\(\int \frac {1}{(a+\frac {b}{x^4})^{5/2}} \, dx\) [566]

Optimal result

Integrand size = 11, antiderivative size = 277 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2}} \, dx=-\frac {7 \sqrt {b} \sqrt {a+\frac {b}{x^4}}}{4 a^3 \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) x}-\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}}-\frac {7 x}{12 a^2 \sqrt {a+\frac {b}{x^4}}}+\frac {7 \sqrt {a+\frac {b}{x^4}} x}{4 a^3}+\frac {7 \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{4 a^{11/4} \sqrt {a+\frac {b}{x^4}}}-\frac {7 \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{8 a^{11/4} \sqrt {a+\frac {b}{x^4}}} \] Output:

-7/4*b^(1/2)*(a+b/x^4)^(1/2)/a^3/(a^(1/2)+b^(1/2)/x^2)/x-1/6*x/a/(a+b/x^4) 
^(3/2)-7/12*x/a^2/(a+b/x^4)^(1/2)+7/4*(a+b/x^4)^(1/2)*x/a^3+7/4*b^(1/4)*(( 
a+b/x^4)/(a^(1/2)+b^(1/2)/x^2)^2)^(1/2)*(a^(1/2)+b^(1/2)/x^2)*EllipticE(si 
n(2*arccot(a^(1/4)*x/b^(1/4))),1/2*2^(1/2))/a^(11/4)/(a+b/x^4)^(1/2)-7/8*b 
^(1/4)*((a+b/x^4)/(a^(1/2)+b^(1/2)/x^2)^2)^(1/2)*(a^(1/2)+b^(1/2)/x^2)*Inv 
erseJacobiAM(2*arccot(a^(1/4)*x/b^(1/4)),1/2*2^(1/2))/a^(11/4)/(a+b/x^4)^( 
1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 10.04 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.29 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2}} \, dx=\frac {7 b x+3 a x^5-7 x \left (b+a x^4\right ) \sqrt {1+\frac {a x^4}{b}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{4},-\frac {a x^4}{b}\right )}{3 a^2 \sqrt {a+\frac {b}{x^4}} \left (b+a x^4\right )} \] Input:

Integrate[(a + b/x^4)^(-5/2),x]
 

Output:

(7*b*x + 3*a*x^5 - 7*x*(b + a*x^4)*Sqrt[1 + (a*x^4)/b]*Hypergeometric2F1[3 
/4, 5/2, 7/4, -((a*x^4)/b)])/(3*a^2*Sqrt[a + b/x^4]*(b + a*x^4))
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {773, 819, 819, 847, 834, 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.

\(\displaystyle \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 773

\(\displaystyle -\int \frac {x^2}{\left (a+\frac {b}{x^4}\right )^{5/2}}d\frac {1}{x}\)

\(\Big \downarrow \) 819

\(\displaystyle -\frac {7 \int \frac {x^2}{\left (a+\frac {b}{x^4}\right )^{3/2}}d\frac {1}{x}}{6 a}-\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}}\)

\(\Big \downarrow \) 819

\(\displaystyle -\frac {7 \left (\frac {3 \int \frac {x^2}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{2 a}+\frac {x}{2 a \sqrt {a+\frac {b}{x^4}}}\right )}{6 a}-\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}}\)

\(\Big \downarrow \) 847

\(\displaystyle -\frac {7 \left (\frac {3 \left (\frac {b \int \frac {1}{\sqrt {a+\frac {b}{x^4}} x^2}d\frac {1}{x}}{a}-\frac {x \sqrt {a+\frac {b}{x^4}}}{a}\right )}{2 a}+\frac {x}{2 a \sqrt {a+\frac {b}{x^4}}}\right )}{6 a}-\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}}\)

\(\Big \downarrow \) 834

\(\displaystyle -\frac {7 \left (\frac {3 \left (\frac {b \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\frac {\sqrt {b}}{x^2}}{\sqrt {a} \sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}\right )}{a}-\frac {x \sqrt {a+\frac {b}{x^4}}}{a}\right )}{2 a}+\frac {x}{2 a \sqrt {a+\frac {b}{x^4}}}\right )}{6 a}-\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {7 \left (\frac {3 \left (\frac {b \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}-\frac {\int \frac {\sqrt {a}-\frac {\sqrt {b}}{x^2}}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}\right )}{a}-\frac {x \sqrt {a+\frac {b}{x^4}}}{a}\right )}{2 a}+\frac {x}{2 a \sqrt {a+\frac {b}{x^4}}}\right )}{6 a}-\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}}\)

\(\Big \downarrow \) 761

\(\displaystyle -\frac {7 \left (\frac {3 \left (\frac {b \left (\frac {\sqrt [4]{a} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b}}{\sqrt [4]{a} x}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {\int \frac {\sqrt {a}-\frac {\sqrt {b}}{x^2}}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}\right )}{a}-\frac {x \sqrt {a+\frac {b}{x^4}}}{a}\right )}{2 a}+\frac {x}{2 a \sqrt {a+\frac {b}{x^4}}}\right )}{6 a}-\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}}\)

