Integrand size = 44, antiderivative size = 313 \[ \int \frac {b^8+a^8 x^8}{\sqrt {-b^4+a^4 x^4} \left (-b^8+a^8 x^8\right )} \, dx=-\frac {x}{2 \sqrt {-b^4+a^4 x^4}}-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \arctan \left (\frac {(1+i) a b x}{i b^2+a^2 x^2+\sqrt {-b^4+a^4 x^4}}\right )}{a b}+\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \text {arctanh}\left (\frac {\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) b}{\sqrt {3-2 \sqrt {2}} a}+\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) a x^2}{\sqrt {3-2 \sqrt {2}} b}+\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {-b^4+a^4 x^4}}{\sqrt {3-2 \sqrt {2}} a b}}{x}\right )}{a b}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \text {arctanh}\left (\frac {\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) b}{\sqrt {3+2 \sqrt {2}} a}+\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) a x^2}{\sqrt {3+2 \sqrt {2}} b}+\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {-b^4+a^4 x^4}}{\sqrt {3+2 \sqrt {2}} a b}}{x}\right )}{a b} \]
-1/2*x/(a^4*x^4-b^4)^(1/2)+(-1/4+1/4*I)*arctan((1+I)*a*b*x/(I*b^2+a^2*x^2+ (a^4*x^4-b^4)^(1/2)))/a/b+(1/8-1/8*I)*arctanh(((1/2+1/2*I)*b/(2^(1/2)-1)/a +(1/2-1/2*I)*a*x^2/(2^(1/2)-1)/b+(1/2-1/2*I)*(a^4*x^4-b^4)^(1/2)/(2^(1/2)- 1)/a/b)/x)/a/b+(-1/8+1/8*I)*arctanh(((1/2+1/2*I)*b/(1+2^(1/2))/a+(1/2-1/2* I)*a*x^2/(1+2^(1/2))/b+(1/2-1/2*I)*(a^4*x^4-b^4)^(1/2)/(1+2^(1/2))/a/b)/x) /a/b
Time = 0.99 (sec) , antiderivative size = 245, normalized size of antiderivative = 0.78 \[ \int \frac {b^8+a^8 x^8}{\sqrt {-b^4+a^4 x^4} \left (-b^8+a^8 x^8\right )} \, dx=\frac {1}{8} \left (-\frac {4 x}{\sqrt {-b^4+a^4 x^4}}-\frac {(2-2 i) \arctan \left (\frac {(1+i) a b x}{i b^2+a^2 x^2+\sqrt {-b^4+a^4 x^4}}\right )}{a b}-\frac {(1+i) \arctan \left (\frac {i b^4+(1-i) a b^3 x-(1+i) a^3 b x^3-i a^4 x^4+\left (b^2-(1+i) a b x-i a^2 x^2\right ) \sqrt {-b^4+a^4 x^4}}{i b^4-(1-i) a b^3 x+(1+i) a^3 b x^3-i a^4 x^4+\left (b^2+(1+i) a b x-i a^2 x^2\right ) \sqrt {-b^4+a^4 x^4}}\right )}{a b}\right ) \]
((-4*x)/Sqrt[-b^4 + a^4*x^4] - ((2 - 2*I)*ArcTan[((1 + I)*a*b*x)/(I*b^2 + a^2*x^2 + Sqrt[-b^4 + a^4*x^4])])/(a*b) - ((1 + I)*ArcTan[(I*b^4 + (1 - I) *a*b^3*x - (1 + I)*a^3*b*x^3 - I*a^4*x^4 + (b^2 - (1 + I)*a*b*x - I*a^2*x^ 2)*Sqrt[-b^4 + a^4*x^4])/(I*b^4 - (1 - I)*a*b^3*x + (1 + I)*a^3*b*x^3 - I* a^4*x^4 + (b^2 + (1 + I)*a*b*x - I*a^2*x^2)*Sqrt[-b^4 + a^4*x^4])])/(a*b)) /8
Time = 0.95 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.43, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {1388, 7276, 2009}
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.
