Integrand size = 28, antiderivative size = 367 \[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^2}{\sqrt {a+c x^4}} \, dx=\frac {e (2 B d+A e) x \sqrt {a+c x^4}}{3 c}+\frac {B e^2 x^3 \sqrt {a+c x^4}}{5 c}+\frac {\left (5 B c d^2+10 A c d e-3 a B e^2\right ) x \sqrt {a+c x^4}}{5 c^{3/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {\sqrt [4]{a} \left (5 B c d^2+10 A c d e-3 a B e^2\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{7/4} \sqrt {a+c x^4}}+\frac {\left (15 A c^{3/2} d^2-9 a^{3/2} B e^2-5 a \sqrt {c} e (2 B d+A e)+15 \sqrt {a} c d (B d+2 A e)\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{30 \sqrt [4]{a} c^{7/4} \sqrt {a+c x^4}} \] Output:
1/3*e*(A*e+2*B*d)*x*(c*x^4+a)^(1/2)/c+1/5*B*e^2*x^3*(c*x^4+a)^(1/2)/c+1/5* (10*A*c*d*e-3*B*a*e^2+5*B*c*d^2)*x*(c*x^4+a)^(1/2)/c^(3/2)/(a^(1/2)+c^(1/2 )*x^2)-1/5*a^(1/4)*(10*A*c*d*e-3*B*a*e^2+5*B*c*d^2)*(a^(1/2)+c^(1/2)*x^2)* ((c*x^4+a)/(a^(1/2)+c^(1/2)*x^2)^2)^(1/2)*EllipticE(sin(2*arctan(c^(1/4)*x /a^(1/4))),1/2*2^(1/2))/c^(7/4)/(c*x^4+a)^(1/2)+1/30*(15*A*c^(3/2)*d^2-9*a ^(3/2)*B*e^2-5*a*c^(1/2)*e*(A*e+2*B*d)+15*a^(1/2)*c*d*(2*A*e+B*d))*(a^(1/2 )+c^(1/2)*x^2)*((c*x^4+a)/(a^(1/2)+c^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2 *arctan(c^(1/4)*x/a^(1/4)),1/2*2^(1/2))/a^(1/4)/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 = 10.19 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.43 \[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^2}{\sqrt {a+c x^4}} \, dx=\frac {e x \left (10 B d+5 A e+3 B e x^2\right ) \left (a+c x^4\right )-5 \left (-3 A c d^2+2 a B d e+a A e^2\right ) x \sqrt {1+\frac {c x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^4}{a}\right )+\left (5 B c d^2+10 A c d e-3 a B e^2\right ) x^3 \sqrt {1+\frac {c x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {c x^4}{a}\right )}{15 c \sqrt {a+c x^4}} \] Input:
Integrate[((A + B*x^2)*(d + e*x^2)^2)/Sqrt[a + c*x^4],x]
Output:
(e*x*(10*B*d + 5*A*e + 3*B*e*x^2)*(a + c*x^4) - 5*(-3*A*c*d^2 + 2*a*B*d*e + a*A*e^2)*x*Sqrt[1 + (c*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^4 )/a)] + (5*B*c*d^2 + 10*A*c*d*e - 3*a*B*e^2)*x^3*Sqrt[1 + (c*x^4)/a]*Hyper geometric2F1[1/2, 3/4, 7/4, -((c*x^4)/a)])/(15*c*Sqrt[a + c*x^4])
Time = 0.76 (sec) , antiderivative size = 706, normalized size of antiderivative = 1.92, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2259, 2009}
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.
