*Sqrt[1 - (c*x^4)/a]*EllipticPi[-((Sqrt[a]*e)/(Sqrt[c]*d)), I*ArcSinh[Sqrt[-(Sqrt[c]/Sqrt[a])]*x], -1] - (3*I)
*c*d^2*e*x^2*Sqrt[1 - (c*x^4)/a]*EllipticPi[-((Sqrt[a]*e)/(Sqrt[c]*d)), I*ArcSinh[Sqrt[-(Sqrt[c]/Sqrt[a])]*x],
-1] + I*a*e^3*x^2*Sqrt[1 - (c*x^4)/a]*EllipticPi[-((Sqrt[a]*e)/(Sqrt[c]*d)), I*ArcSinh[Sqrt[-(Sqrt[c]/Sqrt[a]
)]*x], -1])/(2*Sqrt[-(Sqrt[c]/Sqrt[a])]*d^2*(c*d^2 - a*e^2)*(d + e*x^2)*Sqrt[a - 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)^2/(-c*x^4+a)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.03, size = 523, normalized size = 1.75 \[ \frac {\sqrt {-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, a \,e^{2} \EllipticPi \left (\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, x , -\frac {\sqrt {a}\, e}{\sqrt {c}\, d}, \frac {\sqrt {-\frac {\sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}}\right )}{2 \left (a \,e^{2}-c \,d^{2}\right ) \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}\, d^{2}}+\frac {\sqrt {-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {a}\, \sqrt {c}\, e \EllipticE \left (\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, x , i\right )}{2 \left (a \,e^{2}-c \,d^{2}\right ) \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}\, d}-\frac {\sqrt {-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {a}\, \sqrt {c}\, e \EllipticF \left (\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, x , i\right )}{2 \left (a \,e^{2}-c \,d^{2}\right ) \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}\, d}+\frac {\sqrt {-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, c \EllipticF \left (\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, x , i\right )}{2 \left (a \,e^{2}-c \,d^{2}\right ) \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}}-\frac {3 \sqrt {-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, c \EllipticPi \left (\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, x , -\frac {\sqrt {a}\, e}{\sqrt {c}\, d}, \frac {\sqrt {-\frac {\sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}}\right )}{2 \left (a \,e^{2}-c \,d^{2}\right ) \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}}+\frac {\sqrt {-c \,x^{4}+a}\, e^{2} x}{2 \left (a \,e^{2}-c \,d^{2}\right ) \left (e \,x^{2}+d \right ) d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*e^2/(a*e^2-c*d^2)/d*x*(-c*x^4+a)^(1/2)/(e*x^2+d)+1/2*c/(a*e^2-c*d^2)/(1/a^(1/2)*c^(1/2))^(1/2)*(-1/a^(1/2)
*c^(1/2)*x^2+1)^(1/2)*(1/a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(-c*x^4+a)^(1/2)*EllipticF((1/a^(1/2)*c^(1/2))^(1/2)*x,I
)-1/2*c^(1/2)*e/(a*e^2-c*d^2)/d*a^(1/2)/(1/a^(1/2)*c^(1/2))^(1/2)*(-1/a^(1/2)*c^(1/2)*x^2+1)^(1/2)*(1/a^(1/2)*
c^(1/2)*x^2+1)^(1/2)/(-c*x^4+a)^(1/2)*EllipticF((1/a^(1/2)*c^(1/2))^(1/2)*x,I)+1/2*c^(1/2)*e/(a*e^2-c*d^2)/d*a
^(1/2)/(1/a^(1/2)*c^(1/2))^(1/2)*(-1/a^(1/2)*c^(1/2)*x^2+1)^(1/2)*(1/a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(-c*x^4+a)^(
1/2)*EllipticE((1/a^(1/2)*c^(1/2))^(1/2)*x,I)+1/2/(a*e^2-c*d^2)/d^2*e^2/(1/a^(1/2)*c^(1/2))^(1/2)*(-1/a^(1/2)*
c^(1/2)*x^2+1)^(1/2)*(1/a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(-c*x^4+a)^(1/2)*EllipticPi((1/a^(1/2)*c^(1/2))^(1/2)*x,-
a^(1/2)/c^(1/2)/d*e,(-1/a^(1/2)*c^(1/2))^(1/2)/(1/a^(1/2)*c^(1/2))^(1/2))*a-3/2/(a*e^2-c*d^2)/(1/a^(1/2)*c^(1/
2))^(1/2)*(-1/a^(1/2)*c^(1/2)*x^2+1)^(1/2)*(1/a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(-c*x^4+a)^(1/2)*EllipticPi((1/a^(1
/2)*c^(1/2))^(1/2)*x,-a^(1/2)/c^(1/2)/d*e,(-1/a^(1/2)*c^(1/2))^(1/2)/(1/a^(1/2)*c^(1/2))^(1/2))*c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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