\(\int \frac {d+e x^2}{(9+x^4)^{5/2}} \, dx\) [252]

Optimal result

Integrand size = 17, antiderivative size = 184 \[ \int \frac {d+e x^2}{\left (9+x^4\right )^{5/2}} \, dx=\frac {x \left (d+e x^2\right )}{54 \left (9+x^4\right )^{3/2}}+\frac {x \left (5 d+3 e x^2\right )}{972 \sqrt {9+x^4}}-\frac {e x \sqrt {9+x^4}}{324 \left (3+x^2\right )}+\frac {e \left (3+x^2\right ) \sqrt {\frac {9+x^4}{\left (3+x^2\right )^2}} E\left (2 \arctan \left (\frac {x}{\sqrt {3}}\right )|\frac {1}{2}\right )}{108 \sqrt {3} \sqrt {9+x^4}}+\frac {(5 d-9 e) \left (3+x^2\right ) \sqrt {\frac {9+x^4}{\left (3+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt {3}}\right ),\frac {1}{2}\right )}{1944 \sqrt {3} \sqrt {9+x^4}} \] 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)-e*x*(x^ 
4+9)^(1/2)/(324*x^2+972)+1/324*3^(1/2)*e*(x^2+3)*((x^4+9)/(x^2+3)^2)^(1/2) 
*EllipticE(sin(2*arctan(1/3*x*3^(1/2))),1/2*2^(1/2))/(x^4+9)^(1/2)+1/5832* 
(5*d-9*e)*(x^2+3)*((x^4+9)/(x^2+3)^2)^(1/2)*InverseJacobiAM(2*arctan(1/3*x 
*3^(1/2)),1/2*2^(1/2))*3^(1/2)/(x^4+9)^(1/2)
 

Mathematica [C] (verified)

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

Time = 10.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.37 \[ \int \frac {d+e x^2}{\left (9+x^4\right )^{5/2}} \, dx=\frac {x \left (\frac {3 d \left (63+5 x^4\right )}{\left (9+x^4\right )^{3/2}}+5 d \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {x^4}{9}\right )+4 e x^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{4},-\frac {x^4}{9}\right )\right )}{2916} \] 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, -1/9*x^4] + 4*e*x^2*Hypergeometric2F1[3/4, 5/2, 7/4, -1/9*x^4]))/2916
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {1493, 25, 1493, 25, 1512, 27, 761, 1510}

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.

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

\(\Big \downarrow \) 1493

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1493

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1512

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

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*e*(-((x*Sqrt[9 + x^4])/(3 + x^2)) + (Sqrt[3]*(3 + x^2)*Sqrt[(9 + 
 x^4)/(3 + x^2)^2]*EllipticE[2*ArcTan[x/Sqrt[3]], 1/2])/Sqrt[9 + x^4]) + ( 
(5*d - 9*e)*(3 + x^2)*Sqrt[(9 + x^4)/(3 + x^2)^2]*EllipticF[2*ArcTan[x/Sqr 
t[3]], 1/2])/(2*Sqrt[3]*Sqrt[9 + x^4]))/18)/54
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

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 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

rule 1512

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 2]}, Simp[(e + d*q)/q   Int[1/Sqrt[a + c*x^4], x], x] - Simp[e/q 
 Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a, c 
, d, e}, x] && PosQ[c/a]
 

Maple [A] (verified)

