\(\int \frac {d+e x^2}{(9-x^4)^{5/2}} \, dx\) [225]

Optimal result

Integrand size = 19, antiderivative size = 96 \[ \int \frac {d+e x^2}{\left (9-x^4\right )^{5/2}} \, dx=\frac {x \left (d+e x^2\right )}{54 \left (9-x^4\right )^{3/2}}+\frac {x \left (5 d+3 e x^2\right )}{972 \sqrt {9-x^4}}-\frac {e E\left (\left .\arcsin \left (\frac {x}{\sqrt {3}}\right )\right |-1\right )}{108 \sqrt {3}}+\frac {(5 d+9 e) \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{\sqrt {3}}\right ),-1\right )}{972 \sqrt {3}} \] Output:

1/54*x*(e*x^2+d)/(-x^4+9)^(3/2)+1/972*x*(3*e*x^2+5*d)/(-x^4+9)^(1/2)-1/324 
*3^(1/2)*e*EllipticE(1/3*x*3^(1/2),I)+1/2916*(5*d+9*e)*EllipticF(1/3*x*3^( 
1/2),I)*3^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.73 \[ \int \frac {d+e x^2}{\left (9-x^4\right )^{5/2}} \, dx=\frac {x \left (\frac {3 d \left (63-5 x^4\right )}{\left (9-x^4\right )^{3/2}}+5 d \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {x^4}{9}\right )+4 e x^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{4},\frac {x^4}{9}\right )\right )}{2916} \] Input:

Integrate[(d + e*x^2)/(9 - x^4)^(5/2),x]
 

Output:

(x*((3*d*(63 - 5*x^4))/(9 - x^4)^(3/2) + 5*d*Hypergeometric2F1[1/4, 1/2, 5 
/4, x^4/9] + 4*e*x^2*Hypergeometric2F1[3/4, 5/2, 7/4, x^4/9]))/2916
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {1493, 25, 1493, 25, 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.

\(\displaystyle \int \frac {d+e x^2}{\left (9-x^4\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1493

\(\displaystyle \frac {x \left (d+e x^2\right )}{54 \left (9-x^4\right )^{3/2}}-\frac {1}{54} \int -\frac {3 e x^2+5 d}{\left (9-x^4\right )^{3/2}}dx\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{54} \int \frac {3 e x^2+5 d}{\left (9-x^4\right )^{3/2}}dx+\frac {x \left (d+e x^2\right )}{54 \left (9-x^4\right )^{3/2}}\)

\(\Big \downarrow \) 1493

\(\displaystyle \frac {1}{54} \left (\frac {x \left (5 d+3 e x^2\right )}{18 \sqrt {9-x^4}}-\frac {1}{18} \int -\frac {5 d-3 e x^2}{\sqrt {9-x^4}}dx\right )+\frac {x \left (d+e x^2\right )}{54 \left (9-x^4\right )^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{54} \left (\frac {1}{18} \int \frac {5 d-3 e x^2}{\sqrt {9-x^4}}dx+\frac {x \left (5 d+3 e x^2\right )}{18 \sqrt {9-x^4}}\right )+\frac {x \left (d+e x^2\right )}{54 \left (9-x^4\right )^{3/2}}\)

\(\Big \downarrow \) 1495

\(\displaystyle \frac {1}{54} \left (\frac {1}{18} \int \frac {5 d-3 e x^2}{\sqrt {3-x^2} \sqrt {x^2+3}}dx+\frac {x \left (5 d+3 e x^2\right )}{18 \sqrt {9-x^4}}\right )+\frac {x \left (d+e x^2\right )}{54 \left (9-x^4\right )^{3/2}}\)

\(\Big \downarrow \) 399

\(\displaystyle \frac {1}{54} \left (\frac {1}{18} \left ((5 d+9 e) \int \frac {1}{\sqrt {3-x^2} \sqrt {x^2+3}}dx-3 e \int \frac {\sqrt {x^2+3}}{\sqrt {3-x^2}}dx\right )+\frac {x \left (5 d+3 e x^2\right )}{18 \sqrt {9-x^4}}\right )+\frac {x \left (d+e x^2\right )}{54 \left (9-x^4\right )^{3/2}}\)

