3.380 \(\int \frac{(d+e x^2)^3}{\sqrt{a+b x^2+c x^4}} \, dx\)

Optimal. Leaf size=467 \[ \frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (e \left (-3 c e (3 a e+10 b d)+8 b^2 e^2+45 c^2 d^2\right )+\frac{\sqrt{c} \left (4 a b e^3-15 a c d e^2+15 c^2 d^3\right )}{\sqrt{a}}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{30 c^{11/4} \sqrt{a+b x^2+c x^4}}+\frac{e x \sqrt{a+b x^2+c x^4} \left (-3 c e (3 a e+10 b d)+8 b^2 e^2+45 c^2 d^2\right )}{15 c^{5/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{\sqrt [4]{a} e \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (-3 c e (3 a e+10 b d)+8 b^2 e^2+45 c^2 d^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{15 c^{11/4} \sqrt{a+b x^2+c x^4}}+\frac{e^2 x \sqrt{a+b x^2+c x^4} (15 c d-4 b e)}{15 c^2}+\frac{e^3 x^3 \sqrt{a+b x^2+c x^4}}{5 c} \]

[Out]

(e^2*(15*c*d - 4*b*e)*x*Sqrt[a + b*x^2 + c*x^4])/(15*c^2) + (e^3*x^3*Sqrt[a + b*x^2 + c*x^4])/(5*c) + (e*(45*c
^2*d^2 + 8*b^2*e^2 - 3*c*e*(10*b*d + 3*a*e))*x*Sqrt[a + b*x^2 + c*x^4])/(15*c^(5/2)*(Sqrt[a] + Sqrt[c]*x^2)) -
(a^(1/4)*e*(45*c^2*d^2 + 8*b^2*e^2 - 3*c*e*(10*b*d + 3*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)*((Sqrt[c]*(15*c^2*d^3 - 15*a*c*d*e^2 + 4*a*b*e^3))/Sqrt[a] + e*(45*c^2*d
^2 + 8*b^2*e^2 - 3*c*e*(10*b*d + 3*a*e)))*(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])/(30*c^(11/4)*Sqrt[a + b*x^2 + c
*x^4])

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Rubi [A]  time = 0.424683, antiderivative size = 467, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {1206, 1679, 1197, 1103, 1195} \[ \frac{e x \sqrt{a+b x^2+c x^4} \left (-3 c e (3 a e+10 b d)+8 b^2 e^2+45 c^2 d^2\right )}{15 c^{5/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}+\frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (e \left (-3 c e (3 a e+10 b d)+8 b^2 e^2+45 c^2 d^2\right )+\frac{\sqrt{c} \left (4 a b e^3-15 a c d e^2+15 c^2 d^3\right )}{\sqrt{a}}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{30 c^{11/4} \sqrt{a+b x^2+c x^4}}-\frac{\sqrt [4]{a} e \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (-3 c e (3 a e+10 b d)+8 b^2 e^2+45 c^2 d^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{15 c^{11/4} \sqrt{a+b x^2+c x^4}}+\frac{e^2 x \sqrt{a+b x^2+c x^4} (15 c d-4 b e)}{15 c^2}+\frac{e^3 x^3 \sqrt{a+b x^2+c x^4}}{5 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)^3/Sqrt[a + b*x^2 + c*x^4],x]

[Out]

(e^2*(15*c*d - 4*b*e)*x*Sqrt[a + b*x^2 + c*x^4])/(15*c^2) + (e^3*x^3*Sqrt[a + b*x^2 + c*x^4])/(5*c) + (e*(45*c
^2*d^2 + 8*b^2*e^2 - 3*c*e*(10*b*d + 3*a*e))*x*Sqrt[a + b*x^2 + c*x^4])/(15*c^(5/2)*(Sqrt[a] + Sqrt[c]*x^2)) -
(a^(1/4)*e*(45*c^2*d^2 + 8*b^2*e^2 - 3*c*e*(10*b*d + 3*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)*((Sqrt[c]*(15*c^2*d^3 - 15*a*c*d*e^2 + 4*a*b*e^3))/Sqrt[a] + e*(45*c^2*d
^2 + 8*b^2*e^2 - 3*c*e*(10*b*d + 3*a*e)))*(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])/(30*c^(11/4)*Sqrt[a + b*x^2 + c
*x^4])

Rule 1206

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e^q*x^(2*q - 3)*(
a + b*x^2 + c*x^4)^(p + 1))/(c*(4*p + 2*q + 1)), x] + Dist[1/(c*(4*p + 2*q + 1)), Int[(a + b*x^2 + c*x^4)^p*Ex
pandToSum[c*(4*p + 2*q + 1)*(d + e*x^2)^q - a*(2*q - 3)*e^q*x^(2*q - 4) - b*(2*p + 2*q - 1)*e^q*x^(2*q - 2) -
c*(4*p + 2*q + 1)*e^q*x^(2*q), x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && IGtQ[q, 1]

