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

Optimal. Leaf size=730 \[ -\frac {\sqrt {-a e^4-c d^4} \tan ^{-1}\left (\frac {x \sqrt {-a e^4-c d^4}}{d e \sqrt {a+c x^4}}\right )}{2 e^3}+\frac {\sqrt {c} d^2 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{2 e^3}-\frac {\sqrt [4]{a} \sqrt [4]{c} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\frac {\sqrt {c} d^2}{\sqrt {a}}+e^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 e^4 \sqrt {a+c x^4}}+\frac {\sqrt [4]{c} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (a e^4+c d^4\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} e^4 \sqrt {a+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}-\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \left (a e^4+c d^4\right ) \Pi \left (\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right )^2}{4 \sqrt {a} \sqrt {c} d^2 e^2};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{c} d e^4 \sqrt {a+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}-\frac {\sqrt {a e^4+c d^4} \tanh ^{-1}\left (\frac {a e^2+c d^2 x^2}{\sqrt {a+c x^4} \sqrt {a e^4+c d^4}}\right )}{2 e^3}-\frac {\sqrt {c} d x \sqrt {a+c x^4}}{e^2 \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {\sqrt [4]{a} \sqrt [4]{c} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{e^2 \sqrt {a+c x^4}}+\frac {\sqrt {a+c x^4}}{2 e} \]

[Out]

1/2*d^2*arctanh(x^2*c^(1/2)/(c*x^4+a)^(1/2))*c^(1/2)/e^3-1/2*arctan(x*(-a*e^4-c*d^4)^(1/2)/d/e/(c*x^4+a)^(1/2)
)*(-a*e^4-c*d^4)^(1/2)/e^3-1/2*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)^(1
/2)/e^3+1/2*(c*x^4+a)^(1/2)/e-d*x*c^(1/2)*(c*x^4+a)^(1/2)/e^2/(a^(1/2)+x^2*c^(1/2))+a^(1/4)*c^(1/4)*d*(cos(2*a
rctan(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)/e^2/(c*x^4+a)^(1/2)+1/2*c^(1/4)*
d*(a*e^4+c*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)))*EllipticF(sin(2*ar
ctan(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^4/(e^2*a^(1/2)+d^2*c^(1/2))/(c*x^4+a)^(1/2)-1/4*(a*e^4+c*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^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/e^4/(e^2*a^(1/2)+d^2*c^(1/2))/(c*x^4+a)^(1/2)-1/2*a^(1/4)*c^(1/4)*d*(cos(2*arc
tan(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))*(e^2+d^2*c^(1/2)/a^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/e^4/(c*x
^4+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.73, antiderivative size = 730, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 13, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.684, Rules used = {1729, 1209, 1198, 220, 1196, 1217, 1707, 1248, 735, 844, 217, 206, 725} \[ -\frac {\sqrt {-a e^4-c d^4} \tan ^{-1}\left (\frac {x \sqrt {-a e^4-c d^4}}{d e \sqrt {a+c x^4}}\right )}{2 e^3}+\frac {\sqrt {c} d^2 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{2 e^3}-\frac {\sqrt {a e^4+c d^4} \tanh ^{-1}\left (\frac {a e^2+c d^2 x^2}{\sqrt {a+c x^4} \sqrt {a e^4+c d^4}}\right )}{2 e^3}-\frac {\sqrt [4]{a} \sqrt [4]{c} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\frac {\sqrt {c} d^2}{\sqrt {a}}+e^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 e^4 \sqrt {a+c x^4}}+\frac {\sqrt [4]{c} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (a e^4+c d^4\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} e^4 \sqrt {a+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}-\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \left (a e^4+c d^4\right ) \Pi \left (\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right )^2}{4 \sqrt {a} \sqrt {c} d^2 e^2};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{c} d e^4 \sqrt {a+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}-\frac {\sqrt {c} d x \sqrt {a+c x^4}}{e^2 \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {\sqrt [4]{a} \sqrt [4]{c} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{e^2 \sqrt {a+c x^4}}+\frac {\sqrt {a+c x^4}}{2 e} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + c*x^4]/(d + e*x),x]

[Out]

Sqrt[a + c*x^4]/(2*e) - (Sqrt[c]*d*x*Sqrt[a + c*x^4])/(e^2*(Sqrt[a] + Sqrt[c]*x^2)) - (Sqrt[-(c*d^4) - a*e^4]*
ArcTan[(Sqrt[-(c*d^4) - a*e^4]*x)/(d*e*Sqrt[a + c*x^4])])/(2*e^3) + (Sqrt[c]*d^2*ArcTanh[(Sqrt[c]*x^2)/Sqrt[a
+ c*x^4]])/(2*e^3) - (Sqrt[c*d^4 + a*e^4]*ArcTanh[(a*e^2 + c*d^2*x^2)/(Sqrt[c*d^4 + a*e^4]*Sqrt[a + c*x^4])])/
(2*e^3) + (a^(1/4)*c^(1/4)*d*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*A
rcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(e^2*Sqrt[a + c*x^4]) - (a^(1/4)*c^(1/4)*d*((Sqrt[c]*d^2)/Sqrt[a] + e^2)*(Sq
rt[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*e^4*Sqrt[a + c*x^4]) + (c^(1/4)*d*(c*d^4 + a*e^4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqr
t[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(1/4)*e^4*(Sqrt[c]*d^2 + Sqrt[a]*e^2)*Sqrt[a
+ c*x^4]) - ((Sqrt[c]*d^2 - Sqrt[a]*e^2)*(c*d^4 + a*e^4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + S
qrt[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])/(4*a^(1/4)*c^(1/4)*d*e^4*(Sqrt[c]*d^2 + Sqrt[a]*e^2)*Sqrt[a + c*x^4])

