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

Optimal. Leaf size=553 \[ -\frac {e \left (b-\sqrt {4 a c+b^2}\right ) \sqrt {\sqrt {4 a c+b^2}+b} \sqrt {1-\frac {2 c x^2}{b-\sqrt {4 a c+b^2}}} \sqrt {1-\frac {2 c x^2}{\sqrt {4 a c+b^2}+b}} \left (3 c e (3 a e+10 b d)+8 b^2 e^2+45 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2+4 a c}}}\right )|\frac {b+\sqrt {b^2+4 a c}}{b-\sqrt {b^2+4 a c}}\right )}{30 \sqrt {2} c^{7/2} \sqrt {a+b x^2-c x^4}}+\frac {\left (b-\sqrt {4 a c+b^2}\right ) \sqrt {\sqrt {4 a c+b^2}+b} \sqrt {1-\frac {2 c x^2}{b-\sqrt {4 a c+b^2}}} \sqrt {1-\frac {2 c x^2}{\sqrt {4 a c+b^2}+b}} \left (\frac {2 c \left (4 a b e^3+15 a c d e^2+15 c^2 d^3\right )}{b-\sqrt {4 a c+b^2}}+e \left (3 c e (3 a e+10 b d)+8 b^2 e^2+45 c^2 d^2\right )\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2+4 a c}}}\right )|\frac {b+\sqrt {b^2+4 a c}}{b-\sqrt {b^2+4 a c}}\right )}{30 \sqrt {2} c^{7/2} \sqrt {a+b x^2-c x^4}}-\frac {e^2 x \sqrt {a+b x^2-c x^4} (4 b e+15 c d)}{15 c^2}-\frac {e^3 x^3 \sqrt {a+b x^2-c x^4}}{5 c} \]

[Out]

-1/15*e^2*(4*b*e+15*c*d)*x*(-c*x^4+b*x^2+a)^(1/2)/c^2-1/5*e^3*x^3*(-c*x^4+b*x^2+a)^(1/2)/c-1/60*e*(45*c^2*d^2+
8*b^2*e^2+3*c*e*(3*a*e+10*b*d))*EllipticE(x*2^(1/2)*c^(1/2)/(b+(4*a*c+b^2)^(1/2))^(1/2),((b+(4*a*c+b^2)^(1/2))
/(b-(4*a*c+b^2)^(1/2)))^(1/2))*(b-(4*a*c+b^2)^(1/2))*(1-2*c*x^2/(b-(4*a*c+b^2)^(1/2)))^(1/2)*(b+(4*a*c+b^2)^(1
/2))^(1/2)*(1-2*c*x^2/(b+(4*a*c+b^2)^(1/2)))^(1/2)/c^(7/2)*2^(1/2)/(-c*x^4+b*x^2+a)^(1/2)+1/60*EllipticF(x*2^(
1/2)*c^(1/2)/(b+(4*a*c+b^2)^(1/2))^(1/2),((b+(4*a*c+b^2)^(1/2))/(b-(4*a*c+b^2)^(1/2)))^(1/2))*(e*(45*c^2*d^2+8
*b^2*e^2+3*c*e*(3*a*e+10*b*d))+2*c*(4*a*b*e^3+15*a*c*d*e^2+15*c^2*d^3)/(b-(4*a*c+b^2)^(1/2)))*(b-(4*a*c+b^2)^(
1/2))*(1-2*c*x^2/(b-(4*a*c+b^2)^(1/2)))^(1/2)*(b+(4*a*c+b^2)^(1/2))^(1/2)*(1-2*c*x^2/(b+(4*a*c+b^2)^(1/2)))^(1
/2)/c^(7/2)*2^(1/2)/(-c*x^4+b*x^2+a)^(1/2)

