Integrand size = 19, antiderivative size = 39 \[ \int \frac {d+e x^2}{\sqrt {9-x^4}} \, dx=\sqrt {3} e E\left (\left .\arcsin \left (\frac {x}{\sqrt {3}}\right )\right |-1\right )+\frac {(d-3 e) \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{\sqrt {3}}\right ),-1\right )}{\sqrt {3}} \] Output:
3^(1/2)*e*EllipticE(1/3*x*3^(1/2),I)+1/3*(d-3*e)*EllipticF(1/3*x*3^(1/2),I )*3^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.26 \[ \int \frac {d+e x^2}{\sqrt {9-x^4}} \, dx=\frac {1}{3} d x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {x^4}{9}\right )+\frac {1}{9} e x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},\frac {x^4}{9}\right ) \] Input:
Integrate[(d + e*x^2)/Sqrt[9 - x^4],x]
Output:
(d*x*Hypergeometric2F1[1/4, 1/2, 5/4, x^4/9])/3 + (e*x^3*Hypergeometric2F1 [1/2, 3/4, 7/4, x^4/9])/9
Time = 0.31 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1495, 399, 284, 327, 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.
\(\displaystyle \int \frac {d+e x^2}{\sqrt {9-x^4}} \, dx\) |
\(\Big \downarrow \) 1495 |
\(\displaystyle \int \frac {d+e x^2}{\sqrt {3-x^2} \sqrt {x^2+3}}dx\) |
\(\Big \downarrow \) 399 |
\(\displaystyle (d-3 e) \int \frac {1}{\sqrt {3-x^2} \sqrt {x^2+3}}dx+e \int \frac {\sqrt {x^2+3}}{\sqrt {3-x^2}}dx\) |
\(\Big \downarrow \) 284 |
\(\displaystyle (d-3 e) \int \frac {1}{\sqrt {9-x^4}}dx+e \int \frac {\sqrt {x^2+3}}{\sqrt {3-x^2}}dx\) |
\(\Big \downarrow \) 327 |
\(\displaystyle (d-3 e) \int \frac {1}{\sqrt {9-x^4}}dx+\sqrt {3} e E\left (\left .\arcsin \left (\frac {x}{\sqrt {3}}\right )\right |-1\right )\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {(d-3 e) \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{\sqrt {3}}\right ),-1\right )}{\sqrt {3}}+\sqrt {3} e E\left (\left .\arcsin \left (\frac {x}{\sqrt {3}}\right )\right |-1\right )\) |
Input:
Int[(d + e*x^2)/Sqrt[9 - x^4],x]
Output:
Sqrt[3]*e*EllipticE[ArcSin[x/Sqrt[3]], -1] + ((d - 3*e)*EllipticF[ArcSin[x /Sqrt[3]], -1])/Sqrt[3]
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> I nt[(a*c + b*d*x^4)^p, x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0]))
Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ (Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[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]
Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_) ^2]), x_Symbol] :> Simp[f/b Int[Sqrt[a + b*x^2]/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/b Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; Fr eeQ[{a, b, c, d, e, f}, x] && !((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-b/a, -d/c])))))
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[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Simp[Sqrt[-c] Int[(d + e*x^2)/(Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c, 0]
Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 0.76 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.87
method | result | size |
meijerg | \(\frac {e \,x^{3} \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], \frac {x^{4}}{9}\right )}{9}+\frac {d x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {5}{4}\right ], \frac {x^{4}}{9}\right )}{3}\) | \(34\) |
