Integrand size = 16, antiderivative size = 111 \[ \int \frac {1}{\left (d^2-e^2 x^4\right )^{5/2}} \, dx=\frac {x}{6 d^2 \left (d^2-e^2 x^4\right )^{3/2}}+\frac {5 x}{12 d^4 \sqrt {d^2-e^2 x^4}}+\frac {5 \sqrt {1-\frac {e^2 x^4}{d^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ),-1\right )}{12 d^{7/2} \sqrt {e} \sqrt {d^2-e^2 x^4}} \] Output:
1/6*x/d^2/(-e^2*x^4+d^2)^(3/2)+5/12*x/d^4/(-e^2*x^4+d^2)^(1/2)+5/12*(1-e^2 *x^4/d^2)^(1/2)*EllipticF(e^(1/2)*x/d^(1/2),I)/d^(7/2)/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 = 10.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\left (d^2-e^2 x^4\right )^{5/2}} \, dx=\frac {7 d^2 x-5 e^2 x^5+5 x \left (d^2-e^2 x^4\right ) \sqrt {1-\frac {e^2 x^4}{d^2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {e^2 x^4}{d^2}\right )}{12 d^4 \left (d^2-e^2 x^4\right )^{3/2}} \] Input:
Integrate[(d^2 - e^2*x^4)^(-5/2),x]
Output:
(7*d^2*x - 5*e^2*x^5 + 5*x*(d^2 - e^2*x^4)*Sqrt[1 - (e^2*x^4)/d^2]*Hyperge ometric2F1[1/4, 1/2, 5/4, (e^2*x^4)/d^2])/(12*d^4*(d^2 - e^2*x^4)^(3/2))
Time = 0.37 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {749, 749, 765, 762}
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.
\(\displaystyle \int \frac {1}{\left (d^2-e^2 x^4\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 749 |
\(\displaystyle \frac {5 \int \frac {1}{\left (d^2-e^2 x^4\right )^{3/2}}dx}{6 d^2}+\frac {x}{6 d^2 \left (d^2-e^2 x^4\right )^{3/2}}\) |
\(\Big \downarrow \) 749 |
\(\displaystyle \frac {5 \left (\frac {\int \frac {1}{\sqrt {d^2-e^2 x^4}}dx}{2 d^2}+\frac {x}{2 d^2 \sqrt {d^2-e^2 x^4}}\right )}{6 d^2}+\frac {x}{6 d^2 \left (d^2-e^2 x^4\right )^{3/2}}\) |
\(\Big \downarrow \) 765 |
\(\displaystyle \frac {5 \left (\frac {\sqrt {1-\frac {e^2 x^4}{d^2}} \int \frac {1}{\sqrt {1-\frac {e^2 x^4}{d^2}}}dx}{2 d^2 \sqrt {d^2-e^2 x^4}}+\frac {x}{2 d^2 \sqrt {d^2-e^2 x^4}}\right )}{6 d^2}+\frac {x}{6 d^2 \left (d^2-e^2 x^4\right )^{3/2}}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {5 \left (\frac {\sqrt {1-\frac {e^2 x^4}{d^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ),-1\right )}{2 d^{3/2} \sqrt {e} \sqrt {d^2-e^2 x^4}}+\frac {x}{2 d^2 \sqrt {d^2-e^2 x^4}}\right )}{6 d^2}+\frac {x}{6 d^2 \left (d^2-e^2 x^4\right )^{3/2}}\) |
Input:
Int[(d^2 - e^2*x^4)^(-5/2),x]
Output:
x/(6*d^2*(d^2 - e^2*x^4)^(3/2)) + (5*(x/(2*d^2*Sqrt[d^2 - e^2*x^4]) + (Sqr t[1 - (e^2*x^4)/d^2]*EllipticF[ArcSin[(Sqrt[e]*x)/Sqrt[d]], -1])/(2*d^(3/2 )*Sqrt[e]*Sqrt[d^2 - e^2*x^4])))/(6*d^2)
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Simp[(n*(p + 1) + 1)/(a*n*(p + 1)) Int[(a + b*x^ n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (Inte gerQ[2*p] || Denominator[p + 1/n] < Denominator[p])
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) )*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[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]
Time = 0.76 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.14
method | result | size |
