Integrand size = 45, antiderivative size = 196 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=-\frac {32}{15} \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\frac {8}{5} \sqrt {1+x} \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\frac {32}{15} \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {RootSum}\left [2-8 \text {$\#$1}^4+8 \text {$\#$1}^6+14 \text {$\#$1}^8-32 \text {$\#$1}^{10}+24 \text {$\#$1}^{12}-8 \text {$\#$1}^{14}+\text {$\#$1}^{16}\&,\frac {\log \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {$\#$1}\right )}{\text {$\#$1}-\text {$\#$1}^3-4 \text {$\#$1}^5+8 \text {$\#$1}^7-5 \text {$\#$1}^9+\text {$\#$1}^{11}}\&\right ] \]
Time = 0.01 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.80 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\frac {8}{15} \left (-4+3 \sqrt {1+x}-4 \sqrt {1+\sqrt {1+x}}\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {RootSum}\left [2-8 \text {$\#$1}^4+8 \text {$\#$1}^6+14 \text {$\#$1}^8-32 \text {$\#$1}^{10}+24 \text {$\#$1}^{12}-8 \text {$\#$1}^{14}+\text {$\#$1}^{16}\&,\frac {\log \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {$\#$1}\right )}{\text {$\#$1}-\text {$\#$1}^3-4 \text {$\#$1}^5+8 \text {$\#$1}^7-5 \text {$\#$1}^9+\text {$\#$1}^{11}}\&\right ] \]
(8*(-4 + 3*Sqrt[1 + x] - 4*Sqrt[1 + Sqrt[1 + x]])*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]])/15 - 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 - #1^3 - 4*#1^5 + 8*#1^7 - 5*#1^9 + #1^11) & ]
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.
\(\displaystyle \int \frac {x^2-1}{\left (x^2+1\right ) \sqrt {\sqrt {x+1}+1} \sqrt {\sqrt {\sqrt {x+1}+1}+1}} \, dx\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 2 \int -\frac {(1-x) (x+1)^{3/2}}{\sqrt {\sqrt {x+1}+1} \left ((x+1)^2-2 (x+1)+2\right ) \sqrt {\sqrt {\sqrt {x+1}+1}+1}}d\sqrt {x+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {(1-x) (x+1)^{3/2}}{\sqrt {\sqrt {x+1}+1} \left ((x+1)^2-2 (x+1)+2\right ) \sqrt {\sqrt {\sqrt {x+1}+1}+1}}d\sqrt {x+1}\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 4 \int -\frac {x^3 \left (-(x+1)^2+2 (x+1)+1\right )}{\left ((x+1)^4-4 (x+1)^3+4 (x+1)^2+1\right ) \sqrt {\sqrt {\sqrt {x+1}+1}+1}}d\sqrt {\sqrt {x+1}+1}\) |
\(\Big \downarrow \) 2003 |
\(\displaystyle 4 \int \frac {\left (-(x+1)^2+2 (x+1)+1\right ) \left (1-\sqrt {\sqrt {x+1}+1}\right )^3 \left (\sqrt {\sqrt {x+1}+1}+1\right )^{5/2}}{(x+1)^4-4 (x+1)^3+4 (x+1)^2+1}d\sqrt {\sqrt {x+1}+1}\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 8 \int \frac {(1-x)^3 (x+1)^3 \left (-(x+1)^4+4 (x+1)^3-4 (x+1)^2+2\right )}{x^8-4 x^6+4 x^4+1}d\sqrt {\sqrt {\sqrt {x+1}+1}+1}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 8 \int \left ((x+1)^2-\frac {2 (x-1) (x+1)}{x^8-4 x^6+4 x^4+1}-2 (x+1)\right )d\sqrt {\sqrt {\sqrt {x+1}+1}+1}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 8 \left (4 \int \frac {x+1}{(x+1)^8-8 (x+1)^7+24 (x+1)^6-32 (x+1)^5+14 (x+1)^4+8 (x+1)^3-8 (x+1)^2+2}d\sqrt {\sqrt {\sqrt {x+1}+1}+1}-2 \int \frac {(x+1)^2}{(x+1)^8-8 (x+1)^7+24 (x+1)^6-32 (x+1)^5+14 (x+1)^4+8 (x+1)^3-8 (x+1)^2+2}d\sqrt {\sqrt {\sqrt {x+1}+1}+1}+\frac {1}{5} (x+1)^{5/2}-\frac {2}{3} (x+1)^{3/2}\right )\) |
