3.412 \(\int \frac {f+g x}{(d+e x) \sqrt {a+b x^2+c x^4}} \, dx\)

Optimal. Leaf size=560 \[ \frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right ) \left (\sqrt {a} e g+\sqrt {c} d f\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+b x^2+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}-\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) (e f-d g) \Pi \left (\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right )^2}{4 \sqrt {a} \sqrt {c} d^2 e^2};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{4 \sqrt [4]{a} \sqrt [4]{c} d e \sqrt {a+b x^2+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}+\frac {(e f-d g) \tan ^{-1}\left (\frac {x \sqrt {-a e^4-b d^2 e^2-c d^4}}{d e \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {-e^2 \left (a e^2+b d^2\right )-c d^4}}-\frac {(e f-d g) \tanh ^{-1}\left (\frac {2 a e^2+x^2 \left (b e^2+2 c d^2\right )+b d^2}{2 \sqrt {a+b x^2+c x^4} \sqrt {a e^4+b d^2 e^2+c d^4}}\right )}{2 \sqrt {a e^4+b d^2 e^2+c d^4}} \]

[Out]

1/2*(-d*g+e*f)*arctan(x*(-a*e^4-b*d^2*e^2-c*d^4)^(1/2)/d/e/(c*x^4+b*x^2+a)^(1/2))/(-a*e^4-b*d^2*e^2-c*d^4)^(1/
2)-1/2*(-d*g+e*f)*arctanh(1/2*(b*d^2+2*a*e^2+(b*e^2+2*c*d^2)*x^2)/(a*e^4+b*d^2*e^2+c*d^4)^(1/2)/(c*x^4+b*x^2+a
)^(1/2))/(a*e^4+b*d^2*e^2+c*d^4)^(1/2)-1/4*(-d*g+e*f)*(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-b/a^(1/2)/c^(1/2))^(1/2))*(-e^2*a^(1/2)+d^2*c^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+b*x^2+a)/(
a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(1/4)/c^(1/4)/d/e/(e^2*a^(1/2)+d^2*c^(1/2))/(c*x^4+b*x^2+a)^(1/2)+1/2*(cos(2*a
rctan(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-b/a^(1/2)/c^(1/2))^(1/2))*(e*g*a^(1/2)+d*f*c^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+b*x^2+a)/(a^(1/2)+x^
2*c^(1/2))^2)^(1/2)/a^(1/4)/c^(1/4)/(e^2*a^(1/2)+d^2*c^(1/2))/(c*x^4+b*x^2+a)^(1/2)

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Rubi [A]  time = 0.62, antiderivative size = 560, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {1741, 12, 1247, 724, 206, 1708, 1103, 1706} \[ \frac {(e f-d g) \tan ^{-1}\left (\frac {x \sqrt {-a e^4-b d^2 e^2-c d^4}}{d e \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {-e^2 \left (a e^2+b d^2\right )-c d^4}}-\frac {(e f-d g) \tanh ^{-1}\left (\frac {2 a e^2+x^2 \left (b e^2+2 c d^2\right )+b d^2}{2 \sqrt {a+b x^2+c x^4} \sqrt {a e^4+b d^2 e^2+c d^4}}\right )}{2 \sqrt {a e^4+b d^2 e^2+c d^4}}+\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right ) \left (\sqrt {a} e g+\sqrt {c} d f\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+b x^2+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right )}-\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {c} d^2-\sqrt {a} e^2\right ) (e f-d g) \Pi \left (\frac {\left (\sqrt {c} d^2+\sqrt {a} e^2\right )^2}{4 \sqrt {a} \sqrt {c} d^2 e^2};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{4 \sqrt [4]{a} \sqrt [4]{c} d e \sqrt {a+b x^2+c x^4} \left (\sqrt {a} e^2+\sqrt {c} d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*f - d*g)*ArcTan[(Sqrt[-(c*d^4) - b*d^2*e^2 - a*e^4]*x)/(d*e*Sqrt[a + b*x^2 + c*x^4])])/(2*Sqrt[-(c*d^4) -
e^2*(b*d^2 + a*e^2)]) - ((e*f - d*g)*ArcTanh[(b*d^2 + 2*a*e^2 + (2*c*d^2 + b*e^2)*x^2)/(2*Sqrt[c*d^4 + b*d^2*e
^2 + a*e^4]*Sqrt[a + b*x^2 + c*x^4])])/(2*Sqrt[c*d^4 + b*d^2*e^2 + a*e^4]) + ((Sqrt[c]*d*f + Sqrt[a]*e*g)*(Sqr
t[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)
], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(2*a^(1/4)*c^(1/4)*(Sqrt[c]*d^2 + Sqrt[a]*e^2)*Sqrt[a + b*x^2 + c*x^4]) - ((S
qrt[c]*d^2 - Sqrt[a]*e^2)*(e*f - d*g)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + 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)], (2 -
b/(Sqrt[a]*Sqrt[c]))/4])/(4*a^(1/4)*c^(1/4)*d*e*(Sqrt[c]*d^2 + Sqrt[a]*e^2)*Sqrt[a + b*x^2 + 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 724

