\(\int \frac {\sqrt {a+\frac {b}{x^4}}}{x^4} \, dx\) [528]

Optimal result

Integrand size = 15, antiderivative size = 236 \[ \int \frac {\sqrt {a+\frac {b}{x^4}}}{x^4} \, dx=-\frac {\sqrt {a+\frac {b}{x^4}}}{5 x^3}-\frac {2 a \sqrt {a+\frac {b}{x^4}}}{5 \sqrt {b} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) x}+\frac {2 a^{5/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {a^{5/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+\frac {b}{x^4}}} \] Output:

-1/5*(a+b/x^4)^(1/2)/x^3-2/5*a*(a+b/x^4)^(1/2)/b^(1/2)/(a^(1/2)+b^(1/2)/x^ 
2)/x+2/5*a^(5/4)*((a+b/x^4)/(a^(1/2)+b^(1/2)/x^2)^2)^(1/2)*(a^(1/2)+b^(1/2 
)/x^2)*EllipticE(sin(2*arccot(a^(1/4)*x/b^(1/4))),1/2*2^(1/2))/b^(3/4)/(a+ 
b/x^4)^(1/2)-1/5*a^(5/4)*((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))/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 = 10.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.22 \[ \int \frac {\sqrt {a+\frac {b}{x^4}}}{x^4} \, dx=-\frac {\sqrt {a+\frac {b}{x^4}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},-\frac {1}{2},-\frac {1}{4},-\frac {a x^4}{b}\right )}{5 x^3 \sqrt {1+\frac {a x^4}{b}}} \] Input:

Integrate[Sqrt[a + b/x^4]/x^4,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 243, 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, 811, 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.

\(\displaystyle \int \frac {\sqrt {a+\frac {b}{x^4}}}{x^4} \, dx\)

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 811

\(\displaystyle -\frac {2}{5} a \int \frac {1}{\sqrt {a+\frac {b}{x^4}} x^2}d\frac {1}{x}-\frac {\sqrt {a+\frac {b}{x^4}}}{5 x^3}\)

\(\Big \downarrow \) 834

\(\displaystyle -\frac {2}{5} a \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\frac {\sqrt {b}}{x^2}}{\sqrt {a} \sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}\right )-\frac {\sqrt {a+\frac {b}{x^4}}}{5 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {2}{5} a \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}-\frac {\int \frac {\sqrt {a}-\frac {\sqrt {b}}{x^2}}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}\right )-\frac {\sqrt {a+\frac {b}{x^4}}}{5 x^3}\)

\(\Big \downarrow \) 761

\(\displaystyle -\frac {2}{5} a \left (\frac {\sqrt [4]{a} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b}}{\sqrt [4]{a} x}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {\int \frac {\sqrt {a}-\frac {\sqrt {b}}{x^2}}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}\right )-\frac {\sqrt {a+\frac {b}{x^4}}}{5 x^3}\)

\(\Big \downarrow \) 1510

\(\displaystyle -\frac {2}{5} a \left (\frac {\sqrt [4]{a} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b}}{\sqrt [4]{a} x}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {\frac {\sqrt [4]{a} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{b}}{\sqrt [4]{a} x}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+\frac {b}{x^4}}}-\frac {\sqrt {a+\frac {b}{x^4}}}{x \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )}}{\sqrt {b}}\right )-\frac {\sqrt {a+\frac {b}{x^4}}}{5 x^3}\)

Input:

Int[Sqrt[a + b/x^4]/x^4,x]
 

Output:

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

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 811

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* 
x)^(m + 1)*((a + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 
))   Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I 
GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && 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 = 0.58 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.61

method	result	size
risch	\(-\frac {\left (2 a \,x^{4}+b \right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}}{5 x^{3} b}+\frac {2 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \sqrt {1+\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2}}{5 \sqrt {b}\, \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, \left (a \,x^{4}+b \right )}\)	\(143\)
default	\(-\frac {\sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, \left (-2 i a^{\frac {3}{2}} \sqrt {\frac {-i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, x^{5} b \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )+2 i a^{\frac {3}{2}} \sqrt {\frac {-i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, x^{5} b \operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )+2 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, \sqrt {b}\, a^{2} x^{8}+3 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b^{\frac {3}{2}} a \,x^{4}+\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b^{\frac {5}{2}}\right )}{5 x^{3} \left (a \,x^{4}+b \right ) b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}}\)	\(228\)

Input:

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

Output:

-1/5*(2*a*x^4+b)/x^3/b*((a*x^4+b)/x^4)^(1/2)+2/5*I/b^(1/2)*a^(3/2)/(I*a^(1 
/2)/b^(1/2))^(1/2)*(1-I*a^(1/2)/b^(1/2)*x^2)^(1/2)*(1+I*a^(1/2)/b^(1/2)*x^ 
2)^(1/2)/(a*x^4+b)*(EllipticF(x*(I*a^(1/2)/b^(1/2))^(1/2),I)-EllipticE(x*( 
I*a^(1/2)/b^(1/2))^(1/2),I))*((a*x^4+b)/x^4)^(1/2)*x^2
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.39 \[ \int \frac {\sqrt {a+\frac {b}{x^4}}}{x^4} \, dx=-\frac {2 \, a \sqrt {b} x^{3} \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )\,|\,-1) - 2 \, a \sqrt {b} x^{3} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )\,|\,-1) + {\left (2 \, a x^{4} + b\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{5 \, b x^{3}} \] Input:

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

Output:

-1/5*(2*a*sqrt(b)*x^3*(-a/b)^(3/4)*elliptic_e(arcsin(x*(-a/b)^(1/4)), -1) 
- 2*a*sqrt(b)*x^3*(-a/b)^(3/4)*elliptic_f(arcsin(x*(-a/b)^(1/4)), -1) + (2 
*a*x^4 + b)*sqrt((a*x^4 + b)/x^4))/(b*x^3)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.70 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.17 \[ \int \frac {\sqrt {a+\frac {b}{x^4}}}{x^4} \, dx=- \frac {\sqrt {a} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 x^{3} \Gamma \left (\frac {7}{4}\right )} \] Input:

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

Output:

-sqrt(a)*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), b*exp_polar(I*pi)/(a*x**4)) 
/(4*x**3*gamma(7/4))
 

Maxima [F]

\[ \int \frac {\sqrt {a+\frac {b}{x^4}}}{x^4} \, dx=\int { \frac {\sqrt {a + \frac {b}{x^{4}}}}{x^{4}} \,d x } \] Input:

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

Output:

integrate(sqrt(a + b/x^4)/x^4, x)
 

Giac [F]

\[ \int \frac {\sqrt {a+\frac {b}{x^4}}}{x^4} \, dx=\int { \frac {\sqrt {a + \frac {b}{x^{4}}}}{x^{4}} \,d x } \] Input:

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

Output:

integrate(sqrt(a + b/x^4)/x^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+\frac {b}{x^4}}}{x^4} \, dx=\int \frac {\sqrt {a+\frac {b}{x^4}}}{x^4} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {a+\frac {b}{x^4}}}{x^4} \, dx=\frac {-\sqrt {a \,x^{4}+b}-2 \left (\int \frac {\sqrt {a \,x^{4}+b}}{a \,x^{10}+b \,x^{6}}d x \right ) b \,x^{5}}{3 x^{5}} \] Input:

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

Output:

( - sqrt(a*x**4 + b) - 2*int(sqrt(a*x**4 + b)/(a*x**10 + b*x**6),x)*b*x**5 
)/(3*x**5)