\(\int \frac {(d+e x^2)^3}{\sqrt {a-c x^4}} \, dx\) [157]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 22, antiderivative size = 213 \[ \int \frac {\left (d+e x^2\right )^3}{\sqrt {a-c x^4}} \, dx=-\frac {d e^2 x \sqrt {a-c x^4}}{c}-\frac {e^3 x^3 \sqrt {a-c x^4}}{5 c}+\frac {3 a^{3/4} e \left (5 c d^2+a e^2\right ) \sqrt {1-\frac {c x^4}{a}} E\left (\left .\arcsin \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{5 c^{7/4} \sqrt {a-c x^4}}+\frac {a^{3/4} \left (\frac {5 \sqrt {c} d \left (c d^2+a e^2\right )}{\sqrt {a}}-3 e \left (5 c d^2+a e^2\right )\right ) \sqrt {1-\frac {c x^4}{a}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),-1\right )}{5 c^{7/4} \sqrt {a-c x^4}} \]

[Out]

-d*e^2*x*(-c*x^4+a)^(1/2)/c-1/5*e^3*x^3*(-c*x^4+a)^(1/2)/c+3/5*a^(3/4)*e*(a*e^2+5*c*d^2)*EllipticE(c^(1/4)*x/a
^(1/4),I)*(1-c*x^4/a)^(1/2)/c^(7/4)/(-c*x^4+a)^(1/2)+1/5*a^(3/4)*EllipticF(c^(1/4)*x/a^(1/4),I)*(-3*e*(a*e^2+5
*c*d^2)+5*d*(a*e^2+c*d^2)*c^(1/2)/a^(1/2))*(1-c*x^4/a)^(1/2)/c^(7/4)/(-c*x^4+a)^(1/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {1221, 1902, 1215, 230, 227, 1214, 1213, 435} \[ \int \frac {\left (d+e x^2\right )^3}{\sqrt {a-c x^4}} \, dx=\frac {a^{3/4} \sqrt {1-\frac {c x^4}{a}} \left (\frac {5 \sqrt {c} d \left (a e^2+c d^2\right )}{\sqrt {a}}-3 e \left (a e^2+5 c d^2\right )\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),-1\right )}{5 c^{7/4} \sqrt {a-c x^4}}+\frac {3 a^{3/4} e \sqrt {1-\frac {c x^4}{a}} \left (a e^2+5 c d^2\right ) E\left (\left .\arcsin \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{5 c^{7/4} \sqrt {a-c x^4}}-\frac {d e^2 x \sqrt {a-c x^4}}{c}-\frac {e^3 x^3 \sqrt {a-c x^4}}{5 c} \]

[In]

Int[(d + e*x^2)^3/Sqrt[a - c*x^4],x]

[Out]

-((d*e^2*x*Sqrt[a - c*x^4])/c) - (e^3*x^3*Sqrt[a - c*x^4])/(5*c) + (3*a^(3/4)*e*(5*c*d^2 + a*e^2)*Sqrt[1 - (c*
x^4)/a]*EllipticE[ArcSin[(c^(1/4)*x)/a^(1/4)], -1])/(5*c^(7/4)*Sqrt[a - c*x^4]) + (a^(3/4)*((5*Sqrt[c]*d*(c*d^
2 + a*e^2))/Sqrt[a] - 3*e*(5*c*d^2 + a*e^2))*Sqrt[1 - (c*x^4)/a]*EllipticF[ArcSin[(c^(1/4)*x)/a^(1/4)], -1])/(
5*c^(7/4)*Sqrt[a - c*x^4])

Rule 227

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[Rt[-b, 4]*(x/Rt[a, 4])], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 230

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + b*(x^4/a)]/Sqrt[a + b*x^4], Int[1/Sqrt[1 + b*(x^4/
a)], x], x] /; FreeQ[{a, b}, x] && NegQ[b/a] &&  !GtQ[a, 0]

Rule 435

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

Rule 1213

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + e*(x^2/d)]/Sqrt
[1 - e*(x^2/d)], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

Rule 1214

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + c*(x^4/a)]/Sqrt[a + c*x^4], In
t[(d + e*x^2)/Sqrt[1 + c*(x^4/a)], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] &&
!GtQ[a, 0]

