3.156 \(\int \frac {1}{(d+e x^2)^3 \sqrt {a+c x^4}} \, dx\)

Optimal. Leaf size=729 \[ \frac {3 \sqrt {e} \left (a^2 e^4+2 a c d^2 e^2+5 c^2 d^4\right ) \tan ^{-1}\left (\frac {x \sqrt {a e^2+c d^2}}{\sqrt {d} \sqrt {e} \sqrt {a+c x^4}}\right )}{16 d^{5/2} \left (a e^2+c d^2\right )^{5/2}}-\frac {3 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {a} e+\sqrt {c} d\right ) \left (a^2 e^4+2 a c d^2 e^2+5 c^2 d^4\right ) \Pi \left (-\frac {\left (\sqrt {c} d-\sqrt {a} e\right )^2}{4 \sqrt {a} \sqrt {c} d e};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{32 \sqrt [4]{a} \sqrt [4]{c} d^3 \sqrt {a+c x^4} \left (\sqrt {c} d-\sqrt {a} e\right ) \left (a e^2+c d^2\right )^2}+\frac {3 e^2 x \sqrt {a+c x^4} \left (a e^2+3 c d^2\right )}{8 d^2 \left (d+e x^2\right ) \left (a e^2+c d^2\right )^2}+\frac {e^2 x \sqrt {a+c x^4}}{4 d \left (d+e x^2\right )^2 \left (a e^2+c d^2\right )}-\frac {3 \sqrt {c} e x \sqrt {a+c x^4} \left (a e^2+3 c d^2\right )}{8 d^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \left (a e^2+c d^2\right )^2}+\frac {\sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (-\sqrt {a} \sqrt {c} d e+3 a e^2+4 c d^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{8 \sqrt [4]{a} d^2 \sqrt {a+c x^4} \left (\sqrt {c} d-\sqrt {a} e\right ) \left (a e^2+c d^2\right )}+\frac {3 \sqrt [4]{a} \sqrt [4]{c} e \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (a e^2+3 c d^2\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{8 d^2 \sqrt {a+c x^4} \left (a e^2+c d^2\right )^2} \]

[Out]

3/16*(a^2*e^4+2*a*c*d^2*e^2+5*c^2*d^4)*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
^(5/2)/(a*e^2+c*d^2)^(5/2)+1/4*e^2*x*(c*x^4+a)^(1/2)/d/(a*e^2+c*d^2)/(e*x^2+d)^2+3/8*e^2*(a*e^2+3*c*d^2)*x*(c*
x^4+a)^(1/2)/d^2/(a*e^2+c*d^2)^2/(e*x^2+d)-3/8*e*(a*e^2+3*c*d^2)*x*c^(1/2)*(c*x^4+a)^(1/2)/d^2/(a*e^2+c*d^2)^2
/(a^(1/2)+x^2*c^(1/2))+3/8*a^(1/4)*c^(1/4)*e*(a*e^2+3*c*d^2)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*
arctan(c^(1/4)*x/a^(1/4)))*EllipticE(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)/d^2/(a*e^2+c*d^2)^2/(c*x^4+a)^(1/2)-3/32*(a^2*e^4+2*a*c*d^2*e^2+5*c^2*d^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))*(e*a^(1/2)+d*c^(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)/c^(1/4)/d^3/(a*e^2+c*d^2)^2/(-e*a^(1/2)+d*c^(1/2))/
(c*x^4+a)^(1/2)+1/8*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)))*Ellipt
icF(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*c^(1/2))*(4*c*d^2+3*a*e^2-d*e*a^(1/2)*c^(1/2))*
((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(1/4)/d^2/(a*e^2+c*d^2)/(-e*a^(1/2)+d*c^(1/2))/(c*x^4+a)^(1/2)

