\(\int \frac {1}{(a+b x^8)^{3/2}} \, dx\) [34]

Optimal result

Integrand size = 11, antiderivative size = 404 \[ \int \frac {1}{\left (a+b x^8\right )^{3/2}} \, dx=\frac {x}{4 a \sqrt {a+b x^8}}+\frac {3 \sqrt [4]{b} x^3 \sqrt {\frac {\left (\sqrt [4]{a}+\sqrt [4]{b} x^2\right )^2}{\sqrt [4]{a} \sqrt [4]{b} x^2}} \sqrt {-\frac {a+b x^8}{\sqrt {a} \sqrt {b} x^4}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {-\frac {\sqrt [4]{a} \left (\sqrt {2}-\frac {2 \sqrt [4]{b} x^2}{\sqrt [4]{a}}+\frac {\sqrt {2} \sqrt {b} x^4}{\sqrt {a}}\right )}{\sqrt [4]{b} x^2}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{8 \sqrt {2+\sqrt {2}} a \left (\sqrt [4]{a}+\sqrt [4]{b} x^2\right ) \sqrt {a+b x^8}}-\frac {3 \sqrt [4]{b} x^3 \sqrt {-\frac {\left (\sqrt [4]{a}-\sqrt [4]{b} x^2\right )^2}{\sqrt [4]{a} \sqrt [4]{b} x^2}} \sqrt {-\frac {a+b x^8}{\sqrt {a} \sqrt {b} x^4}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {\frac {\sqrt [4]{a} \left (\sqrt {2}+\frac {2 \sqrt [4]{b} x^2}{\sqrt [4]{a}}+\frac {\sqrt {2} \sqrt {b} x^4}{\sqrt {a}}\right )}{\sqrt [4]{b} x^2}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{8 \sqrt {2+\sqrt {2}} a \left (\sqrt [4]{a}-\sqrt [4]{b} x^2\right ) \sqrt {a+b x^8}} \] Output:

1/4*x/a/(b*x^8+a)^(1/2)+3/8*b^(1/4)*x^3*((a^(1/4)+b^(1/4)*x^2)^2/a^(1/4)/b 
^(1/4)/x^2)^(1/2)*(-(b*x^8+a)/a^(1/2)/b^(1/2)/x^4)^(1/2)*EllipticF(1/2*(-a 
^(1/4)*(2^(1/2)-2*b^(1/4)*x^2/a^(1/4)+2^(1/2)*b^(1/2)*x^4/a^(1/2))/b^(1/4) 
/x^2)^(1/2),(-2+2*2^(1/2))^(1/2))/(2+2^(1/2))^(1/2)/a/(a^(1/4)+b^(1/4)*x^2 
)/(b*x^8+a)^(1/2)-3/8*b^(1/4)*x^3*(-(a^(1/4)-b^(1/4)*x^2)^2/a^(1/4)/b^(1/4 
)/x^2)^(1/2)*(-(b*x^8+a)/a^(1/2)/b^(1/2)/x^4)^(1/2)*EllipticF(1/2*(a^(1/4) 
*(2^(1/2)+2*b^(1/4)*x^2/a^(1/4)+2^(1/2)*b^(1/2)*x^4/a^(1/2))/b^(1/4)/x^2)^ 
(1/2),(-2+2*2^(1/2))^(1/2))/(2+2^(1/2))^(1/2)/a/(a^(1/4)-b^(1/4)*x^2)/(b*x 
^8+a)^(1/2)
 

Mathematica [C] (verified)

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

Time = 10.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.14 \[ \int \frac {1}{\left (a+b x^8\right )^{3/2}} \, dx=\frac {x+3 x \sqrt {1+\frac {b x^8}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{2},\frac {9}{8},-\frac {b x^8}{a}\right )}{4 a \sqrt {a+b x^8}} \] Input:

Integrate[(a + b*x^8)^(-3/2),x]
 

Output:

(x + 3*x*Sqrt[1 + (b*x^8)/a]*Hypergeometric2F1[1/8, 1/2, 9/8, -((b*x^8)/a) 
])/(4*a*Sqrt[a + b*x^8])
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 408, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {749, 767, 27, 2422}

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.