\(\Big \downarrow \) 1510

\(\displaystyle -\frac {7 \left (\frac {3 \left (\frac {b \left (\frac {\sqrt [4]{a} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b}}{\sqrt [4]{a} x}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {\frac {\sqrt [4]{a} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{b}}{\sqrt [4]{a} x}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+\frac {b}{x^4}}}-\frac {\sqrt {a+\frac {b}{x^4}}}{x \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )}}{\sqrt {b}}\right )}{a}-\frac {x \sqrt {a+\frac {b}{x^4}}}{a}\right )}{2 a}+\frac {x}{2 a \sqrt {a+\frac {b}{x^4}}}\right )}{6 a}-\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}}\)

Input:

Int[(a + b/x^4)^(-5/2),x]
 

Output:

-1/6*x/(a*(a + b/x^4)^(3/2)) - (7*(x/(2*a*Sqrt[a + b/x^4]) + (3*(-((Sqrt[a 
 + b/x^4]*x)/a) + (b*(-((-(Sqrt[a + b/x^4]/((Sqrt[a] + Sqrt[b]/x^2)*x)) + 
(a^(1/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^ 
2)*EllipticE[2*ArcTan[b^(1/4)/(a^(1/4)*x)], 1/2])/(b^(1/4)*Sqrt[a + b/x^4] 
))/Sqrt[b]) + (a^(1/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a 
] + Sqrt[b]/x^2)*EllipticF[2*ArcTan[b^(1/4)/(a^(1/4)*x)], 1/2])/(2*b^(3/4) 
*Sqrt[a + b/x^4])))/a))/(2*a)))/(6*a)
 

Defintions of rubi rules used

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 773

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^ 
2, x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] &&  !IntegerQ[p]
 

rule 819

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-( 
c*x)^(m + 1))*((a + b*x^n)^(p + 1)/(a*c*n*(p + 1))), x] + Simp[(m + n*(p + 
1) + 1)/(a*n*(p + 1))   Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a 
, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p 
, x]
 

rule 834

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

rule 847

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x 
)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) 
+ 1)/(a*c^n*(m + 1)))   Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a 
, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p 
, x]
 

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]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.64 (sec) , antiderivative size = 503, normalized size of antiderivative = 1.82

method	result	size
default	\(\frac {-9 a^{\frac {9}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x^{11}+21 i \sqrt {b}\, \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) a^{4} x^{8}-21 i \sqrt {b}\, \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) a^{4} x^{8}-16 a^{\frac {7}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b \,x^{7}+42 i b^{\frac {3}{2}} \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) a^{3} x^{4}-42 i b^{\frac {3}{2}} \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) a^{3} x^{4}+21 i b^{\frac {5}{2}} \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) a^{2}-21 i b^{\frac {5}{2}} \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) a^{2}-7 a^{\frac {5}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b^{2} x^{3}}{12 a^{\frac {9}{2}} \left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}} x^{10} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}}\)	\(503\)

Input:

int(1/(a+b/x^4)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/12*(-9*a^(9/2)*(I*a^(1/2)/b^(1/2))^(1/2)*x^11+21*I*b^(1/2)*(-(I*a^(1/2)* 
x^2-b^(1/2))/b^(1/2))^(1/2)*((I*a^(1/2)*x^2+b^(1/2))/b^(1/2))^(1/2)*Ellipt 
icF(x*(I*a^(1/2)/b^(1/2))^(1/2),I)*a^4*x^8-21*I*b^(1/2)*(-(I*a^(1/2)*x^2-b 
^(1/2))/b^(1/2))^(1/2)*((I*a^(1/2)*x^2+b^(1/2))/b^(1/2))^(1/2)*EllipticE(x 
*(I*a^(1/2)/b^(1/2))^(1/2),I)*a^4*x^8-16*a^(7/2)*(I*a^(1/2)/b^(1/2))^(1/2) 
*b*x^7+42*I*b^(3/2)*(-(I*a^(1/2)*x^2-b^(1/2))/b^(1/2))^(1/2)*((I*a^(1/2)*x 
^2+b^(1/2))/b^(1/2))^(1/2)*EllipticF(x*(I*a^(1/2)/b^(1/2))^(1/2),I)*a^3*x^ 
4-42*I*b^(3/2)*(-(I*a^(1/2)*x^2-b^(1/2))/b^(1/2))^(1/2)*((I*a^(1/2)*x^2+b^ 
(1/2))/b^(1/2))^(1/2)*EllipticE(x*(I*a^(1/2)/b^(1/2))^(1/2),I)*a^3*x^4+21* 
I*b^(5/2)*(-(I*a^(1/2)*x^2-b^(1/2))/b^(1/2))^(1/2)*((I*a^(1/2)*x^2+b^(1/2) 
)/b^(1/2))^(1/2)*EllipticF(x*(I*a^(1/2)/b^(1/2))^(1/2),I)*a^2-21*I*b^(5/2) 
*(-(I*a^(1/2)*x^2-b^(1/2))/b^(1/2))^(1/2)*((I*a^(1/2)*x^2+b^(1/2))/b^(1/2) 
)^(1/2)*EllipticE(x*(I*a^(1/2)/b^(1/2))^(1/2),I)*a^2-7*a^(5/2)*(I*a^(1/2)/ 
b^(1/2))^(1/2)*b^2*x^3)/a^(9/2)/((a*x^4+b)/x^4)^(5/2)/x^10/(I*a^(1/2)/b^(1 
/2))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.57 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2}} \, dx=\frac {21 \, {\left (a^{2} x^{8} + 2 \, a b x^{4} + b^{2}\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {b}{a}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 21 \, {\left (a^{2} x^{8} + 2 \, a b x^{4} + b^{2}\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {b}{a}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (12 \, a^{2} x^{9} + 35 \, a b x^{5} + 21 \, b^{2} x\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{12 \, {\left (a^{5} x^{8} + 2 \, a^{4} b x^{4} + a^{3} b^{2}\right )}} \] Input:

