\(\int \frac {1}{x^6 \sqrt {a+b x^4}} \, dx\) [373]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [C] (verified)
Fricas [A] (verification not implemented)
Sympy [C] (verification not implemented)
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 15, antiderivative size = 261 \[ \int \frac {1}{x^6 \sqrt {a+b x^4}} \, dx=-\frac {\sqrt {a+b x^4}}{5 a x^5}+\frac {3 b \sqrt {a+b x^4}}{5 a^2 x}-\frac {3 b^{3/2} x \sqrt {a+b x^4}}{5 a^2 \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {3 b^{5/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{7/4} \sqrt {a+b x^4}}-\frac {3 b^{5/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{10 a^{7/4} \sqrt {a+b x^4}} \] Output:

-1/5*(b*x^4+a)^(1/2)/a/x^5+3/5*b*(b*x^4+a)^(1/2)/a^2/x-3/5*b^(3/2)*x*(b*x^ 
4+a)^(1/2)/a^2/(a^(1/2)+b^(1/2)*x^2)+3/5*b^(5/4)*(a^(1/2)+b^(1/2)*x^2)*((b 
*x^4+a)/(a^(1/2)+b^(1/2)*x^2)^2)^(1/2)*EllipticE(sin(2*arctan(b^(1/4)*x/a^ 
(1/4))),1/2*2^(1/2))/a^(7/4)/(b*x^4+a)^(1/2)-3/10*b^(5/4)*(a^(1/2)+b^(1/2) 
*x^2)*((b*x^4+a)/(a^(1/2)+b^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(b 
^(1/4)*x/a^(1/4)),1/2*2^(1/2))/a^(7/4)/(b*x^4+a)^(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 = 0.20 \[ \int \frac {1}{x^6 \sqrt {a+b x^4}} \, dx=-\frac {\sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{2},-\frac {1}{4},-\frac {b x^4}{a}\right )}{5 x^5 \sqrt {a+b x^4}} \] Input:

Integrate[1/(x^6*Sqrt[a + b*x^4]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {847, 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.

\(\displaystyle \int \frac {1}{x^6 \sqrt {a+b x^4}} \, dx\)

\(\Big \downarrow \) 847

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

\(\Big \downarrow \) 847

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

Int[1/(x^6*Sqrt[a + b*x^4]),x]
 

Output:

-1/5*Sqrt[a + b*x^4]/(a*x^5) - (3*b*(-(Sqrt[a + b*x^4]/(a*x)) + (b*(-((-(( 
x*Sqrt[a + b*x^4])/(Sqrt[a] + Sqrt[b]*x^2)) + (a^(1/4)*(Sqrt[a] + Sqrt[b]* 
x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticE[2*ArcTan[(b^(1/ 
4)*x)/a^(1/4)], 1/2])/(b^(1/4)*Sqrt[a + b*x^4]))/Sqrt[b]) + (a^(1/4)*(Sqrt 
[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2 
*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(2*b^(3/4)*Sqrt[a + b*x^4])))/a))/(5*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 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 = 0.80 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.47

method result size
risch \(-\frac {\sqrt {b \,x^{4}+a}\, \left (-3 b \,x^{4}+a \right )}{5 a^{2} x^{5}}-\frac {3 i b^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 a^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) \(123\)
default \(-\frac {\sqrt {b \,x^{4}+a}}{5 a \,x^{5}}+\frac {3 b \sqrt {b \,x^{4}+a}}{5 a^{2} x}-\frac {3 i b^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 a^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) \(133\)
elliptic \(-\frac {\sqrt {b \,x^{4}+a}}{5 a \,x^{5}}+\frac {3 b \sqrt {b \,x^{4}+a}}{5 a^{2} x}-\frac {3 i b^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 a^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) \(133\)

Input:

int(1/x^6/(b*x^4+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.34 \[ \int \frac {1}{x^6 \sqrt {a+b x^4}} \, dx=\frac {3 \, \sqrt {a} b x^{5} \left (-\frac {b}{a}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - 3 \, \sqrt {a} b x^{5} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) + {\left (3 \, b x^{4} - a\right )} \sqrt {b x^{4} + a}}{5 \, a^{2} x^{5}} \] Input:

integrate(1/x^6/(b*x^4+a)^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.60 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.17 \[ \int \frac {1}{x^6 \sqrt {a+b x^4}} \, dx=\frac {\Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} x^{5} \Gamma \left (- \frac {1}{4}\right )} \] Input:

integrate(1/x**6/(b*x**4+a)**(1/2),x)
 

Output:

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

Maxima [F]

\[ \int \frac {1}{x^6 \sqrt {a+b x^4}} \, dx=\int { \frac {1}{\sqrt {b x^{4} + a} x^{6}} \,d x } \] Input:

integrate(1/x^6/(b*x^4+a)^(1/2),x, algorithm="maxima")
 

Output:

integrate(1/(sqrt(b*x^4 + a)*x^6), x)
 

Giac [F]

\[ \int \frac {1}{x^6 \sqrt {a+b x^4}} \, dx=\int { \frac {1}{\sqrt {b x^{4} + a} x^{6}} \,d x } \] Input:

integrate(1/x^6/(b*x^4+a)^(1/2),x, algorithm="giac")
 

Output:

integrate(1/(sqrt(b*x^4 + a)*x^6), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^6 \sqrt {a+b x^4}} \, dx=\int \frac {1}{x^6\,\sqrt {b\,x^4+a}} \,d x \] Input:

int(1/(x^6*(a + b*x^4)^(1/2)),x)
 

Output:

int(1/(x^6*(a + b*x^4)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {1}{x^6 \sqrt {a+b x^4}} \, dx=\int \frac {\sqrt {b \,x^{4}+a}}{b \,x^{10}+a \,x^{6}}d x \] Input:

int(1/x^6/(b*x^4+a)^(1/2),x)
 

Output:

int(sqrt(a + b*x**4)/(a*x**6 + b*x**10),x)