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

Optimal. Leaf size=528 \[ \frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right ) \left (-\frac{\sqrt{c} \left (2 a B e (5 c d-2 b e)-5 A c \left (3 c d^2-a e^2\right )\right )}{\sqrt{a}}+B \left (-c e (9 a e+20 b d)+8 b^2 e^2+15 c^2 d^2\right )+10 A c e (3 c d-b e)\right )}{30 c^{11/4} \sqrt{a+b x^2+c x^4}}+\frac{x \sqrt{a+b x^2+c x^4} \left (B \left (-c e (9 a e+20 b d)+8 b^2 e^2+15 c^2 d^2\right )+10 A c e (3 c d-b e)\right )}{15 c^{5/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right ) \left (B \left (-c e (9 a e+20 b d)+8 b^2 e^2+15 c^2 d^2\right )+10 A c e (3 c d-b e)\right )}{15 c^{11/4} \sqrt{a+b x^2+c x^4}}+\frac{e x \sqrt{a+b x^2+c x^4} (5 A c e-4 b B e+10 B c d)}{15 c^2}+\frac{B e^2 x^3 \sqrt{a+b x^2+c x^4}}{5 c} \]

[Out]

(e*(10*B*c*d - 4*b*B*e + 5*A*c*e)*x*Sqrt[a + b*x^2 + c*x^4])/(15*c^2) + (B*e^2*x^3*Sqrt[a + b*x^2 + c*x^4])/(5
*c) + ((10*A*c*e*(3*c*d - b*e) + B*(15*c^2*d^2 + 8*b^2*e^2 - c*e*(20*b*d + 9*a*e)))*x*Sqrt[a + b*x^2 + c*x^4])
/(15*c^(5/2)*(Sqrt[a] + Sqrt[c]*x^2)) - (a^(1/4)*(10*A*c*e*(3*c*d - b*e) + B*(15*c^2*d^2 + 8*b^2*e^2 - c*e*(20
*b*d + 9*a*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])/(15*c^(11/4)*Sqrt[a + b*x^2 + c*x^4]) + (a^(1/4)*(10*A*c*
e*(3*c*d - b*e) + B*(15*c^2*d^2 + 8*b^2*e^2 - c*e*(20*b*d + 9*a*e)) - (Sqrt[c]*(2*a*B*e*(5*c*d - 2*b*e) - 5*A*
c*(3*c*d^2 - a*e^2)))/Sqrt[a])*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*Ell
ipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(30*c^(11/4)*Sqrt[a + b*x^2 + c*x^4])

