Integrand size = 11, antiderivative size = 379 \[ \int \frac {1}{\sqrt {a+b x^8}} \, dx=\frac {\sqrt [4]{b} x^3 \sqrt {\frac {\left (\sqrt [4]{a}+\sqrt [4]{b} x^2\right )^2}{\sqrt [4]{a} \sqrt [4]{b} x^2}} \sqrt {-\frac {a+b x^8}{\sqrt {a} \sqrt {b} x^4}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {-\frac {\sqrt [4]{a} \left (\sqrt {2}-\frac {2 \sqrt [4]{b} x^2}{\sqrt [4]{a}}+\frac {\sqrt {2} \sqrt {b} x^4}{\sqrt {a}}\right )}{\sqrt [4]{b} x^2}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{2 \sqrt {2+\sqrt {2}} \left (\sqrt [4]{a}+\sqrt [4]{b} x^2\right ) \sqrt {a+b x^8}}-\frac {\sqrt [4]{b} x^3 \sqrt {-\frac {\left (\sqrt [4]{a}-\sqrt [4]{b} x^2\right )^2}{\sqrt [4]{a} \sqrt [4]{b} x^2}} \sqrt {-\frac {a+b x^8}{\sqrt {a} \sqrt {b} x^4}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {\frac {\sqrt [4]{a} \left (\sqrt {2}+\frac {2 \sqrt [4]{b} x^2}{\sqrt [4]{a}}+\frac {\sqrt {2} \sqrt {b} x^4}{\sqrt {a}}\right )}{\sqrt [4]{b} x^2}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{2 \sqrt {2+\sqrt {2}} \left (\sqrt [4]{a}-\sqrt [4]{b} x^2\right ) \sqrt {a+b x^8}} \] Output:
1/2*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^(1/4)+b^(1/4)*x^2)/(b*x^8+a)^(1/2)-1/2*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^(1/4)-b^(1/4)*x^2)/(b*x^8+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.12 \[ \int \frac {1}{\sqrt {a+b x^8}} \, dx=\frac {x \sqrt {1+\frac {b x^8}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{2},\frac {9}{8},-\frac {b x^8}{a}\right )}{\sqrt {a+b x^8}} \] Input:
Integrate[1/Sqrt[a + b*x^8],x]
Output:
(x*Sqrt[1 + (b*x^8)/a]*Hypergeometric2F1[1/8, 1/2, 9/8, -((b*x^8)/a)])/Sqr t[a + b*x^8]
Time = 0.58 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {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.
\(\displaystyle \int \frac {1}{\sqrt {a+b x^8}} \, dx\) |
\(\Big \downarrow \) 767 |
\(\displaystyle \frac {1}{2} \int \frac {\sqrt [4]{a}-\sqrt [4]{b} x^2}{\sqrt [4]{a} \sqrt {b x^8+a}}dx+\frac {1}{2} \int \frac {\sqrt [4]{b} x^2+\sqrt [4]{a}}{\sqrt [4]{a} \sqrt {b x^8+a}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sqrt [4]{a}-\sqrt [4]{b} x^2}{\sqrt {b x^8+a}}dx}{2 \sqrt [4]{a}}+\frac {\int \frac {\sqrt [4]{b} x^2+\sqrt [4]{a}}{\sqrt {b x^8+a}}dx}{2 \sqrt [4]{a}}\) |
\(\Big \downarrow \) 2422 |
\(\displaystyle \frac {\sqrt [4]{b} x^3 \sqrt {\frac {\left (\sqrt [4]{a}+\sqrt [4]{b} x^2\right )^2}{\sqrt [4]{a} \sqrt [4]{b} x^2}} \sqrt {-\frac {a+b x^8}{\sqrt {a} \sqrt {b} x^4}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {-\frac {\sqrt {2} \sqrt {b} x^4-2 \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {2} \sqrt {a}}{\sqrt [4]{a} \sqrt [4]{b} x^2}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{2 \sqrt {2+\sqrt {2}} \left (\sqrt [4]{a}+\sqrt [4]{b} x^2\right ) \sqrt {a+b x^8}}-\frac {\sqrt [4]{b} x^3 \sqrt {-\frac {\left (\sqrt [4]{a}-\sqrt [4]{b} x^2\right )^2}{\sqrt [4]{a} \sqrt [4]{b} x^2}} \sqrt {-\frac {a+b x^8}{\sqrt {a} \sqrt {b} x^4}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {\frac {\sqrt {2} \sqrt {b} x^4+2 \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {2} \sqrt {a}}{\sqrt [4]{a} \sqrt [4]{b} x^2}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{2 \sqrt {2+\sqrt {2}} \left (\sqrt [4]{a}-\sqrt [4]{b} x^2\right ) \sqrt {a+b x^8}}\) |
Input:
Int[1/Sqrt[a + b*x^8],x]
Output:
(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 + 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 + 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])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
\[\int \frac {1}{\sqrt {b \,x^{8}+a}}d x\]
Input:
int(1/(b*x^8+a)^(1/2),x)
Output:
int(1/(b*x^8+a)^(1/2),x)
\[ \int \frac {1}{\sqrt {a+b x^8}} \, dx=\int { \frac {1}{\sqrt {b x^{8} + a}} \,d x } \] Input:
integrate(1/(b*x^8+a)^(1/2),x, algorithm="fricas")
Output:
integral(1/sqrt(b*x^8 + a), x)
Result contains complex when optimal does not.
Time = 0.45 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.09 \[ \int \frac {1}{\sqrt {a+b x^8}} \, dx=\frac {x \Gamma \left (\frac {1}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{8}, \frac {1}{2} \\ \frac {9}{8} \end {matrix}\middle | {\frac {b x^{8} e^{i \pi }}{a}} \right )}}{8 \sqrt {a} \Gamma \left (\frac {9}{8}\right )} \] Input:
integrate(1/(b*x**8+a)**(1/2),x)
Output:
x*gamma(1/8)*hyper((1/8, 1/2), (9/8,), b*x**8*exp_polar(I*pi)/a)/(8*sqrt(a )*gamma(9/8))
\[ \int \frac {1}{\sqrt {a+b x^8}} \, dx=\int { \frac {1}{\sqrt {b x^{8} + a}} \,d x } \] Input:
integrate(1/(b*x^8+a)^(1/2),x, algorithm="maxima")
Output:
integrate(1/sqrt(b*x^8 + a), x)
\[ \int \frac {1}{\sqrt {a+b x^8}} \, dx=\int { \frac {1}{\sqrt {b x^{8} + a}} \,d x } \] Input:
integrate(1/(b*x^8+a)^(1/2),x, algorithm="giac")
Output:
integrate(1/sqrt(b*x^8 + a), x)
Time = 0.15 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.10 \[ \int \frac {1}{\sqrt {a+b x^8}} \, dx=\frac {x\,\sqrt {\frac {b\,x^8}{a}+1}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{8},\frac {1}{2};\ \frac {9}{8};\ -\frac {b\,x^8}{a}\right )}{\sqrt {b\,x^8+a}} \] Input:
int(1/(a + b*x^8)^(1/2),x)
Output:
(x*((b*x^8)/a + 1)^(1/2)*hypergeom([1/8, 1/2], 9/8, -(b*x^8)/a))/(a + b*x^ 8)^(1/2)
\[ \int \frac {1}{\sqrt {a+b x^8}} \, dx=\int \frac {\sqrt {b \,x^{8}+a}}{b \,x^{8}+a}d x \] Input:
int(1/(b*x^8+a)^(1/2),x)
Output:
int(sqrt(a + b*x**8)/(a + b*x**8),x)