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

Optimal. Leaf size=818 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {-c d^4-a e^4} x}{d e \sqrt {c x^4+a}}\right ) e^5}{2 \left (-c d^4-a e^4\right )^{3/2}}-\frac {\tanh ^{-1}\left (\frac {a e^2+c d^2 x^2}{\sqrt {c d^4+a e^4} \sqrt {c x^4+a}}\right ) e^5}{2 \left (c d^4+a e^4\right )^{3/2}}+\frac {\sqrt [4]{c} d \left (\sqrt {c} x^2+\sqrt {a}\right ) \sqrt {\frac {c x^4+a}{\left (\sqrt {c} x^2+\sqrt {a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right ) e^4}{2 \sqrt [4]{a} \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \left (c d^4+a e^4\right ) \sqrt {c x^4+a}}-\frac {\left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \left (\sqrt {c} x^2+\sqrt {a}\right ) \sqrt {\frac {c x^4+a}{\left (\sqrt {c} x^2+\sqrt {a}\right )^2}} \Pi \left (\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right )^2}{4 \sqrt {a} \sqrt {c} d^2 e^2};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right ) e^4}{4 \sqrt [4]{a} \sqrt [4]{c} d \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \left (c d^4+a e^4\right ) \sqrt {c x^4+a}}+\frac {\sqrt [4]{c} d \left (\sqrt {c} x^2+\sqrt {a}\right ) \sqrt {\frac {c x^4+a}{\left (\sqrt {c} x^2+\sqrt {a}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right ) e^2}{2 a^{3/4} \left (c d^4+a e^4\right ) \sqrt {c x^4+a}}-\frac {\sqrt {c} d x \sqrt {c x^4+a} e^2}{2 a \left (c d^4+a e^4\right ) \left (\sqrt {c} x^2+\sqrt {a}\right )}+\frac {\left (a e^2-c d^2 x^2\right ) e}{2 a \left (c d^4+a e^4\right ) \sqrt {c x^4+a}}+\frac {\sqrt [4]{c} d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \left (\sqrt {c} x^2+\sqrt {a}\right ) \sqrt {\frac {c x^4+a}{\left (\sqrt {c} x^2+\sqrt {a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 a^{5/4} \left (c d^4+a e^4\right ) \sqrt {c x^4+a}}+\frac {c d x \left (d^2+e^2 x^2\right )}{2 a \left (c d^4+a e^4\right ) \sqrt {c x^4+a}} \]

[Out]