Rule 1215

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

Rule 1221

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

Rule 1902

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, With[{Pqq = Coeff[Pq, x, q]}, D
ist[1/(b*(q + n*p + 1)), Int[ExpandToSum[b*(q + n*p + 1)*(Pq - Pqq*x^q) - a*Pqq*(q - n + 1)*x^(q - n), x]*(a +
 b*x^n)^p, x], x] + Simp[Pqq*x^(q - n + 1)*((a + b*x^n)^(p + 1)/(b*(q + n*p + 1))), x]] /; NeQ[q + n*p + 1, 0]
 && q - n >= 0 && (IntegerQ[2*p] || IntegerQ[p + (q + 1)/(2*n)])] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && IG
tQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {e^3 x^3 \sqrt {a-c x^4}}{5 c}-\frac {\int \frac {-5 c d^3-3 e \left (5 c d^2+a e^2\right ) x^2-15 c d e^2 x^4}{\sqrt {a-c x^4}} \, dx}{5 c} \\ & = -\frac {d e^2 x \sqrt {a-c x^4}}{c}-\frac {e^3 x^3 \sqrt {a-c x^4}}{5 c}+\frac {\int \frac {15 c d \left (c d^2+a e^2\right )+9 c e \left (5 c d^2+a e^2\right ) x^2}{\sqrt {a-c x^4}} \, dx}{15 c^2} \\ & = -\frac {d e^2 x \sqrt {a-c x^4}}{c}-\frac {e^3 x^3 \sqrt {a-c x^4}}{5 c}+\frac {\left (3 \sqrt {a} e \left (5 c d^2+a e^2\right )\right ) \int \frac {1+\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a-c x^4}} \, dx}{5 c^{3/2}}+\frac {\left (5 \sqrt {c} d \left (c d^2+a e^2\right )-3 \sqrt {a} e \left (5 c d^2+a e^2\right )\right ) \int \frac {1}{\sqrt {a-c x^4}} \, dx}{5 c^{3/2}} \\ & = -\frac {d e^2 x \sqrt {a-c x^4}}{c}-\frac {e^3 x^3 \sqrt {a-c x^4}}{5 c}+\frac {\left (3 \sqrt {a} e \left (5 c d^2+a e^2\right ) \sqrt {1-\frac {c x^4}{a}}\right ) \int \frac {1+\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {1-\frac {c x^4}{a}}} \, dx}{5 c^{3/2} \sqrt {a-c x^4}}+\frac {\left (\left (5 \sqrt {c} d \left (c d^2+a e^2\right )-3 \sqrt {a} e \left (5 c d^2+a e^2\right )\right ) \sqrt {1-\frac {c x^4}{a}}\right ) \int \frac {1}{\sqrt {1-\frac {c x^4}{a}}} \, dx}{5 c^{3/2} \sqrt {a-c x^4}} \\ & = -\frac {d e^2 x \sqrt {a-c x^4}}{c}-\frac {e^3 x^3 \sqrt {a-c x^4}}{5 c}+\frac {\sqrt [4]{a} \left (5 \sqrt {c} d \left (c d^2+a e^2\right )-3 \sqrt {a} e \left (5 c d^2+a e^2\right )\right ) \sqrt {1-\frac {c x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{5 c^{7/4} \sqrt {a-c x^4}}+\frac {\left (3 \sqrt {a} e \left (5 c d^2+a e^2\right ) \sqrt {1-\frac {c x^4}{a}}\right ) \int \frac {\sqrt {1+\frac {\sqrt {c} x^2}{\sqrt {a}}}}{\sqrt {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}} \, dx}{5 c^{3/2} \sqrt {a-c x^4}} \\ & = -\frac {d e^2 x \sqrt {a-c x^4}}{c}-\frac {e^3 x^3 \sqrt {a-c x^4}}{5 c}+\frac {3 a^{3/4} e \left (5 c d^2+a e^2\right ) \sqrt {1-\frac {c x^4}{a}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{5 c^{7/4} \sqrt {a-c x^4}}+\frac {\sqrt [4]{a} \left (5 \sqrt {c} d \left (c d^2+a e^2\right )-3 \sqrt {a} e \left (5 c d^2+a e^2\right )\right ) \sqrt {1-\frac {c x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{5 c^{7/4} \sqrt {a-c x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.12 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.66 \[ \int \frac {\left (d+e x^2\right )^3}{\sqrt {a-c x^4}} \, dx=\frac {5 d \left (c d^2+a e^2\right ) x \sqrt {1-\frac {c x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {c x^4}{a}\right )+e x \left (e \left (5 d+e x^2\right ) \left (-a+c x^4\right )+\left (5 c d^2+a e^2\right ) x^2 \sqrt {1-\frac {c x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},\frac {c x^4}{a}\right )\right )}{5 c \sqrt {a-c x^4}} \]

[In]

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

[Out]

(5*d*(c*d^2 + a*e^2)*x*Sqrt[1 - (c*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, (c*x^4)/a] + e*x*(e*(5*d + e*x^2)*
(-a + c*x^4) + (5*c*d^2 + a*e^2)*x^2*Sqrt[1 - (c*x^4)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, (c*x^4)/a]))/(5*c*Sq
rt[a - c*x^4])

Maple [A] (verified)

