Integrand size = 52, antiderivative size = 138 \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=8 \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {RootSum}\left [2-8 \text {$\#$1}^4+8 \text {$\#$1}^6+14 \text {$\#$1}^8-32 \text {$\#$1}^{10}+24 \text {$\#$1}^{12}-8 \text {$\#$1}^{14}+\text {$\#$1}^{16}\&,\frac {\log \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {$\#$1}\right )}{-2 \text {$\#$1}^3+3 \text {$\#$1}^5+7 \text {$\#$1}^7-20 \text {$\#$1}^9+18 \text {$\#$1}^{11}-7 \text {$\#$1}^{13}+\text {$\#$1}^{15}}\&\right ] \]
Time = 0.00 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00 \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=8 \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {RootSum}\left [2-8 \text {$\#$1}^4+8 \text {$\#$1}^6+14 \text {$\#$1}^8-32 \text {$\#$1}^{10}+24 \text {$\#$1}^{12}-8 \text {$\#$1}^{14}+\text {$\#$1}^{16}\&,\frac {\log \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {$\#$1}\right )}{-2 \text {$\#$1}^3+3 \text {$\#$1}^5+7 \text {$\#$1}^7-20 \text {$\#$1}^9+18 \text {$\#$1}^{11}-7 \text {$\#$1}^{13}+\text {$\#$1}^{15}}\&\right ] \]
Integrate[(-1 + x^2)/(Sqrt[1 + x]*(1 + x^2)*Sqrt[1 + Sqrt[1 + x]]*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]),x]
8*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]] - 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]/(-2*#1^3 + 3*#1^5 + 7*#1^7 - 20*#1^9 + 18*#1^11 - 7*#1^13 + #1 ^15) & ]
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.
\(\displaystyle \int \frac {x^2-1}{\sqrt {x+1} \left (x^2+1\right ) \sqrt {\sqrt {x+1}+1} \sqrt {\sqrt {\sqrt {x+1}+1}+1}} \, dx\) |
\(\Big \downarrow \) 2003 |
\(\displaystyle \int \frac {(x-1) \sqrt {x+1}}{\left (x^2+1\right ) \sqrt {\sqrt {x+1}+1} \sqrt {\sqrt {\sqrt {x+1}+1}+1}}dx\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 2 \int -\frac {(1-x) (x+1)}{\left (x^2+1\right ) \sqrt {\sqrt {x+1}+1} \sqrt {\sqrt {\sqrt {x+1}+1}+1}}d\sqrt {x+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {(1-x) (x+1)}{\left (x^2+1\right ) \sqrt {\sqrt {x+1}+1} \sqrt {\sqrt {\sqrt {x+1}+1}+1}}d\sqrt {x+1}\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle -4 \int \frac {-(x+1)^4+4 (x+1)^3-4 (x+1)^2+1}{\left ((x+1)^4-4 (x+1)^3+4 (x+1)^2+1\right ) \sqrt {\sqrt {\sqrt {x+1}+1}+1}}d\sqrt {\sqrt {x+1}+1}\) |
\(\Big \downarrow \) 2019 |
\(\displaystyle -4 \int \frac {\left (-(x+1)^{7/2}+(x+1)^3+3 (x+1)^{5/2}-3 (x+1)^2-(x+1)^{3/2}+x-\sqrt {\sqrt {x+1}+1}+2\right ) \sqrt {\sqrt {\sqrt {x+1}+1}+1}}{(x+1)^4-4 (x+1)^3+4 (x+1)^2+1}d\sqrt {\sqrt {x+1}+1}\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 8 \int -\frac {(1-x)^2 (x+1)^2 \left (-(x+1)^4+4 (x+1)^3-4 (x+1)^2+2\right )}{(x+1)^8-8 (x+1)^7+24 (x+1)^6-32 (x+1)^5+14 (x+1)^4+8 (x+1)^3-8 (x+1)^2+2}d\sqrt {\sqrt {\sqrt {x+1}+1}+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -8 \int \frac {(1-x)^2 (x+1)^2 \left (-(x+1)^4+4 (x+1)^3-4 (x+1)^2+2\right )}{(x+1)^8-8 (x+1)^7+24 (x+1)^6-32 (x+1)^5+14 (x+1)^4+8 (x+1)^3-8 (x+1)^2+2}d\sqrt {\sqrt {\sqrt {x+1}+1}+1}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -8 \int \left (\frac {2}{(x+1)^8-8 (x+1)^7+24 (x+1)^6-32 (x+1)^5+14 (x+1)^4+8 (x+1)^3-8 (x+1)^2+2}-1\right )d\sqrt {\sqrt {\sqrt {x+1}+1}+1}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 8 \left (\sqrt {\sqrt {\sqrt {x+1}+1}+1}-2 \int \frac {1}{(x+1)^8-8 (x+1)^7+24 (x+1)^6-32 (x+1)^5+14 (x+1)^4+8 (x+1)^3-8 (x+1)^2+2}d\sqrt {\sqrt {\sqrt {x+1}+1}+1}\right )\) |
