3.6 \(\int \frac{A+B x^2}{(d+e x^2)^2 \sqrt{a+c x^4}} \, dx\)

Optimal. Leaf size=641 \[ \frac{A \sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{2 \sqrt [4]{a} d \sqrt{a+c x^4} \left (\sqrt{c} d-\sqrt{a} e\right )}+\frac{\sqrt{c} x \sqrt{a+c x^4} (B d-A e)}{2 d \left (\sqrt{a}+\sqrt{c} x^2\right ) \left (a e^2+c d^2\right )}-\frac{e x \sqrt{a+c x^4} (B d-A e)}{2 d \left (d+e x^2\right ) \left (a e^2+c d^2\right )}-\frac{\left (-a A e^3-a B d e^2-3 A c d^2 e+B c d^3\right ) \tan ^{-1}\left (\frac{x \sqrt{a e^2+c d^2}}{\sqrt{d} \sqrt{e} \sqrt{a+c x^4}}\right )}{4 d^{3/2} \sqrt{e} \left (a e^2+c d^2\right )^{3/2}}-\frac{\sqrt [4]{a} \sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} (B d-A e) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 d \sqrt{a+c x^4} \left (a e^2+c d^2\right )}+\frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\sqrt{a} e+\sqrt{c} d\right ) \left (-a A e^3-a B d e^2-3 A c d^2 e+B c d^3\right ) \Pi \left (-\frac{\left (\sqrt{c} d-\sqrt{a} e\right )^2}{4 \sqrt{a} \sqrt{c} d e};2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{8 \sqrt [4]{a} \sqrt [4]{c} d^2 e \sqrt{a+c x^4} \left (\sqrt{c} d-\sqrt{a} e\right ) \left (a e^2+c d^2\right )} \]

[Out]

(Sqrt[c]*(B*d - A*e)*x*Sqrt[a + c*x^4])/(2*d*(c*d^2 + a*e^2)*(Sqrt[a] + Sqrt[c]*x^2)) - (e*(B*d - A*e)*x*Sqrt[
a + c*x^4])/(2*d*(c*d^2 + a*e^2)*(d + e*x^2)) - ((B*c*d^3 - 3*A*c*d^2*e - a*B*d*e^2 - a*A*e^3)*ArcTan[(Sqrt[c*
d^2 + a*e^2]*x)/(Sqrt[d]*Sqrt[e]*Sqrt[a + c*x^4])])/(4*d^(3/2)*Sqrt[e]*(c*d^2 + a*e^2)^(3/2)) - (a^(1/4)*c^(1/
4)*(B*d - A*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)
*x)/a^(1/4)], 1/2])/(2*d*(c*d^2 + a*e^2)*Sqrt[a + c*x^4]) + (A*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)*d*(Sqrt[c]*d - Sqrt[a]*
e)*Sqrt[a + c*x^4]) + ((Sqrt[c]*d + Sqrt[a]*e)*(B*c*d^3 - 3*A*c*d^2*e - a*B*d*e^2 - a*A*e^3)*(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])/(8*a^(1/4)*c^(1/4)*d^2*e*(Sqrt[c]*d - Sqrt[a]*e)*(c*d^2 + a*e^2)*Sqrt
[a + c*x^4])

