Integrand size = 31, antiderivative size = 242 \[ \int \sqrt {1+x^2} \sqrt {1+\sqrt {x+\sqrt {1+x^2}}} \, dx=\frac {\left (78032-1254 x+193024 x^2-3072 x^3+35840 x^4\right ) \sqrt {1+\sqrt {x+\sqrt {1+x^2}}}+\left (184+345 x+2048 x^2+2560 x^3\right ) \sqrt {x+\sqrt {1+x^2}} \sqrt {1+\sqrt {x+\sqrt {1+x^2}}}+\sqrt {1+x^2} \left (\left (282+175104 x-3072 x^2+35840 x^3\right ) \sqrt {1+\sqrt {x+\sqrt {1+x^2}}}+\left (-935+2048 x+2560 x^2\right ) \sqrt {x+\sqrt {1+x^2}} \sqrt {1+\sqrt {x+\sqrt {1+x^2}}}\right )}{80640 x \sqrt {1+x^2}+40320 \left (1+2 x^2\right )}-\frac {251}{128} \text {arctanh}\left (\sqrt {1+\sqrt {x+\sqrt {1+x^2}}}\right ) \]
((35840*x^4-3072*x^3+193024*x^2-1254*x+78032)*(1+(x+(x^2+1)^(1/2))^(1/2))^ (1/2)+(2560*x^3+2048*x^2+345*x+184)*(x+(x^2+1)^(1/2))^(1/2)*(1+(x+(x^2+1)^ (1/2))^(1/2))^(1/2)+(x^2+1)^(1/2)*((35840*x^3-3072*x^2+175104*x+282)*(1+(x +(x^2+1)^(1/2))^(1/2))^(1/2)+(2560*x^2+2048*x-935)*(x+(x^2+1)^(1/2))^(1/2) *(1+(x+(x^2+1)^(1/2))^(1/2))^(1/2)))/(80640*x*(x^2+1)^(1/2)+80640*x^2+4032 0)-251/128*arctanh((1+(x+(x^2+1)^(1/2))^(1/2))^(1/2))
Time = 0.32 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.73 \[ \int \sqrt {1+x^2} \sqrt {1+\sqrt {x+\sqrt {1+x^2}}} \, dx=\frac {\frac {\sqrt {1+\sqrt {x+\sqrt {1+x^2}}} \left (2 \left (39016-627 x+96512 x^2-1536 x^3+17920 x^4\right )+\left (184+345 x+2048 x^2+2560 x^3\right ) \sqrt {x+\sqrt {1+x^2}}+\sqrt {1+x^2} \left (282+175104 x-3072 x^2+35840 x^3+\left (-935+2048 x+2560 x^2\right ) \sqrt {x+\sqrt {1+x^2}}\right )\right )}{1+2 x^2+2 x \sqrt {1+x^2}}-79065 \text {arctanh}\left (\sqrt {1+\sqrt {x+\sqrt {1+x^2}}}\right )}{40320} \]
((Sqrt[1 + Sqrt[x + Sqrt[1 + x^2]]]*(2*(39016 - 627*x + 96512*x^2 - 1536*x ^3 + 17920*x^4) + (184 + 345*x + 2048*x^2 + 2560*x^3)*Sqrt[x + Sqrt[1 + x^ 2]] + Sqrt[1 + x^2]*(282 + 175104*x - 3072*x^2 + 35840*x^3 + (-935 + 2048* x + 2560*x^2)*Sqrt[x + Sqrt[1 + x^2]])))/(1 + 2*x^2 + 2*x*Sqrt[1 + x^2]) - 79065*ArcTanh[Sqrt[1 + Sqrt[x + Sqrt[1 + x^2]]]])/40320
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.
\(\displaystyle \int \sqrt {x^2+1} \sqrt {\sqrt {\sqrt {x^2+1}+x}+1} \, dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \sqrt {x^2+1} \sqrt {\sqrt {\sqrt {x^2+1}+x}+1}dx\) |
3.27.81.3.1 Defintions of rubi rules used
\[\int \sqrt {x^{2}+1}\, \sqrt {1+\sqrt {x +\sqrt {x^{2}+1}}}d x\]
Time = 0.31 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.49 \[ \int \sqrt {1+x^2} \sqrt {1+\sqrt {x+\sqrt {1+x^2}}} \, dx=-\frac {1}{40320} \, {\left (1120 \, x^{2} - 2 \, \sqrt {x^{2} + 1} {\left (9520 \, x + 141\right )} + {\left (1680 \, x^{2} - 5 \, \sqrt {x^{2} + 1} {\left (336 \, x - 187\right )} - 2215 \, x - 184\right )} \sqrt {x + \sqrt {x^{2} + 1}} + 1818 \, x - 78032\right )} \sqrt {\sqrt {x + \sqrt {x^{2} + 1}} + 1} - \frac {251}{256} \, \log \left (\sqrt {\sqrt {x + \sqrt {x^{2} + 1}} + 1} + 1\right ) + \frac {251}{256} \, \log \left (\sqrt {\sqrt {x + \sqrt {x^{2} + 1}} + 1} - 1\right ) \]
-1/40320*(1120*x^2 - 2*sqrt(x^2 + 1)*(9520*x + 141) + (1680*x^2 - 5*sqrt(x ^2 + 1)*(336*x - 187) - 2215*x - 184)*sqrt(x + sqrt(x^2 + 1)) + 1818*x - 7 8032)*sqrt(sqrt(x + sqrt(x^2 + 1)) + 1) - 251/256*log(sqrt(sqrt(x + sqrt(x ^2 + 1)) + 1) + 1) + 251/256*log(sqrt(sqrt(x + sqrt(x^2 + 1)) + 1) - 1)
\[ \int \sqrt {1+x^2} \sqrt {1+\sqrt {x+\sqrt {1+x^2}}} \, dx=\int \sqrt {x^{2} + 1} \sqrt {\sqrt {x + \sqrt {x^{2} + 1}} + 1}\, dx \]
\[ \int \sqrt {1+x^2} \sqrt {1+\sqrt {x+\sqrt {1+x^2}}} \, dx=\int { \sqrt {x^{2} + 1} \sqrt {\sqrt {x + \sqrt {x^{2} + 1}} + 1} \,d x } \]
\[ \int \sqrt {1+x^2} \sqrt {1+\sqrt {x+\sqrt {1+x^2}}} \, dx=\int { \sqrt {x^{2} + 1} \sqrt {\sqrt {x + \sqrt {x^{2} + 1}} + 1} \,d x } \]
Timed out. \[ \int \sqrt {1+x^2} \sqrt {1+\sqrt {x+\sqrt {1+x^2}}} \, dx=\int \sqrt {\sqrt {x+\sqrt {x^2+1}}+1}\,\sqrt {x^2+1} \,d x \]