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Rubi [A]  time = 1.25, antiderivative size = 729, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1224, 1697, 1715, 1196, 1709, 220, 1707} \[ \frac {3 \sqrt {e} \left (a^2 e^4+2 a c d^2 e^2+5 c^2 d^4\right ) \tan ^{-1}\left (\frac {x \sqrt {a e^2+c d^2}}{\sqrt {d} \sqrt {e} \sqrt {a+c x^4}}\right )}{16 d^{5/2} \left (a e^2+c d^2\right )^{5/2}}-\frac {3 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {a} e+\sqrt {c} d\right ) \left (a^2 e^4+2 a c d^2 e^2+5 c^2 d^4\right ) \Pi \left (-\frac {\left (\sqrt {c} d-\sqrt {a} e\right )^2}{4 \sqrt {a} \sqrt {c} d e};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{32 \sqrt [4]{a} \sqrt [4]{c} d^3 \sqrt {a+c x^4} \left (\sqrt {c} d-\sqrt {a} e\right ) \left (a e^2+c d^2\right )^2}+\frac {3 e^2 x \sqrt {a+c x^4} \left (a e^2+3 c d^2\right )}{8 d^2 \left (d+e x^2\right ) \left (a e^2+c d^2\right )^2}+\frac {e^2 x \sqrt {a+c x^4}}{4 d \left (d+e x^2\right )^2 \left (a e^2+c d^2\right )}-\frac {3 \sqrt {c} e x \sqrt {a+c x^4} \left (a e^2+3 c d^2\right )}{8 d^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \left (a e^2+c d^2\right )^2}+\frac {\sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (-\sqrt {a} \sqrt {c} d e+3 a e^2+4 c d^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{8 \sqrt [4]{a} d^2 \sqrt {a+c x^4} \left (\sqrt {c} d-\sqrt {a} e\right ) \left (a e^2+c d^2\right )}+\frac {3 \sqrt [4]{a} \sqrt [4]{c} e \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (a e^2+3 c d^2\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{8 d^2 \sqrt {a+c x^4} \left (a e^2+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x^2)^3*Sqrt[a + c*x^4]),x]

[Out]

(-3*Sqrt[c]*e*(3*c*d^2 + a*e^2)*x*Sqrt[a + c*x^4])/(8*d^2*(c*d^2 + a*e^2)^2*(Sqrt[a] + Sqrt[c]*x^2)) + (e^2*x*
Sqrt[a + c*x^4])/(4*d*(c*d^2 + a*e^2)*(d + e*x^2)^2) + (3*e^2*(3*c*d^2 + a*e^2)*x*Sqrt[a + c*x^4])/(8*d^2*(c*d
^2 + a*e^2)^2*(d + e*x^2)) + (3*Sqrt[e]*(5*c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4)*ArcTan[(Sqrt[c*d^2 + a*e^2]*x)/(
Sqrt[d]*Sqrt[e]*Sqrt[a + c*x^4])])/(16*d^(5/2)*(c*d^2 + a*e^2)^(5/2)) + (3*a^(1/4)*c^(1/4)*e*(3*c*d^2 + a*e^2)
*(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])/(8*d^2*(c*d^2 + a*e^2)^2*Sqrt[a + c*x^4]) + (c^(1/4)*(4*c*d^2 - Sqrt[a]*Sqrt[c]*d*e + 3*a*e^2)*(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])/(8*a
^(1/4)*d^2*(Sqrt[c]*d - Sqrt[a]*e)*(c*d^2 + a*e^2)*Sqrt[a + c*x^4]) - (3*(Sqrt[c]*d + Sqrt[a]*e)*(5*c^2*d^4 +
2*a*c*d^2*e^2 + a^2*e^4)*(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])/(32*a^(1/4)*c^(1/4)*d^3*(Sq
rt[c]*d - Sqrt[a]*e)*(c*d^2 + a*e^2)^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 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 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 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 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]

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 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]