| \(\displaystyle \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^2}{\sqrt {a+c x^4}} \, dx\) |
| \(\Big \downarrow \) 2259 |
| \(\displaystyle \int \left (\frac {e x^4 (A e+2 B d)}{\sqrt {a+c x^4}}+\frac {d x^2 (2 A e+B d)}{\sqrt {a+c x^4}}+\frac {A d^2}{\sqrt {a+c x^4}}+\frac {B e^2 x^6}{\sqrt {a+c x^4}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^{3/4} e \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} (A e+2 B d) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{6 c^{5/4} \sqrt {a+c x^4}}-\frac {3 a^{5/4} B e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{10 c^{7/4} \sqrt {a+c x^4}}+\frac {3 a^{5/4} B e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{7/4} \sqrt {a+c x^4}}+\frac {\sqrt [4]{a} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} (2 A e+B d) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {a+c x^4}}-\frac {\sqrt [4]{a} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} (2 A e+B d) E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{c^{3/4} \sqrt {a+c x^4}}+\frac {A d^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^4}}+\frac {e x \sqrt {a+c x^4} (A e+2 B d)}{3 c}+\frac {d x \sqrt {a+c x^4} (2 A e+B d)}{\sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {3 a B e^2 x \sqrt {a+c x^4}}{5 c^{3/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {B e^2 x^3 \sqrt {a+c x^4}}{5 c}\) |
Input:
Int[((A + B*x^2)*(d + e*x^2)^2)/Sqrt[a + c*x^4],x]
Output:
(e*(2*B*d + A*e)*x*Sqrt[a + c*x^4])/(3*c) + (B*e^2*x^3*Sqrt[a + c*x^4])/(5 *c) - (3*a*B*e^2*x*Sqrt[a + c*x^4])/(5*c^(3/2)*(Sqrt[a] + Sqrt[c]*x^2)) + (d*(B*d + 2*A*e)*x*Sqrt[a + c*x^4])/(Sqrt[c]*(Sqrt[a] + Sqrt[c]*x^2)) + (3 *a^(5/4)*B*e^2*(Sqrt[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])/(5*c^(7/4)*Sqrt[a + c*x^4]) - (a^(1/4)*d*(B*d + 2*A*e)*(Sqrt[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^(3/4)*Sqrt[a + c*x^4]) + (A*d^2*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c *x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(1/4)*c^(1/4)*Sqrt[a + c*x^4]) - (3*a^(5/4)*B*e^2*(Sqrt[a] + Sq rt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[ (c^(1/4)*x)/a^(1/4)], 1/2])/(10*c^(7/4)*Sqrt[a + c*x^4]) - (a^(3/4)*e*(2*B *d + A*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2) ^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(6*c^(5/4)*Sqrt[a + c*x ^4]) + (a^(1/4)*d*(B*d + 2*A*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/( Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/( 2*c^(3/4)*Sqrt[a + c*x^4])
Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[ExpandIntegrand[1/Sqrt[a + c*x^4], Px*(d + e*x^2)^q*(a + c*x^4)^(p + 1/2), x], x] /; FreeQ[{a, c, d, e}, x] && PolyQ[Px, x] && IntegerQ[p + 1/ 2] && IntegerQ[q]
Result contains complex when optimal does not.
