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

Optimal. Leaf size=610 \[ -\frac {c^{3/4} d^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \Pi \left (\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right )^2}{4 \sqrt {a} \sqrt {c} d^2 e^2};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt {a+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right ) \left (a e^4+c d^4\right )}-\frac {e^3 \sqrt {a+c x^4}}{(d+e x) \left (a e^4+c d^4\right )}+\frac {\sqrt {c} e^2 x \sqrt {a+c x^4}}{\left (\sqrt {a}+\sqrt {c} x^2\right ) \left (a e^4+c d^4\right )}-\frac {\sqrt [4]{a} \sqrt [4]{c} e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt {a+c x^4} \left (a e^4+c d^4\right )}+\frac {\sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt {a+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}-\frac {c d^3 e \tan ^{-1}\left (\frac {x \sqrt {-a e^4-c d^4}}{d e \sqrt {a+c x^4}}\right )}{\left (-a e^4-c d^4\right )^{3/2}}-\frac {c d^3 e \tanh ^{-1}\left (\frac {a e^2+c d^2 x^2}{\sqrt {a+c x^4} \sqrt {a e^4+c d^4}}\right )}{\left (a e^4+c d^4\right )^{3/2}} \]

[Out]

-c*d^3*e*arctan(x*(-a*e^4-c*d^4)^(1/2)/d/e/(c*x^4+a)^(1/2))/(-a*e^4-c*d^4)^(3/2)-c*d^3*e*arctanh((c*d^2*x^2+a*
e^2)/(a*e^4+c*d^4)^(1/2)/(c*x^4+a)^(1/2))/(a*e^4+c*d^4)^(3/2)-e^3*(c*x^4+a)^(1/2)/(a*e^4+c*d^4)/(e*x+d)+e^2*x*
c^(1/2)*(c*x^4+a)^(1/2)/(a*e^4+c*d^4)/(a^(1/2)+x^2*c^(1/2))-a^(1/4)*c^(1/4)*e^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)/(a*e^4+c*d^4)/(c*x^4+a)^(1/2)+1/2*c^(1/4)*(cos(2*arcta
n(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^2*a^(1/2)+d^2*c^(1/2))/(c
*x^4+a)^(1/2)-1/2*c^(3/4)*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)))*Elli
pticPi(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/4*(e^2*a^(1/2)+d^2*c^(1/2))^2/d^2/e^2/a^(1/2)/c^(1/2),1/2*2^(1/2))*(
-e^2*a^(1/2)+d^2*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)/(a*e^4+c*d^4
)/(e^2*a^(1/2)+d^2*c^(1/2))/(c*x^4+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.76, antiderivative size = 610, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.579, Rules used = {1727, 1742, 12, 1248, 725, 206, 1715, 1196, 1709, 220, 1707} \[ -\frac {c^{3/4} d^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \Pi \left (\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right )^2}{4 \sqrt {a} \sqrt {c} d^2 e^2};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt {a+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right ) \left (a e^4+c d^4\right )}-\frac {e^3 \sqrt {a+c x^4}}{(d+e x) \left (a e^4+c d^4\right )}+\frac {\sqrt {c} e^2 x \sqrt {a+c x^4}}{\left (\sqrt {a}+\sqrt {c} x^2\right ) \left (a e^4+c d^4\right )}-\frac {c d^3 e \tan ^{-1}\left (\frac {x \sqrt {-a e^4-c d^4}}{d e \sqrt {a+c x^4}}\right )}{\left (-a e^4-c d^4\right )^{3/2}}-\frac {c d^3 e \tanh ^{-1}\left (\frac {a e^2+c d^2 x^2}{\sqrt {a+c x^4} \sqrt {a e^4+c d^4}}\right )}{\left (a e^4+c d^4\right )^{3/2}}+\frac {\sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt {a+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}-\frac {\sqrt [4]{a} \sqrt [4]{c} e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt {a+c x^4} \left (a e^4+c d^4\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((e^3*Sqrt[a + c*x^4])/((c*d^4 + a*e^4)*(d + e*x))) + (Sqrt[c]*e^2*x*Sqrt[a + c*x^4])/((c*d^4 + a*e^4)*(Sqrt[
a] + Sqrt[c]*x^2)) - (c*d^3*e*ArcTan[(Sqrt[-(c*d^4) - a*e^4]*x)/(d*e*Sqrt[a + c*x^4])])/(-(c*d^4) - a*e^4)^(3/
2) - (c*d^3*e*ArcTanh[(a*e^2 + c*d^2*x^2)/(Sqrt[c*d^4 + a*e^4]*Sqrt[a + c*x^4])])/(c*d^4 + a*e^4)^(3/2) - (a^(
1/4)*c^(1/4)*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])/((c*d^4 + a*e^4)*Sqrt[a + c*x^4]) + (c^(1/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(S
qrt[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^2 + Sqrt[a]*e^2)
*Sqrt[a + c*x^4]) - (c^(3/4)*d^2*(Sqrt[c]*d^2 - 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^2 + Sqrt[a]*e^2)^2/(4*Sqrt[a]*Sqrt[c]*d^2*e^2), 2*ArcTan[(c^(1/4)*x)/
a^(1/4)], 1/2])/(2*a^(1/4)*(Sqrt[c]*d^2 + Sqrt[a]*e^2)*(c*d^4 + a*e^4)*Sqrt[a + c*x^4])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 725

