Integrand size = 24, antiderivative size = 190 \[ \int \frac {d+e x^2}{\left (d^2-e^2 x^4\right )^{5/2}} \, dx=\frac {x \left (d+e x^2\right )}{6 d^2 \left (d^2-e^2 x^4\right )^{3/2}}+\frac {x \left (5 d+3 e x^2\right )}{12 d^4 \sqrt {d^2-e^2 x^4}}-\frac {\sqrt {1-\frac {e^2 x^4}{d^2}} E\left (\left .\arcsin \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )\right |-1\right )}{4 d^{5/2} \sqrt {e} \sqrt {d^2-e^2 x^4}}+\frac {2 \sqrt {1-\frac {e^2 x^4}{d^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ),-1\right )}{3 d^{5/2} \sqrt {e} \sqrt {d^2-e^2 x^4}} \] Output:
1/6*x*(e*x^2+d)/d^2/(-e^2*x^4+d^2)^(3/2)+1/12*x*(3*e*x^2+5*d)/d^4/(-e^2*x^ 4+d^2)^(1/2)-1/4*(1-e^2*x^4/d^2)^(1/2)*EllipticE(e^(1/2)*x/d^(1/2),I)/d^(5 /2)/e^(1/2)/(-e^2*x^4+d^2)^(1/2)+2/3*(1-e^2*x^4/d^2)^(1/2)*EllipticF(e^(1/ 2)*x/d^(1/2),I)/d^(5/2)/e^(1/2)/(-e^2*x^4+d^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.09 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.77 \[ \int \frac {d+e x^2}{\left (d^2-e^2 x^4\right )^{5/2}} \, dx=\frac {7 d^3 x-5 d e^2 x^5+5 d x \left (d^2-e^2 x^4\right ) \sqrt {1-\frac {e^2 x^4}{d^2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {e^2 x^4}{d^2}\right )+4 e x^3 \left (d^2-e^2 x^4\right ) \sqrt {1-\frac {e^2 x^4}{d^2}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{4},\frac {e^2 x^4}{d^2}\right )}{12 d^4 \left (d^2-e^2 x^4\right )^{3/2}} \] Input:
Integrate[(d + e*x^2)/(d^2 - e^2*x^4)^(5/2),x]
Output:
(7*d^3*x - 5*d*e^2*x^5 + 5*d*x*(d^2 - e^2*x^4)*Sqrt[1 - (e^2*x^4)/d^2]*Hyp ergeometric2F1[1/4, 1/2, 5/4, (e^2*x^4)/d^2] + 4*e*x^3*(d^2 - e^2*x^4)*Sqr t[1 - (e^2*x^4)/d^2]*Hypergeometric2F1[3/4, 5/2, 7/4, (e^2*x^4)/d^2])/(12* d^4*(d^2 - e^2*x^4)^(3/2))
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.
| \(\displaystyle \int \frac {d+e x^2}{\left (d^2-e^2 x^4\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1571 |
| \(\displaystyle \int \frac {d+e x^2}{\left (d^2-e^2 x^4\right )^{5/2}}dx\) |
Input:
Int[(d + e*x^2)/(d^2 - e^2*x^4)^(5/2),x]
Output:
$Aborted
Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> U nintegrable[(d + e*x^2)^q*(a + c*x^4)^p, x] /; FreeQ[{a, c, d, e, p, q}, x]
Time = 2.25 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.43
| method | result | size |