Time = 2.12 (sec) , antiderivative size = 258, normalized size of antiderivative = 0.70
| method | result | size |
| elliptic | \(\frac {B \,e^{2} x^{3} \sqrt {c \,x^{4}+a}}{5 c}+\frac {\left (A \,e^{2}+2 B d e \right ) x \sqrt {c \,x^{4}+a}}{3 c}+\frac {\left (A \,d^{2}-\frac {\left (A \,e^{2}+2 B d e \right ) a}{3 c}\right ) \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {i \left (2 d e A +B \,d^{2}-\frac {3 B \,e^{2} a}{5 c}\right ) \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}\) | \(258\) |
| risch | \(\frac {e x \left (3 B e \,x^{2}+5 A e +10 B d \right ) \sqrt {c \,x^{4}+a}}{15 c}-\frac {-\frac {i \left (30 A c d e -9 B a \,e^{2}+15 B c \,d^{2}\right ) \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}+\frac {5 A a \,e^{2} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}-\frac {15 A c \,d^{2} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {10 a B d e \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}}{15 c}\) | \(381\) |
| default | \(\frac {A \,d^{2} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+e \left (A e +2 B d \right ) \left (\frac {x \sqrt {c \,x^{4}+a}}{3 c}-\frac {a \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{3 c \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\right )+\frac {i d \left (2 A e +B d \right ) \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}+B \,e^{2} \left (\frac {x^{3} \sqrt {c \,x^{4}+a}}{5 c}-\frac {3 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 c^{\frac {3}{2}} \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\right )\) | \(399\) |
Input:
int((B*x^2+A)*(e*x^2+d)^2/(c*x^4+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/5*B*e^2*x^3*(c*x^4+a)^(1/2)/c+1/3*(A*e^2+2*B*d*e)/c*x*(c*x^4+a)^(1/2)+(A *d^2-1/3*(A*e^2+2*B*d*e)/c*a)/(I*c^(1/2)/a^(1/2))^(1/2)*(1-I*c^(1/2)*x^2/a ^(1/2))^(1/2)*(1+I*c^(1/2)*x^2/a^(1/2))^(1/2)/(c*x^4+a)^(1/2)*EllipticF(x* (I*c^(1/2)/a^(1/2))^(1/2),I)+I*(2*d*e*A+B*d^2-3/5*B*e^2/c*a)*a^(1/2)/(I*c^ (1/2)/a^(1/2))^(1/2)*(1-I*c^(1/2)*x^2/a^(1/2))^(1/2)*(1+I*c^(1/2)*x^2/a^(1 /2))^(1/2)/(c*x^4+a)^(1/2)/c^(1/2)*(EllipticF(x*(I*c^(1/2)/a^(1/2))^(1/2), I)-EllipticE(x*(I*c^(1/2)/a^(1/2))^(1/2),I))
Time = 0.08 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.56 \[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^2}{\sqrt {a+c x^4}} \, dx=\frac {3 \, {\left (5 \, B a c d^{2} + 10 \, A a c d e - 3 \, B a^{2} e^{2}\right )} \sqrt {c} x \left (-\frac {a}{c}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - {\left (10 \, {\left (3 \, A + B\right )} a c d e + 15 \, {\left (B a c - A c^{2}\right )} d^{2} - {\left (9 \, B a^{2} - 5 \, A a c\right )} e^{2}\right )} \sqrt {c} x \left (-\frac {a}{c}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (3 \, B a c e^{2} x^{4} + 15 \, B a c d^{2} + 30 \, A a c d e - 9 \, B a^{2} e^{2} + 5 \, {\left (2 \, B a c d e + A a c e^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + a}}{15 \, a c^{2} x} \] Input:
integrate((B*x^2+A)*(e*x^2+d)^2/(c*x^4+a)^(1/2),x, algorithm="fricas")
Output:
1/15*(3*(5*B*a*c*d^2 + 10*A*a*c*d*e - 3*B*a^2*e^2)*sqrt(c)*x*(-a/c)^(3/4)* elliptic_e(arcsin((-a/c)^(1/4)/x), -1) - (10*(3*A + B)*a*c*d*e + 15*(B*a*c - A*c^2)*d^2 - (9*B*a^2 - 5*A*a*c)*e^2)*sqrt(c)*x*(-a/c)^(3/4)*elliptic_f (arcsin((-a/c)^(1/4)/x), -1) + (3*B*a*c*e^2*x^4 + 15*B*a*c*d^2 + 30*A*a*c* d*e - 9*B*a^2*e^2 + 5*(2*B*a*c*d*e + A*a*c*e^2)*x^2)*sqrt(c*x^4 + a))/(a*c ^2*x)
Result contains complex when optimal does not.
