Integrand size = 32, antiderivative size = 224 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\frac {32}{105} \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\frac {32}{105} \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\sqrt {1+x} \left (-\frac {48}{35} \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\frac {8}{7} \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )-\text {RootSum}\left [2-8 \text {$\#$1}^4+8 \text {$\#$1}^6+14 \text {$\#$1}^8-32 \text {$\#$1}^{10}+24 \text {$\#$1}^{12}-8 \text {$\#$1}^{14}+\text {$\#$1}^{16}\&,\frac {\log \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {$\#$1}\right )}{-\text {$\#$1}+4 \text {$\#$1}^5-4 \text {$\#$1}^7+\text {$\#$1}^9}\&\right ] \]
Time = 0.00 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.75 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\frac {8}{105} \sqrt {1+\sqrt {1+\sqrt {1+x}}} \left (4-18 \sqrt {1+x}+4 \sqrt {1+\sqrt {1+x}}+15 \sqrt {1+x} \sqrt {1+\sqrt {1+x}}\right )-\text {RootSum}\left [2-8 \text {$\#$1}^4+8 \text {$\#$1}^6+14 \text {$\#$1}^8-32 \text {$\#$1}^{10}+24 \text {$\#$1}^{12}-8 \text {$\#$1}^{14}+\text {$\#$1}^{16}\&,\frac {\log \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {$\#$1}\right )}{-\text {$\#$1}+4 \text {$\#$1}^5-4 \text {$\#$1}^7+\text {$\#$1}^9}\&\right ] \]
(8*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]*(4 - 18*Sqrt[1 + x] + 4*Sqrt[1 + Sqrt[1 + x]] + 15*Sqrt[1 + x]*Sqrt[1 + Sqrt[1 + x]]))/105 - RootSum[2 - 8*#1^4 + 8*#1^6 + 14*#1^8 - 32*#1^10 + 24*#1^12 - 8*#1^14 + #1^16 & , Log[Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]] - #1]/(-#1 + 4*#1^5 - 4*#1^7 + #1^9) & ]
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.
\(\displaystyle \int \frac {x^2-1}{\left (x^2+1\right ) \sqrt {\sqrt {\sqrt {x+1}+1}+1}} \, dx\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 2 \int -\frac {(1-x) (x+1)^{3/2}}{\left ((x+1)^2-2 (x+1)+2\right ) \sqrt {\sqrt {\sqrt {x+1}+1}+1}}d\sqrt {x+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {(1-x) (x+1)^{3/2}}{\left ((x+1)^2-2 (x+1)+2\right ) \sqrt {\sqrt {\sqrt {x+1}+1}+1}}d\sqrt {x+1}\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 4 \int -\frac {x^3 \sqrt {\sqrt {x+1}+1} \left (-(x+1)^2+2 (x+1)+1\right )}{\left ((x+1)^4-4 (x+1)^3+4 (x+1)^2+1\right ) \sqrt {\sqrt {\sqrt {x+1}+1}+1}}d\sqrt {\sqrt {x+1}+1}\) |
\(\Big \downarrow \) 2003 |
\(\displaystyle 4 \int \frac {\sqrt {\sqrt {x+1}+1} \left (-(x+1)^2+2 (x+1)+1\right ) \left (1-\sqrt {\sqrt {x+1}+1}\right )^3 \left (\sqrt {\sqrt {x+1}+1}+1\right )^{5/2}}{(x+1)^4-4 (x+1)^3+4 (x+1)^2+1}d\sqrt {\sqrt {x+1}+1}\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 8 \int \frac {(1-x)^3 x (x+1)^3 \left (-(x+1)^4+4 (x+1)^3-4 (x+1)^2+2\right )}{x^8-4 x^6+4 x^4+1}d\sqrt {\sqrt {\sqrt {x+1}+1}+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -8 \int -\frac {(1-x)^3 x (x+1)^3 \left (-(x+1)^4+4 (x+1)^3-4 (x+1)^2+2\right )}{x^8-4 x^6+4 x^4+1}d\sqrt {\sqrt {\sqrt {x+1}+1}+1}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -8 \int \left (-(x+1)^3+3 (x+1)^2+\frac {2 \left ((x+1)^2-3 (x+1)+2\right ) (x+1)}{x^8-4 x^6+4 x^4+1}-2 (x+1)\right )d\sqrt {\sqrt {\sqrt {x+1}+1}+1}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 8 \left (-4 \int \frac {x+1}{(x+1)^8-8 (x+1)^7+24 (x+1)^6-32 (x+1)^5+14 (x+1)^4+8 (x+1)^3-8 (x+1)^2+2}d\sqrt {\sqrt {\sqrt {x+1}+1}+1}+6 \int \frac {(x+1)^2}{(x+1)^8-8 (x+1)^7+24 (x+1)^6-32 (x+1)^5+14 (x+1)^4+8 (x+1)^3-8 (x+1)^2+2}d\sqrt {\sqrt {\sqrt {x+1}+1}+1}-2 \int \frac {(x+1)^3}{(x+1)^8-8 (x+1)^7+24 (x+1)^6-32 (x+1)^5+14 (x+1)^4+8 (x+1)^3-8 (x+1)^2+2}d\sqrt {\sqrt {\sqrt {x+1}+1}+1}+\frac {1}{7} (x+1)^{7/2}-\frac {3}{5} (x+1)^{5/2}+\frac {2}{3} (x+1)^{3/2}\right )\) |