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

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 1248

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

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]

Rule 1727

Int[((d_) + (e_.)*(x_))^(q_)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Simp[(e^3*(d + e*x)^(q + 1)*Sqrt[a + c*x^
4])/((q + 1)*(c*d^4 + a*e^4)), x] + Dist[c/((q + 1)*(c*d^4 + a*e^4)), Int[((d + e*x)^(q + 1)*Simp[d^3*(q + 1)
- d^2*e*(q + 1)*x + d*e^2*(q + 1)*x^2 - e^3*(q + 3)*x^3, x])/Sqrt[a + c*x^4], x], x] /; FreeQ[{a, c, d, e}, x]
&& NeQ[c*d^4 + a*e^4, 0] && ILtQ[q, -1]

Rule 1742

Int[(Px_)/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{A = Coeff[Px, x, 0], B = Coeff[P
x, x, 1], C = Coeff[Px, x, 2], D = Coeff[Px, x, 3]}, Int[(x*(B*d - A*e + (d*D - C*e)*x^2))/((d^2 - e^2*x^2)*Sq
rt[a + c*x^4]), x] + Int[(A*d + (C*d - B*e)*x^2 - D*e*x^4)/((d^2 - e^2*x^2)*Sqrt[a + c*x^4]), x]] /; FreeQ[{a,
c, d, e}, x] && PolyQ[Px, x] && LeQ[Expon[Px, x], 3] && NeQ[c*d^4 + a*e^4, 0]

