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

Optimal. Leaf size=369 \[ -\frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\sqrt{a} B-A \sqrt{c}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt{a+c x^4} \left (\sqrt{c} d-\sqrt{a} e\right )}+\frac{a^{3/4} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\frac{\sqrt{c} d}{\sqrt{a}}+e\right )^2 (B d-A e) \Pi \left (-\frac{\left (\sqrt{c} d-\sqrt{a} e\right )^2}{4 \sqrt{a} \sqrt{c} d e};2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 \sqrt [4]{c} d e \sqrt{a+c x^4} \left (c d^2-a e^2\right )}-\frac{(B d-A e) \tan ^{-1}\left (\frac{x \sqrt{a e^2+c d^2}}{\sqrt{d} \sqrt{e} \sqrt{a+c x^4}}\right )}{2 \sqrt{d} \sqrt{e} \sqrt{a e^2+c d^2}} \]

[Out]

-((B*d - A*e)*ArcTan[(Sqrt[c*d^2 + a*e^2]*x)/(Sqrt[d]*Sqrt[e]*Sqrt[a + c*x^4])])/(2*Sqrt[d]*Sqrt[e]*Sqrt[c*d^2
+ a*e^2]) - ((Sqrt[a]*B - A*Sqrt[c])*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*Elli
pticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(1/4)*c^(1/4)*(Sqrt[c]*d - Sqrt[a]*e)*Sqrt[a + c*x^4]) + (a^(3
/4)*((Sqrt[c]*d)/Sqrt[a] + e)^2*(B*d - A*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + 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)], 1/2])/(4*c^(1/
4)*d*e*(c*d^2 - a*e^2)*Sqrt[a + c*x^4])

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Rubi [A]  time = 0.360138, antiderivative size = 369, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {1709, 220, 1707} \[ \frac{a^{3/4} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\frac{\sqrt{c} d}{\sqrt{a}}+e\right )^2 (B d-A e) \Pi \left (-\frac{\left (\sqrt{c} d-\sqrt{a} e\right )^2}{4 \sqrt{a} \sqrt{c} d e};2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 \sqrt [4]{c} d e \sqrt{a+c x^4} \left (c d^2-a e^2\right )}-\frac{(B d-A e) \tan ^{-1}\left (\frac{x \sqrt{a e^2+c d^2}}{\sqrt{d} \sqrt{e} \sqrt{a+c x^4}}\right )}{2 \sqrt{d} \sqrt{e} \sqrt{a e^2+c d^2}}-\frac{\left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\sqrt{a} B-A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt{a+c x^4} \left (\sqrt{c} d-\sqrt{a} e\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((B*d - A*e)*ArcTan[(Sqrt[c*d^2 + a*e^2]*x)/(Sqrt[d]*Sqrt[e]*Sqrt[a + c*x^4])])/(2*Sqrt[d]*Sqrt[e]*Sqrt[c*d^2
+ a*e^2]) - ((Sqrt[a]*B - A*Sqrt[c])*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*Elli
pticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(1/4)*c^(1/4)*(Sqrt[c]*d - Sqrt[a]*e)*Sqrt[a + c*x^4]) + (a^(3
/4)*((Sqrt[c]*d)/Sqrt[a] + e)^2*(B*d - A*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + 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)], 1/2])/(4*c^(1/
4)*d*e*(c*d^2 - a*e^2)*Sqrt[a + c*x^4])

Rule 1709

Int[((A_.) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (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 + 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 + c*x^4]), x], 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] && NeQ[c*A^2 - a*B^2, 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 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]

Rubi steps

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

Mathematica [C]  time = 0.304231, size = 138, normalized size = 0.37 \[ -\frac{i \sqrt{\frac{c x^4}{a}+1} \left (B d \text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{i \sqrt{c}}{\sqrt{a}}}\right ),-1\right )+(A e-B d) \Pi \left (-\frac{i \sqrt{a} e}{\sqrt{c} d};\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{c}}{\sqrt{a}}} x\right )\right |-1\right )\right )}{d e \sqrt{\frac{i \sqrt{c}}{\sqrt{a}}} \sqrt{a+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[1 + (c*x^4)/a]*(B*d*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[a]]*x], -1] + (-(B*d) + A*e)*Elliptic
Pi[((-I)*Sqrt[a]*e)/(Sqrt[c]*d), I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[a]]*x], -1]))/(Sqrt[(I*Sqrt[c])/Sqrt[a]]*d*e*
Sqrt[a + c*x^4])

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Maple [C]  time = 0.033, size = 192, normalized size = 0.5 \begin{align*}{\frac{B}{e}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{c{x}^{4}+a}}}}+{\frac{Ae-Bd}{ed}\sqrt{1-{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\it EllipticPi} \left ( x\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}},{\frac{ie}{d}\sqrt{a}{\frac{1}{\sqrt{c}}}},{\sqrt{{-i\sqrt{c}{\frac{1}{\sqrt{a}}}}}{\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}} \right ){\frac{1}{\sqrt{{i\sqrt{c}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{c{x}^{4}+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+a)^(1/2),x)

[Out]

B/e/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)*
EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I)+(A*e-B*d)/e/d/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)*x^2)^(1/
2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)*EllipticPi(x*(I/a^(1/2)*c^(1/2))^(1/2),I*a^(1/2)/c^(1/2)*e/
d,(-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., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{2} + A}{\sqrt{c x^{4} + 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+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)/(sqrt(c*x^4 + 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+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}}{\sqrt{a + c x^{4}} \left (d + e x^{2}\right )}\, 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+a)**(1/2),x)

[Out]

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

[Out]

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