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 (d+e x^2\right )^3}{\sqrt{a+b x^2+c x^4}} \, dx &=\frac{e^3 x^3 \sqrt{a+b x^2+c x^4}}{5 c}+\frac{\int \frac{5 c d^3+3 e \left (5 c d^2-a e^2\right ) x^2+e^2 (15 c d-4 b e) x^4}{\sqrt{a+b x^2+c x^4}} \, dx}{5 c}\\ &=\frac{e^2 (15 c d-4 b e) x \sqrt{a+b x^2+c x^4}}{15 c^2}+\frac{e^3 x^3 \sqrt{a+b x^2+c x^4}}{5 c}+\frac{\int \frac{15 c^2 d^3-15 a c d e^2+4 a b e^3+e \left (45 c^2 d^2+8 b^2 e^2-3 c e (10 b d+3 a e)\right ) x^2}{\sqrt{a+b x^2+c x^4}} \, dx}{15 c^2}\\ &=\frac{e^2 (15 c d-4 b e) x \sqrt{a+b x^2+c x^4}}{15 c^2}+\frac{e^3 x^3 \sqrt{a+b x^2+c x^4}}{5 c}-\frac{\left (\sqrt{a} e \left (45 c^2 d^2+8 b^2 e^2-3 c e (10 b d+3 a e)\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 (15 c^2 d^3-15 a c d e^2+4 a b e^3+\frac{\sqrt{a} e \left (45 c^2 d^2+8 b^2 e^2-3 c e (10 b d+3 a e)\right )}{\sqrt{c}}\right ) \int \frac{1}{\sqrt{a+b x^2+c x^4}} \, dx}{15 c^2}\\ &=\frac{e^2 (15 c d-4 b e) x \sqrt{a+b x^2+c x^4}}{15 c^2}+\frac{e^3 x^3 \sqrt{a+b x^2+c x^4}}{5 c}+\frac{e \left (45 c^2 d^2+8 b^2 e^2-3 c e (10 b d+3 a e)\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} e \left (45 c^2 d^2+8 b^2 e^2-3 c e (10 b d+3 a e)\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{\left (15 c^2 d^3-15 a c d e^2+4 a b e^3+\frac{\sqrt{a} e \left (45 c^2 d^2+8 b^2 e^2-3 c e (10 b d+3 a e)\right )}{\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 )}{30 \sqrt [4]{a} c^{9/4} \sqrt{a+b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 2.90362, size = 584, normalized size = 1.25 \[ \frac{-i \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{-2 \sqrt{b^2-4 a c}+2 b+4 c x^2}{b-\sqrt{b^2-4 a c}}} \left (15 c^2 d e \left (3 d \sqrt{b^2-4 a c}-2 a e-3 b d\right )+c e^2 \left (-30 b d \sqrt{b^2-4 a c}-9 a e \sqrt{b^2-4 a c}+17 a b e+30 b^2 d\right )+8 b^2 e^3 \left (\sqrt{b^2-4 a c}-b\right )+30 c^3 d^3\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} x \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}}\right ),\frac{\sqrt{b^2-4 a c}+b}{b-\sqrt{b^2-4 a c}}\right )+i e \left (\sqrt{b^2-4 a c}-b\right ) \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{-2 \sqrt{b^2-4 a c}+2 b+4 c x^2}{b-\sqrt{b^2-4 a c}}} \left (-3 c e (3 a e+10 b d)+8 b^2 e^2+45 c^2 d^2\right ) E\left (i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{c}{b+\sqrt{b^2-4 a c}}} x\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )+4 c e^2 x \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \left (a+b x^2+c x^4\right ) \left (3 c \left (5 d+e x^2\right )-4 b e\right )}{60 c^3 \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \sqrt{a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x^2)^3/Sqrt[a + b*x^2 + c*x^4],x]

[Out]

(4*c*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*e^2*x*(a + b*x^2 + c*x^4)*(-4*b*e + 3*c*(5*d + e*x^2)) + I*(-b + Sqrt[b^2
- 4*a*c])*e*(45*c^2*d^2 + 8*b^2*e^2 - 3*c*e*(10*b*d + 3*a*e))*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqr
t[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*(30*c^3*d^3 + 8*b^
2*(-b + Sqrt[b^2 - 4*a*c])*e^3 + 15*c^2*d*e*(-3*b*d + 3*Sqrt[b^2 - 4*a*c]*d - 2*a*e) + c*e^2*(30*b^2*d - 30*b*
Sqrt[b^2 - 4*a*c]*d + 17*a*b*e - 9*a*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*ArcSinh[Sqrt[2]*S
qrt[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 = 0.066, size = 1186, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^3/(c*x^4+b*x^2+a)^(1/2),x)

[Out]

e^3*(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/c*a+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))))+3*d*e^2*(1/3/c*x*(c*x^4+b*x
^2+a)^(1/2)-1/12/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)*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))))-3
/2*d^2*e*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*d^3*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)*Elliptic
F(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 (e x^{2} + d\right )}^{3}}{\sqrt{c x^{4} + b x^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3/(c*x^4+b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((e*x^2 + d)^3/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{e^{3} x^{6} + 3 \, d e^{2} x^{4} + 3 \, d^{2} e x^{2} + d^{3}}{\sqrt{c x^{4} + b x^{2} + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3/(c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

integral((e^3*x^6 + 3*d*e^2*x^4 + 3*d^2*e*x^2 + d^3)/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 (d + e x^{2}\right )^{3}}{\sqrt{a + b x^{2} + c x^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**3/(c*x**4+b*x**2+a)**(1/2),x)

[Out]

Integral((d + e*x**2)**3/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 (e x^{2} + d\right )}^{3}}{\sqrt{c x^{4} + b x^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3/(c*x^4+b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^3/sqrt(c*x^4 + b*x^2 + a), x)