Integrand size = 21, antiderivative size = 326 \[ \int \frac {\left (d+e x^2\right )^3}{\sqrt {a+c x^4}} \, dx=\frac {d e^2 x \sqrt {a+c x^4}}{c}+\frac {e^3 x^3 \sqrt {a+c x^4}}{5 c}+\frac {3 e \left (5 c d^2-a e^2\right ) x \sqrt {a+c x^4}}{5 c^{3/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {3 \sqrt [4]{a} e \left (5 c d^2-a e^2\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{7/4} \sqrt {a+c x^4}}+\frac {\sqrt [4]{a} \left (15 c d^2 e-3 a e^3+\frac {5 \sqrt {c} d \left (c d^2-a e^2\right )}{\sqrt {a}}\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{10 c^{7/4} \sqrt {a+c x^4}} \]
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*e*(-a*e^2+5*c* d^2)*x*(c*x^4+a)^(1/2)/c^(3/2)/(a^(1/2)+x^2*c^(1/2))-3/5*a^(1/4)*e*(-a*e^2 +5*c*d^2)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)* x/a^(1/4)))*EllipticE(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/ 2)+x^2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/c^(7/4)/(c*x^4+a )^(1/2)+1/10*a^(1/4)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arct an(c^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*2^(1 /2))*(a^(1/2)+x^2*c^(1/2))*(15*c*d^2*e-3*a*e^3+5*d*(-a*e^2+c*d^2)*c^(1/2)/ a^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/c^(7/4)/(c*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.10 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.43 \[ \int \frac {\left (d+e x^2\right )^3}{\sqrt {a+c x^4}} \, dx=\frac {5 d \left (c d^2-a e^2\right ) x \sqrt {1+\frac {c x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^4}{a}\right )+e x \left (e \left (5 d+e x^2\right ) \left (a+c x^4\right )+\left (5 c d^2-a e^2\right ) x^2 \sqrt {1+\frac {c x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {c x^4}{a}\right )\right )}{5 c \sqrt {a+c x^4}} \]
(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 *Sqrt[a + c*x^4])
Time = 0.51 (sec) , antiderivative size = 322, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1519, 2427, 27, 1512, 27, 761, 1510}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\left (d+e x^2\right )^3}{\sqrt {a+c x^4}} \, dx\) |
| \(\Big \downarrow \) 1519 |
| \(\displaystyle \frac {\int \frac {15 c d e^2 x^4+3 e \left (5 c d^2-a e^2\right ) x^2+5 c d^3}{\sqrt {c x^4+a}}dx}{5 c}+\frac {e^3 x^3 \sqrt {a+c x^4}}{5 c}\) |
| \(\Big \downarrow \) 2427 |
| \(\displaystyle \frac {\frac {\int \frac {3 c \left (3 e \left (5 c d^2-a e^2\right ) x^2+5 d \left (c d^2-a e^2\right )\right )}{\sqrt {c x^4+a}}dx}{3 c}+5 d e^2 x \sqrt {a+c x^4}}{5 c}+\frac {e^3 x^3 \sqrt {a+c x^4}}{5 c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {3 e \left (5 c d^2-a e^2\right ) x^2+5 d \left (c d^2-a e^2\right )}{\sqrt {c x^4+a}}dx+5 d e^2 x \sqrt {a+c x^4}}{5 c}+\frac {e^3 x^3 \sqrt {a+c x^4}}{5 c}\) |
| \(\Big \downarrow \) 1512 |
| \(\displaystyle \frac {\frac {\left (-3 a^{3/2} e^3+15 \sqrt {a} c d^2 e-5 a \sqrt {c} d e^2+5 c^{3/2} d^3\right ) \int \frac {1}{\sqrt {c x^4+a}}dx}{\sqrt {c}}-\frac {3 \sqrt {a} e \left (5 c d^2-a e^2\right ) \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {a} \sqrt {c x^4+a}}dx}{\sqrt {c}}+5 d e^2 x \sqrt {a+c x^4}}{5 c}+\frac {e^3 x^3 \sqrt {a+c x^4}}{5 c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\left (-3 a^{3/2} e^3+15 \sqrt {a} c d^2 e-5 a \sqrt {c} d e^2+5 c^{3/2} d^3\right ) \int \frac {1}{\sqrt {c x^4+a}}dx}{\sqrt {c}}-\frac {3 e \left (5 c d^2-a e^2\right ) \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+a}}dx}{\sqrt {c}}+5 d e^2 x \sqrt {a+c x^4}}{5 c}+\frac {e^3 x^3 \sqrt {a+c x^4}}{5 c}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {-\frac {3 e \left (5 c d^2-a e^2\right ) \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+a}}dx}{\sqrt {c}}+\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (-3 a^{3/2} e^3+15 \sqrt {a} c d^2 e-5 a \sqrt {c} d e^2+5 c^{3/2} d^3\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} c^{3/4} \sqrt {a+c x^4}}+5 d e^2 x \sqrt {a+c x^4}}{5 c}+\frac {e^3 x^3 \sqrt {a+c x^4}}{5 c}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle \frac {\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (-3 a^{3/2} e^3+15 \sqrt {a} c d^2 e-5 a \sqrt {c} d e^2+5 c^{3/2} d^3\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} c^{3/4} \sqrt {a+c x^4}}-\frac {3 e \left (5 c d^2-a e^2\right ) \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt [4]{c} \sqrt {a+c x^4}}-\frac {x \sqrt {a+c x^4}}{\sqrt {a}+\sqrt {c} x^2}\right )}{\sqrt {c}}+5 d e^2 x \sqrt {a+c x^4}}{5 c}+\frac {e^3 x^3 \sqrt {a+c x^4}}{5 c}\) |