| \(\displaystyle \int \frac {a^8 x^8+b^8}{\sqrt {a^4 x^4-b^4} \left (a^8 x^8-b^8\right )} \, dx\) |
| \(\Big \downarrow \) 1388 |
| \(\displaystyle \int \frac {a^8 x^8+b^8}{\left (a^4 x^4-b^4\right )^{3/2} \left (a^4 x^4+b^4\right )}dx\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle \int \left (-\frac {b^4}{\left (a^4 x^4-b^4\right )^{3/2}}+\frac {a^4 x^4}{\left (a^4 x^4-b^4\right )^{3/2}}+\frac {2 b^8}{\left (a^4 x^4-b^4\right )^{3/2} \left (a^4 x^4+b^4\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{-a^4} b x}{\sqrt {a^4 x^4-b^4}}\right )}{4 \sqrt {2} \sqrt [4]{-a^4} b}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{-a^4} b x}{\sqrt {a^4 x^4-b^4}}\right )}{4 \sqrt {2} \sqrt [4]{-a^4} b}-\frac {x}{2 \sqrt {a^4 x^4-b^4}}\) |
-1/2*x/Sqrt[-b^4 + a^4*x^4] - ArcTan[(Sqrt[2]*(-a^4)^(1/4)*b*x)/Sqrt[-b^4 + a^4*x^4]]/(4*Sqrt[2]*(-a^4)^(1/4)*b) - ArcTanh[(Sqrt[2]*(-a^4)^(1/4)*b*x )/Sqrt[-b^4 + a^4*x^4]]/(4*Sqrt[2]*(-a^4)^(1/4)*b)
3.29.87.3.1 Defintions of rubi rules used
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 5.00 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.72
| method | result | size |
| elliptic | \(\frac {\left (-\frac {\sqrt {2}\, x}{2 \sqrt {a^{4} x^{4}-b^{4}}}+\frac {\sqrt {2}\, \left (\ln \left (\frac {\frac {a^{4} x^{4}-b^{4}}{2 x^{2}}-\frac {\left (a^{4} b^{4}\right )^{\frac {1}{4}} \sqrt {a^{4} x^{4}-b^{4}}}{x}+\sqrt {a^{4} b^{4}}}{\frac {a^{4} x^{4}-b^{4}}{2 x^{2}}+\frac {\left (a^{4} b^{4}\right )^{\frac {1}{4}} \sqrt {a^{4} x^{4}-b^{4}}}{x}+\sqrt {a^{4} b^{4}}}\right )+2 \arctan \left (\frac {\sqrt {a^{4} x^{4}-b^{4}}}{\left (a^{4} b^{4}\right )^{\frac {1}{4}} x}+1\right )+2 \arctan \left (\frac {\sqrt {a^{4} x^{4}-b^{4}}}{\left (a^{4} b^{4}\right )^{\frac {1}{4}} x}-1\right )\right )}{16 \left (a^{4} b^{4}\right )^{\frac {1}{4}}}\right ) \sqrt {2}}{2}\) | \(226\) |
| default | \(-\frac {i a^{2} \left (\left (\sqrt {i a^{2} b^{2}}\, \sqrt {a^{4} x^{4}-b^{4}}\, x +\frac {\left (\ln \left (\frac {a^{2} \left (-2 i a^{2} b^{2} x +2 \sqrt {i a^{2} b^{2}}\, a^{2} x^{2}+2 i \sqrt {i a^{2} b^{2}}\, b^{2}+\sqrt {2}\, \sqrt {i a^{2} b^{2}}\, \sqrt {a^{4} x^{4}-b^{4}}\right )}{a^{2} x^{2}+i b^{2}-2 \sqrt {i a^{2} b^{2}}\, x}\right )+\ln \left (-\frac {2 a^{2} \left (-\frac {\sqrt {2}\, \sqrt {i a^{2} b^{2}}\, \sqrt {a^{4} x^{4}-b^{4}}}{2}+\left (a^{2} x^{2}+i b^{2}\right ) \sqrt {i a^{2} b^{2}}+i a^{2} b^{2} x \right )}{a^{2} x^{2}+i b^{2}+2 \sqrt {i a^{2} b^{2}}\, x}\right )+2 \ln \left (2\right )\right ) \left (a^{2} x^{2}+b^{2}\right ) \left (a x -b \right ) \sqrt {2}\, \left (a x +b \right )}{8}\right ) \sqrt {-i a^{2} b^{2}}+\frac {\left (a^{2} x^{2}+b^{2}\right ) \left (a x -b \right ) \left (\ln \left (\frac {\left (-2 i a^{2} b^{2} x +\sqrt {2}\, \sqrt {-i a^{2} b^{2}}\, \sqrt {a^{4} x^{4}-b^{4}}\right ) a^{2}}{a^{2} x^{2}+i b^{2}}\right )+\ln \left (2\right )\right ) \sqrt {2}\, \left (a x +b \right ) \sqrt {i a^{2} b^{2}}}{4}\right ) b^{2}}{\sqrt {i a^{2} b^{2}}\, \sqrt {-i a^{2} b^{2}}\, \left (-a x +i b \right ) \left (-2 \sqrt {i a^{2} b^{2}}+\left (1+i\right ) a b \right ) \left (a x -b \right ) \left (a x +b \right ) \left (a x +i b \right ) \left (2 \sqrt {i a^{2} b^{2}}+\left (1+i\right ) a b \right )}\) | \(508\) |
| pseudoelliptic | \(-\frac {i a^{2} \left (\left (\sqrt {i a^{2} b^{2}}\, \sqrt {a^{4} x^{4}-b^{4}}\, x +\frac {\left (\ln \left (\frac {a^{2} \left (-2 i a^{2} b^{2} x +2 \sqrt {i a^{2} b^{2}}\, a^{2} x^{2}+2 i \sqrt {i a^{2} b^{2}}\, b^{2}+\sqrt {2}\, \sqrt {i a^{2} b^{2}}\, \sqrt {a^{4} x^{4}-b^{4}}\right )}{a^{2} x^{2}+i b^{2}-2 \sqrt {i a^{2} b^{2}}\, x}\right )+\ln \left (-\frac {2 a^{2} \left (-\frac {\sqrt {2}\, \sqrt {i a^{2} b^{2}}\, \sqrt {a^{4} x^{4}-b^{4}}}{2}+\left (a^{2} x^{2}+i b^{2}\right ) \sqrt {i a^{2} b^{2}}+i a^{2} b^{2} x \right )}{a^{2} x^{2}+i b^{2}+2 \sqrt {i a^{2} b^{2}}\, x}\right )+2 \ln \left (2\right )\right ) \left (a^{2} x^{2}+b^{2}\right ) \left (a x -b \right ) \sqrt {2}\, \left (a x +b \right )}{8}\right ) \sqrt {-i a^{2} b^{2}}+\frac {\left (a^{2} x^{2}+b^{2}\right ) \left (a x -b \right ) \left (\ln \left (\frac {\left (-2 i a^{2} b^{2} x +\sqrt {2}\, \sqrt {-i a^{2} b^{2}}\, \sqrt {a^{4} x^{4}-b^{4}}\right ) a^{2}}{a^{2} x^{2}+i b^{2}}\right )+\ln \left (2\right )\right ) \sqrt {2}\, \left (a x +b \right ) \sqrt {i a^{2} b^{2}}}{4}\right ) b^{2}}{\sqrt {i a^{2} b^{2}}\, \sqrt {-i a^{2} b^{2}}\, \left (-a x +i b \right ) \left (-2 \sqrt {i a^{2} b^{2}}+\left (1+i\right ) a b \right ) \left (a x -b \right ) \left (a x +b \right ) \left (a x +i b \right ) \left (2 \sqrt {i a^{2} b^{2}}+\left (1+i\right ) a b \right )}\) | \(508\) |