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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 735

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 2*p + 1)), x] + Dist[(2*p)/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[a*e - c*d*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] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !Ration
alQ[m] || LtQ[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

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 1209

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

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

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Mathematica [C]  time = 0.84, size = 405, normalized size = 0.55 \[ \frac {2 c^{3/4} d^2 \sqrt {\frac {c x^4}{a}+1} \left (\sqrt {a} e^2+i \sqrt {c} d^2\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )-2 \sqrt {a} c^{3/4} d^2 e^2 \sqrt {\frac {c x^4}{a}+1} E\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )+\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} \left (\sqrt [4]{c} d e \left (\sqrt {c} d^2 \sqrt {a+c x^4} \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )-\sqrt {a+c x^4} \sqrt {a e^4+c d^4} \tanh ^{-1}\left (\frac {a e^2+c d^2 x^2}{\sqrt {a+c x^4} \sqrt {a e^4+c d^4}}\right )+e^2 \left (a+c x^4\right )\right )-2 \sqrt [4]{-1} \sqrt [4]{a} \sqrt {\frac {c x^4}{a}+1} \left (a e^4+c d^4\right ) \Pi \left (\frac {i \sqrt {a} e^2}{\sqrt {c} d^2};\left .\sin ^{-1}\left (\frac {(-1)^{3/4} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )\right )}{2 \sqrt [4]{c} d e^4 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} \sqrt {a+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x^{4} + a}}{e x + d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 565, normalized size = 0.77 \[ \frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, a \EllipticPi \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , -\frac {i \sqrt {a}\, e^{2}}{\sqrt {c}\, d^{2}}, \frac {\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, d}-\frac {a \arctanh \left (\frac {\frac {2 c \,d^{2} x^{2}}{e^{2}}+2 a}{2 \sqrt {a +\frac {c \,d^{4}}{e^{4}}}\, \sqrt {c \,x^{4}+a}}\right )}{2 \sqrt {a +\frac {c \,d^{4}}{e^{4}}}\, e}-\frac {c \,d^{4} \arctanh \left (\frac {\frac {2 c \,d^{2} x^{2}}{e^{2}}+2 a}{2 \sqrt {a +\frac {c \,d^{4}}{e^{4}}}\, \sqrt {c \,x^{4}+a}}\right )}{2 \sqrt {a +\frac {c \,d^{4}}{e^{4}}}\, e^{5}}-\frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, c \,d^{3} \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, e^{4}}+\frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, c \,d^{3} \EllipticPi \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , -\frac {i \sqrt {a}\, e^{2}}{\sqrt {c}\, d^{2}}, \frac {\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, e^{4}}+\frac {\sqrt {c}\, d^{2} \ln \left (2 \sqrt {c}\, x^{2}+2 \sqrt {c \,x^{4}+a}\right )}{2 e^{3}}-\frac {i \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )\right ) \sqrt {a}\, \sqrt {c}\, d}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, e^{2}}+\frac {\sqrt {c \,x^{4}+a}}{2 e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(c*x^4+a)^(1/2)/e-c*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)*EllipticF((I/a^(1/2)*c^(1/2))^(1/2)*x,I)+1/2*c^(1/2)*d^2/e^3*ln(2*c^(1/2)*x^2+2*(
c*x^4+a)^(1/2))-I*c^(1/2)*d/e^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)-EllipticE((I/a^(1/2)*c^(1/2))^(
1/2)*x,I))-1/2/e/(c*d^4/e^4+a)^(1/2)*arctanh(1/2*(2*c*x^2*d^2/e^2+2*a)/(c*d^4/e^4+a)^(1/2)/(c*x^4+a)^(1/2))*a-
1/2/e^5/(c*d^4/e^4+a)^(1/2)*arctanh(1/2*(2*c*x^2*d^2/e^2+2*a)/(c*d^4/e^4+a)^(1/2)/(c*x^4+a)^(1/2))*c*d^4+1/(I/
a^(1/2)*c^(1/2))^(1/2)/d*(-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)*Elli
pticPi((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))*a+1/e^4/(I/a^(1/2)*c^(1/2))^(1/2)*d^3*(-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))*c

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x^{4} + a}}{e x + d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^4)^(1/2)/(d + e*x),x)

[Out]

int((a + c*x^4)^(1/2)/(d + e*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + c x^{4}}}{d + e x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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