default | \(\frac {d \sqrt {3}\, \sqrt {-3 x^{2}+9}\, \sqrt {3 x^{2}+9}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, x}{3}, i\right )}{9 \sqrt {-x^{4}+9}}-\frac {e \sqrt {3}\, \sqrt {-3 x^{2}+9}\, \sqrt {3 x^{2}+9}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {3}\, x}{3}, i\right )-\operatorname {EllipticE}\left (\frac {\sqrt {3}\, x}{3}, i\right )\right )}{3 \sqrt {-x^{4}+9}}\) | \(98\) |
elliptic | \(\frac {d \sqrt {3}\, \sqrt {-3 x^{2}+9}\, \sqrt {3 x^{2}+9}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, x}{3}, i\right )}{9 \sqrt {-x^{4}+9}}-\frac {e \sqrt {3}\, \sqrt {-3 x^{2}+9}\, \sqrt {3 x^{2}+9}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {3}\, x}{3}, i\right )-\operatorname {EllipticE}\left (\frac {\sqrt {3}\, x}{3}, i\right )\right )}{3 \sqrt {-x^{4}+9}}\) | \(98\) |
Input:
int((e*x^2+d)/(-x^4+9)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/9*e*x^3*hypergeom([1/2,3/4],[7/4],1/9*x^4)+1/3*d*x*hypergeom([1/4,1/2],[ 5/4],1/9*x^4)
Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.44 \[ \int \frac {d+e x^2}{\sqrt {9-x^4}} \, dx=\frac {-9 i \, \sqrt {3} e x E(\arcsin \left (\frac {\sqrt {3}}{x}\right )\,|\,-1) + i \, \sqrt {3} {\left (d + 9 \, e\right )} x F(\arcsin \left (\frac {\sqrt {3}}{x}\right )\,|\,-1) - 3 \, \sqrt {-x^{4} + 9} e}{3 \, x} \] Input:
integrate((e*x^2+d)/(-x^4+9)^(1/2),x, algorithm="fricas")
Output:
1/3*(-9*I*sqrt(3)*e*x*elliptic_e(arcsin(sqrt(3)/x), -1) + I*sqrt(3)*(d + 9 *e)*x*elliptic_f(arcsin(sqrt(3)/x), -1) - 3*sqrt(-x^4 + 9)*e)/x
Time = 0.72 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.74 \[ \int \frac {d+e x^2}{\sqrt {9-x^4}} \, dx=\frac {d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {x^{4} e^{2 i \pi }}{9}} \right )}}{12 \Gamma \left (\frac {5}{4}\right )} + \frac {e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {x^{4} e^{2 i \pi }}{9}} \right )}}{12 \Gamma \left (\frac {7}{4}\right )} \] Input:
integrate((e*x**2+d)/(-x**4+9)**(1/2),x)
Output:
d*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), x**4*exp_polar(2*I*pi)/9)/(12*gam ma(5/4)) + e*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), x**4*exp_polar(2*I* pi)/9)/(12*gamma(7/4))
\[ \int \frac {d+e x^2}{\sqrt {9-x^4}} \, dx=\int { \frac {e x^{2} + d}{\sqrt {-x^{4} + 9}} \,d x } \] Input:
integrate((e*x^2+d)/(-x^4+9)^(1/2),x, algorithm="maxima")
Output:
integrate((e*x^2 + d)/sqrt(-x^4 + 9), x)
\[ \int \frac {d+e x^2}{\sqrt {9-x^4}} \, dx=\int { \frac {e x^{2} + d}{\sqrt {-x^{4} + 9}} \,d x } \] Input:
integrate((e*x^2+d)/(-x^4+9)^(1/2),x, algorithm="giac")
Output:
integrate((e*x^2 + d)/sqrt(-x^4 + 9), x)
Timed out. \[ \int \frac {d+e x^2}{\sqrt {9-x^4}} \, dx=\int \frac {e\,x^2+d}{\sqrt {9-x^4}} \,d x \] Input:
int((d + e*x^2)/(9 - x^4)^(1/2),x)
Output:
int((d + e*x^2)/(9 - x^4)^(1/2), x)
\[ \int \frac {d+e x^2}{\sqrt {9-x^4}} \, dx=-\left (\int \frac {\sqrt {-x^{4}+9}}{x^{4}-9}d x \right ) d -\left (\int \frac {\sqrt {-x^{4}+9}\, x^{2}}{x^{4}-9}d x \right ) e \] Input:
int((e*x^2+d)/(-x^4+9)^(1/2),x)
Output:
- (int(sqrt( - x**4 + 9)/(x**4 - 9),x)*d + int((sqrt( - x**4 + 9)*x**2)/( x**4 - 9),x)*e)