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Rubi [A]  time = 1.18937, antiderivative size = 641, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {1697, 1715, 1196, 1709, 220, 1707} \[ \frac{\sqrt{c} x \sqrt{a+c x^4} (B d-A e)}{2 d \left (\sqrt{a}+\sqrt{c} x^2\right ) \left (a e^2+c d^2\right )}-\frac{e x \sqrt{a+c x^4} (B d-A e)}{2 d \left (d+e x^2\right ) \left (a e^2+c d^2\right )}-\frac{\left (-a A e^3-a B d e^2-3 A c d^2 e+B c d^3\right ) \tan ^{-1}\left (\frac{x \sqrt{a e^2+c d^2}}{\sqrt{d} \sqrt{e} \sqrt{a+c x^4}}\right )}{4 d^{3/2} \sqrt{e} \left (a e^2+c d^2\right )^{3/2}}-\frac{\sqrt [4]{a} \sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} (B d-A e) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 d \sqrt{a+c x^4} \left (a e^2+c d^2\right )}+\frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\sqrt{a} e+\sqrt{c} d\right ) \left (-a A e^3-a B d e^2-3 A c d^2 e+B c d^3\right ) \Pi \left (-\frac{\left (\sqrt{c} d-\sqrt{a} e\right )^2}{4 \sqrt{a} \sqrt{c} d e};2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{8 \sqrt [4]{a} \sqrt [4]{c} d^2 e \sqrt{a+c x^4} \left (\sqrt{c} d-\sqrt{a} e\right ) \left (a e^2+c d^2\right )}+\frac{A \sqrt [4]{c} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} d \sqrt{a+c x^4} \left (\sqrt{c} d-\sqrt{a} e\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*(B*d - A*e)*x*Sqrt[a + c*x^4])/(2*d*(c*d^2 + a*e^2)*(Sqrt[a] + Sqrt[c]*x^2)) - (e*(B*d - A*e)*x*Sqrt[
a + c*x^4])/(2*d*(c*d^2 + a*e^2)*(d + e*x^2)) - ((B*c*d^3 - 3*A*c*d^2*e - a*B*d*e^2 - a*A*e^3)*ArcTan[(Sqrt[c*
d^2 + a*e^2]*x)/(Sqrt[d]*Sqrt[e]*Sqrt[a + c*x^4])])/(4*d^(3/2)*Sqrt[e]*(c*d^2 + a*e^2)^(3/2)) - (a^(1/4)*c^(1/
4)*(B*d - A*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)
*x)/a^(1/4)], 1/2])/(2*d*(c*d^2 + a*e^2)*Sqrt[a + c*x^4]) + (A*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)*d*(Sqrt[c]*d - Sqrt[a]*
e)*Sqrt[a + c*x^4]) + ((Sqrt[c]*d + Sqrt[a]*e)*(B*c*d^3 - 3*A*c*d^2*e - a*B*d*e^2 - a*A*e^3)*(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])/(8*a^(1/4)*c^(1/4)*d^2*e*(Sqrt[c]*d - Sqrt[a]*e)*(c*d^2 + a*e^2)*Sqrt
[a + c*x^4])

Rule 1697

Int[((P4x_)*((d_) + (e_.)*(x_)^2)^(q_))/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{A = Coeff[P4x, x, 0], B
= Coeff[P4x, x, 2], C = Coeff[P4x, x, 4]}, -Simp[((C*d^2 - B*d*e + A*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*d
*(C*d - B*e) + A*(a*e^2*(2*q + 3) + 2*c*d^2*(q + 1)) + 2*d*(B*c*d - A*c*e + a*C*e)*(q + 1)*x^2 + c*(C*d^2 - B*
d*e + A*e^2)*(2*q + 5)*x^4, x])/Sqrt[a + c*x^4], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[P4x, x^2] && LeQ[E
xpon[P4x, x], 4] && NeQ[c*d^2 + a*e^2, 0] && ILtQ[q, -1]