Time = 3.88 (sec) , antiderivative size = 262, normalized size of antiderivative = 0.71 \[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^2}{\sqrt {a+c x^4}} \, dx=\frac {A d^{2} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} + \frac {A d e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} + \frac {A e^{2} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {9}{4}\right )} + \frac {B d^{2} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} + \frac {B d e x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {9}{4}\right )} + \frac {B e^{2} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {11}{4}\right )} \] Input:
integrate((B*x**2+A)*(e*x**2+d)**2/(c*x**4+a)**(1/2),x)
Output:
A*d**2*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), c*x**4*exp_polar(I*pi)/a)/(4 *sqrt(a)*gamma(5/4)) + A*d*e*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), c*x **4*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(7/4)) + A*e**2*x**5*gamma(5/4)*hyp er((1/2, 5/4), (9/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(9/4)) + B*d**2*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)) + B*d*e*x**5*gamma(5/4)*hyper((1/2, 5/4), (9/4,), c*x**4*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(9/4)) + B*e**2*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 ))
\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^2}{\sqrt {a+c x^4}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{2}}{\sqrt {c x^{4} + a}} \,d x } \] Input:
integrate((B*x^2+A)*(e*x^2+d)^2/(c*x^4+a)^(1/2),x, algorithm="maxima")
Output:
integrate((B*x^2 + A)*(e*x^2 + d)^2/sqrt(c*x^4 + a), x)
\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^2}{\sqrt {a+c x^4}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{2}}{\sqrt {c x^{4} + a}} \,d x } \] Input:
integrate((B*x^2+A)*(e*x^2+d)^2/(c*x^4+a)^(1/2),x, algorithm="giac")
Output:
integrate((B*x^2 + A)*(e*x^2 + d)^2/sqrt(c*x^4 + a), x)
Timed out. \[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^2}{\sqrt {a+c x^4}} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (e\,x^2+d\right )}^2}{\sqrt {c\,x^4+a}} \,d x \] Input:
int(((A + B*x^2)*(d + e*x^2)^2)/(a + c*x^4)^(1/2),x)
Output:
int(((A + B*x^2)*(d + e*x^2)^2)/(a + c*x^4)^(1/2), x)
\[ \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^2}{\sqrt {a+c x^4}} \, dx=\frac {5 \sqrt {c \,x^{4}+a}\, a \,e^{2} x +10 \sqrt {c \,x^{4}+a}\, b d e x +3 \sqrt {c \,x^{4}+a}\, b \,e^{2} x^{3}-5 \left (\int \frac {\sqrt {c \,x^{4}+a}}{c \,x^{4}+a}d x \right ) a^{2} e^{2}-10 \left (\int \frac {\sqrt {c \,x^{4}+a}}{c \,x^{4}+a}d x \right ) a b d e +15 \left (\int \frac {\sqrt {c \,x^{4}+a}}{c \,x^{4}+a}d x \right ) a c \,d^{2}-9 \left (\int \frac {\sqrt {c \,x^{4}+a}\, x^{2}}{c \,x^{4}+a}d x \right ) a b \,e^{2}+30 \left (\int \frac {\sqrt {c \,x^{4}+a}\, x^{2}}{c \,x^{4}+a}d x \right ) a c d e +15 \left (\int \frac {\sqrt {c \,x^{4}+a}\, x^{2}}{c \,x^{4}+a}d x \right ) b c \,d^{2}}{15 c} \] Input:
int((B*x^2+A)*(e*x^2+d)^2/(c*x^4+a)^(1/2),x)
Output:
(5*sqrt(a + c*x**4)*a*e**2*x + 10*sqrt(a + c*x**4)*b*d*e*x + 3*sqrt(a + c* x**4)*b*e**2*x**3 - 5*int(sqrt(a + c*x**4)/(a + c*x**4),x)*a**2*e**2 - 10* int(sqrt(a + c*x**4)/(a + c*x**4),x)*a*b*d*e + 15*int(sqrt(a + c*x**4)/(a + c*x**4),x)*a*c*d**2 - 9*int((sqrt(a + c*x**4)*x**2)/(a + c*x**4),x)*a*b* e**2 + 30*int((sqrt(a + c*x**4)*x**2)/(a + c*x**4),x)*a*c*d*e + 15*int((sq rt(a + c*x**4)*x**2)/(a + c*x**4),x)*b*c*d**2)/(15*c)