1/2*(-1/2/(a^4*x^4-b^4)^(1/2)*2^(1/2)*x+1/16/(a^4*b^4)^(1/4)*2^(1/2)*(ln(( 1/2*(a^4*x^4-b^4)/x^2-(a^4*b^4)^(1/4)*(a^4*x^4-b^4)^(1/2)/x+(a^4*b^4)^(1/2 ))/(1/2*(a^4*x^4-b^4)/x^2+(a^4*b^4)^(1/4)*(a^4*x^4-b^4)^(1/2)/x+(a^4*b^4)^ (1/2)))+2*arctan(1/(a^4*b^4)^(1/4)*(a^4*x^4-b^4)^(1/2)/x+1)+2*arctan(1/(a^ 4*b^4)^(1/4)*(a^4*x^4-b^4)^(1/2)/x-1)))*2^(1/2)
Time = 0.50 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.52 \[ \int \frac {b^8+a^8 x^8}{\sqrt {-b^4+a^4 x^4} \left (-b^8+a^8 x^8\right )} \, dx=-\frac {4 \, \sqrt {a^{4} x^{4} - b^{4}} a b x - 2 \, {\left (a^{4} x^{4} - b^{4}\right )} \arctan \left (\frac {\sqrt {a^{4} x^{4} - b^{4}} a x}{a^{2} b x^{2} + b^{3}}\right ) - {\left (a^{4} x^{4} - b^{4}\right )} \log \left (\frac {a^{4} x^{4} + 2 \, a^{2} b^{2} x^{2} - b^{4} - 2 \, \sqrt {a^{4} x^{4} - b^{4}} a b x}{a^{4} x^{4} + b^{4}}\right )}{8 \, {\left (a^{5} b x^{4} - a b^{5}\right )}} \]
-1/8*(4*sqrt(a^4*x^4 - b^4)*a*b*x - 2*(a^4*x^4 - b^4)*arctan(sqrt(a^4*x^4 - b^4)*a*x/(a^2*b*x^2 + b^3)) - (a^4*x^4 - b^4)*log((a^4*x^4 + 2*a^2*b^2*x ^2 - b^4 - 2*sqrt(a^4*x^4 - b^4)*a*b*x)/(a^4*x^4 + b^4)))/(a^5*b*x^4 - a*b ^5)
\[ \int \frac {b^8+a^8 x^8}{\sqrt {-b^4+a^4 x^4} \left (-b^8+a^8 x^8\right )} \, dx=\int \frac {a^{8} x^{8} + b^{8}}{\sqrt {\left (a x - b\right ) \left (a x + b\right ) \left (a^{2} x^{2} + b^{2}\right )} \left (a x - b\right ) \left (a x + b\right ) \left (a^{2} x^{2} + b^{2}\right ) \left (a^{4} x^{4} + b^{4}\right )}\, dx \]
Integral((a**8*x**8 + b**8)/(sqrt((a*x - b)*(a*x + b)*(a**2*x**2 + b**2))* (a*x - b)*(a*x + b)*(a**2*x**2 + b**2)*(a**4*x**4 + b**4)), x)
\[ \int \frac {b^8+a^8 x^8}{\sqrt {-b^4+a^4 x^4} \left (-b^8+a^8 x^8\right )} \, dx=\int { \frac {a^{8} x^{8} + b^{8}}{{\left (a^{8} x^{8} - b^{8}\right )} \sqrt {a^{4} x^{4} - b^{4}}} \,d x } \]
\[ \int \frac {b^8+a^8 x^8}{\sqrt {-b^4+a^4 x^4} \left (-b^8+a^8 x^8\right )} \, dx=\int { \frac {a^{8} x^{8} + b^{8}}{{\left (a^{8} x^{8} - b^{8}\right )} \sqrt {a^{4} x^{4} - b^{4}}} \,d x } \]
Timed out. \[ \int \frac {b^8+a^8 x^8}{\sqrt {-b^4+a^4 x^4} \left (-b^8+a^8 x^8\right )} \, dx=\int -\frac {a^8\,x^8+b^8}{\sqrt {a^4\,x^4-b^4}\,\left (b^8-a^8\,x^8\right )} \,d x \]