Int[(-1 + x^2)/(Sqrt[1 + x]*(1 + x^2)*Sqrt[1 + Sqrt[1 + x]]*Sqrt[1 + Sqrt[ 1 + Sqrt[1 + x]]]),x]
3.20.62.3.1 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_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px , Qx, x]^p*Qx^(p + q), x] /; FreeQ[q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]
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 = 0.28 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(8 \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{16}-8 \textit {\_Z}^{14}+24 \textit {\_Z}^{12}-32 \textit {\_Z}^{10}+14 \textit {\_Z}^{8}+8 \textit {\_Z}^{6}-8 \textit {\_Z}^{4}+2\right )}{\sum }\frac {\ln \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\textit {\_R} \right )}{\textit {\_R}^{15}-7 \textit {\_R}^{13}+18 \textit {\_R}^{11}-20 \textit {\_R}^{9}+7 \textit {\_R}^{7}+3 \textit {\_R}^{5}-2 \textit {\_R}^{3}}\right )\) | \(113\) |
default | \(8 \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{16}-8 \textit {\_Z}^{14}+24 \textit {\_Z}^{12}-32 \textit {\_Z}^{10}+14 \textit {\_Z}^{8}+8 \textit {\_Z}^{6}-8 \textit {\_Z}^{4}+2\right )}{\sum }\frac {\ln \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\textit {\_R} \right )}{\textit {\_R}^{15}-7 \textit {\_R}^{13}+18 \textit {\_R}^{11}-20 \textit {\_R}^{9}+7 \textit {\_R}^{7}+3 \textit {\_R}^{5}-2 \textit {\_R}^{3}}\right )\) | \(113\) |
int((x^2-1)/(1+x)^(1/2)/(x^2+1)/(1+(1+x)^(1/2))^(1/2)/(1+(1+(1+x)^(1/2))^( 1/2))^(1/2),x,method=_RETURNVERBOSE)
8*(1+(1+(1+x)^(1/2))^(1/2))^(1/2)-sum(1/(_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. \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\text {Timed out} \]
integrate((x^2-1)/(1+x)^(1/2)/(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. \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\text {Timed out} \]
integrate((x**2-1)/(1+x)**(1/2)/(x**2+1)/(1+(1+x)**(1/2))**(1/2)/(1+(1+(1+ x)**(1/2))**(1/2))**(1/2),x)
Not integrable
Time = 0.63 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.30 \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int { \frac {x^{2} - 1}{{\left (x^{2} + 1\right )} \sqrt {x + 1} \sqrt {\sqrt {x + 1} + 1} \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}} \,d x } \]
integrate((x^2-1)/(1+x)^(1/2)/(x^2+1)/(1+(1+x)^(1/2))^(1/2)/(1+(1+(1+x)^(1 /2))^(1/2))^(1/2),x, algorithm="maxima")
integrate((x^2 - 1)/((x^2 + 1)*sqrt(x + 1)*sqrt(sqrt(x + 1) + 1)*sqrt(sqrt (sqrt(x + 1) + 1) + 1)), x)
Timed out. \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\text {Timed out} \]
integrate((x^2-1)/(1+x)^(1/2)/(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 = 42, normalized size of antiderivative = 0.30 \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int \frac {x^2-1}{\left (x^2+1\right )\,\sqrt {\sqrt {\sqrt {x+1}+1}+1}\,\sqrt {\sqrt {x+1}+1}\,\sqrt {x+1}} \,d x \]
int((x^2 - 1)/((x^2 + 1)*(((x + 1)^(1/2) + 1)^(1/2) + 1)^(1/2)*((x + 1)^(1 /2) + 1)^(1/2)*(x + 1)^(1/2)),x)