-1/2*e^5*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)-1/2*e^5*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)+1/2*e*(-c*d^2*x^2+a*e^2)/a/(a*e^4+c*d^4)/(c*x^4+
a)^(1/2)+1/2*c*d*x*(e^2*x^2+d^2)/a/(a*e^4+c*d^4)/(c*x^4+a)^(1/2)-1/2*d*e^2*x*c^(1/2)*(c*x^4+a)^(1/2)/a/(a*e^4+
c*d^4)/(a^(1/2)+x^2*c^(1/2))+1/2*c^(1/4)*d*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^(3/4)/(a*e^4+c*d^4)/(c*x^4+a)^(1/2)+1/4*c^(1/4)*d*(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))*(-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^(5/4)/(a*e^4+c*d^4)/(c*x^4+a)^
(1/2)+1/2*c^(1/4)*d*e^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)/(a*e^4+c*d^4)/(e^2*a^(1/2)+d^2*c^(1/2))/(c*x^4+a)^(1/2)-1/4*e^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^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)/c^(1/4)/d/(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 = 0.60, antiderivative size = 818, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 13, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.684, Rules used = {1729, 1222, 1179, 1198, 220, 1196, 1217, 1707, 1248, 741, 12, 725, 206} \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {-c d^4-a e^4} x}{d e \sqrt {c x^4+a}}\right ) e^5}{2 \left (-c d^4-a e^4\right )^{3/2}}-\frac {\tanh ^{-1}\left (\frac {a e^2+c d^2 x^2}{\sqrt {c d^4+a e^4} \sqrt {c x^4+a}}\right ) e^5}{2 \left (c d^4+a e^4\right )^{3/2}}+\frac {\sqrt [4]{c} d \left (\sqrt {c} x^2+\sqrt {a}\right ) \sqrt {\frac {c x^4+a}{\left (\sqrt {c} x^2+\sqrt {a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right ) e^4}{2 \sqrt [4]{a} \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \left (c d^4+a e^4\right ) \sqrt {c x^4+a}}-\frac {\left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \left (\sqrt {c} x^2+\sqrt {a}\right ) \sqrt {\frac {c x^4+a}{\left (\sqrt {c} x^2+\sqrt {a}\right )^2}} \Pi \left (\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right )^2}{4 \sqrt {a} \sqrt {c} d^2 e^2};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right ) e^4}{4 \sqrt [4]{a} \sqrt [4]{c} d \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \left (c d^4+a e^4\right ) \sqrt {c x^4+a}}+\frac {\sqrt [4]{c} d \left (\sqrt {c} x^2+\sqrt {a}\right ) \sqrt {\frac {c x^4+a}{\left (\sqrt {c} x^2+\sqrt {a}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right ) e^2}{2 a^{3/4} \left (c d^4+a e^4\right ) \sqrt {c x^4+a}}-\frac {\sqrt {c} d x \sqrt {c x^4+a} e^2}{2 a \left (c d^4+a e^4\right ) \left (\sqrt {c} x^2+\sqrt {a}\right )}+\frac {\left (a e^2-c d^2 x^2\right ) e}{2 a \left (c d^4+a e^4\right ) \sqrt {c x^4+a}}+\frac {\sqrt [4]{c} d \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \left (\sqrt {c} x^2+\sqrt {a}\right ) \sqrt {\frac {c x^4+a}{\left (\sqrt {c} x^2+\sqrt {a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 a^{5/4} \left (c d^4+a e^4\right ) \sqrt {c x^4+a}}+\frac {c d x \left (d^2+e^2 x^2\right )}{2 a \left (c d^4+a e^4\right ) \sqrt {c x^4+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(a*e^2 - c*d^2*x^2))/(2*a*(c*d^4 + a*e^4)*Sqrt[a + c*x^4]) + (c*d*x*(d^2 + e^2*x^2))/(2*a*(c*d^4 + a*e^4)*S
qrt[a + c*x^4]) - (Sqrt[c]*d*e^2*x*Sqrt[a + c*x^4])/(2*a*(c*d^4 + a*e^4)*(Sqrt[a] + Sqrt[c]*x^2)) - (e^5*ArcTa
n[(Sqrt[-(c*d^4) - a*e^4]*x)/(d*e*Sqrt[a + c*x^4])])/(2*(-(c*d^4) - a*e^4)^(3/2)) - (e^5*ArcTanh[(a*e^2 + c*d^
2*x^2)/(Sqrt[c*d^4 + a*e^4]*Sqrt[a + c*x^4])])/(2*(c*d^4 + a*e^4)^(3/2)) + (c^(1/4)*d*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])/(2*a^(3/4)*(c*d
^4 + a*e^4)*Sqrt[a + c*x^4]) + (c^(1/4)*d*(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]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(4*a^(5/4)*(c*d^4 + a*e^4)*Sqrt[a +
c*x^4]) + (c^(1/4)*d*e^4*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcT
an[(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]) - (e^4*
(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[(Sq
rt[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])/(4*a^(1/4)*c^(1/4
)*d*(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 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)*(a + c*x^4)^(p + 1))/(
4*a*(p + 1)), x] + Dist[1/(4*a*(p + 1)), Int[Simp[d*(4*p + 5) + e*(4*p + 7)*x^2, x]*(a + c*x^4)^(p + 1), x], x
] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntegerQ[2*p]

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 1198

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(e + d*q)/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] /; NeQ[e + d*q, 0]] /; FreeQ[{a
, c, d, e}, x] && PosQ[c/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 1222