3.26.92.3.1 Defintions of rubi rules used
Int[(u_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] : > Int[u*(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, n, p} , x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && !IntegerQ[n]))
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
Time = 0.00 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.70
method | result | size |
derivativedivides | \(\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {7}{2}}}{7}-\frac {24 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {5}{2}}}{5}+\frac {16 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {3}{2}}}{3}-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{16}-8 \textit {\_Z}^{14}+24 \textit {\_Z}^{12}-32 \textit {\_Z}^{10}+14 \textit {\_Z}^{8}+8 \textit {\_Z}^{6}-8 \textit {\_Z}^{4}+2\right )}{\sum }\frac {\left (\textit {\_R}^{6}-3 \textit {\_R}^{4}+2 \textit {\_R}^{2}\right ) \ln \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\textit {\_R} \right )}{\textit {\_R}^{15}-7 \textit {\_R}^{13}+18 \textit {\_R}^{11}-20 \textit {\_R}^{9}+7 \textit {\_R}^{7}+3 \textit {\_R}^{5}-2 \textit {\_R}^{3}}\right )\) | \(157\) |
default | \(\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {7}{2}}}{7}-\frac {24 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {5}{2}}}{5}+\frac {16 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {3}{2}}}{3}-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{16}-8 \textit {\_Z}^{14}+24 \textit {\_Z}^{12}-32 \textit {\_Z}^{10}+14 \textit {\_Z}^{8}+8 \textit {\_Z}^{6}-8 \textit {\_Z}^{4}+2\right )}{\sum }\frac {\left (\textit {\_R}^{6}-3 \textit {\_R}^{4}+2 \textit {\_R}^{2}\right ) \ln \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\textit {\_R} \right )}{\textit {\_R}^{15}-7 \textit {\_R}^{13}+18 \textit {\_R}^{11}-20 \textit {\_R}^{9}+7 \textit {\_R}^{7}+3 \textit {\_R}^{5}-2 \textit {\_R}^{3}}\right )\) | \(157\) |
8/7*(1+(1+(1+x)^(1/2))^(1/2))^(7/2)-24/5*(1+(1+(1+x)^(1/2))^(1/2))^(5/2)+1 6/3*(1+(1+(1+x)^(1/2))^(1/2))^(3/2)-sum((_R^6-3*_R^4+2*_R^2)/(_R^15-7*_R^1 3+18*_R^11-20*_R^9+7*_R^7+3*_R^5-2*_R^3)*ln((1+(1+(1+x)^(1/2))^(1/2))^(1/2 )-_R),_R=RootOf(_Z^16-8*_Z^14+24*_Z^12-32*_Z^10+14*_Z^8+8*_Z^6-8*_Z^4+2))
Timed out. \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\text {Timed out} \]
Not integrable
Time = 0.58 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.12 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int { \frac {x^{2} - 1}{{\left (x^{2} + 1\right )} \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}} \,d x } \]
Not integrable
Time = 1.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.12 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int { \frac {x^{2} - 1}{{\left (x^{2} + 1\right )} \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}} \,d x } \]
Not integrable
Time = 0.00 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.12 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int \frac {x^2-1}{\left (x^2+1\right )\,\sqrt {\sqrt {\sqrt {x+1}+1}+1}} \,d x \]