Time = 4.50 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.04

method result size
elliptic \(-\frac {e^{3} x^{3} \sqrt {-c \,x^{4}+a}}{5 c}-\frac {d \,e^{2} x \sqrt {-c \,x^{4}+a}}{c}+\frac {\left (d^{3}+\frac {d \,e^{2} a}{c}\right ) \sqrt {1-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}}-\frac {\left (3 d^{2} e +\frac {3 e^{3} a}{5 c}\right ) \sqrt {a}\, \sqrt {1-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}\, \sqrt {c}}\) \(222\)
risch \(-\frac {e^{2} x \left (e \,x^{2}+5 d \right ) \sqrt {-c \,x^{4}+a}}{5 c}+\frac {\frac {5 d^{3} c \sqrt {1-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}}+\frac {5 d \,e^{2} a \sqrt {1-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}}-\frac {\left (3 a \,e^{3}+15 c \,d^{2} e \right ) \sqrt {a}\, \sqrt {1-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}\, \sqrt {c}}}{5 c}\) \(274\)
default \(\frac {d^{3} \sqrt {1-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}}+e^{3} \left (-\frac {x^{3} \sqrt {-c \,x^{4}+a}}{5 c}-\frac {3 a^{\frac {3}{2}} \sqrt {1-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 c^{\frac {3}{2}} \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}}\right )+3 d \,e^{2} \left (-\frac {x \sqrt {-c \,x^{4}+a}}{3 c}+\frac {a \sqrt {1-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )}{3 c \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}}\right )-\frac {3 d^{2} e \sqrt {a}\, \sqrt {1-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}\, \sqrt {c}}\) \(360\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.10 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.78 \[ \int \frac {\left (d+e x^2\right )^3}{\sqrt {a-c x^4}} \, dx=-\frac {3 \, {\left (5 \, a c d^{2} e + a^{2} e^{3}\right )} \sqrt {-c} x \left (\frac {a}{c}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - {\left (5 \, c^{2} d^{3} + 15 \, a c d^{2} e + 5 \, a c d e^{2} + 3 \, a^{2} e^{3}\right )} \sqrt {-c} x \left (\frac {a}{c}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (a c e^{3} x^{4} + 5 \, a c d e^{2} x^{2} + 15 \, a c d^{2} e + 3 \, a^{2} e^{3}\right )} \sqrt {-c x^{4} + a}}{5 \, a c^{2} x} \]

[In]

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

[Out]

-1/5*(3*(5*a*c*d^2*e + a^2*e^3)*sqrt(-c)*x*(a/c)^(3/4)*elliptic_e(arcsin((a/c)^(1/4)/x), -1) - (5*c^2*d^3 + 15
*a*c*d^2*e + 5*a*c*d*e^2 + 3*a^2*e^3)*sqrt(-c)*x*(a/c)^(3/4)*elliptic_f(arcsin((a/c)^(1/4)/x), -1) + (a*c*e^3*
x^4 + 5*a*c*d*e^2*x^2 + 15*a*c*d^2*e + 3*a^2*e^3)*sqrt(-c*x^4 + a))/(a*c^2*x)

Sympy [A] (verification not implemented)

Time = 1.88 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.85 \[ \int \frac {\left (d+e x^2\right )^3}{\sqrt {a-c x^4}} \, dx=\frac {d^{3} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{2 i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} + \frac {3 d^{2} e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{2 i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} + \frac {3 d e^{2} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {c x^{4} e^{2 i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {9}{4}\right )} + \frac {e^{3} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {c x^{4} e^{2 i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {11}{4}\right )} \]

[In]

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

[Out]

d**3*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), c*x**4*exp_polar(2*I*pi)/a)/(4*sqrt(a)*gamma(5/4)) + 3*d**2*e*x**3
*gamma(3/4)*hyper((1/2, 3/4), (7/4,), c*x**4*exp_polar(2*I*pi)/a)/(4*sqrt(a)*gamma(7/4)) + 3*d*e**2*x**5*gamma
(5/4)*hyper((1/2, 5/4), (9/4,), c*x**4*exp_polar(2*I*pi)/a)/(4*sqrt(a)*gamma(9/4)) + e**3*x**7*gamma(7/4)*hype
r((1/2, 7/4), (11/4,), c*x**4*exp_polar(2*I*pi)/a)/(4*sqrt(a)*gamma(11/4))

Maxima [F]

\[ \int \frac {\left (d+e x^2\right )^3}{\sqrt {a-c x^4}} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{3}}{\sqrt {-c x^{4} + a}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\left (d+e x^2\right )^3}{\sqrt {a-c x^4}} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{3}}{\sqrt {-c x^{4} + a}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right )^3}{\sqrt {a-c x^4}} \, dx=\int \frac {{\left (e\,x^2+d\right )}^3}{\sqrt {a-c\,x^4}} \,d x \]

[In]

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

[Out]

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