3.25.30.3.1 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 = 0.17 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.70
method | result | size |
derivativedivides | \(\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {5}{2}}}{5}-\frac {16 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {3}{2}}}{3}-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{16}-8 \textit {\_Z}^{14}+24 \textit {\_Z}^{12}-32 \textit {\_Z}^{10}+14 \textit {\_Z}^{8}+8 \textit {\_Z}^{6}-8 \textit {\_Z}^{4}+2\right )}{\sum }\frac {\left (\textit {\_R}^{4}-2 \textit {\_R}^{2}\right ) \ln \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\textit {\_R} \right )}{\textit {\_R}^{15}-7 \textit {\_R}^{13}+18 \textit {\_R}^{11}-20 \textit {\_R}^{9}+7 \textit {\_R}^{7}+3 \textit {\_R}^{5}-2 \textit {\_R}^{3}}\right )\) | \(137\) |
default | \(\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {5}{2}}}{5}-\frac {16 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {3}{2}}}{3}-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{16}-8 \textit {\_Z}^{14}+24 \textit {\_Z}^{12}-32 \textit {\_Z}^{10}+14 \textit {\_Z}^{8}+8 \textit {\_Z}^{6}-8 \textit {\_Z}^{4}+2\right )}{\sum }\frac {\left (\textit {\_R}^{4}-2 \textit {\_R}^{2}\right ) \ln \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\textit {\_R} \right )}{\textit {\_R}^{15}-7 \textit {\_R}^{13}+18 \textit {\_R}^{11}-20 \textit {\_R}^{9}+7 \textit {\_R}^{7}+3 \textit {\_R}^{5}-2 \textit {\_R}^{3}}\right )\) | \(137\) |
int((x^2-1)/(x^2+1)/(1+(1+x)^(1/2))^(1/2)/(1+(1+(1+x)^(1/2))^(1/2))^(1/2), x,method=_RETURNVERBOSE)
8/5*(1+(1+(1+x)^(1/2))^(1/2))^(5/2)-16/3*(1+(1+(1+x)^(1/2))^(1/2))^(3/2)-s um((_R^4-2*_R^2)/(_R^15-7*_R^13+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. \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\text {Timed out} \]
integrate((x^2-1)/(x^2+1)/(1+(1+x)^(1/2))^(1/2)/(1+(1+(1+x)^(1/2))^(1/2))^ (1/2),x, algorithm="fricas")
Timed out. \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\text {Timed out} \]
Not integrable
Time = 0.61 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.19 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int { \frac {x^{2} - 1}{{\left (x^{2} + 1\right )} \sqrt {\sqrt {x + 1} + 1} \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}} \,d x } \]
integrate((x^2-1)/(x^2+1)/(1+(1+x)^(1/2))^(1/2)/(1+(1+(1+x)^(1/2))^(1/2))^ (1/2),x, algorithm="maxima")
Not integrable
Time = 1.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.19 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int { \frac {x^{2} - 1}{{\left (x^{2} + 1\right )} \sqrt {\sqrt {x + 1} + 1} \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}} \,d x } \]
integrate((x^2-1)/(x^2+1)/(1+(1+x)^(1/2))^(1/2)/(1+(1+(1+x)^(1/2))^(1/2))^ (1/2),x, algorithm="giac")
Not integrable
Time = 0.00 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.19 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int \frac {x^2-1}{\left (x^2+1\right )\,\sqrt {\sqrt {\sqrt {x+1}+1}+1}\,\sqrt {\sqrt {x+1}+1}} \,d x \]