(e^3*x^3*Sqrt[a + c*x^4])/(5*c) + (5*d*e^2*x*Sqrt[a + c*x^4] - (3*e*(5*c*d ^2 - a*e^2)*(-((x*Sqrt[a + c*x^4])/(Sqrt[a] + Sqrt[c]*x^2)) + (a^(1/4)*(Sq rt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE [2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(c^(1/4)*Sqrt[a + c*x^4])))/Sqrt[c] + ((5*c^(3/2)*d^3 + 15*Sqrt[a]*c*d^2*e - 5*a*Sqrt[c]*d*e^2 - 3*a^(3/2)*e^3 )*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*Elli pticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(1/4)*c^(3/4)*Sqrt[a + c*x ^4]))/(5*c)
3.2.51.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Simp[(e + d*q)/q Int[1/Sqrt[a + c*x^4], x], x] - Simp[e/q Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a, c , d, e}, x] && PosQ[c/a]
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Sim p[e^q*x^(2*q - 3)*((a + c*x^4)^(p + 1)/(c*(4*p + 2*q + 1))), x] + Simp[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]
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{q = Expon[Pq, x ]}, With[{Pqq = Coeff[Pq, x, q]}, Simp[Pqq*x^(q - n + 1)*((a + b*x^n)^(p + 1)/(b*(q + n*p + 1))), x] + Simp[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]] /; 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] && IGtQ[n, 0]
Result contains complex when optimal does not.
Time = 2.70 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.72
| method | result | size |
| elliptic | \(\frac {e^{3} x^{3} \sqrt {c \,x^{4}+a}}{5 c}+\frac {d \,e^{2} x \sqrt {c \,x^{4}+a}}{c}+\frac {\left (d^{3}-\frac {d \,e^{2} a}{c}\right ) \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {i \left (3 d^{2} e -\frac {3 e^{3} a}{5 c}\right ) \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}\) | \(235\) |
| risch | \(\frac {e^{2} x \left (e \,x^{2}+5 d \right ) \sqrt {c \,x^{4}+a}}{5 c}-\frac {-\frac {5 d^{3} c \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {5 d \,e^{2} a \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {i \left (3 a \,e^{3}-15 c \,d^{2} e \right ) \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}}{5 c}\) | \(294\) |
| default | \(\frac {d^{3} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+e^{3} \left (\frac {x^{3} \sqrt {c \,x^{4}+a}}{5 c}-\frac {3 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 c^{\frac {3}{2}} \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\right )+3 d \,e^{2} \left (\frac {x \sqrt {c \,x^{4}+a}}{3 c}-\frac {a \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{3 c \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\right )+\frac {3 i d^{2} e \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}\) | \(388\) |
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)/(I /a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/ 2)*x^2)^(1/2)/(c*x^4+a)^(1/2)*EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I)+I*( 3*d^2*e-3/5*e^3/c*a)*a^(1/2)/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2 )*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)/c^(1/2)*(Elli pticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1/2)*c^(1/2))^(1/2) ,I))
Time = 0.09 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.51 \[ \int \frac {\left (d+e x^2\right )^3}{\sqrt {a+c x^4}} \, dx=\frac {3 \, {\left (5 \, a c d^{2} e - a^{2} e^{3}\right )} \sqrt {c} x \left (-\frac {a}{c}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (5 \, c^{2} d^{3} - 15 \, a c d^{2} e - 5 \, a c d e^{2} + 3 \, a^{2} e^{3}\right )} \sqrt {c} x \left (-\frac {a}{c}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (a c e^{3} x^{4} + 5 \, a c d e^{2} x^{2} + 15 \, a c d^{2} e - 3 \, a^{2} e^{3}\right )} \sqrt {c x^{4} + a}}{5 \, a c^{2} x} \]
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 )
Result contains complex when optimal does not.
Time = 1.72 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.53 \[ \int \frac {\left (d+e x^2\right )^3}{\sqrt {a+c x^4}} \, dx=\frac {d^{3} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} + \frac {3 d^{2} e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} + \frac {3 d e^{2} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {9}{4}\right )} + \frac {e^{3} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {11}{4}\right )} \]
d**3*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), c*x**4*exp_polar(I*pi)/a)/(4*s qrt(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(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(I*pi)/a)/(4*sqrt(a)*gamma(9/4)) + e**3*x**7*gamma(7/4)*hyper((1/2, 7/4), (11/4,), c*x**4*exp_polar(I*pi)/ a)/(4*sqrt(a)*gamma(11/4))
\[ \int \frac {\left (d+e x^2\right )^3}{\sqrt {a+c x^4}} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{3}}{\sqrt {c x^{4} + a}} \,d x } \]
\[ \int \frac {\left (d+e x^2\right )^3}{\sqrt {a+c x^4}} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{3}}{\sqrt {c x^{4} + a}} \,d x } \]
Timed out. \[ \int \frac {\left (d+e x^2\right )^3}{\sqrt {a+c x^4}} \, dx=\int \frac {{\left (e\,x^2+d\right )}^3}{\sqrt {c\,x^4+a}} \,d x \]