\(\int \frac {1}{(a+\frac {b}{x^4})^{3/2} x^4} \, dx\) [560]

Optimal result

Integrand size = 15, antiderivative size = 241 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2} x^4} \, dx=-\frac {1}{2 a \sqrt {a+\frac {b}{x^4}} x^3}+\frac {\sqrt {a+\frac {b}{x^4}}}{2 a \sqrt {b} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) x}-\frac {\sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{2 a^{3/4} b^{3/4} \sqrt {a+\frac {b}{x^4}}}+\frac {\sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{4 a^{3/4} b^{3/4} \sqrt {a+\frac {b}{x^4}}} \] Output:

-1/2/a/(a+b/x^4)^(1/2)/x^3+1/2*(a+b/x^4)^(1/2)/a/b^(1/2)/(a^(1/2)+b^(1/2)/ 
x^2)/x-1/2*((a+b/x^4)/(a^(1/2)+b^(1/2)/x^2)^2)^(1/2)*(a^(1/2)+b^(1/2)/x^2) 
*EllipticE(sin(2*arccot(a^(1/4)*x/b^(1/4))),1/2*2^(1/2))/a^(3/4)/b^(3/4)/( 
a+b/x^4)^(1/2)+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)*InverseJacobiAM(2*arccot(a^(1/4)*x/b^(1/4)),1/2*2^(1/2))/a^(3/4)/ 
b^(3/4)/(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 = 5.23 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.22 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2} x^4} \, dx=\frac {x \sqrt {1+\frac {a x^4}{b}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {a x^4}{b}\right )}{3 b \sqrt {a+\frac {b}{x^4}}} \] Input:

Integrate[1/((a + b/x^4)^(3/2)*x^4),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {858, 819, 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.

\(\displaystyle \int \frac {1}{x^4 \left (a+\frac {b}{x^4}\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 819

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

Int[1/((a + b/x^4)^(3/2)*x^4),x]
 

Output:

-1/2*1/(a*Sqrt[a + b/x^4]*x^3) + (-((-(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)*Sqr 
t[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]))/(2*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 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 858

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

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 = 0.32 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.78

method	result	size
default	\(\frac {\left (a \,x^{4}+b \right ) \left (x^{3} \sqrt {b}\, \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, \sqrt {a}-i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, b \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )+i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, b \operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )\right )}{2 \left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {3}{2}} x^{6} b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, \sqrt {a}}\)	\(187\)

Input:

int(1/(a+b/x^4)^(3/2)/x^4,x,method=_RETURNVERBOSE)
 

Output:

1/2*(a*x^4+b)*(x^3*b^(1/2)*(I*a^(1/2)/b^(1/2))^(1/2)*a^(1/2)-I*(-(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)*b*El 
lipticF(x*(I*a^(1/2)/b^(1/2))^(1/2),I)+I*(-(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)*b*EllipticE(x*(I*a^(1/2)/b 
^(1/2))^(1/2),I))/((a*x^4+b)/x^4)^(3/2)/x^6/b^(3/2)/(I*a^(1/2)/b^(1/2))^(1 
/2)/a^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.42 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2} x^4} \, dx=\frac {a x^{5} \sqrt {\frac {a x^{4} + b}{x^{4}}} + {\left (a x^{4} + b\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )\,|\,-1) - {\left (a x^{4} + b\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )\,|\,-1)}{2 \, {\left (a^{2} b x^{4} + a b^{2}\right )}} \] Input:

integrate(1/(a+b/x^4)^(3/2)/x^4,x, algorithm="fricas")
 

Output:

1/2*(a*x^5*sqrt((a*x^4 + b)/x^4) + (a*x^4 + b)*sqrt(b)*(-a/b)^(3/4)*ellipt 
ic_e(arcsin(x*(-a/b)^(1/4)), -1) - (a*x^4 + b)*sqrt(b)*(-a/b)^(3/4)*ellipt 
ic_f(arcsin(x*(-a/b)^(1/4)), -1))/(a^2*b*x^4 + a*b^2)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.16 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2} x^4} \, dx=- \frac {\Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 a^{\frac {3}{2}} x^{3} \Gamma \left (\frac {7}{4}\right )} \] Input:

integrate(1/(a+b/x**4)**(3/2)/x**4,x)
 

Output:

-gamma(3/4)*hyper((3/4, 3/2), (7/4,), b*exp_polar(I*pi)/(a*x**4))/(4*a**(3 
/2)*x**3*gamma(7/4))
 

Maxima [F]

\[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2} x^4} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} x^{4}} \,d x } \] Input:

integrate(1/(a+b/x^4)^(3/2)/x^4,x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

integrate(1/((a + b/x^4)^(3/2)*x^4), x)
 

Giac [F]

\[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2} x^4} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} x^{4}} \,d x } \] Input:

integrate(1/(a+b/x^4)^(3/2)/x^4,x, algorithm="giac")
 

Output:

integrate(1/((a + b/x^4)^(3/2)*x^4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2} x^4} \, dx=\int \frac {1}{x^4\,{\left (a+\frac {b}{x^4}\right )}^{3/2}} \,d x \] Input:

int(1/(x^4*(a + b/x^4)^(3/2)),x)
 

Output:

int(1/(x^4*(a + b/x^4)^(3/2)), x)
 

Reduce [F]

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

int(1/(a+b/x^4)^(3/2)/x^4,x)
 

Output:

int((sqrt(a*x**4 + b)*x**2)/(a**2*x**8 + 2*a*b*x**4 + b**2),x)