Rubi steps

\begin {align*} \int \frac {1}{\left (d+e x^2\right )^3 \sqrt {a+c x^4}} \, dx &=\frac {e^2 x \sqrt {a+c x^4}}{4 d \left (c d^2+a e^2\right ) \left (d+e x^2\right )^2}-\frac {\int \frac {-4 c d^2-3 a e^2+4 c d e x^2-c e^2 x^4}{\left (d+e x^2\right )^2 \sqrt {a+c x^4}} \, dx}{4 d \left (c d^2+a e^2\right )}\\ &=\frac {e^2 x \sqrt {a+c x^4}}{4 d \left (c d^2+a e^2\right ) \left (d+e x^2\right )^2}+\frac {3 e^2 \left (3 c d^2+a e^2\right ) x \sqrt {a+c x^4}}{8 d^2 \left (c d^2+a e^2\right )^2 \left (d+e x^2\right )}+\frac {\int \frac {8 c^2 d^4+5 a c d^2 e^2+3 a^2 e^4-4 c d e \left (4 c d^2+a e^2\right ) x^2-3 c e^2 \left (3 c d^2+a e^2\right ) x^4}{\left (d+e x^2\right ) \sqrt {a+c x^4}} \, dx}{8 d^2 \left (c d^2+a e^2\right )^2}\\ &=\frac {e^2 x \sqrt {a+c x^4}}{4 d \left (c d^2+a e^2\right ) \left (d+e x^2\right )^2}+\frac {3 e^2 \left (3 c d^2+a e^2\right ) x \sqrt {a+c x^4}}{8 d^2 \left (c d^2+a e^2\right )^2 \left (d+e x^2\right )}+\frac {\int \frac {-3 \sqrt {a} c^{3/2} d e^2 \left (3 c d^2+a e^2\right )+c e \left (8 c^2 d^4+5 a c d^2 e^2+3 a^2 e^4\right )+\left (3 c e^2 \left (c d-\sqrt {a} \sqrt {c} e\right ) \left (3 c d^2+a e^2\right )-4 c^2 d e^2 \left (4 c d^2+a e^2\right )\right ) x^2}{\left (d+e x^2\right ) \sqrt {a+c x^4}} \, dx}{8 c d^2 e \left (c d^2+a e^2\right )^2}+\frac {\left (3 \sqrt {a} \sqrt {c} e \left (3 c d^2+a e^2\right )\right ) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+c x^4}} \, dx}{8 d^2 \left (c d^2+a e^2\right )^2}\\ &=-\frac {3 \sqrt {c} e \left (3 c d^2+a e^2\right ) x \sqrt {a+c x^4}}{8 d^2 \left (c d^2+a e^2\right )^2 \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {e^2 x \sqrt {a+c x^4}}{4 d \left (c d^2+a e^2\right ) \left (d+e x^2\right )^2}+\frac {3 e^2 \left (3 c d^2+a e^2\right ) x \sqrt {a+c x^4}}{8 d^2 \left (c d^2+a e^2\right )^2 \left (d+e x^2\right )}+\frac {3 \sqrt [4]{a} \sqrt [4]{c} e \left (3 c d^2+a e^2\right ) \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 )}{8 d^2 \left (c d^2+a e^2\right )^2 \sqrt {a+c x^4}}+\frac {\left (\sqrt {c} \left (4 c d^2-\sqrt {a} \sqrt {c} d e+3 a e^2\right )\right ) \int \frac {1}{\sqrt {a+c x^4}} \, dx}{4 d^2 \left (\sqrt {c} d-\sqrt {a} e\right ) \left (c d^2+a e^2\right )}-\frac {\left (3 \sqrt {a} e \left (5 c^2 d^4+2 a c d^2 e^2+a^2 e^4\right )\right ) \int \frac {1+\frac {\sqrt {c} x^2}{\sqrt {a}}}{\left (d+e x^2\right ) \sqrt {a+c x^4}} \, dx}{8 d^2 \left (\sqrt {c} d-\sqrt {a} e\right ) \left (c d^2+a e^2\right )^2}\\ &=-\frac {3 \sqrt {c} e \left (3 c d^2+a e^2\right ) x \sqrt {a+c x^4}}{8 d^2 \left (c d^2+a e^2\right )^2 \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {e^2 x \sqrt {a+c x^4}}{4 d \left (c d^2+a e^2\right ) \left (d+e x^2\right )^2}+\frac {3 e^2 \left (3 c d^2+a e^2\right ) x \sqrt {a+c x^4}}{8 d^2 \left (c d^2+a e^2\right )^2 \left (d+e x^2\right )}+\frac {3 \sqrt {e} \left (5 c^2 d^4+2 a c d^2 e^2+a^2 e^4\right ) \tan ^{-1}\left (\frac {\sqrt {c d^2+a e^2} x}{\sqrt {d} \sqrt {e} \sqrt {a+c x^4}}\right )}{16 d^{5/2} \left (c d^2+a e^2\right )^{5/2}}+\frac {3 \sqrt [4]{a} \sqrt [4]{c} e \left (3 c d^2+a e^2\right ) \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 )}{8 d^2 \left (c d^2+a e^2\right )^2 \sqrt {a+c x^4}}+\frac {\sqrt [4]{c} \left (4 c d^2-\sqrt {a} \sqrt {c} d e+3 a e^2\right ) \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 )}{8 \sqrt [4]{a} d^2 \left (\sqrt {c} d-\sqrt {a} e\right ) \left (c d^2+a e^2\right ) \sqrt {a+c x^4}}-\frac {3 \sqrt [4]{a} \left (\frac {\sqrt {c} d}{\sqrt {a}}+e\right ) \left (5 c^2 d^4+2 a c d^2 e^2+a^2 e^4\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 )}{32 \sqrt [4]{c} d^3 \left (\sqrt {c} d-\sqrt {a} e\right ) \left (c d^2+a e^2\right )^2 \sqrt {a+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 1.10, size = 332, normalized size = 0.46 \[ \frac {\frac {d e^2 x \left (a+c x^4\right ) \left (a e^2 \left (5 d+3 e x^2\right )+c d^2 \left (11 d+9 e x^2\right )\right )}{\left (d+e x^2\right )^2}+\frac {\sqrt {\frac {c x^4}{a}+1} \left (i \left (\sqrt {c} d \left (-3 i a^{3/2} e^3-9 i \sqrt {a} c d^2 e+a \sqrt {c} d e^2+7 c^{3/2} d^3\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )-3 \left (a^2 e^4+2 a c d^2 e^2+5 c^2 d^4\right ) \Pi \left (-\frac {i \sqrt {a} e}{\sqrt {c} d};\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )\right )-3 \sqrt {a} \sqrt {c} d e \left (a e^2+3 c d^2\right ) E\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}}{8 d^3 \sqrt {a+c x^4} \left (a e^2+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x^2)^3*Sqrt[a + c*x^4]),x]