Rubi steps

\begin {align*} \int \frac {1}{(d+e x)^2 \sqrt {a+c x^4}} \, dx &=-\frac {e^3 \sqrt {a+c x^4}}{\left (c d^4+a e^4\right ) (d+e x)}-\frac {c \int \frac {-d^3+d^2 e x-d e^2 x^2-e^3 x^3}{(d+e x) \sqrt {a+c x^4}} \, dx}{c d^4+a e^4}\\ &=-\frac {e^3 \sqrt {a+c x^4}}{\left (c d^4+a e^4\right ) (d+e x)}-\frac {c \int \frac {2 d^3 e x}{\left (d^2-e^2 x^2\right ) \sqrt {a+c x^4}} \, dx}{c d^4+a e^4}-\frac {c \int \frac {-d^4-2 d^2 e^2 x^2+e^4 x^4}{\left (d^2-e^2 x^2\right ) \sqrt {a+c x^4}} \, dx}{c d^4+a e^4}\\ &=-\frac {e^3 \sqrt {a+c x^4}}{\left (c d^4+a e^4\right ) (d+e x)}+\frac {\int \frac {c d^4 e^2+\sqrt {a} \sqrt {c} d^2 e^4+\left (2 c d^2 e^4-e^4 \left (c d^2+\sqrt {a} \sqrt {c} e^2\right )\right ) x^2}{\left (d^2-e^2 x^2\right ) \sqrt {a+c x^4}} \, dx}{e^2 \left (c d^4+a e^4\right )}-\frac {\left (2 c d^3 e\right ) \int \frac {x}{\left (d^2-e^2 x^2\right ) \sqrt {a+c x^4}} \, dx}{c d^4+a e^4}-\frac {\left (\sqrt {a} \sqrt {c} e^2\right ) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+c x^4}} \, dx}{c d^4+a e^4}\\ &=-\frac {e^3 \sqrt {a+c x^4}}{\left (c d^4+a e^4\right ) (d+e x)}+\frac {\sqrt {c} e^2 x \sqrt {a+c x^4}}{\left (c d^4+a e^4\right ) \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {\sqrt [4]{a} \sqrt [4]{c} e^2 \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 )}{\left (c d^4+a e^4\right ) \sqrt {a+c x^4}}+\frac {\sqrt {c} \int \frac {1}{\sqrt {a+c x^4}} \, dx}{\sqrt {c} d^2+\sqrt {a} e^2}-\frac {\left (c d^3 e\right ) \operatorname {Subst}\left (\int \frac {1}{\left (d^2-e^2 x\right ) \sqrt {a+c x^2}} \, dx,x,x^2\right )}{c d^4+a e^4}+\frac {\left (2 \sqrt {a} c d^4 e^2\right ) \int \frac {1+\frac {\sqrt {c} x^2}{\sqrt {a}}}{\left (d^2-e^2 x^2\right ) \sqrt {a+c x^4}} \, dx}{\left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \left (c d^4+a e^4\right )}\\ &=-\frac {e^3 \sqrt {a+c x^4}}{\left (c d^4+a e^4\right ) (d+e x)}+\frac {\sqrt {c} e^2 x \sqrt {a+c x^4}}{\left (c d^4+a e^4\right ) \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {c d^3 e \tan ^{-1}\left (\frac {\sqrt {-c d^4-a e^4} x}{d e \sqrt {a+c x^4}}\right )}{\left (-c d^4-a e^4\right )^{3/2}}-\frac {\sqrt [4]{a} \sqrt [4]{c} e^2 \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 )}{\left (c d^4+a e^4\right ) \sqrt {a+c x^4}}+\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^2+\sqrt {a} e^2\right ) \sqrt {a+c x^4}}-\frac {c^{3/4} d^2 \left (\sqrt {c} d^2-\sqrt {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}} \Pi \left (\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right )^2}{4 \sqrt {a} \sqrt {c} d^2 e^2};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^2+\sqrt {a} e^2\right ) \left (c d^4+a e^4\right ) \sqrt {a+c x^4}}+\frac {\left (c d^3 e\right ) \operatorname {Subst}\left (\int \frac {1}{c d^4+a e^4-x^2} \, dx,x,\frac {-a e^2-c d^2 x^2}{\sqrt {a+c x^4}}\right )}{c d^4+a e^4}\\ &=-\frac {e^3 \sqrt {a+c x^4}}{\left (c d^4+a e^4\right ) (d+e x)}+\frac {\sqrt {c} e^2 x \sqrt {a+c x^4}}{\left (c d^4+a e^4\right ) \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {c d^3 e \tan ^{-1}\left (\frac {\sqrt {-c d^4-a e^4} x}{d e \sqrt {a+c x^4}}\right )}{\left (-c d^4-a e^4\right )^{3/2}}-\frac {c d^3 e \tanh ^{-1}\left (\frac {a e^2+c d^2 x^2}{\sqrt {c d^4+a e^4} \sqrt {a+c x^4}}\right )}{\left (c d^4+a e^4\right )^{3/2}}-\frac {\sqrt [4]{a} \sqrt [4]{c} e^2 \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 )}{\left (c d^4+a e^4\right ) \sqrt {a+c x^4}}+\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^2+\sqrt {a} e^2\right ) \sqrt {a+c x^4}}-\frac {c^{3/4} d^2 \left (\sqrt {c} d^2-\sqrt {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}} \Pi \left (\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right )^2}{4 \sqrt {a} \sqrt {c} d^2 e^2};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^2+\sqrt {a} e^2\right ) \left (c d^4+a e^4\right ) \sqrt {a+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 1.11, size = 425, normalized size = 0.70 \[ \frac {-\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} \left (2 \sqrt [4]{-1} \sqrt [4]{a} c^{3/4} d^2 \sqrt {\frac {c x^4}{a}+1} (d+e x) \sqrt {a e^4+c d^4} \Pi \left (\frac {i \sqrt {a} e^2}{\sqrt {c} d^2};\left .