Integrand size = 31, antiderivative size = 171 \[ \int \frac {\sqrt {b+\sqrt {b^2+a x^2}}}{\left (b^2+a x^2\right )^4} \, dx=\frac {11 x \left (611 b^4+798 a b^2 x^2+315 a^2 x^4\right )}{7680 b^6 \left (b^2+a x^2\right )^{5/2} \sqrt {b+\sqrt {b^2+a x^2}}}+\frac {x \left (2587 b^4+3102 a b^2 x^2+1155 a^2 x^4\right )}{3840 b^5 \left (b^2+a x^2\right )^3 \sqrt {b+\sqrt {b^2+a x^2}}}+\frac {231 \arctan \left (\frac {\sqrt {a} x}{\sqrt {b} \sqrt {b+\sqrt {b^2+a x^2}}}\right )}{512 \sqrt {a} b^{13/2}} \]
11/7680*x*(315*a^2*x^4+798*a*b^2*x^2+611*b^4)/b^6/(a*x^2+b^2)^(5/2)/(b+(a* x^2+b^2)^(1/2))^(1/2)+1/3840*x*(1155*a^2*x^4+3102*a*b^2*x^2+2587*b^4)/b^5/ (a*x^2+b^2)^3/(b+(a*x^2+b^2)^(1/2))^(1/2)+231/512*arctan(a^(1/2)*x/b^(1/2) /(b+(a*x^2+b^2)^(1/2))^(1/2))/a^(1/2)/b^(13/2)
Time = 0.33 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {b+\sqrt {b^2+a x^2}}}{\left (b^2+a x^2\right )^4} \, dx=\frac {11 x \left (611 b^4+798 a b^2 x^2+315 a^2 x^4\right )}{7680 b^6 \left (b^2+a x^2\right )^{5/2} \sqrt {b+\sqrt {b^2+a x^2}}}+\frac {x \left (2587 b^4+3102 a b^2 x^2+1155 a^2 x^4\right )}{3840 b^5 \left (b^2+a x^2\right )^3 \sqrt {b+\sqrt {b^2+a x^2}}}+\frac {231 \arctan \left (\frac {\sqrt {a} x}{\sqrt {b} \sqrt {b+\sqrt {b^2+a x^2}}}\right )}{512 \sqrt {a} b^{13/2}} \]
(11*x*(611*b^4 + 798*a*b^2*x^2 + 315*a^2*x^4))/(7680*b^6*(b^2 + a*x^2)^(5/ 2)*Sqrt[b + Sqrt[b^2 + a*x^2]]) + (x*(2587*b^4 + 3102*a*b^2*x^2 + 1155*a^2 *x^4))/(3840*b^5*(b^2 + a*x^2)^3*Sqrt[b + Sqrt[b^2 + a*x^2]]) + (231*ArcTa n[(Sqrt[a]*x)/(Sqrt[b]*Sqrt[b + Sqrt[b^2 + a*x^2]])])/(512*Sqrt[a]*b^(13/2 ))
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.
\(\displaystyle \int \frac {\sqrt {\sqrt {a x^2+b^2}+b}}{\left (a x^2+b^2\right )^4} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {a^4 \sqrt {\sqrt {a x^2+b^2}+b}}{8 (-a)^{5/2} b^5 \left (\sqrt {-a} b-a x\right )^3}+\frac {a^4 \sqrt {\sqrt {a x^2+b^2}+b}}{8 (-a)^{5/2} b^5 \left (\sqrt {-a} b+a x\right )^3}-\frac {5 a \sqrt {\sqrt {a x^2+b^2}+b}}{16 b^6 \left (-a^2 x^2-a b^2\right )}+\frac {a^2 \sqrt {\sqrt {a x^2+b^2}+b}}{16 b^4 \left (\sqrt {-a} b-a x\right )^4}+\frac {a^2 \sqrt {\sqrt {a x^2+b^2}+b}}{16 b^4 \left (\sqrt {-a} b+a x\right )^4}-\frac {5 a \sqrt {\sqrt {a x^2+b^2}+b}}{32 b^6 \left (\sqrt {-a} b-a x\right )^2}-\frac {5 a \sqrt {\sqrt {a x^2+b^2}+b}}{32 b^6 \left (\sqrt {-a} b+a x\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \int \frac {\sqrt {b+\sqrt {b^2+a x^2}}}{\left (\sqrt {-a} b-a x\right )^4}dx}{16 b^4}+\frac {a^2 \int \frac {\sqrt {b+\sqrt {b^2+a x^2}}}{\left (\sqrt {-a} b+a x\right )^4}dx}{16 b^4}+\frac {5 \int \frac {\sqrt {b+\sqrt {b^2+a x^2}}}{b-\sqrt {-a} x}dx}{32 b^7}+\frac {5 \int \frac {\sqrt {b+\sqrt {b^2+a x^2}}}{b+\sqrt {-a} x}dx}{32 b^7}-\frac {5 a \int \frac {\sqrt {b+\sqrt {b^2+a x^2}}}{\left (\sqrt {-a} b-a x\right )^2}dx}{32 b^6}-\frac {5 a \int \frac {\sqrt {b+\sqrt {b^2+a x^2}}}{\left (\sqrt {-a} b+a x\right )^2}dx}{32 b^6}+\frac {(-a)^{3/2} \int \frac {\sqrt {b+\sqrt {b^2+a x^2}}}{\left (\sqrt {-a} b-a x\right )^3}dx}{8 b^5}+\frac {(-a)^{3/2} \int \frac {\sqrt {b+\sqrt {b^2+a x^2}}}{\left (\sqrt {-a} b+a x\right )^3}dx}{8 b^5}\) |
3.23.66.3.1 Defintions of rubi rules used
\[\int \frac {\sqrt {b +\sqrt {a \,x^{2}+b^{2}}}}{\left (a \,x^{2}+b^{2}\right )^{4}}d x\]
Timed out. \[ \int \frac {\sqrt {b+\sqrt {b^2+a x^2}}}{\left (b^2+a x^2\right )^4} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {b+\sqrt {b^2+a x^2}}}{\left (b^2+a x^2\right )^4} \, dx=\int \frac {\sqrt {b + \sqrt {a x^{2} + b^{2}}}}{\left (a x^{2} + b^{2}\right )^{4}}\, dx \]
\[ \int \frac {\sqrt {b+\sqrt {b^2+a x^2}}}{\left (b^2+a x^2\right )^4} \, dx=\int { \frac {\sqrt {b + \sqrt {a x^{2} + b^{2}}}}{{\left (a x^{2} + b^{2}\right )}^{4}} \,d x } \]
\[ \int \frac {\sqrt {b+\sqrt {b^2+a x^2}}}{\left (b^2+a x^2\right )^4} \, dx=\int { \frac {\sqrt {b + \sqrt {a x^{2} + b^{2}}}}{{\left (a x^{2} + b^{2}\right )}^{4}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {b+\sqrt {b^2+a x^2}}}{\left (b^2+a x^2\right )^4} \, dx=\int \frac {\sqrt {b+\sqrt {b^2+a\,x^2}}}{{\left (b^2+a\,x^2\right )}^4} \,d x \]