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Rubi [A]  time = 1.28, antiderivative size = 553, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1206, 1679, 1202, 524, 424, 419} \[ \frac {\left (b-\sqrt {4 a c+b^2}\right ) \sqrt {\sqrt {4 a c+b^2}+b} \sqrt {1-\frac {2 c x^2}{b-\sqrt {4 a c+b^2}}} \sqrt {1-\frac {2 c x^2}{\sqrt {4 a c+b^2}+b}} \left (\frac {2 c \left (4 a b e^3+15 a c d e^2+15 c^2 d^3\right )}{b-\sqrt {4 a c+b^2}}+e \left (3 c e (3 a e+10 b d)+8 b^2 e^2+45 c^2 d^2\right )\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2+4 a c}}}\right )|\frac {b+\sqrt {b^2+4 a c}}{b-\sqrt {b^2+4 a c}}\right )}{30 \sqrt {2} c^{7/2} \sqrt {a+b x^2-c x^4}}-\frac {e \left (b-\sqrt {4 a c+b^2}\right ) \sqrt {\sqrt {4 a c+b^2}+b} \sqrt {1-\frac {2 c x^2}{b-\sqrt {4 a c+b^2}}} \sqrt {1-\frac {2 c x^2}{\sqrt {4 a c+b^2}+b}} \left (3 c e (3 a e+10 b d)+8 b^2 e^2+45 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2+4 a c}}}\right )|\frac {b+\sqrt {b^2+4 a c}}{b-\sqrt {b^2+4 a c}}\right )}{30 \sqrt {2} c^{7/2} \sqrt {a+b x^2-c x^4}}-\frac {e^2 x \sqrt {a+b x^2-c x^4} (4 b e+15 c d)}{15 c^2}-\frac {e^3 x^3 \sqrt {a+b x^2-c x^4}}{5 c} \]

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) - ((b - S
qrt[b^2 + 4*a*c])*Sqrt[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[1 - (2*
c*x^2)/(b - Sqrt[b^2 + 4*a*c])]*Sqrt[1 - (2*c*x^2)/(b + Sqrt[b^2 + 4*a*c])]*EllipticE[ArcSin[(Sqrt[2]*Sqrt[c]*
x)/Sqrt[b + Sqrt[b^2 + 4*a*c]]], (b + Sqrt[b^2 + 4*a*c])/(b - Sqrt[b^2 + 4*a*c])])/(30*Sqrt[2]*c^(7/2)*Sqrt[a
+ b*x^2 - c*x^4]) + ((b - Sqrt[b^2 + 4*a*c])*Sqrt[b + Sqrt[b^2 + 4*a*c]]*((2*c*(15*c^2*d^3 + 15*a*c*d*e^2 + 4*
a*b*e^3))/(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[1 - (2*c*x^2)/(b
- Sqrt[b^2 + 4*a*c])]*Sqrt[1 - (2*c*x^2)/(b + Sqrt[b^2 + 4*a*c])]*EllipticF[ArcSin[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b
+ Sqrt[b^2 + 4*a*c]]], (b + Sqrt[b^2 + 4*a*c])/(b - Sqrt[b^2 + 4*a*c])])/(30*Sqrt[2]*c^(7/2)*Sqrt[a + b*x^2 -
c*x^4])

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 524

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-(b/a), -(d/c)]))))))

Rule 1202

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

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]