\sin ^{-1}\left (\frac {(-1)^{3/4} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )+e^3 \left (a+c x^4\right ) \sqrt {a e^4+c d^4}+c d^3 e \sqrt {a+c x^4} (d+e x) \tanh ^{-1}\left (\frac {a e^2+c d^2 x^2}{\sqrt {a+c x^4} \sqrt {a e^4+c d^4}}\right )\right )+\sqrt {a} \sqrt {c} e^2 \sqrt {\frac {c x^4}{a}+1} (d+e x) \sqrt {a e^4+c d^4} E\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )+i \sqrt {c} \sqrt {\frac {c x^4}{a}+1} (d+e x) \left (\sqrt {c} d^2+i \sqrt {a} e^2\right ) \sqrt {a e^4+c d^4} F\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} \sqrt {a+c x^4} (d+e x) \left (a e^4+c d^4\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a]*Sqrt[c]*e^2*Sqrt[c*d^4 + a*e^4]*(d + e*x)*Sqrt[1 + (c*x^4)/a]*EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sq
rt[a]]*x], -1] + I*Sqrt[c]*(Sqrt[c]*d^2 + I*Sqrt[a]*e^2)*Sqrt[c*d^4 + a*e^4]*(d + e*x)*Sqrt[1 + (c*x^4)/a]*Ell
ipticF[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[a]]*x], -1] - Sqrt[(I*Sqrt[c])/Sqrt[a]]*(e^3*Sqrt[c*d^4 + a*e^4]*(a + c
*x^4) + c*d^3*e*(d + e*x)*Sqrt[a + c*x^4]*ArcTanh[(a*e^2 + c*d^2*x^2)/(Sqrt[c*d^4 + a*e^4]*Sqrt[a + c*x^4])] +
2*(-1)^(1/4)*a^(1/4)*c^(3/4)*d^2*Sqrt[c*d^4 + a*e^4]*(d + e*x)*Sqrt[1 + (c*x^4)/a]*EllipticPi[(I*Sqrt[a]*e^2)
/(Sqrt[c]*d^2), ArcSin[((-1)^(3/4)*c^(1/4)*x)/a^(1/4)], -1]))/(Sqrt[(I*Sqrt[c])/Sqrt[a]]*(c*d^4 + a*e^4)^(3/2)
*(d + e*x)*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+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 \[ \int \frac {1}{\sqrt {c x^{4} + a} {\left (e x + d\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.03, size = 421, normalized size = 0.69 \[ -\frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, c \,d^{2} \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{\left (a \,e^{4}+c \,d^{4}\right ) \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {i \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )\right ) \sqrt {a}\, \sqrt {c}\, e^{2}}{\left (a \,e^{4}+c \,d^{4}\right ) \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {2 \left (\frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, e \EllipticPi \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , -\frac {i \sqrt {a}\, e^{2}}{\sqrt {c}\, d^{2}}, \frac {\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, d}-\frac {\arctanh \left (\frac {\frac {2 c \,d^{2} x^{2}}{e^{2}}+2 a}{2 \sqrt {a +\frac {c \,d^{4}}{e^{4}}}\, \sqrt {c \,x^{4}+a}}\right )}{2 \sqrt {a +\frac {c \,d^{4}}{e^{4}}}}\right ) c \,d^{3}}{\left (a \,e^{4}+c \,d^{4}\right ) e}-\frac {\sqrt {c \,x^{4}+a}\, e^{3}}{\left (a \,e^{4}+c \,d^{4}\right ) \left (e x +d \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^3*(c*x^4+a)^(1/2)/(a*e^4+c*d^4)/(e*x+d)-c*d^2/(a*e^4+c*d^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)*EllipticF((I/a^(1/2)*c^(1/2))^(1/2)*x,I)+I*c^(1/2)
*e^2/(a*e^4+c*d^4)*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)-EllipticE((I/a^(1/2)*c^(1/2))^(1/2)*x,I))+2*
c*d^3/(a*e^4+c*d^4)/e*(-1/2/(a+c*d^4/e^4)^(1/2)*arctanh(1/2*(2*c*d^2/e^2*x^2+2*a)/(a+c*d^4/e^4)^(1/2)/(c*x^4+a
)^(1/2))+1/(I/a^(1/2)*c^(1/2))^(1/2)/d*e*(-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^2*e^2,(-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 \[ \int \frac {1}{\sqrt {c x^{4} + a} {\left (e x + d\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + c*x^4)^(1/2)*(d + e*x)^2),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 \[ \int \frac {1}{\sqrt {a + c x^{4}} \left (d + e x\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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