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

Optimal. Leaf size=368 \[ \frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right ) \left (\frac{\sqrt{c} (3 A c d-a B e)}{\sqrt{a}}+3 A c e-2 b B e+3 B c d\right )}{6 c^{7/4} \sqrt{a+b x^2+c x^4}}+\frac{x \sqrt{a+b x^2+c x^4} (3 A c e-2 b B e+3 B c d)}{3 c^{3/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right ) (3 A c e-2 b B e+3 B c d)}{3 c^{7/4} \sqrt{a+b x^2+c x^4}}+\frac{B e x \sqrt{a+b x^2+c x^4}}{3 c} \]

[Out]

(B*e*x*Sqrt[a + b*x^2 + c*x^4])/(3*c) + ((3*B*c*d - 2*b*B*e + 3*A*c*e)*x*Sqrt[a + b*x^2 + c*x^4])/(3*c^(3/2)*(
Sqrt[a] + Sqrt[c]*x^2)) - (a^(1/4)*(3*B*c*d - 2*b*B*e + 3*A*c*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x
^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(3*c^(7/
4)*Sqrt[a + b*x^2 + c*x^4]) + (a^(1/4)*(3*B*c*d - 2*b*B*e + 3*A*c*e + (Sqrt[c]*(3*A*c*d - a*B*e))/Sqrt[a])*(Sq
rt[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])/(6*c^(7/4)*Sqrt[a + b*x^2 + c*x^4])

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Rubi [A]  time = 0.25398, antiderivative size = 368, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {1679, 1197, 1103, 1195} \[ \frac{x \sqrt{a+b x^2+c x^4} (3 A c e-2 b B e+3 B c d)}{3 c^{3/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right ) \left (\frac{\sqrt{c} (3 A c d-a B e)}{\sqrt{a}}+3 A c e-2 b B e+3 B c d\right )}{6 c^{7/4} \sqrt{a+b x^2+c x^4}}-\frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right ) (3 A c e-2 b B e+3 B c d)}{3 c^{7/4} \sqrt{a+b x^2+c x^4}}+\frac{B e x \sqrt{a+b x^2+c x^4}}{3 c} \]

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*e*x*Sqrt[a + b*x^2 + c*x^4])/(3*c) + ((3*B*c*d - 2*b*B*e + 3*A*c*e)*x*Sqrt[a + b*x^2 + c*x^4])/(3*c^(3/2)*(
Sqrt[a] + Sqrt[c]*x^2)) - (a^(1/4)*(3*B*c*d - 2*b*B*e + 3*A*c*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x
^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(3*c^(7/
4)*Sqrt[a + b*x^2 + c*x^4]) + (a^(1/4)*(3*B*c*d - 2*b*B*e + 3*A*c*e + (Sqrt[c]*(3*A*c*d - a*B*e))/Sqrt[a])*(Sq
rt[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])/(6*c^(7/4)*Sqrt[a + b*x^2 + c*x^4])

Rule 1679

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{q = Expon[Pq, x^2], e = Coeff[Pq, x^2,
Expon[Pq, x^2]]}, Simp[(e*x^(2*q - 3)*(a + b*x^2 + c*x^4)^(p + 1))/(c*(2*q + 4*p + 1)), x] + Dist[1/(c*(2*q +
4*p + 1)), Int[(a + b*x^2 + c*x^4)^p*ExpandToSum[c*(2*q + 4*p + 1)*Pq - a*e*(2*q - 3)*x^(2*q - 4) - b*e*(2*q
+ 2*p - 1)*x^(2*q - 2) - c*e*(2*q + 4*p + 1)*x^(2*q), x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2]
&& Expon[Pq, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] &&  !LtQ[p, -1]