default | \(\frac {x \sqrt {-e^{2} x^{4}+d^{2}}}{6 d^{2} e^{4} \left (x^{4}-\frac {d^{2}}{e^{2}}\right )^{2}}+\frac {5 x}{12 d^{4} \sqrt {-\left (x^{4}-\frac {d^{2}}{e^{2}}\right ) e^{2}}}+\frac {5 \sqrt {1-\frac {e \,x^{2}}{d}}\, \sqrt {1+\frac {e \,x^{2}}{d}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {e}{d}}, i\right )}{12 d^{4} \sqrt {\frac {e}{d}}\, \sqrt {-e^{2} x^{4}+d^{2}}}\) | \(127\) |
elliptic | \(\frac {x \sqrt {-e^{2} x^{4}+d^{2}}}{6 d^{2} e^{4} \left (x^{4}-\frac {d^{2}}{e^{2}}\right )^{2}}+\frac {5 x}{12 d^{4} \sqrt {-\left (x^{4}-\frac {d^{2}}{e^{2}}\right ) e^{2}}}+\frac {5 \sqrt {1-\frac {e \,x^{2}}{d}}\, \sqrt {1+\frac {e \,x^{2}}{d}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {e}{d}}, i\right )}{12 d^{4} \sqrt {\frac {e}{d}}\, \sqrt {-e^{2} x^{4}+d^{2}}}\) | \(127\) |
Input:
int(1/(-e^2*x^4+d^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/6/d^2*x/e^4*(-e^2*x^4+d^2)^(1/2)/(x^4-d^2/e^2)^2+5/12/d^4*x/(-(x^4-d^2/e ^2)*e^2)^(1/2)+5/12/d^4/(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)
Time = 0.07 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (d^2-e^2 x^4\right )^{5/2}} \, dx=\frac {5 \, {\left (e^{4} x^{8} - 2 \, d^{2} e^{2} x^{4} + d^{4}\right )} \sqrt {\frac {e}{d}} F(\arcsin \left (x \sqrt {\frac {e}{d}}\right )\,|\,-1) - {\left (5 \, e^{3} x^{5} - 7 \, d^{2} e x\right )} \sqrt {-e^{2} x^{4} + d^{2}}}{12 \, {\left (d^{4} e^{5} x^{8} - 2 \, d^{6} e^{3} x^{4} + d^{8} e\right )}} \] Input:
integrate(1/(-e^2*x^4+d^2)^(5/2),x, algorithm="fricas")
Output:
1/12*(5*(e^4*x^8 - 2*d^2*e^2*x^4 + d^4)*sqrt(e/d)*elliptic_f(arcsin(x*sqrt (e/d)), -1) - (5*e^3*x^5 - 7*d^2*e*x)*sqrt(-e^2*x^4 + d^2))/(d^4*e^5*x^8 - 2*d^6*e^3*x^4 + d^8*e)
Time = 0.53 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.35 \[ \int \frac {1}{\left (d^2-e^2 x^4\right )^{5/2}} \, dx=\frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {e^{2} x^{4} e^{2 i \pi }}{d^{2}}} \right )}}{4 d^{5} \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate(1/(-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**5*gamma(5/4))
\[ \int \frac {1}{\left (d^2-e^2 x^4\right )^{5/2}} \, dx=\int { \frac {1}{{\left (-e^{2} x^{4} + d^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/(-e^2*x^4+d^2)^(5/2),x, algorithm="maxima")
Output:
integrate((-e^2*x^4 + d^2)^(-5/2), x)
\[ \int \frac {1}{\left (d^2-e^2 x^4\right )^{5/2}} \, dx=\int { \frac {1}{{\left (-e^{2} x^{4} + d^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/(-e^2*x^4+d^2)^(5/2),x, algorithm="giac")
Output:
integrate((-e^2*x^4 + d^2)^(-5/2), x)
Time = 17.93 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.41 \[ \int \frac {1}{\left (d^2-e^2 x^4\right )^{5/2}} \, dx=\frac {x\,{\left (1-\frac {e^2\,x^4}{d^2}\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {5}{2};\ \frac {5}{4};\ \frac {e^2\,x^4}{d^2}\right )}{{\left (d^2-e^2\,x^4\right )}^{5/2}} \] Input:
int(1/(d^2 - e^2*x^4)^(5/2),x)
Output:
(x*(1 - (e^2*x^4)/d^2)^(5/2)*hypergeom([1/4, 5/2], 5/4, (e^2*x^4)/d^2))/(d ^2 - e^2*x^4)^(5/2)
\[ \int \frac {1}{\left (d^2-e^2 x^4\right )^{5/2}} \, dx=\int \frac {\sqrt {-e^{2} x^{4}+d^{2}}}{-e^{6} x^{12}+3 d^{2} e^{4} x^{8}-3 d^{4} e^{2} x^{4}+d^{6}}d x \] Input:
int(1/(-e^2*x^4+d^2)^(5/2),x)
Output:
int(sqrt(d**2 - e**2*x**4)/(d**6 - 3*d**4*e**2*x**4 + 3*d**2*e**4*x**8 - e **6*x**12),x)