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 {1-\frac {2 c x^2}{b-\sqrt {b^2+4 a c}}} \sqrt {1-\frac {2 c x^2}{b+\sqrt {b^2+4 a c}}}\right ) \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 {1-\frac {2 c x^2}{b-\sqrt {b^2+4 a c}}} \sqrt {1-\frac {2 c x^2}{b+\sqrt {b^2+4 a c}}}} \, dx}{15 c^2 \sqrt {a+b x^2-c x^4}}\\ &=-\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 (\left (b-\sqrt {b^2+4 a c}\right ) e \left (45 c^2 d^2+8 b^2 e^2+3 c e (10 b d+3 a e)\right ) \sqrt {1-\frac {2 c x^2}{b-\sqrt {b^2+4 a c}}} \sqrt {1-\frac {2 c x^2}{b+\sqrt {b^2+4 a c}}}\right ) \int \frac {\sqrt {1-\frac {2 c x^2}{b-\sqrt {b^2+4 a c}}}}{\sqrt {1-\frac {2 c x^2}{b+\sqrt {b^2+4 a c}}}} \, dx}{30 c^3 \sqrt {a+b x^2-c x^4}}+\frac {\left (\left (b-\sqrt {b^2+4 a c}\right ) \left (\frac {2 c \left (15 c^2 d^3+15 a c d e^2+4 a b e^3\right )}{b-\sqrt {b^2+4 a c}}+e \left (45 c^2 d^2+8 b^2 e^2+3 c e (10 b d+3 a e)\right )\right ) \sqrt {1-\frac {2 c x^2}{b-\sqrt {b^2+4 a c}}} \sqrt {1-\frac {2 c x^2}{b+\sqrt {b^2+4 a c}}}\right ) \int \frac {1}{\sqrt {1-\frac {2 c x^2}{b-\sqrt {b^2+4 a c}}} \sqrt {1-\frac {2 c x^2}{b+\sqrt {b^2+4 a c}}}} \, dx}{30 c^3 \sqrt {a+b x^2-c x^4}}\\ &=-\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 (b-\sqrt {b^2+4 a c}\right ) \sqrt {b+\sqrt {b^2+4 a c}} e \left (45 c^2 d^2+8 b^2 e^2+3 c e (10 b d+3 a e)\right ) \sqrt {1-\frac {2 c x^2}{b-\sqrt {b^2+4 a c}}} \sqrt {1-\frac {2 c x^2}{b+\sqrt {b^2+4 a c}}} E\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2+4 a c}}}\right )|\frac {b+\sqrt {b^2+4 a c}}{b-\sqrt {b^2+4 a c}}\right )}{30 \sqrt {2} c^{7/2} \sqrt {a+b x^2-c x^4}}+\frac {\left (b-\sqrt {b^2+4 a c}\right ) \sqrt {b+\sqrt {b^2+4 a c}} \left (\frac {2 c \left (15 c^2 d^3+15 a c d e^2+4 a b e^3\right )}{b-\sqrt {b^2+4 a c}}+e \left (45 c^2 d^2+8 b^2 e^2+3 c e (10 b d+3 a e)\right )\right ) \sqrt {1-\frac {2 c x^2}{b-\sqrt {b^2+4 a c}}} \sqrt {1-\frac {2 c x^2}{b+\sqrt {b^2+4 a c}}} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2+4 a c}}}\right )|\frac {b+\sqrt {b^2+4 a c}}{b-\sqrt {b^2+4 a c}}\right )}{30 \sqrt {2} c^{7/2} \sqrt {a+b x^2-c x^4}}\\ \end {align*}

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Mathematica [C]  time = 2.47, size = 596, normalized size = 1.08 \[ \frac {-i \sqrt {2} e \left (\sqrt {4 a c+b^2}-b\right ) \sqrt {\frac {\sqrt {4 a c+b^2}+b-2 c x^2}{\sqrt {4 a c+b^2}+b}} \sqrt {\frac {\sqrt {4 a c+b^2}-b+2 c x^2}{\sqrt {4 a c+b^2}-b}} \left (3 c e (3 a e+10 b d)+8 b^2 e^2+45 c^2 d^2\right ) E\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {-\frac {c}{b+\sqrt {b^2+4 a c}}} x\right )|\frac {b+\sqrt {b^2+4 a c}}{b-\sqrt {b^2+4 a c}}\right )+i \sqrt {2} \sqrt {\frac {\sqrt {4 a c+b^2}+b-2 c x^2}{\sqrt {4 a c+b^2}+b}} \sqrt {\frac {\sqrt {4 a c+b^2}-b+2 c x^2}{\sqrt {4 a c+b^2}-b}} \left (15 c^2 d e \left (3 d \sqrt {4 a c+b^2}-2 a e-3 b d\right )+c e^2 \left (30 b d \sqrt {4 a c+b^2}+9 a e \sqrt {4 a c+b^2}-17 a b e-30 b^2 d\right )+8 b^2 e^3 \left (\sqrt {4 a c+b^2}-b\right )-30 c^3 d^3\right ) F\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {-\frac {c}{b+\sqrt {b^2+4 a c}}} x\right )|\frac {b+\sqrt {b^2+4 a c}}{b-\sqrt {b^2+4 a c}}\right )-4 c e^2 x \sqrt {-\frac {c}{\sqrt {4 a c+b^2}+b}} \left (a+b x^2-c x^4\right ) \left (4 b e+3 c \left (5 d+e x^2\right )\right )}{60 c^3 \sqrt {-\frac {c}{\sqrt {4 a c+b^2}+b}} \sqrt {a+b x^2-c x^4}} \]