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

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1247

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

Rule 1706

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

Rule 1708

Int[((A_.) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (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 + b*x^2 + 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 + b*x^2 + c*x^4]), x]
, x]] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[c*d^2
- a*e^2, 0] && PosQ[c/a] && NeQ[c*A^2 - a*B^2, 0]

Rule 1741

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

Rubi steps

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

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Mathematica [C]  time = 7.87, size = 3652, normalized size = 6.52 \[ \text {Result too large to show} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*g*Sqrt[1 - (2*c*x^2)/(-b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 - (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])]*EllipticF[I*
ArcSinh[Sqrt[2]*Sqrt[-(c/(-b - Sqrt[b^2 - 4*a*c]))]*x], (-b - Sqrt[b^2 - 4*a*c])/(-b + Sqrt[b^2 - 4*a*c])])/(S
qrt[2]*Sqrt[-(c/(-b - Sqrt[b^2 - 4*a*c]))]*e*Sqrt[a + b*x^2 + c*x^4]) + (2*(Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]
/Sqrt[2] + Sqrt[-(b/c) + Sqrt[b^2 - 4*a*c]/c]/Sqrt[2])*f*(-(Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/Sqrt[2]) + x)^2
*Sqrt[(Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c]*(-(Sqrt[-(b/c) + Sqrt[b^2 - 4*a*c]/c]/Sqrt[2]) + x))/((Sqrt[-(b/c) - S
qrt[b^2 - 4*a*c]/c]/Sqrt[2] + Sqrt[-(b/c) + Sqrt[b^2 - 4*a*c]/c]/Sqrt[2])*(-(Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c
]/Sqrt[2]) + x))]*Sqrt[(Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c]*(Sqrt[-(b/c) + Sqrt[b^2 - 4*a*c]/c]/Sqrt[2] + x))/((S
qrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/Sqrt[2] - Sqrt[-(b/c) + Sqrt[b^2 - 4*a*c]/c]/Sqrt[2])*(-(Sqrt[-(b/c) - Sqrt[
b^2 - 4*a*c]/c]/Sqrt[2]) + x))]*Sqrt[((Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] - Sqrt[(-b + Sqrt[b^2 - 4*a*c])/c])*(S
qrt[2]*Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] + 2*x))/((Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] + Sqrt[(-b + Sqrt[b^2 - 4*a
*c])/c])*(Sqrt[2]*Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] - 2*x))]*((-d + (Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]*e)/Sqrt
[2])*EllipticF[ArcSin[Sqrt[((Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] - Sqrt[(-b + Sqrt[b^2 - 4*a*c])/c])*(Sqrt[2]*Sqr
t[(-b - Sqrt[b^2 - 4*a*c])/c] + 2*x))/((Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] + Sqrt[(-b + Sqrt[b^2 - 4*a*c])/c])*(
Sqrt[2]*Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] - 2*x))]], (Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] + Sqrt[(-b + Sqrt[b^2 -
4*a*c])/c])^2/(Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] - Sqrt[(-b + Sqrt[b^2 - 4*a*c])/c])^2] - Sqrt[2]*Sqrt[(-b - Sq
rt[b^2 - 4*a*c])/c]*e*EllipticPi[((Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/Sqrt[2] + Sqrt[-(b/c) + Sqrt[b^2 - 4*a*c
]/c]/Sqrt[2])*(d + (Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]*e)/Sqrt[2]))/((-(Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/Sqr