\(\displaystyle \int \frac {1}{\left (a+b x^8\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 749

\(\displaystyle \frac {3 \int \frac {1}{\sqrt {b x^8+a}}dx}{4 a}+\frac {x}{4 a \sqrt {a+b x^8}}\)

\(\Big \downarrow \) 767

\(\displaystyle \frac {3 \left (\frac {1}{2} \int \frac {\sqrt [4]{a}-\sqrt [4]{b} x^2}{\sqrt [4]{a} \sqrt {b x^8+a}}dx+\frac {1}{2} \int \frac {\sqrt [4]{b} x^2+\sqrt [4]{a}}{\sqrt [4]{a} \sqrt {b x^8+a}}dx\right )}{4 a}+\frac {x}{4 a \sqrt {a+b x^8}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3 \left (\frac {\int \frac {\sqrt [4]{a}-\sqrt [4]{b} x^2}{\sqrt {b x^8+a}}dx}{2 \sqrt [4]{a}}+\frac {\int \frac {\sqrt [4]{b} x^2+\sqrt [4]{a}}{\sqrt {b x^8+a}}dx}{2 \sqrt [4]{a}}\right )}{4 a}+\frac {x}{4 a \sqrt {a+b x^8}}\)

\(\Big \downarrow \) 2422

\(\displaystyle \frac {3 \left (\frac {\sqrt [4]{b} x^3 \sqrt {\frac {\left (\sqrt [4]{a}+\sqrt [4]{b} x^2\right )^2}{\sqrt [4]{a} \sqrt [4]{b} x^2}} \sqrt {-\frac {a+b x^8}{\sqrt {a} \sqrt {b} x^4}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {-\frac {\sqrt {2} \sqrt {b} x^4-2 \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {2} \sqrt {a}}{\sqrt [4]{a} \sqrt [4]{b} x^2}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{2 \sqrt {2+\sqrt {2}} \left (\sqrt [4]{a}+\sqrt [4]{b} x^2\right ) \sqrt {a+b x^8}}-\frac {\sqrt [4]{b} x^3 \sqrt {-\frac {\left (\sqrt [4]{a}-\sqrt [4]{b} x^2\right )^2}{\sqrt [4]{a} \sqrt [4]{b} x^2}} \sqrt {-\frac {a+b x^8}{\sqrt {a} \sqrt {b} x^4}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {\frac {\sqrt {2} \sqrt {b} x^4+2 \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {2} \sqrt {a}}{\sqrt [4]{a} \sqrt [4]{b} x^2}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{2 \sqrt {2+\sqrt {2}} \left (\sqrt [4]{a}-\sqrt [4]{b} x^2\right ) \sqrt {a+b x^8}}\right )}{4 a}+\frac {x}{4 a \sqrt {a+b x^8}}\)

Input:

Int[(a + b*x^8)^(-3/2),x]
 

Output:

x/(4*a*Sqrt[a + b*x^8]) + (3*((b^(1/4)*x^3*Sqrt[(a^(1/4) + b^(1/4)*x^2)^2/ 
(a^(1/4)*b^(1/4)*x^2)]*Sqrt[-((a + b*x^8)/(Sqrt[a]*Sqrt[b]*x^4))]*Elliptic 
F[ArcSin[Sqrt[-((Sqrt[2]*Sqrt[a] - 2*a^(1/4)*b^(1/4)*x^2 + Sqrt[2]*Sqrt[b] 
*x^4)/(a^(1/4)*b^(1/4)*x^2))]/2], -2*(1 - Sqrt[2])])/(2*Sqrt[2 + Sqrt[2]]* 
(a^(1/4) + b^(1/4)*x^2)*Sqrt[a + b*x^8]) - (b^(1/4)*x^3*Sqrt[-((a^(1/4) - 
b^(1/4)*x^2)^2/(a^(1/4)*b^(1/4)*x^2))]*Sqrt[-((a + b*x^8)/(Sqrt[a]*Sqrt[b] 
*x^4))]*EllipticF[ArcSin[Sqrt[(Sqrt[2]*Sqrt[a] + 2*a^(1/4)*b^(1/4)*x^2 + S 
qrt[2]*Sqrt[b]*x^4)/(a^(1/4)*b^(1/4)*x^2)]/2], -2*(1 - Sqrt[2])])/(2*Sqrt[ 
2 + Sqrt[2]]*(a^(1/4) - b^(1/4)*x^2)*Sqrt[a + b*x^8])))/(4*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 749