integrate(1/(a+b/x^4)^(5/2),x, algorithm="fricas")
 

Output:

1/12*(21*(a^2*x^8 + 2*a*b*x^4 + b^2)*sqrt(a)*(-b/a)^(3/4)*elliptic_e(arcsi 
n((-b/a)^(1/4)/x), -1) - 21*(a^2*x^8 + 2*a*b*x^4 + b^2)*sqrt(a)*(-b/a)^(3/ 
4)*elliptic_f(arcsin((-b/a)^(1/4)/x), -1) + (12*a^2*x^9 + 35*a*b*x^5 + 21* 
b^2*x)*sqrt((a*x^4 + b)/x^4))/(a^5*x^8 + 2*a^4*b*x^4 + a^3*b^2)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.95 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.15 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2}} \, dx=- \frac {x \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 a^{\frac {5}{2}} \Gamma \left (\frac {3}{4}\right )} \] Input:

integrate(1/(a+b/x**4)**(5/2),x)
 

Output:

-x*gamma(-1/4)*hyper((-1/4, 5/2), (3/4,), b*exp_polar(I*pi)/(a*x**4))/(4*a 
**(5/2)*gamma(3/4))
 

Maxima [F]

\[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2}} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(a+b/x^4)^(5/2),x, algorithm="maxima")
 

Output:

integrate((a + b/x^4)^(-5/2), x)
 

Giac [F]

\[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2}} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(a+b/x^4)^(5/2),x, algorithm="giac")
 

Output:

integrate((a + b/x^4)^(-5/2), x)
 

Mupad [B] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.16 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2}} \, dx=\frac {x\,{\left (\frac {a\,x^4}{b}+1\right )}^{5/2}\,\sqrt {x^{20}}\,{{}}_2{\mathrm {F}}_1\left (\frac {5}{2},\frac {11}{4};\ \frac {15}{4};\ -\frac {a\,x^4}{b}\right )}{11\,{\left (a\,x^4+b\right )}^{5/2}} \] Input:

int(1/(a + b/x^4)^(5/2),x)
 

Output:

(x*((a*x^4)/b + 1)^(5/2)*(x^20)^(1/2)*hypergeom([5/2, 11/4], 15/4, -(a*x^4 
)/b))/(11*(b + a*x^4)^(5/2))
 

Reduce [F]

\[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2}} \, dx=\frac {3 \sqrt {a \,x^{4}+b}\, a \,x^{7}+7 \sqrt {a \,x^{4}+b}\, b \,x^{3}-21 \left (\int \frac {\sqrt {a \,x^{4}+b}\, x^{2}}{a^{3} x^{12}+3 a^{2} b \,x^{8}+3 a \,b^{2} x^{4}+b^{3}}d x \right ) a^{2} b^{2} x^{8}-42 \left (\int \frac {\sqrt {a \,x^{4}+b}\, x^{2}}{a^{3} x^{12}+3 a^{2} b \,x^{8}+3 a \,b^{2} x^{4}+b^{3}}d x \right ) a \,b^{3} x^{4}-21 \left (\int \frac {\sqrt {a \,x^{4}+b}\, x^{2}}{a^{3} x^{12}+3 a^{2} b \,x^{8}+3 a \,b^{2} x^{4}+b^{3}}d x \right ) b^{4}}{3 a^{2} \left (a^{2} x^{8}+2 a b \,x^{4}+b^{2}\right )} \] Input:

int(1/(a+b/x^4)^(5/2),x)
 

Output:

(3*sqrt(a*x**4 + b)*a*x**7 + 7*sqrt(a*x**4 + b)*b*x**3 - 21*int((sqrt(a*x* 
*4 + b)*x**2)/(a**3*x**12 + 3*a**2*b*x**8 + 3*a*b**2*x**4 + b**3),x)*a**2* 
b**2*x**8 - 42*int((sqrt(a*x**4 + b)*x**2)/(a**3*x**12 + 3*a**2*b*x**8 + 3 
*a*b**2*x**4 + b**3),x)*a*b**3*x**4 - 21*int((sqrt(a*x**4 + b)*x**2)/(a**3 
*x**12 + 3*a**2*b*x**8 + 3*a*b**2*x**4 + b**3),x)*b**4)/(3*a**2*(a**2*x**8 
 + 2*a*b*x**4 + b**2))