\(\Big \downarrow \) 284

\(\displaystyle \frac {1}{54} \left (\frac {1}{18} \left ((5 d+9 e) \int \frac {1}{\sqrt {9-x^4}}dx-3 e \int \frac {\sqrt {x^2+3}}{\sqrt {3-x^2}}dx\right )+\frac {x \left (5 d+3 e x^2\right )}{18 \sqrt {9-x^4}}\right )+\frac {x \left (d+e x^2\right )}{54 \left (9-x^4\right )^{3/2}}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {1}{54} \left (\frac {1}{18} \left ((5 d+9 e) \int \frac {1}{\sqrt {9-x^4}}dx-3 \sqrt {3} e E\left (\left .\arcsin \left (\frac {x}{\sqrt {3}}\right )\right |-1\right )\right )+\frac {x \left (5 d+3 e x^2\right )}{18 \sqrt {9-x^4}}\right )+\frac {x \left (d+e x^2\right )}{54 \left (9-x^4\right )^{3/2}}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {1}{54} \left (\frac {1}{18} \left (\frac {(5 d+9 e) \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{\sqrt {3}}\right ),-1\right )}{\sqrt {3}}-3 \sqrt {3} e E\left (\left .\arcsin \left (\frac {x}{\sqrt {3}}\right )\right |-1\right )\right )+\frac {x \left (5 d+3 e x^2\right )}{18 \sqrt {9-x^4}}\right )+\frac {x \left (d+e x^2\right )}{54 \left (9-x^4\right )^{3/2}}\)

Input:

Int[(d + e*x^2)/(9 - x^4)^(5/2),x]
 

Output:

(x*(d + e*x^2))/(54*(9 - x^4)^(3/2)) + ((x*(5*d + 3*e*x^2))/(18*Sqrt[9 - x 
^4]) + (-3*Sqrt[3]*e*EllipticE[ArcSin[x/Sqrt[3]], -1] + ((5*d + 9*e)*Ellip 
ticF[ArcSin[x/Sqrt[3]], -1])/Sqrt[3])/18)/54
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 284

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]))
 

rule 327

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]
 

rule 399

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])))))
 

rule 762

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]
 

rule 1493

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(-x 
)*(d + e*x^2)*((a + c*x^4)^(p + 1)/(4*a*(p + 1))), x] + Simp[1/(4*a*(p + 1) 
)   Int[Simp[d*(4*p + 5) + e*(4*p + 7)*x^2, x]*(a + c*x^4)^(p + 1), x], x] 
/; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && Integer 
Q[2*p]
 

rule 1495

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]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 1.42 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.35

method	result	size
meijerg	\(\frac {e \,x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {5}{2}\right ], \left [\frac {7}{4}\right ], \frac {x^{4}}{9}\right )}{729}+\frac {d x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {5}{2}\right ], \left [\frac {5}{4}\right ], \frac {x^{4}}{9}\right )}{243}\)	\(34\)
risch	\(\frac {x \left (3 e \,x^{6}+5 d \,x^{4}-45 e \,x^{2}-63 d \right )}{972 \left (x^{4}-9\right ) \sqrt {-x^{4}+9}}+\frac {5 d \sqrt {3}\, \sqrt {-3 x^{2}+9}\, \sqrt {3 x^{2}+9}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, x}{3}, i\right )}{8748 \sqrt {-x^{4}+9}}+\frac {e \sqrt {3}\, \sqrt {-3 x^{2}+9}\, \sqrt {3 x^{2}+9}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {3}\, x}{3}, i\right )-\operatorname {EllipticE}\left (\frac {\sqrt {3}\, x}{3}, i\right )\right )}{972 \sqrt {-x^{4}+9}}\)	\(139\)
elliptic	\(\frac {\left (\frac {1}{54} e \,x^{3}+\frac {1}{54} d x \right ) \sqrt {-x^{4}+9}}{\left (x^{4}-9\right )^{2}}+\frac {\frac {1}{324} e \,x^{3}+\frac {5}{972} d x}{\sqrt {-x^{4}+9}}+\frac {5 d \sqrt {3}\, \sqrt {-3 x^{2}+9}\, \sqrt {3 x^{2}+9}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, x}{3}, i\right )}{8748 \sqrt {-x^{4}+9}}+\frac {e \sqrt {3}\, \sqrt {-3 x^{2}+9}\, \sqrt {3 x^{2}+9}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {3}\, x}{3}, i\right )-\operatorname {EllipticE}\left (\frac {\sqrt {3}\, x}{3}, i\right )\right )}{972 \sqrt {-x^{4}+9}}\)	\(148\)
default	\(d \left (\frac {x \sqrt {-x^{4}+9}}{54 \left (x^{4}-9\right )^{2}}+\frac {5 x}{972 \sqrt {-x^{4}+9}}+\frac {5 \sqrt {3}\, \sqrt {-3 x^{2}+9}\, \sqrt {3 x^{2}+9}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, x}{3}, i\right )}{8748 \sqrt {-x^{4}+9}}\right )+e \left (\frac {x^{3} \sqrt {-x^{4}+9}}{54 \left (x^{4}-9\right )^{2}}+\frac {x^{3}}{324 \sqrt {-x^{4}+9}}+\frac {\sqrt {3}\, \sqrt {-3 x^{2}+9}\, \sqrt {3 x^{2}+9}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {3}\, x}{3}, i\right )-\operatorname {EllipticE}\left (\frac {\sqrt {3}\, x}{3}, i\right )\right )}{972 \sqrt {-x^{4}+9}}\right )\)	\(168\)