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Rubi [A]  time = 0.726367, antiderivative size = 528, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {1679, 1197, 1103, 1195} \[ \frac{x \sqrt{a+b x^2+c x^4} \left (B \left (-c e (9 a e+20 b d)+8 b^2 e^2+15 c^2 d^2\right )+10 A c e (3 c d-b e)\right )}{15 c^{5/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right ) \left (-\frac{\sqrt{c} \left (2 a B e (5 c d-2 b e)-5 A c \left (3 c d^2-a e^2\right )\right )}{\sqrt{a}}+B \left (-c e (9 a e+20 b d)+8 b^2 e^2+15 c^2 d^2\right )+10 A c e (3 c d-b e)\right )}{30 c^{11/4} \sqrt{a+b x^2+c x^4}}-\frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right ) \left (B \left (-c e (9 a e+20 b d)+8 b^2 e^2+15 c^2 d^2\right )+10 A c e (3 c d-b e)\right )}{15 c^{11/4} \sqrt{a+b x^2+c x^4}}+\frac{e x \sqrt{a+b x^2+c x^4} (5 A c e-4 b B e+10 B c d)}{15 c^2}+\frac{B e^2 x^3 \sqrt{a+b x^2+c x^4}}{5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(10*B*c*d - 4*b*B*e + 5*A*c*e)*x*Sqrt[a + b*x^2 + c*x^4])/(15*c^2) + (B*e^2*x^3*Sqrt[a + b*x^2 + c*x^4])/(5
*c) + ((10*A*c*e*(3*c*d - b*e) + B*(15*c^2*d^2 + 8*b^2*e^2 - c*e*(20*b*d + 9*a*e)))*x*Sqrt[a + b*x^2 + c*x^4])
/(15*c^(5/2)*(Sqrt[a] + Sqrt[c]*x^2)) - (a^(1/4)*(10*A*c*e*(3*c*d - b*e) + B*(15*c^2*d^2 + 8*b^2*e^2 - c*e*(20
*b*d + 9*a*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])/(15*c^(11/4)*Sqrt[a + b*x^2 + c*x^4]) + (a^(1/4)*(10*A*c*
e*(3*c*d - b*e) + B*(15*c^2*d^2 + 8*b^2*e^2 - c*e*(20*b*d + 9*a*e)) - (Sqrt[c]*(2*a*B*e*(5*c*d - 2*b*e) - 5*A*
c*(3*c*d^2 - a*e^2)))/Sqrt[a])*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*Ell
ipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(30*c^(11/4)*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 )^2}{\sqrt{a+b x^2+c x^4}} \, dx &=\frac{B e^2 x^3 \sqrt{a+b x^2+c x^4}}{5 c}+\frac{\int \frac{5 A c d^2+\left (5 B c d^2+10 A c d e-3 a B e^2\right ) x^2+e (10 B c d-4 b B e+5 A c e) x^4}{\sqrt{a+b x^2+c x^4}} \, dx}{5 c}\\ &=\frac{e (10 B c d-4 b B e+5 A c e) x \sqrt{a+b x^2+c x^4}}{15 c^2}+\frac{B e^2 x^3 \sqrt{a+b x^2+c x^4}}{5 c}+\frac{\int \frac{-2 a B e (5 c d-2 b e)+5 A c \left (3 c d^2-a e^2\right )+\left (10 A c e (3 c d-b e)+B \left (15 c^2 d^2+8 b^2 e^2-c e (20 b d+9 a e)\right )\right ) x^2}{\sqrt{a+b x^2+c x^4}} \, dx}{15 c^2}\\ &=\frac{e (10 B c d-4 b B e+5 A c e) x \sqrt{a+b x^2+c x^4}}{15 c^2}+\frac{B e^2 x^3 \sqrt{a+b x^2+c x^4}}{5 c}-\frac{\left (\sqrt{a} \left (10 A c e (3 c d-b e)+B \left (15 c^2 d^2+8 b^2 e^2-c e (20 b d+9 a e)\right )\right )\right ) \int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+b x^2+c x^4}} \, dx}{15 c^{5/2}}+\frac{\left (\sqrt{a} \left (10 A c e (3 c d-b e)+B \left (15 c^2 d^2+8 b^2 e^2-c e (20 b d+9 a e)\right )+\frac{\sqrt{c} \left (-2 a B e (5 c d-2 b e)+5 A c \left (3 c d^2-a e^2\right )\right )}{\sqrt{a}}\right )\right ) \int \frac{1}{\sqrt{a+b x^2+c x^4}} \, dx}{15 c^{5/2}}\\ &=\frac{e (10 B c d-4 b B e+5 A c e) x \sqrt{a+b x^2+c x^4}}{15 c^2}+\frac{B e^2 x^3 \sqrt{a+b x^2+c x^4}}{5 c}+\frac{\left (10 A c e (3 c d-b e)+B \left (15 c^2 d^2+8 b^2 e^2-c e (20 b d+9 a e)\right )\right ) x \sqrt{a+b x^2+c x^4}}{15 c^{5/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{\sqrt [4]{a} \left (10 A c e (3 c d-b e)+B \left (15 c^2 d^2+8 b^2 e^2-c e (20 b d+9 a e)\right )\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}} 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 )}{15 c^{11/4} \sqrt{a+b x^2+c x^4}}+\frac{\sqrt [4]{a} \left (10 A c e (3 c d-b e)+B \left (15 c^2 d^2+8 b^2 e^2-c e (20 b d+9 a e)\right )-\frac{\sqrt{c} \left (2 a B e (5 c d-2 b e)-5 A c \left (3 c d^2-a e^2\right )\right )}{\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 )}{30 