3.21 \(\int \frac {A+B x^2}{(d+e x^2) \sqrt {a+b x^2+c x^4}} \, dx\)

Optimal. Leaf size=436 \[ \frac {a^{3/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\frac {\sqrt {c} d}{\sqrt {a}}+e\right )^2 (B d-A e) \Pi \left (-\frac {\left (\sqrt {c} d-\sqrt {a} e\right )^2}{4 \sqrt {a} \sqrt {c} d e};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{4 \sqrt [4]{c} d e \sqrt {a+b x^2+c x^4} \left (c d^2-a e^2\right )}-\frac {(B d-A e) \tan ^{-1}\left (\frac {x \sqrt {a e^2-b d e+c d^2}}{\sqrt {d} \sqrt {e} \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {d} \sqrt {e} \sqrt {a e^2-b d e+c d^2}}-\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \left (\sqrt {a} B-A \sqrt {c}\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+b x^2+c x^4} \left (\sqrt {c} d-\sqrt {a} e\right )} \]

[Out]

-1/2*(-A*e+B*d)*arctan(x*(a*e^2-b*d*e+c*d^2)^(1/2)/d^(1/2)/e^(1/2)/(c*x^4+b*x^2+a)^(1/2))/d^(1/2)/e^(1/2)/(a*e
^2-b*d*e+c*d^2)^(1/2)-1/2*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*Elliptic
F(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*(2-b/a^(1/2)/c^(1/2))^(1/2))*(B*a^(1/2)-A*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*a^(1/2)+d*c^(1/2))/(c*x^4+b*x^2+a)^(1/2)
+1/4*a^(3/4)*(-A*e+B*d)*(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-b/a^(1/2)/c^(1/2))^
(1/2))*(a^(1/2)+x^2*c^(1/2))*(e+d*c^(1/2)/a^(1/2))^2*((c*x^4+b*x^2+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/c^(1/4)/d
/e/(-a*e^2+c*d^2)/(c*x^4+b*x^2+a)^(1/2)

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Rubi [A]  time = 0.41, antiderivative size = 436, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1708, 1103, 1706} \[ \frac {a^{3/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\frac {\sqrt {c} d}{\sqrt {a}}+e\right )^2 (B d-A e) \Pi \left (-\frac {\left (\sqrt {c} d-\sqrt {a} e\right )^2}{4 \sqrt {a} \sqrt {c} d e};2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{4 \sqrt [4]{c} d e \sqrt {a+b x^2+c x^4} \left (c d^2-a e^2\right )}-\frac {(B d-A e) \tan ^{-1}\left (\frac {x \sqrt {a e^2-b d e+c d^2}}{\sqrt {d} \sqrt {e} \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {d} \sqrt {e} \sqrt {a e^2-b d e+c d^2}}-\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \left (\sqrt {a} B-A \sqrt {c}\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+b x^2+c x^4} \left (\sqrt {c} d-\sqrt {a} e\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((B*d - A*e)*ArcTan[(Sqrt[c*d^2 - b*d*e + a*e^2]*x)/(Sqrt[d]*Sqrt[e]*Sqrt[a + b*x^2 + c*x^4])])/(2*Sqrt[d]*Sq
rt[e]*Sqrt[c*d^2 - b*d*e + a*e^2]) - ((Sqrt[a]*B - A*Sqrt[c])*(Sqrt[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 - Sqrt[a]*e)*Sqrt[a + b*x^2 + c*x^4]) + (a^(3/4)*((Sqrt[c]*d)/Sqrt[a] + e)^2*(B*d - A*e)*(S
qrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + 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)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(4*c^(1/4)*d*e*(c*d^2
- a*e^2)*Sqrt[a + b*x^2 + c*x^4])

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 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]

Rubi steps

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

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Mathematica [C]  time = 0.49, size = 298, normalized size = 0.68 \[ -\frac {i \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}+b}} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \left ((A e-B d) \Pi \left (\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c d};i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )+B d F\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )\right )}{\sqrt {2} d e \sqrt {\frac {c}{\sqrt {b^2-4 a c}+b}} \sqrt {a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c
])]*(B*d*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])] + (-(B*d) + A*e)*EllipticPi[((b + Sqrt[b^2 - 4*a*c])*e)/(2*c*d), I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqr
t[b^2 - 4*a*c])]*x], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])]))/(Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c]
)]*d*e*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((B*x^2+A)/(e*x^2+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 \[ \int \frac {B x^{2} + A}{\sqrt {c x^{4} + b x^{2} + a} {\left (e x^{2} + d\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 359, normalized size = 0.82 \[ \frac {\sqrt {2}\, \sqrt {-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}+4}\, \sqrt {\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}+4}\, B \EllipticF \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, x}{2}, \frac {\sqrt {\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) b}{a c}-4}}{2}\right )}{4 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, e}+\frac {\left (A e -B d \right ) \sqrt {2}\, \sqrt {\frac {b \,x^{2}}{2 a}-\frac {\sqrt {-4 a c +b^{2}}\, x^{2}}{2 a}+1}\, \sqrt {\frac {b \,x^{2}}{2 a}+\frac {\sqrt {-4 a c +b^{2}}\, x^{2}}{2 a}+1}\, \EllipticPi \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, x}{2}, -\frac {2 a e}{\left (-b +\sqrt {-4 a c +b^{2}}\right ) d}, \frac {\sqrt {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 a}}\, \sqrt {2}}{\sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}\right )}{\sqrt {-\frac {b}{a}+\frac {\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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