| elliptic | \(\frac {\left (-e^{2} x^{2}+d e \right ) x}{8 e \,d^{4} \sqrt {\left (x^{2}+\frac {d}{e}\right ) \left (-e^{2} x^{2}+d e \right )}}+\frac {x \sqrt {-e^{2} x^{4}+d^{2}}}{12 e^{2} d^{3} \left (x^{2}-\frac {d}{e}\right )^{2}}-\frac {3 \left (-e^{2} x^{2}-d e \right ) x}{8 d^{4} e \sqrt {\left (x^{2}-\frac {d}{e}\right ) \left (-e^{2} x^{2}-d e \right )}}+\frac {5 \sqrt {1-\frac {e \,x^{2}}{d}}\, \sqrt {1+\frac {e \,x^{2}}{d}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {e}{d}}, i\right )}{12 d^{3} \sqrt {\frac {e}{d}}\, \sqrt {-e^{2} x^{4}+d^{2}}}+\frac {\sqrt {1-\frac {e \,x^{2}}{d}}\, \sqrt {1+\frac {e \,x^{2}}{d}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {e}{d}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {e}{d}}, i\right )\right )}{4 d^{3} \sqrt {\frac {e}{d}}\, \sqrt {-e^{2} x^{4}+d^{2}}}\) | \(271\) |
| default | \(d \left (\frac {x \sqrt {-e^{2} x^{4}+d^{2}}}{6 d^{2} e^{4} \left (x^{4}-\frac {d^{2}}{e^{2}}\right )^{2}}+\frac {5 x}{12 d^{4} \sqrt {-\left (x^{4}-\frac {d^{2}}{e^{2}}\right ) e^{2}}}+\frac {5 \sqrt {1-\frac {e \,x^{2}}{d}}\, \sqrt {1+\frac {e \,x^{2}}{d}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {e}{d}}, i\right )}{12 d^{4} \sqrt {\frac {e}{d}}\, \sqrt {-e^{2} x^{4}+d^{2}}}\right )+e \left (\frac {x^{3} \sqrt {-e^{2} x^{4}+d^{2}}}{6 d^{2} e^{4} \left (x^{4}-\frac {d^{2}}{e^{2}}\right )^{2}}+\frac {x^{3}}{4 d^{4} \sqrt {-\left (x^{4}-\frac {d^{2}}{e^{2}}\right ) e^{2}}}+\frac {\sqrt {1-\frac {e \,x^{2}}{d}}\, \sqrt {1+\frac {e \,x^{2}}{d}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {e}{d}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {e}{d}}, i\right )\right )}{4 d^{3} \sqrt {\frac {e}{d}}\, \sqrt {-e^{2} x^{4}+d^{2}}\, e}\right )\) | \(280\) |
Input:
int((e*x^2+d)/(-e^2*x^4+d^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/8*(-e^2*x^2+d*e)/e/d^4*x/((x^2+d/e)*(-e^2*x^2+d*e))^(1/2)+1/12/e^2/d^3*x *(-e^2*x^4+d^2)^(1/2)/(x^2-d/e)^2-3/8*(-e^2*x^2-d*e)/d^4*x/e/((x^2-d/e)*(- e^2*x^2-d*e))^(1/2)+5/12/d^3/(e/d)^(1/2)*(1-e*x^2/d)^(1/2)*(1+e*x^2/d)^(1/ 2)/(-e^2*x^4+d^2)^(1/2)*EllipticF(x*(e/d)^(1/2),I)+1/4/d^3/(e/d)^(1/2)*(1- e*x^2/d)^(1/2)*(1+e*x^2/d)^(1/2)/(-e^2*x^4+d^2)^(1/2)*(EllipticF(x*(e/d)^( 1/2),I)-EllipticE(x*(e/d)^(1/2),I))
Time = 0.08 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.19 \[ \int \frac {d+e x^2}{\left (d^2-e^2 x^4\right )^{5/2}} \, dx=-\frac {3 \, {\left (e^{4} x^{6} - d e^{3} x^{4} - d^{2} e^{2} x^{2} + d^{3} e\right )} \sqrt {\frac {e}{d}} E(\arcsin \left (x \sqrt {\frac {e}{d}}\right )\,|\,-1) - {\left ({\left (5 \, d e^{3} + 3 \, e^{4}\right )} x^{6} - {\left (5 \, d^{2} e^{2} + 3 \, d e^{3}\right )} x^{4} + 5 \, d^{4} + 3 \, d^{3} e - {\left (5 \, d^{3} e + 3 \, d^{2} e^{2}\right )} x^{2}\right )} \sqrt {\frac {e}{d}} F(\arcsin \left (x \sqrt {\frac {e}{d}}\right )\,|\,-1) + {\left (3 \, e^{3} x^{5} + 2 \, d e^{2} x^{3} - 7 \, d^{2} e x\right )} \sqrt {-e^{2} x^{4} + d^{2}}}{12 \, {\left (d^{4} e^{4} x^{6} - d^{5} e^{3} x^{4} - d^{6} e^{2} x^{2} + d^{7} e\right )}} \] Input:
integrate((e*x^2+d)/(-e^2*x^4+d^2)^(5/2),x, algorithm="fricas")
Output:
-1/12*(3*(e^4*x^6 - d*e^3*x^4 - d^2*e^2*x^2 + d^3*e)*sqrt(e/d)*elliptic_e( arcsin(x*sqrt(e/d)), -1) - ((5*d*e^3 + 3*e^4)*x^6 - (5*d^2*e^2 + 3*d*e^3)* x^4 + 5*d^4 + 3*d^3*e - (5*d^3*e + 3*d^2*e^2)*x^2)*sqrt(e/d)*elliptic_f(ar csin(x*sqrt(e/d)), -1) + (3*e^3*x^5 + 2*d*e^2*x^3 - 7*d^2*e*x)*sqrt(-e^2*x ^4 + d^2))/(d^4*e^4*x^6 - d^5*e^3*x^4 - d^6*e^2*x^2 + d^7*e)
Time = 6.45 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.44 \[ \int \frac {d+e x^2}{\left (d^2-e^2 x^4\right )^{5/2}} \, dx=\frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {e^{2} x^{4} e^{2 i \pi }}{d^{2}}} \right )}}{4 d^{4} \Gamma \left (\frac {5}{4}\right )} + \frac {e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {e^{2} x^{4} e^{2 i \pi }}{d^{2}}} \right )}}{4 d^{5} \Gamma \left (\frac {7}{4}\right )} \] Input:
integrate((e*x**2+d)/(-e**2*x**4+d**2)**(5/2),x)
Output:
x*gamma(1/4)*hyper((1/4, 5/2), (5/4,), e**2*x**4*exp_polar(2*I*pi)/d**2)/( 4*d**4*gamma(5/4)) + e*x**3*gamma(3/4)*hyper((3/4, 5/2), (7/4,), e**2*x**4 *exp_polar(2*I*pi)/d**2)/(4*d**5*gamma(7/4))
\[ \int \frac {d+e x^2}{\left (d^2-e^2 x^4\right )^{5/2}} \, dx=\int { \frac {e x^{2} + d}{{\left (-e^{2} x^{4} + d^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((e*x^2+d)/(-e^2*x^4+d^2)^(5/2),x, algorithm="maxima")
Output:
integrate((e*x^2 + d)/(-e^2*x^4 + d^2)^(5/2), x)
\[ \int \frac {d+e x^2}{\left (d^2-e^2 x^4\right )^{5/2}} \, dx=\int { \frac {e x^{2} + d}{{\left (-e^{2} x^{4} + d^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((e*x^2+d)/(-e^2*x^4+d^2)^(5/2),x, algorithm="giac")
Output:
integrate((e*x^2 + d)/(-e^2*x^4 + d^2)^(5/2), x)
Timed out. \[ \int \frac {d+e x^2}{\left (d^2-e^2 x^4\right )^{5/2}} \, dx=\int \frac {e\,x^2+d}{{\left (d^2-e^2\,x^4\right )}^{5/2}} \,d x \] Input:
int((d + e*x^2)/(d^2 - e^2*x^4)^(5/2),x)
Output:
int((d + e*x^2)/(d^2 - e^2*x^4)^(5/2), x)
\[ \int \frac {d+e x^2}{\left (d^2-e^2 x^4\right )^{5/2}} \, dx=\int \frac {\sqrt {-e^{2} x^{4}+d^{2}}}{-e^{5} x^{10}+d \,e^{4} x^{8}+2 d^{2} e^{3} x^{6}-2 d^{3} e^{2} x^{4}-d^{4} e \,x^{2}+d^{5}}d x \] Input:
int((e*x^2+d)/(-e^2*x^4+d^2)^(5/2),x)
Output:
int(sqrt(d**2 - e**2*x**4)/(d**5 - d**4*e*x**2 - 2*d**3*e**2*x**4 + 2*d**2 *e**3*x**6 + d*e**4*x**8 - e**5*x**10),x)