t[2]) + Sqrt[-(b/c) + Sqrt[b^2 - 4*a*c]/c]/Sqrt[2])*(d - (Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]*e)/Sqrt[2])), Arc
Sin[Sqrt[((Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] - Sqrt[(-b + Sqrt[b^2 - 4*a*c])/c])*(Sqrt[2]*Sqrt[(-b - Sqrt[b^2 -
4*a*c])/c] + 2*x))/((Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] + Sqrt[(-b + Sqrt[b^2 - 4*a*c])/c])*(Sqrt[2]*Sqrt[(-b -
Sqrt[b^2 - 4*a*c])/c] - 2*x))]], (Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] + Sqrt[(-b + Sqrt[b^2 - 4*a*c])/c])^2/(Sqr
t[(-b - Sqrt[b^2 - 4*a*c])/c] - Sqrt[(-b + Sqrt[b^2 - 4*a*c])/c])^2]))/(Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c]*(Sqrt
[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/Sqrt[2] - Sqrt[-(b/c) + Sqrt[b^2 - 4*a*c]/c]/Sqrt[2])*(-d - (Sqrt[-(b/c) - Sqrt
[b^2 - 4*a*c]/c]*e)/Sqrt[2])*(d - (Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]*e)/Sqrt[2])*Sqrt[a + b*x^2 + c*x^4]) - (
2*(Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/Sqrt[2] + Sqrt[-(b/c) + Sqrt[b^2 - 4*a*c]/c]/Sqrt[2])*d*g*(-(Sqrt[-(b/c)
- Sqrt[b^2 - 4*a*c]/c]/Sqrt[2]) + x)^2*Sqrt[(Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c]*(-(Sqrt[-(b/c) + Sqrt[b^2 - 4*a
*c]/c]/Sqrt[2]) + x))/((Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/Sqrt[2] + Sqrt[-(b/c) + Sqrt[b^2 - 4*a*c]/c]/Sqrt[2
])*(-(Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/Sqrt[2]) + x))]*Sqrt[(Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c]*(Sqrt[-(b/c) +
Sqrt[b^2 - 4*a*c]/c]/Sqrt[2] + x))/((Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/Sqrt[2] - Sqrt[-(b/c) + Sqrt[b^2 - 4*
a*c]/c]/Sqrt[2])*(-(Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/Sqrt[2]) + x))]*Sqrt[((Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c]
- Sqrt[(-b + Sqrt[b^2 - 4*a*c])/c])*(Sqrt[2]*Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] + 2*x))/((Sqrt[(-b - Sqrt[b^2 -
4*a*c])/c] + Sqrt[(-b + Sqrt[b^2 - 4*a*c])/c])*(Sqrt[2]*Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] - 2*x))]*((-d + (Sqr
t[-(b/c) - Sqrt[b^2 - 4*a*c]/c]*e)/Sqrt[2])*EllipticF[ArcSin[Sqrt[((Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] - Sqrt[(-
b + Sqrt[b^2 - 4*a*c])/c])*(Sqrt[2]*Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] + 2*x))/((Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c
] + Sqrt[(-b + Sqrt[b^2 - 4*a*c])/c])*(Sqrt[2]*Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] - 2*x))]], (Sqrt[(-b - Sqrt[b^
2 - 4*a*c])/c] + Sqrt[(-b + Sqrt[b^2 - 4*a*c])/c])^2/(Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] - Sqrt[(-b + Sqrt[b^2 -
4*a*c])/c])^2] - Sqrt[2]*Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c]*e*EllipticPi[((Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/S
qrt[2] + Sqrt[-(b/c) + Sqrt[b^2 - 4*a*c]/c]/Sqrt[2])*(d + (Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]*e)/Sqrt[2]))/((-
(Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/Sqrt[2]) + Sqrt[-(b/c) + Sqrt[b^2 - 4*a*c]/c]/Sqrt[2])*(d - (Sqrt[-(b/c) -
Sqrt[b^2 - 4*a*c]/c]*e)/Sqrt[2])), ArcSin[Sqrt[((Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] - Sqrt[(-b + Sqrt[b^2 - 4*a