Time = 1.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.18

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	\(\frac {x \left (3 e \,x^{6}+5 d \,x^{4}+45 e \,x^{2}+63 d \right )}{972 \left (x^{4}+9\right )^{\frac {3}{2}}}+\frac {5 d \sqrt {-3 i x^{2}+9}\, \sqrt {3 i x^{2}+9}\, \operatorname {EllipticF}\left (x \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ), i\right )}{8748 \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ) \sqrt {x^{4}+9}}-\frac {i e \sqrt {-3 i x^{2}+9}\, \sqrt {3 i x^{2}+9}\, \left (\operatorname {EllipticF}\left (x \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ), i\right )-\operatorname {EllipticE}\left (x \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ), i\right )\right )}{972 \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ) \sqrt {x^{4}+9}}\)	\(177\)
elliptic	\(\frac {\frac {1}{54} e \,x^{3}+\frac {1}{54} d x}{\left (x^{4}+9\right )^{\frac {3}{2}}}-\frac {2 \left (-\frac {1}{648} e \,x^{3}-\frac {5}{1944} d x \right )}{\sqrt {x^{4}+9}}+\frac {5 d \sqrt {-3 i x^{2}+9}\, \sqrt {3 i x^{2}+9}\, \operatorname {EllipticF}\left (x \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ), i\right )}{8748 \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ) \sqrt {x^{4}+9}}-\frac {i e \sqrt {-3 i x^{2}+9}\, \sqrt {3 i x^{2}+9}\, \left (\operatorname {EllipticF}\left (x \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ), i\right )-\operatorname {EllipticE}\left (x \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ), i\right )\right )}{972 \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ) \sqrt {x^{4}+9}}\)	\(184\)
default	\(d \left (\frac {x}{54 \left (x^{4}+9\right )^{\frac {3}{2}}}+\frac {5 x}{972 \sqrt {x^{4}+9}}+\frac {5 \sqrt {-3 i x^{2}+9}\, \sqrt {3 i x^{2}+9}\, \operatorname {EllipticF}\left (x \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ), i\right )}{8748 \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ) \sqrt {x^{4}+9}}\right )+e \left (\frac {x^{3}}{54 \left (x^{4}+9\right )^{\frac {3}{2}}}+\frac {x^{3}}{324 \sqrt {x^{4}+9}}-\frac {i \sqrt {-3 i x^{2}+9}\, \sqrt {3 i x^{2}+9}\, \left (\operatorname {EllipticF}\left (x \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ), i\right )-\operatorname {EllipticE}\left (x \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ), i\right )\right )}{972 \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ) \sqrt {x^{4}+9}}\right )\)	\(193\)

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 [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.17 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.65 \[ \int \frac {d+e x^2}{\left (9+x^4\right )^{5/2}} \, dx=-\frac {3 \, \sqrt {\frac {1}{3} i} {\left (-i \, e x^{8} - 18 i \, e x^{4} - 81 i \, e\right )} E(\arcsin \left (\sqrt {\frac {1}{3} i} x\right )\,|\,-1) - \sqrt {\frac {1}{3} i} {\left (-i \, {\left (5 \, d + 3 \, e\right )} x^{8} - 18 i \, {\left (5 \, d + 3 \, e\right )} x^{4} - 405 i \, d - 243 i \, e\right )} F(\arcsin \left (\sqrt {\frac {1}{3} i} x\right )\,|\,-1) - {\left (3 \, e x^{7} + 5 \, d x^{5} + 45 \, e x^{3} + 63 \, d x\right )} \sqrt {x^{4} + 9}}{972 \, {\left (x^{8} + 18 \, x^{4} + 81\right )}} \] Input:

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

Output:

-1/972*(3*sqrt(1/3*I)*(-I*e*x^8 - 18*I*e*x^4 - 81*I*e)*elliptic_e(arcsin(s 
qrt(1/3*I)*x), -1) - sqrt(1/3*I)*(-I*(5*d + 3*e)*x^8 - 18*I*(5*d + 3*e)*x^ 
4 - 405*I*d - 243*I*e)*elliptic_f(arcsin(sqrt(1/3*I)*x), -1) - (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 [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.83 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.35 \[ \int \frac {d+e x^2}{\left (9+x^4\right )^{5/2}} \, dx=\frac {d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{9}} \right )}}{972 \Gamma \left (\frac {5}{4}\right )} + \frac {e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{9}} \right )}}{972 \Gamma \left (\frac {7}{4}\right )} \] 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(I*pi)/9)/(972*gamm 
a(5/4)) + e*x**3*gamma(3/4)*hyper((3/4, 5/2), (7/4,), x**4*exp_polar(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. \[ \int \frac {d+e x^2}{\left (9+x^4\right )^{5/2}} \, dx=\int \frac {e\,x^2+d}{{\left (x^4+9\right )}^{5/2}} \,d x \] Input:

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

Output:

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