Input:

int((e*x^2+d)/(-x^4+9)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/729*e*x^3*hypergeom([3/4,5/2],[7/4],1/9*x^4)+1/243*d*x*hypergeom([1/4,5/ 
2],[5/4],1/9*x^4)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.27 \[ \int \frac {d+e x^2}{\left (9-x^4\right )^{5/2}} \, dx=-\frac {3 \, \sqrt {3} {\left (e x^{8} - 18 \, e x^{4} + 81 \, e\right )} E(\arcsin \left (\frac {1}{3} \, \sqrt {3} x\right )\,|\,-1) - \sqrt {3} {\left ({\left (5 \, d + 3 \, e\right )} x^{8} - 18 \, {\left (5 \, d + 3 \, e\right )} x^{4} + 405 \, d + 243 \, e\right )} F(\arcsin \left (\frac {1}{3} \, \sqrt {3} x\right )\,|\,-1) + 3 \, {\left (3 \, e x^{7} + 5 \, d x^{5} - 45 \, e x^{3} - 63 \, d x\right )} \sqrt {-x^{4} + 9}}{2916 \, {\left (x^{8} - 18 \, x^{4} + 81\right )}} \] Input:

integrate((e*x^2+d)/(-x^4+9)^(5/2),x, algorithm="fricas")
 

Output:

-1/2916*(3*sqrt(3)*(e*x^8 - 18*e*x^4 + 81*e)*elliptic_e(arcsin(1/3*sqrt(3) 
*x), -1) - sqrt(3)*((5*d + 3*e)*x^8 - 18*(5*d + 3*e)*x^4 + 405*d + 243*e)* 
elliptic_f(arcsin(1/3*sqrt(3)*x), -1) + 3*(3*e*x^7 + 5*d*x^5 - 45*e*x^3 - 
63*d*x)*sqrt(-x^4 + 9))/(x^8 - 18*x^4 + 81)
 

Sympy [A] (verification not implemented)

Time = 3.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.71 \[ \int \frac {d+e x^2}{\left (9-x^4\right )^{5/2}} \, dx=\frac {d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {x^{4} e^{2 i \pi }}{9}} \right )}}{972 \Gamma \left (\frac {5}{4}\right )} + \frac {e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {x^{4} e^{2 i \pi }}{9}} \right )}}{972 \Gamma \left (\frac {7}{4}\right )} \] Input:

integrate((e*x**2+d)/(-x**4+9)**(5/2),x)
 

Output:

d*x*gamma(1/4)*hyper((1/4, 5/2), (5/4,), x**4*exp_polar(2*I*pi)/9)/(972*ga 
mma(5/4)) + e*x**3*gamma(3/4)*hyper((3/4, 5/2), (7/4,), x**4*exp_polar(2*I 
*pi)/9)/(972*gamma(7/4))
 

Maxima [F]

\[ \int \frac {d+e x^2}{\left (9-x^4\right )^{5/2}} \, dx=\int { \frac {e x^{2} + d}{{\left (-x^{4} + 9\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((e*x^2+d)/(-x^4+9)^(5/2),x, algorithm="maxima")
 

Output:

integrate((e*x^2 + d)/(-x^4 + 9)^(5/2), x)
 

Giac [F]

\[ \int \frac {d+e x^2}{\left (9-x^4\right )^{5/2}} \, dx=\int { \frac {e x^{2} + d}{{\left (-x^{4} + 9\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((e*x^2+d)/(-x^4+9)^(5/2),x, algorithm="giac")
 

Output:

integrate((e*x^2 + d)/(-x^4 + 9)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {d+e x^2}{\left (9-x^4\right )^{5/2}} \, dx=\int \frac {e\,x^2+d}{{\left (9-x^4\right )}^{5/2}} \,d x \] Input:

int((d + e*x^2)/(9 - x^4)^(5/2),x)
                                                                                    
                                                                                    
 

Output:

int((d + e*x^2)/(9 - x^4)^(5/2), x)
 

Reduce [F]

\[ \int \frac {d+e x^2}{\left (9-x^4\right )^{5/2}} \, dx=-\left (\int \frac {\sqrt {-x^{4}+9}}{x^{12}-27 x^{8}+243 x^{4}-729}d x \right ) d -\left (\int \frac {\sqrt {-x^{4}+9}\, x^{2}}{x^{12}-27 x^{8}+243 x^{4}-729}d x \right ) e \] Input:

int((e*x^2+d)/(-x^4+9)^(5/2),x)
 

Output:

 - (int(sqrt( - x**4 + 9)/(x**12 - 27*x**8 + 243*x**4 - 729),x)*d + int((s 
qrt( - x**4 + 9)*x**2)/(x**12 - 27*x**8 + 243*x**4 - 729),x)*e)