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*Sqrt[2]*(-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 + Sqrt[b^2 + 4*a*c])]*Sqrt[(-b + Sqrt[b^2 + 4*a*c] + 2*c*x^2)/(-b + Sqrt[b^2 + 4*a*c])]*EllipticE[I*ArcS
inh[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*Sqrt[2
]*(-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[(-b + Sqrt[b^2 + 4*a*c] + 2*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])])/(6
0*c^3*Sqrt[-(c/(b + Sqrt[b^2 + 4*a*c]))]*Sqrt[a + b*x^2 - c*x^4])

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (e^{3} x^{6} + 3 \, d e^{2} x^{4} + 3 \, d^{2} e x^{2} + d^{3}\right )} \sqrt {-c x^{4} + b x^{2} + a}}{c x^{4} - b x^{2} - a}, x\right ) \]

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)/(c*x^4 - b*x^2 - a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )}^{3}}{\sqrt {-c x^{4} + b x^{2} + a}}\,{d x} \]

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)

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maple [B]  time = 0.02, size = 1195, normalized size = 2.16 \[ \text {result too large to display} \]

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)/((-b+(4*a*c+b^
2)^(1/2))/a)^(1/2)*(-2*(-b+(4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)*(2*(b+(4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)/(-c*x^4+b
*x^2+a)^(1/2)*EllipticF(1/2*2^(1/2)*((-b+(4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(-2*(b+(4*a*c+b^2)^(1/2))/a*b/c-4)^
(1/2))-1/2*(3/5*a/c+8/15*b^2/c^2)*a*2^(1/2)/((-b+(4*a*c+b^2)^(1/2))/a)^(1/2)*(-2*(-b+(4*a*c+b^2)^(1/2))/a*x^2+
4)^(1/2)*(2*(b+(4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)/(-c*x^4+b*x^2+a)^(1/2)/(b+(4*a*c+b^2)^(1/2))*(EllipticF(1/2*2
^(1/2)*((-b+(4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(-2*(b+(4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))-EllipticE(1/2*2^(1/2)*
((-b+(4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(-2*(b+(4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))))+3*d*e^2*(-1/3*(-c*x^4+b*x^2
+a)^(1/2)/c*x+1/12/c*a*2^(1/2)/((-b+(4*a*c+b^2)^(1/2))/a)^(1/2)*(-2*(-b+(4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)*(2*(
b+(4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)/(-c*x^4+b*x^2+a)^(1/2)*EllipticF(1/2*2^(1/2)*((-b+(4*a*c+b^2)^(1/2))/a)^(1
/2)*x,1/2*(-2*(b+(4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))-1/3*b/c*a*2^(1/2)/((-b+(4*a*c+b^2)^(1/2))/a)^(1/2)*(-2*(-b
+(4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)*(2*(b+(4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)/(-c*x^4+b*x^2+a)^(1/2)/(b+(4*a*c+b^
2)^(1/2))*(EllipticF(1/2*2^(1/2)*((-b+(4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(-2*(b+(4*a*c+b^2)^(1/2))/a*b/c-4)^(1/
2))-EllipticE(1/2*2^(1/2)*((-b+(4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(-2*(b+(4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))))-3
/2*d^2*e*a*2^(1/2)/((-b+(4*a*c+b^2)^(1/2))/a)^(1/2)*(-2*(-b+(4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)*(2*(b+(4*a*c+b^2
)^(1/2))/a*x^2+4)^(1/2)/(-c*x^4+b*x^2+a)^(1/2)/(b+(4*a*c+b^2)^(1/2))*(EllipticF(1/2*2^(1/2)*((-b+(4*a*c+b^2)^(
1/2))/a)^(1/2)*x,1/2*(-2*(b+(4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))-EllipticE(1/2*2^(1/2)*((-b+(4*a*c+b^2)^(1/2))/a
)^(1/2)*x,1/2*(-2*(b+(4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2)))+1/4*d^3*2^(1/2)/((-b+(4*a*c+b^2)^(1/2))/a)^(1/2)*(-2*
(-b+(4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)*(2*(b+(4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)/(-c*x^4+b*x^2+a)^(1/2)*EllipticF
(1/2*2^(1/2)*((-b+(4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(-2*(b+(4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )}^{3}}{\sqrt {-c x^{4} + b x^{2} + a}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (e\,x^2+d\right )}^3}{\sqrt {-c\,x^4+b\,x^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x^{2}\right )^{3}}{\sqrt {a + b x^{2} - c x^{4}}}\, dx \]

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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