c^{11/4} \sqrt{a+b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 4.57953, size = 674, normalized size = 1.28 \[ \frac{-i \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{-2 \sqrt{b^2-4 a c}+2 b+4 c x^2}{b-\sqrt{b^2-4 a c}}} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} x \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}}\right ),\frac{\sqrt{b^2-4 a c}+b}{b-\sqrt{b^2-4 a c}}\right ) \left (c \left (10 A c \left (e \left (3 d \sqrt{b^2-4 a c}-a e\right )+3 c d^2\right )+B \left (5 c d \left (3 d \sqrt{b^2-4 a c}-4 a e\right )-9 a e^2 \sqrt{b^2-4 a c}\right )\right )-b c \left (10 A e \left (e \sqrt{b^2-4 a c}+3 c d\right )+B e \left (20 d \sqrt{b^2-4 a c}-17 a e\right )+15 B c d^2\right )+2 b^2 e \left (4 B e \sqrt{b^2-4 a c}+5 A c e+10 B c d\right )-8 b^3 B e^2\right )+i \left (\sqrt{b^2-4 a c}-b\right ) \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{-2 \sqrt{b^2-4 a c}+2 b+4 c x^2}{b-\sqrt{b^2-4 a c}}} E\left (i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{c}{b+\sqrt{b^2-4 a c}}} x\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right ) \left (B \left (-c e (9 a e+20 b d)+8 b^2 e^2+15 c^2 d^2\right )+10 A c e (3 c d-b e)\right )+4 c e x \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \left (a+b x^2+c x^4\right ) \left (5 A c e+B \left (-4 b e+10 c d+3 c e x^2\right )\right )}{60 c^3 \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \sqrt{a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*c*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*e*x*(a + b*x^2 + c*x^4)*(5*A*c*e + B*(10*c*d - 4*b*e + 3*c*e*x^2)) + I*(-
b + Sqrt[b^2 - 4*a*c])*(10*A*c*e*(3*c*d - b*e) + B*(15*c^2*d^2 + 8*b^2*e^2 - c*e*(20*b*d + 9*a*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*(-8*b^3*B*e^2 + 2*b^2*e*(10*B*c*d + 5*A*c*e + 4*B*Sqrt[b^2 - 4*a*c]*e) - b*c*(15*B*c*d^2 +
B*e*(20*Sqrt[b^2 - 4*a*c]*d - 17*a*e) + 10*A*e*(3*c*d + Sqrt[b^2 - 4*a*c]*e)) + c*(B*(-9*a*Sqrt[b^2 - 4*a*c]*e
^2 + 5*c*d*(3*Sqrt[b^2 - 4*a*c]*d - 4*a*e)) + 10*A*c*(3*c*d^2 + e*(3*Sqrt[b^2 - 4*a*c]*d - a*e))))*Sqrt[(b + S
qrt[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*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])])/(60*c^3*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[a + b*x^2 + c*x^4])

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Maple [B]  time = 0.009, size = 1201, normalized size = 2.3 \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)^2/(c*x^4+b*x^2+a)^(1/2),x)

[Out]

B*e^2*(1/5/c*x^3*(c*x^4+b*x^2+a)^(1/2)-4/15*b/c^2*x*(c*x^4+b*x^2+a)^(1/2)+1/15*b/c^2*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)*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))-1/2*(-3/5*a/c+8/15*b^2/c^2)*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))))+(A*e^2+2*B*d*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*(2*A*d*e+B*d^2)*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*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 )}^{2}}{\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)^2/(c*x^4+b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*(e*x^2 + d)^2/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^{2} x^{6} +{\left (2 \, B d e + A e^{2}\right )} x^{4} + A d^{2} +{\left (B d^{2} + 2 \, A d e\right )} x^{2}}{\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)^2/(c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

integral((B*e^2*x^6 + (2*B*d*e + A*e^2)*x^4 + A*d^2 + (B*d^2 + 2*A*d*e)*x^2)/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 )^{2}}{\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)**2/(c*x**4+b*x**2+a)**(1/2),x)

[Out]

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

[Out]

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