Rule 1197

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

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 1195

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

Rubi steps

\begin{align*} \int \frac{\left (A+B x^2\right ) \left (d+e x^2\right )}{\sqrt{a+b x^2+c x^4}} \, dx &=\frac{B e x \sqrt{a+b x^2+c x^4}}{3 c}+\frac{\int \frac{3 A c d-a B e+(3 B c d-2 b B e+3 A c e) x^2}{\sqrt{a+b x^2+c x^4}} \, dx}{3 c}\\ &=\frac{B e x \sqrt{a+b x^2+c x^4}}{3 c}-\frac{\left (\sqrt{a} (3 B c d-2 b B e+3 A c e)\right ) \int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+b x^2+c x^4}} \, dx}{3 c^{3/2}}+\frac{\left (\sqrt{a} \left (3 B c d-2 b B e+3 A c e+\frac{\sqrt{c} (3 A c d-a B e)}{\sqrt{a}}\right )\right ) \int \frac{1}{\sqrt{a+b x^2+c x^4}} \, dx}{3 c^{3/2}}\\ &=\frac{B e x \sqrt{a+b x^2+c x^4}}{3 c}+\frac{(3 B c d-2 b B e+3 A c e) x \sqrt{a+b x^2+c x^4}}{3 c^{3/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{\sqrt [4]{a} (3 B c d-2 b B e+3 A c 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}} E\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 )}{3 c^{7/4} \sqrt{a+b x^2+c x^4}}+\frac{\sqrt [4]{a} \left (3 B c d-2 b B e+3 A c e+\frac{\sqrt{c} (3 A c d-a B e)}{\sqrt{a}}\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 )}{6 c^{7/4} \sqrt{a+b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 2.23374, size = 521, normalized size = 1.42 \[ \frac{i \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{-2 \sqrt{b^2-4 a c}+2 b+4 c x^2}{b-\sqrt{b^2-4 a c}}} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} x \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}}\right ),\frac{\sqrt{b^2-4 a c}+b}{b-\sqrt{b^2-4 a c}}\right ) \left (b \left (2 B e \sqrt{b^2-4 a c}+3 A c e+3 B c d\right )-c \left (3 A e \sqrt{b^2-4 a c}+3 B d \sqrt{b^2-4 a c}-2 a B e+6 A c d\right )-2 b^2 B e\right )-i \left (\sqrt{b^2-4 a c}-b\right ) \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{-2 \sqrt{b^2-4 a c}+2 b+4 c x^2}{b-\sqrt{b^2-4 a c}}} E\left (i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{c}{b+\sqrt{b^2-4 a c}}} x\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right ) (-3 A c e+2 b B e-3 B c d)+4 B c e x \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \left (a+b x^2+c x^4\right )}{12 c^2 \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \sqrt{a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*B*c*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*e*x*(a + b*x^2 + c*x^4) - I*(-b + Sqrt[b^2 - 4*a*c])*(-3*B*c*d + 2*b*B*
e - 3*A*c*e)*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[(2*b - 2*Sqrt[b^2 - 4*a*c] +
4*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*EllipticE[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])] + I*(-2*b^2*B*e - c*(6*A*c*d + 3*B*Sqrt[b^2 - 4*a*c]*d - 2*a*B*e + 3*A*S
qrt[b^2 - 4*a*c]*e) + b*(3*B*c*d + 3*A*c*e + 2*B*Sqrt[b^2 - 4*a*c]*e))*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/
(b + Sqrt[b^2 - 4*a*c])]*Sqrt[(2*b - 2*Sqrt[b^2 - 4*a*c] + 4*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*EllipticF[I*ArcSi
nh[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])])/(12*c^2*Sqrt[
c/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[a + b*x^2 + c*x^4])

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Maple [B]  time = 0.007, size = 759, normalized size = 2.1 \begin{align*} \text{result too large to display} \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]

B*e*(1/3/c*x*(c*x^4+b*x^2+a)^(1/2)-1/12*a/c*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))+1/3*b/c*a*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)/(b+(-4*a*c+b^2)^(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))-EllipticE(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))))-1/2*(A*e+B*d)*a*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)/(b+(-4*a*c+b^2)^(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))-EllipticE(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)))+1/4*A*d*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))

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

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

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

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