Integrand size = 37, antiderivative size = 118 \[ \int \frac {\sqrt {b+a^2 x^4}}{\sqrt {a x^2+\sqrt {b+a^2 x^4}}} \, dx=\frac {b x}{8 \left (a x^2+\sqrt {b+a^2 x^4}\right )^{3/2}}+\frac {1}{4} x \sqrt {a x^2+\sqrt {b+a^2 x^4}}+\frac {5 \sqrt {b} \arctan \left (\frac {\sqrt {2} \sqrt {a} x \sqrt {a x^2+\sqrt {b+a^2 x^4}}}{\sqrt {b}}\right )}{8 \sqrt {2} \sqrt {a}} \]
1/8*b*x/(a*x^2+(a^2*x^4+b)^(1/2))^(3/2)+1/4*x*(a*x^2+(a^2*x^4+b)^(1/2))^(1 /2)+5/16*b^(1/2)*arctan(2^(1/2)*a^(1/2)*x*(a*x^2+(a^2*x^4+b)^(1/2))^(1/2)/ b^(1/2))*2^(1/2)/a^(1/2)
Time = 0.28 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.97 \[ \int \frac {\sqrt {b+a^2 x^4}}{\sqrt {a x^2+\sqrt {b+a^2 x^4}}} \, dx=\frac {1}{16} \left (\frac {2 x \left (b+2 \left (a x^2+\sqrt {b+a^2 x^4}\right )^2\right )}{\left (a x^2+\sqrt {b+a^2 x^4}\right )^{3/2}}+\frac {5 \sqrt {2} \sqrt {b} \arctan \left (\frac {\sqrt {2} \sqrt {a} x \sqrt {a x^2+\sqrt {b+a^2 x^4}}}{\sqrt {b}}\right )}{\sqrt {a}}\right ) \]
((2*x*(b + 2*(a*x^2 + Sqrt[b + a^2*x^4])^2))/(a*x^2 + Sqrt[b + a^2*x^4])^( 3/2) + (5*Sqrt[2]*Sqrt[b]*ArcTan[(Sqrt[2]*Sqrt[a]*x*Sqrt[a*x^2 + Sqrt[b + a^2*x^4]])/Sqrt[b]])/Sqrt[a])/16
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.
\(\displaystyle \int \frac {\sqrt {a^2 x^4+b}}{\sqrt {\sqrt {a^2 x^4+b}+a x^2}} \, dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \frac {\sqrt {a^2 x^4+b}}{\sqrt {\sqrt {a^2 x^4+b}+a x^2}}dx\) |
3.18.59.3.1 Defintions of rubi rules used
\[\int \frac {\sqrt {a^{2} x^{4}+b}}{\sqrt {a \,x^{2}+\sqrt {a^{2} x^{4}+b}}}d x\]
Time = 2.61 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.70 \[ \int \frac {\sqrt {b+a^2 x^4}}{\sqrt {a x^2+\sqrt {b+a^2 x^4}}} \, dx=\left [\frac {5 \, \sqrt {\frac {1}{2}} b \sqrt {-\frac {b}{a}} \log \left (4 \, a^{2} b x^{4} - 4 \, \sqrt {a^{2} x^{4} + b} a b x^{2} + b^{2} + 4 \, {\left (2 \, \sqrt {\frac {1}{2}} \sqrt {a^{2} x^{4} + b} a^{2} x^{3} \sqrt {-\frac {b}{a}} - \sqrt {\frac {1}{2}} {\left (2 \, a^{3} x^{5} + a b x\right )} \sqrt {-\frac {b}{a}}\right )} \sqrt {a x^{2} + \sqrt {a^{2} x^{4} + b}}\right ) + 2 \, {\left (2 \, a^{2} x^{5} - 2 \, \sqrt {a^{2} x^{4} + b} a x^{3} + 3 \, b x\right )} \sqrt {a x^{2} + \sqrt {a^{2} x^{4} + b}}}{16 \, b}, -\frac {5 \, \sqrt {\frac {1}{2}} b \sqrt {\frac {b}{a}} \arctan \left (-\frac {{\left (\sqrt {\frac {1}{2}} a x^{2} \sqrt {\frac {b}{a}} - \sqrt {\frac {1}{2}} \sqrt {a^{2} x^{4} + b} \sqrt {\frac {b}{a}}\right )} \sqrt {a x^{2} + \sqrt {a^{2} x^{4} + b}}}{b x}\right ) - {\left (2 \, a^{2} x^{5} - 2 \, \sqrt {a^{2} x^{4} + b} a x^{3} + 3 \, b x\right )} \sqrt {a x^{2} + \sqrt {a^{2} x^{4} + b}}}{8 \, b}\right ] \]
[1/16*(5*sqrt(1/2)*b*sqrt(-b/a)*log(4*a^2*b*x^4 - 4*sqrt(a^2*x^4 + b)*a*b* x^2 + b^2 + 4*(2*sqrt(1/2)*sqrt(a^2*x^4 + b)*a^2*x^3*sqrt(-b/a) - sqrt(1/2 )*(2*a^3*x^5 + a*b*x)*sqrt(-b/a))*sqrt(a*x^2 + sqrt(a^2*x^4 + b))) + 2*(2* a^2*x^5 - 2*sqrt(a^2*x^4 + b)*a*x^3 + 3*b*x)*sqrt(a*x^2 + sqrt(a^2*x^4 + b )))/b, -1/8*(5*sqrt(1/2)*b*sqrt(b/a)*arctan(-(sqrt(1/2)*a*x^2*sqrt(b/a) - sqrt(1/2)*sqrt(a^2*x^4 + b)*sqrt(b/a))*sqrt(a*x^2 + sqrt(a^2*x^4 + b))/(b* x)) - (2*a^2*x^5 - 2*sqrt(a^2*x^4 + b)*a*x^3 + 3*b*x)*sqrt(a*x^2 + sqrt(a^ 2*x^4 + b)))/b]
\[ \int \frac {\sqrt {b+a^2 x^4}}{\sqrt {a x^2+\sqrt {b+a^2 x^4}}} \, dx=\int \frac {\sqrt {a^{2} x^{4} + b}}{\sqrt {a x^{2} + \sqrt {a^{2} x^{4} + b}}}\, dx \]
\[ \int \frac {\sqrt {b+a^2 x^4}}{\sqrt {a x^2+\sqrt {b+a^2 x^4}}} \, dx=\int { \frac {\sqrt {a^{2} x^{4} + b}}{\sqrt {a x^{2} + \sqrt {a^{2} x^{4} + b}}} \,d x } \]
\[ \int \frac {\sqrt {b+a^2 x^4}}{\sqrt {a x^2+\sqrt {b+a^2 x^4}}} \, dx=\int { \frac {\sqrt {a^{2} x^{4} + b}}{\sqrt {a x^{2} + \sqrt {a^{2} x^{4} + b}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {b+a^2 x^4}}{\sqrt {a x^2+\sqrt {b+a^2 x^4}}} \, dx=\int \frac {\sqrt {a^2\,x^4+b}}{\sqrt {\sqrt {a^2\,x^4+b}+a\,x^2}} \,d x \]