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

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 1729

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

Rubi steps

\begin {align*} \int \frac {1}{(d+e x) \left (a+c x^4\right )^{3/2}} \, dx &=d \int \frac {1}{\left (d^2-e^2 x^2\right ) \left (a+c x^4\right )^{3/2}} \, dx-e \int \frac {x}{\left (d^2-e^2 x^2\right ) \left (a+c x^4\right )^{3/2}} \, dx\\ &=-\left (\frac {1}{2} e \operatorname {Subst}\left (\int \frac {1}{\left (d^2-e^2 x\right ) \left (a+c x^2\right )^{3/2}} \, dx,x,x^2\right )\right )+\frac {d \int \frac {c d^2+c e^2 x^2}{\left (a+c x^4\right )^{3/2}} \, dx}{c d^4+a e^4}+\frac {\left (d e^4\right ) \int \frac {1}{\left (d^2-e^2 x^2\right ) \sqrt {a+c x^4}} \, dx}{c d^4+a e^4}\\ &=\frac {e \left (a e^2-c d^2 x^2\right )}{2 a \left (c d^4+a e^4\right ) \sqrt {a+c x^4}}+\frac {c d x \left (d^2+e^2 x^2\right )}{2 a \left (c d^4+a e^4\right ) \sqrt {a+c x^4}}-\frac {d \int \frac {-c d^2+c e^2 x^2}{\sqrt {a+c x^4}} \, dx}{2 a \left (c d^4+a e^4\right )}-\frac {e \operatorname {Subst}\left (\int \frac {a e^4}{\left (d^2-e^2 x\right ) \sqrt {a+c x^2}} \, dx,x,x^2\right )}{2 a \left (c d^4+a e^4\right )}+\frac {\left (\sqrt {c} d e^4\right ) \int \frac {1}{\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 {\left (\sqrt {a} d e^6\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 \left (a e^2-c d^2 x^2\right )}{2 a \left (c d^4+a e^4\right ) \sqrt {a+c x^4}}+\frac {c d x \left (d^2+e^2 x^2\right )}{2 a \left (c d^4+a e^4\right ) \sqrt {a+c x^4}}-\frac {e^5 \tan ^{-1}\left (\frac {\sqrt {-c d^4-a e^4} x}{d e \sqrt {a+c x^4}}\right )}{2 \left (-c d^4-a e^4\right )^{3/2}}+\frac {\sqrt [4]{c} d e^4 \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 ) \left (c d^4+a e^4\right ) \sqrt {a+c x^4}}+\frac {d \left (\frac {\sqrt {a}}{d^2}-\frac {\sqrt {c}}{e^2}\right ) e^6 \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 )}{4 \sqrt [4]{a} \sqrt [4]{c} \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 (\sqrt {c} d e^2\right ) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+c x^4}} \, dx}{2 \sqrt {a} \left (c d^4+a e^4\right )}-\frac {e^5 \operatorname {Subst}\left (\int \frac {1}{\left (d^2-e^2 x\right ) \sqrt {a+c x^2}} \, dx,x,x^2\right )}{2 \left (c d^4+a e^4\right )}+\frac {\left (\sqrt {c} d \left (\sqrt {c} d^2-\sqrt {a} e^2\right )\right ) \int \frac {1}{\sqrt {a+c x^4}} \, dx}{2 a \left (c d^4+a e^4\right )}\\ &=\frac {e \left (a e^2-c d^2 x^2\right )}{2 a \left (c d^4+a e^4\right ) \sqrt {a+c x^4}}+\frac {c d x \left (d^2+e^2 x^2\right )}{2 a \left (c d^4+a e^4\right ) \sqrt {a+c x^4}}-\frac {\sqrt {c} d e^2 x \sqrt {a+c x^4}}{2 a \left (c d^4+a e^4\right ) \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {e^5 \tan ^{-1}\left (\frac {\sqrt {-c d^4-a e^4} x}{d e \sqrt {a+c x^4}}\right )}{2 \left (-c d^4-a e^4\right )^{3/2}}+\frac {\sqrt [4]{c} d 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 )}{2 a^{3/4} \left (c d^4+a e^4\right ) \sqrt {a+c x^4}}+\frac {\sqrt [4]{c} d \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}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 a^{5/4} \left (c d^4+a e^4\right ) \sqrt {a+c x^4}}+\frac {\sqrt [4]{c} d e^4 \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 ) \left (c d^4+a e^4\right ) \sqrt {a+c x^4}}+\frac {d \left (\frac {\sqrt {a}}{d^2}-\frac {\sqrt {c}}{e^2}\right ) e^6 \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 )}{4 \sqrt [4]{a} \sqrt [4]{c} \left (\sqrt {c} d^2+\sqrt {a} e^2\right ) \left (c d^4+a e^4\right ) \sqrt {a+c x^4}}+\frac {e^5 \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 )}{2 \left (c d^4+a e^4\right )}\\ &=\frac {e \left (a e^2-c d^2 x^2\right )}{2 a \left (c d^4+a e^4\right ) \sqrt {a+c x^4}}+\frac {c d x \left (d^2+e^2 x^2\right )}{2 a \left (c d^4+a e^4\right ) \sqrt {a+c x^4}}-\frac {\sqrt {c} d e^2 x \sqrt {a+c x^4}}{2 a \left (c d^4+a e^4\right ) \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {e^5 \tan ^{-1}\left (\frac {\sqrt {-c d^4-a e^4} x}{d e \sqrt {a+c x^4}}\right )}{2 \left (-c d^4-a e^4\right )^{3/2}}-\frac {e^5 \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 )}{2 \left (c d^4+a e^4\right )^{3/2}}+\frac {\sqrt [4]{c} d 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 )}{2 a^{3/4} \left (c d^4+a e^4\right ) \sqrt {a+c x^4}}+\frac {\sqrt [4]{c} d \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}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 a^{5/4} \left (c d^4+a e^4\right ) \sqrt {a+c x^4}}+\frac {\sqrt [4]{c} d e^4 \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 ) \left (c d^4+a e^4\right ) \sqrt {a+c x^4}}+\frac {d \left (\frac {\sqrt {a}}{d^2}-\frac {\sqrt {c}}{e^2}\right ) e^6 \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 )}{4 \sqrt [4]{a} \sqrt [4]{c} \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 = 0.88, size = 434, normalized size = 0.53 \[ \frac {\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} \left (\sqrt [4]{c} d \left (\sqrt {a e^4+c d^4} \left (a e^3+c d x \left (d^2-d e x+e^2 x^2\right )\right )-a e^5 \sqrt {a+c x^4} \tanh ^{-1}\left (\frac {a e^2+c d^2 x^2}{\sqrt {a+c x^4} \sqrt {a e^4+c d^4}}\right )\right )-2 \sqrt [4]{-1} a^{5/4} e^4 \sqrt {\frac {c x^4}{a}+1} \sqrt {a e^4+c d^4} \Pi \left (\frac {i \sqrt {a} e^2}{\sqrt {c} d^2};\left .\sin ^{-1}\left (\frac {(-1)^{3/4} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )\right )+c^{3/4} d^2 \sqrt {\frac {c x^4}{a}+1} \left (\sqrt {a} e^2-i \sqrt {c} d^2\right ) \sqrt {a e^4+c d^4} F\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )-\sqrt {a} c^{3/4} d^2 e^2 \sqrt {\frac {c x^4}{a}+1} \sqrt {a e^4+c d^4} E\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )}{2 a \sqrt [4]{c} d \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} \sqrt {a+c x^4} \left (a e^4+c d^4\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 496, normalized size = 0.61 \[ \frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, c \,d^{3} \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{2 \left (a \,e^{4}+c \,d^{4}\right ) \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, a}-\frac {i \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )\right ) \sqrt {c}\, d \,e^{2}}{2 \left (a \,e^{4}+c \,d^{4}\right ) \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {a}}+\frac {\left (\frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, e \EllipticPi \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , -\frac {i \sqrt {a}\, e^{2}}{\sqrt {c}\, d^{2}}, \frac {\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, d}-\frac {\arctanh \left (\frac {\frac {2 c \,d^{2} x^{2}}{e^{2}}+2 a}{2 \sqrt {a +\frac {c \,d^{4}}{e^{4}}}\, \sqrt {c \,x^{4}+a}}\right )}{2 \sqrt {a +\frac {c \,d^{4}}{e^{4}}}}\right ) e^{3}}{a \,e^{4}+c \,d^{4}}-\frac {2 \left (-\frac {d \,e^{2} x^{3}}{4 \left (a \,e^{4}+c \,d^{4}\right ) a}+\frac {d^{2} e \,x^{2}}{4 \left (a \,e^{4}+c \,d^{4}\right ) a}-\frac {d^{3} x}{4 \left (a \,e^{4}+c \,d^{4}\right ) a}-\frac {e^{3}}{4 \left (a \,e^{4}+c \,d^{4}\right ) c}\right ) c}{\sqrt {\left (x^{4}+\frac {a}{c}\right ) c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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