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 \[ -\frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \left (\sqrt{a} B-A \sqrt{c}\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt{a+b x^2+c x^4} \left (\sqrt{c} d-\sqrt{a} e\right )}+\frac{a^{3/4} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\frac{\sqrt{c} d}{\sqrt{a}}+e\right )^2 (B d-A e) \Pi \left (-\frac{\left (\sqrt{c} d-\sqrt{a} e\right )^2}{4 \sqrt{a} \sqrt{c} d e};2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{4 \sqrt [4]{c} d e \sqrt{a+b x^2+c x^4} \left (c d^2-a e^2\right )}-\frac{(B d-A e) \tan ^{-1}\left (\frac{x \sqrt{a e^2-b d e+c d^2}}{\sqrt{d} \sqrt{e} \sqrt{a+b x^2+c x^4}}\right )}{2 \sqrt{d} \sqrt{e} \sqrt{a e^2-b d e+c d^2}} \]

[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])

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Rubi [A]  time = 0.408582, antiderivative size = 436, normalized size of antiderivative = 1., 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 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 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]

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*}

Mathematica [C]  time = 0.513092, size = 298, normalized size = 0.68 \[ -\frac{i \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \left (B d \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} x \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}}\right ),\frac{\sqrt{b^2-4 a c}+b}{b-\sqrt{b^2-4 a c}}\right )+(A e-B d) \Pi \left (\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{2 c d};i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{c}{b+\sqrt{b^2-4 a c}}} x\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )\right )}{\sqrt{2} d e \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \sqrt{a+b x^2+c x^4}} \]

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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Maple [A]  time = 0.029, size = 359, normalized size = 0.8 \begin{align*}{\frac{B\sqrt{2}}{4\,e}\sqrt{4-2\,{\frac{ \left ( \sqrt{-4\,ac+{b}^{2}}-b \right ){x}^{2}}{a}}}\sqrt{4+2\,{\frac{ \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ){x}^{2}}{a}}}{\it EllipticF} \left ({\frac{x\sqrt{2}}{2}\sqrt{{\frac{1}{a} \left ( \sqrt{-4\,ac+{b}^{2}}-b \right ) }}},{\frac{1}{2}\sqrt{-4+2\,{\frac{b \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) }{ac}}}} \right ){\frac{1}{\sqrt{{\frac{1}{a} \left ( \sqrt{-4\,ac+{b}^{2}}-b \right ) }}}}{\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}+a}}}}+{\frac{ \left ( Ae-Bd \right ) \sqrt{2}}{ed}\sqrt{1-{\frac{{x}^{2}}{2\,a}\sqrt{-4\,ac+{b}^{2}}}+{\frac{b{x}^{2}}{2\,a}}}\sqrt{1+{\frac{b{x}^{2}}{2\,a}}+{\frac{{x}^{2}}{2\,a}\sqrt{-4\,ac+{b}^{2}}}}{\it EllipticPi} \left ({\frac{x\sqrt{2}}{2}\sqrt{{\frac{1}{a} \left ( \sqrt{-4\,ac+{b}^{2}}-b \right ) }}},-2\,{\frac{ae}{ \left ( \sqrt{-4\,ac+{b}^{2}}-b \right ) d}},{\sqrt{2}\sqrt{-{\frac{1}{2\,a} \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) }}{\frac{1}{\sqrt{{\frac{1}{a} \left ( \sqrt{-4\,ac+{b}^{2}}-b \right ) }}}}} \right ){\frac{1}{\sqrt{{\frac{1}{a}\sqrt{-4\,ac+{b}^{2}}}-{\frac{b}{a}}}}}{\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}+a}}}} \end{align*}

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{2} + A}{\sqrt{c x^{4} + b x^{2} + a}{\left (e x^{2} + d\right )}}\,{d x} \end{align*}

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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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B x^{2}}{\left (d + e x^{2}\right ) \sqrt{a + b x^{2} + c x^{4}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{2} + A}{\sqrt{c x^{4} + b x^{2} + a}{\left (e x^{2} + d\right )}}\,{d x} \end{align*}

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)