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 
 1)/(a*n*(p + 1))), x] + Simp[(n*(p + 1) + 1)/(a*n*(p + 1))   Int[(a + b*x^ 
n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (Inte 
gerQ[2*p] || Denominator[p + 1/n] < Denominator[p])
 

rule 767

Int[1/Sqrt[(a_) + (b_.)*(x_)^8], x_Symbol] :> Simp[1/2   Int[(1 - Rt[b/a, 4 
]*x^2)/Sqrt[a + b*x^8], x], x] + Simp[1/2   Int[(1 + Rt[b/a, 4]*x^2)/Sqrt[a 
 + b*x^8], x], x] /; FreeQ[{a, b}, x]
 

rule 2422

Int[((c_) + (d_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^8], x_Symbol] :> Simp[(-c) 
*d*x^3*Sqrt[-(c - d*x^2)^2/(c*d*x^2)]*(Sqrt[(-d^2)*((a + b*x^8)/(b*c^2*x^4) 
)]/(Sqrt[2 + Sqrt[2]]*(c - d*x^2)*Sqrt[a + b*x^8]))*EllipticF[ArcSin[(1/2)* 
Sqrt[(Sqrt[2]*c^2 + 2*c*d*x^2 + Sqrt[2]*d^2*x^4)/(c*d*x^2)]], -2*(1 - Sqrt[ 
2])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^4 - a*d^4, 0]
 

Maple [F]

\[\int \frac {1}{\left (b \,x^{8}+a \right )^{\frac {3}{2}}}d x\]

Input:

int(1/(b*x^8+a)^(3/2),x)
 

Output:

int(1/(b*x^8+a)^(3/2),x)
 

Fricas [F]

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

integrate(1/(b*x^8+a)^(3/2),x, algorithm="fricas")
 

Output:

integral(sqrt(b*x^8 + a)/(b^2*x^16 + 2*a*b*x^8 + a^2), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.52 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.09 \[ \int \frac {1}{\left (a+b x^8\right )^{3/2}} \, dx=\frac {x \Gamma \left (\frac {1}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{8}, \frac {3}{2} \\ \frac {9}{8} \end {matrix}\middle | {\frac {b x^{8} e^{i \pi }}{a}} \right )}}{8 a^{\frac {3}{2}} \Gamma \left (\frac {9}{8}\right )} \] Input:

integrate(1/(b*x**8+a)**(3/2),x)
 

Output:

x*gamma(1/8)*hyper((1/8, 3/2), (9/8,), b*x**8*exp_polar(I*pi)/a)/(8*a**(3/ 
2)*gamma(9/8))
 

Maxima [F]

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

integrate(1/(b*x^8+a)^(3/2),x, algorithm="maxima")
 

Output:

integrate((b*x^8 + a)^(-3/2), x)
 

Giac [F]

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

integrate(1/(b*x^8+a)^(3/2),x, algorithm="giac")
 

Output:

integrate((b*x^8 + a)^(-3/2), x)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.09 \[ \int \frac {1}{\left (a+b x^8\right )^{3/2}} \, dx=\frac {x\,{\left (\frac {b\,x^8}{a}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{8},\frac {3}{2};\ \frac {9}{8};\ -\frac {b\,x^8}{a}\right )}{{\left (b\,x^8+a\right )}^{3/2}} \] Input:

int(1/(a + b*x^8)^(3/2),x)
 

Output:

(x*((b*x^8)/a + 1)^(3/2)*hypergeom([1/8, 3/2], 9/8, -(b*x^8)/a))/(a + b*x^ 
8)^(3/2)
 

Reduce [F]

\[ \int \frac {1}{\left (a+b x^8\right )^{3/2}} \, dx=\int \frac {\sqrt {b \,x^{8}+a}}{b^{2} x^{16}+2 a b \,x^{8}+a^{2}}d x \] Input:

int(1/(b*x^8+a)^(3/2),x)
 

Output:

int(sqrt(a + b*x**8)/(a**2 + 2*a*b*x**8 + b**2*x**16),x)