*c])/c])*(Sqrt[2]*Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] + 2*x))/((Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] + Sqrt[(-b + Sqr
t[b^2 - 4*a*c])/c])*(Sqrt[2]*Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] - 2*x))]], (Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] + S
qrt[(-b + Sqrt[b^2 - 4*a*c])/c])^2/(Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c] - Sqrt[(-b + Sqrt[b^2 - 4*a*c])/c])^2]))/
(Sqrt[(-b - Sqrt[b^2 - 4*a*c])/c]*(Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]/Sqrt[2] - Sqrt[-(b/c) + Sqrt[b^2 - 4*a*c
]/c]/Sqrt[2])*e*(-d - (Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]*e)/Sqrt[2])*(d - (Sqrt[-(b/c) - Sqrt[b^2 - 4*a*c]/c]
*e)/Sqrt[2])*Sqrt[a + b*x^2 + 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((g*x+f)/(e*x+d)/(c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 437, normalized size = 0.78 \[ \frac {\sqrt {2}\, \sqrt {-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}+4}\, \sqrt {\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}+4}\, g \EllipticF \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, x}{2}, \frac {\sqrt {\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) b}{a c}-4}}{2}\right )}{4 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, e}+\frac {\left (-d g +e f \right ) \left (\frac {\sqrt {2}\, \sqrt {-\frac {\left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{2 a}+1}\, \sqrt {\frac {\left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{2 a}+1}\, e \EllipticPi \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, x}{2}, \frac {2 a \,e^{2}}{\left (-b +\sqrt {-4 a c +b^{2}}\right ) d^{2}}, \frac {\sqrt {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 a}}\, \sqrt {2}}{\sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}\right )}{\sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, d}-\frac {\arctanh \left (\frac {b \,x^{2}+\frac {2 c \,d^{2} x^{2}}{e^{2}}+2 a +\frac {b \,d^{2}}{e^{2}}}{2 \sqrt {a +\frac {b \,d^{2}}{e^{2}}+\frac {c \,d^{4}}{e^{4}}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}\right )}{2 \sqrt {a +\frac {b \,d^{2}}{e^{2}}+\frac {c \,d^{4}}{e^{4}}}}\right )}{e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*g/e*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(-2*(-b+(-4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)*(2*(b+(-4*a*c+b^2
)^(1/2))/a*x^2+4)^(1/2)/(c*x^4+b*x^2+a)^(1/2)*EllipticF(1/2*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(2
*(b+(-4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))+(-d*g+e*f)/e^2*(-1/2/(c*d^4/e^4+b*d^2/e^2+a)^(1/2)*arctanh(1/2*(2*c*d^
2/e^2*x^2+b*x^2+b*d^2/e^2+2*a)/(c*d^4/e^4+b*d^2/e^2+a)^(1/2)/(c*x^4+b*x^2+a)^(1/2))+2^(1/2)/((-b+(-4*a*c+b^2)^
(1/2))/a)^(1/2)/d*e*(1-1/2*(-b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)*(1+1/2*(b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)/(c*
x^4+b*x^2+a)^(1/2)*EllipticPi(1/2*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x,2/(-b+(-4*a*c+b^2)^(1/2))*a/d^2*
e^2,(-1/2*(b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((f + g*x)/((d + e*x)*sqrt(a + b*x**2 + c*x**4)), x)

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