Rule 1715

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

Rule 1196

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

Rule 1709

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

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{A+B x^2}{\left (d+e x^2\right )^2 \sqrt{a+c x^4}} \, dx &=-\frac{e (B d-A e) 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 A c d^2-a B d e-a A e^2-2 c d (B d-A e) x^2-c e (B d-A e) 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 (B d-A e) 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{-\sqrt{a} c^{3/2} d e (B d-A e)+c e \left (-2 A c d^2-a B d e-a A e^2\right )+\left (-2 c^2 d e (B d-A e)+c e (B d-A e) \left (c d-\sqrt{a} \sqrt{c} e\right )\right ) x^2}{\left (d+e x^2\right ) \sqrt{a+c x^4}} \, dx}{2 c d e \left (c d^2+a e^2\right )}-\frac{\left (\sqrt{a} \sqrt{c} (B d-A 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{\sqrt{c} (B d-A e) x \sqrt{a+c x^4}}{2 d \left (c d^2+a e^2\right ) \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{e (B d-A e) 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} (B d-A e) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 d \left (c d^2+a e^2\right ) \sqrt{a+c x^4}}+\frac{\left (A \sqrt{c}\right ) \int \frac{1}{\sqrt{a+c x^4}} \, dx}{d \left (\sqrt{c} d-\sqrt{a} e\right )}+\frac{\left (\sqrt{a} \left (B c d^3-3 A c d^2 e-a B d e^2-a A e^3\right )\right ) \int \frac{1+\frac{\sqrt{c} x^2}{\sqrt{a}}}{\left (d+e x^2\right ) \sqrt{a+c x^4}} \, dx}{2 d \left (\sqrt{c} d-\sqrt{a} e\right ) \left (c d^2+a e^2\right )}\\ &=\frac{\sqrt{c} (B d-A e) x \sqrt{a+c x^4}}{2 d \left (c d^2+a e^2\right ) \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{e (B d-A e) x \sqrt{a+c x^4}}{2 d \left (c d^2+a e^2\right ) \left (d+e x^2\right )}-\frac{\left (B c d^3-3 A c d^2 e-a B d e^2-a A e^3\right ) \tan ^{-1}\left (\frac{\sqrt{c d^2+a e^2} x}{\sqrt{d} \sqrt{e} \sqrt{a+c x^4}}\right )}{4 d^{3/2} \sqrt{e} \left (c d^2+a e^2\right )^{3/2}}-\frac{\sqrt [4]{a} \sqrt [4]{c} (B d-A e) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 d \left (c d^2+a e^2\right ) \sqrt{a+c x^4}}+\frac{A \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} d \left (\sqrt{c} d-\sqrt{a} e\right ) \sqrt{a+c x^4}}+\frac{\left (\sqrt{c} d+\sqrt{a} e\right ) \left (B c d^3-3 A c d^2 e-a B d e^2-a A e^3\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 )}{8 \sqrt [4]{a} \sqrt [4]{c} d^2 e \left (\sqrt{c} d-\sqrt{a} e\right ) \left (c d^2+a e^2\right ) \sqrt{a+c x^4}}\\ \end{align*}

Mathematica [C]  time = 1.23753, size = 297, normalized size = 0.46 \[ \frac{\frac{d e x \left (a+c x^4\right ) (A e-B d)}{\left (d+e x^2\right ) \left (a e^2+c d^2\right )}-\frac{i \sqrt{\frac{c x^4}{a}+1} \left (\sqrt{c} d \left (\sqrt{c} d-i \sqrt{a} e\right ) (B d-A e) \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{i \sqrt{c}}{\sqrt{a}}}\right ),-1\right )+\left (a A e^3+a B d e^2+3 A c d^2 e-B c d^3\right ) \Pi \left (-\frac{i \sqrt{a} e}{\sqrt{c} d};\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{c}}{\sqrt{a}}} x\right )\right |-1\right )+i \sqrt{a} \sqrt{c} d e (B d-A e) E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{c}}{\sqrt{a}}} x\right )\right |-1\right )\right )}{\sqrt{\frac{i \sqrt{c}}{\sqrt{a}}} \left (a e^3+c d^2 e\right )}}{2 d^2 \sqrt{a+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.025, size = 679, normalized size = 1.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*e-B*d)/e*(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)/(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)*EllipticF(x*(I/a^(1/2)*c^(1/2
))^(1/2),I)-1/2*I*c^(1/2)*e/(a*e^2+c*d^2)/d*a^(1/2)/(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)*EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I)+1/2*I*c^(1/2)*e/(a*e^
2+c*d^2)/d*a^(1/2)/(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)*EllipticE(x*(I/a^(1/2)*c^(1/2))^(1/2),I)+1/2/(a*e^2+c*d^2)/d^2*e^2/(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))*a+3/2/(a*e^2+c*d^2)/(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))*c)+B/e
/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)*E
llipticPi(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{B x^{2} + A}{\sqrt{c x^{4} + a}{\left (e x^{2} + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x^2 + A)/(sqrt(c*x^4 + a)*(e*x^2 + d)^2), 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((B*x^2+A)/(e*x^2+d)^2/(c*x^4+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{A + B x^{2}}{\sqrt{a + c x^{4}} \left (d + e x^{2}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*x**2)/(sqrt(a + c*x**4)*(d + e*x**2)**2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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