[Out]

((d*e^2*x*(a + c*x^4)*(a*e^2*(5*d + 3*e*x^2) + c*d^2*(11*d + 9*e*x^2)))/(d + e*x^2)^2 + (Sqrt[1 + (c*x^4)/a]*(
-3*Sqrt[a]*Sqrt[c]*d*e*(3*c*d^2 + a*e^2)*EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[a]]*x], -1] + I*(Sqrt[c]*d*
(7*c^(3/2)*d^3 - (9*I)*Sqrt[a]*c*d^2*e + a*Sqrt[c]*d*e^2 - (3*I)*a^(3/2)*e^3)*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt
[c])/Sqrt[a]]*x], -1] - 3*(5*c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4)*EllipticPi[((-I)*Sqrt[a]*e)/(Sqrt[c]*d), I*Arc
Sinh[Sqrt[(I*Sqrt[c])/Sqrt[a]]*x], -1])))/Sqrt[(I*Sqrt[c])/Sqrt[a]])/(8*d^3*(c*d^2 + a*e^2)^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)^3/(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 \[ \int \frac {1}{\sqrt {c x^{4} + a} {\left (e x^{2} + d\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x^2+d)^3/(c*x^4+a)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(c*x^4 + a)*(e*x^2 + d)^3), x)

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maple [C]  time = 0.03, size = 1018, normalized size = 1.40 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x^2+d)^3/(c*x^4+a)^(1/2),x)

[Out]

1/4*e^2*x*(c*x^4+a)^(1/2)/d/(a*e^2+c*d^2)/(e*x^2+d)^2+3/8*e^2*(a*e^2+3*c*d^2)*x*(c*x^4+a)^(1/2)/d^2/(a*e^2+c*d
^2)^2/(e*x^2+d)-1/8*c/d/(a*e^2+c*d^2)^2/(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)*EllipticF((I/a^(1/2)*c^(1/2))^(1/2)*x,I)*a*e^2-7/8*c^2*d/(a*e^2+c*d^2)^2/
(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)*Ell
ipticF((I/a^(1/2)*c^(1/2))^(1/2)*x,I)-9/8*I*c^(3/2)*e/(a*e^2+c*d^2)^2*a^(1/2)/(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)*EllipticF((I/a^(1/2)*c^(1/2))^(1/2)
*x,I)-3/8*I*c^(1/2)*e^3/(a*e^2+c*d^2)^2/d^2*a^(3/2)/(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)*EllipticF((I/a^(1/2)*c^(1/2))^(1/2)*x,I)+9/8*I*c^(3/2)*e/(a*e
^2+c*d^2)^2*a^(1/2)/(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)*EllipticE((I/a^(1/2)*c^(1/2))^(1/2)*x,I)+3/8*I*c^(1/2)*e^3/(a*e^2+c*d^2)^2/d^2*a^(3/2)/(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)*EllipticE
((I/a^(1/2)*c^(1/2))^(1/2)*x,I)+3/8/(a*e^2+c*d^2)^2/d^3*e^4/(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)/d*e,(-I/a^(1/2)*c^(1/2))^(1/2)/(I/a^(1/2)*c^(1/2))^(1/2))*a^2+3/4/(a*e^2+c*d^2)^2/d*e^2/(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)/d*e,(-I/a^(1/2)*c^(1/2))^(1/2)/(I/a^(1/2)*c^(1/2))^(1/2))*a*c+15/8/(a*e^2+
c*d^2)^2*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)/d*e,(-I/a^(1/2)*c^(1/2))^(1/2)/(I/a^(1/2)*c^(
1/2))^(1/2))*c^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c x^{4} + a} {\left (e x^{2} + d\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x^2+d)^3/(c*x^4+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(c*x^4 + a)*(e*x^2 + d)^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{\sqrt {c\,x^4+a}\,{\left (e\,x^2+d\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + c*x^4)^(1/2)*(d + e*x^2)^3),x)

[Out]

int(1/((a + c*x^4)^(1/2)*(d + e*x^2)^3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a + c x^{4}} \left (d + e x^{2}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x**2+d)**3/(c*x**4+a)**(1/2),x)

[Out]

Integral(1/(sqrt(a + c*x**4)*(d + e*x**2)**3), x)

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