3.23.57 \(\int \frac {-1+x^2}{\sqrt {1+x} (1+x^2) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx\) [2257]

3.23.57.1 Optimal result

Integrand size = 39, antiderivative size = 170 \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=-\frac {16}{3} \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\frac {8}{3} \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {RootSum}\left [2-8 \text {$\#$1}^4+8 \text {$\#$1}^6+14 \text {$\#$1}^8-32 \text {$\#$1}^{10}+24 \text {$\#$1}^{12}-8 \text {$\#$1}^{14}+\text {$\#$1}^{16}\&,\frac {\log \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {$\#$1}\right )}{2 \text {$\#$1}^3-\text {$\#$1}^5-8 \text {$\#$1}^7+12 \text {$\#$1}^9-6 \text {$\#$1}^{11}+\text {$\#$1}^{13}}\&\right ] \]

output

Unintegrable
 

3.23.57.2 Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.88 \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\frac {8}{3} \left (-2+\sqrt {1+\sqrt {1+x}}\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {RootSum}\left [2-8 \text {$\#$1}^4+8 \text {$\#$1}^6+14 \text {$\#$1}^8-32 \text {$\#$1}^{10}+24 \text {$\#$1}^{12}-8 \text {$\#$1}^{14}+\text {$\#$1}^{16}\&,\frac {\log \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\text {$\#$1}\right )}{2 \text {$\#$1}^3-\text {$\#$1}^5-8 \text {$\#$1}^7+12 \text {$\#$1}^9-6 \text {$\#$1}^{11}+\text {$\#$1}^{13}}\&\right ] \]

input

Integrate[(-1 + x^2)/(Sqrt[1 + x]*(1 + x^2)*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]] 
]),x]
 

output

(8*(-2 + Sqrt[1 + Sqrt[1 + x]])*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]])/3 - RootS 
um[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 - #1^5 - 8*#1^7 + 12* 
#1^9 - 6*#1^11 + #1^13) & ]
 

3.23.57.3 Rubi [F]

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.

\(\displaystyle \int \frac {x^2-1}{\sqrt {x+1} \left (x^2+1\right ) \sqrt {\sqrt {\sqrt {x+1}+1}+1}} \, dx\)

\(\Big \downarrow \) 2003

\(\displaystyle \int \frac {(x-1) \sqrt {x+1}}{\left (x^2+1\right ) \sqrt {\sqrt {\sqrt {x+1}+1}+1}}dx\)

\(\Big \downarrow \) 7267

\(\displaystyle 2 \int -\frac {(1-x) (x+1)}{\left (x^2+1\right ) \sqrt {\sqrt {\sqrt {x+1}+1}+1}}d\sqrt {x+1}\)

\(\Big \downarrow \) 25

\(\displaystyle -2 \int \frac {(1-x) (x+1)}{\left (x^2+1\right ) \sqrt {\sqrt {\sqrt {x+1}+1}+1}}d\sqrt {x+1}\)

\(\Big \downarrow \) 7267

\(\displaystyle -4 \int \frac {-(x+1)^{9/2}+4 (x+1)^{7/2}-4 (x+1)^{5/2}+\sqrt {\sqrt {x+1}+1}}{\left ((x+1)^4-4 (x+1)^3+4 (x+1)^2+1\right ) \sqrt {\sqrt {\sqrt {x+1}+1}+1}}d\sqrt {\sqrt {x+1}+1}\)

\(\Big \downarrow \) 2019

\(\displaystyle -4 \int \frac {\sqrt {\sqrt {\sqrt {x+1}+1}+1} \left (-(x+1)^4+(x+1)^{7/2}+3 (x+1)^3-3 (x+1)^{5/2}-(x+1)^2+(x+1)^{3/2}-x+\sqrt {\sqrt {x+1}+1}-1\right )}{(x+1)^4-4 (x+1)^3+4 (x+1)^2+1}d\sqrt {\sqrt {x+1}+1}\)

\(\Big \downarrow \) 7267

\(\displaystyle 8 \int \frac {(1-x)^2 (x+1)^2 \left ((x+1)^5-5 (x+1)^4+8 (x+1)^3-4 (x+1)^2-2 (x+1)+2\right )}{(x+1)^8-8 (x+1)^7+24 (x+1)^6-32 (x+1)^5+14 (x+1)^4+8 (x+1)^3-8 (x+1)^2+2}d\sqrt {\sqrt {\sqrt {x+1}+1}+1}\)

\(\Big \downarrow \) 7293

\(\displaystyle 8 \int \left (x-\frac {2 x}{(x+1)^8-8 (x+1)^7+24 (x+1)^6-32 (x+1)^5+14 (x+1)^4+8 (x+1)^3-8 (x+1)^2+2}\right )d\sqrt {\sqrt {\sqrt {x+1}+1}+1}\)

\(\Big \downarrow \) 2009

\(\displaystyle 8 \left (2 \int \frac {1}{(x+1)^8-8 (x+1)^7+24 (x+1)^6-32 (x+1)^5+14 (x+1)^4+8 (x+1)^3-8 (x+1)^2+2}d\sqrt {\sqrt {\sqrt {x+1}+1}+1}-2 \int \frac {x+1}{(x+1)^8-8 (x+1)^7+24 (x+1)^6-32 (x+1)^5+14 (x+1)^4+8 (x+1)^3-8 (x+1)^2+2}d\sqrt {\sqrt {\sqrt {x+1}+1}+1}+\frac {1}{3} (x+1)^{3/2}-\sqrt {\sqrt {\sqrt {x+1}+1}+1}\right )\)

input

Int[(-1 + x^2)/(Sqrt[1 + x]*(1 + x^2)*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]),x]
 

output

$Aborted
 

3.23.57.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2003

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]))
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2019

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]
 

rule 7267

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]]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.23.57.4 Maple [N/A] (verified)

Time = 0.15 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {3}{2}}}{3}-8 \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{16}-8 \textit {\_Z}^{14}+24 \textit {\_Z}^{12}-32 \textit {\_Z}^{10}+14 \textit {\_Z}^{8}+8 \textit {\_Z}^{6}-8 \textit {\_Z}^{4}+2\right )}{\sum }\frac {\left (-\textit {\_R}^{2}+1\right ) \ln \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\textit {\_R} \right )}{\textit {\_R}^{15}-7 \textit {\_R}^{13}+18 \textit {\_R}^{11}-20 \textit {\_R}^{9}+7 \textit {\_R}^{7}+3 \textit {\_R}^{5}-2 \textit {\_R}^{3}}\right )\)	\(133\)
default	\(\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {3}{2}}}{3}-8 \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{16}-8 \textit {\_Z}^{14}+24 \textit {\_Z}^{12}-32 \textit {\_Z}^{10}+14 \textit {\_Z}^{8}+8 \textit {\_Z}^{6}-8 \textit {\_Z}^{4}+2\right )}{\sum }\frac {\left (-\textit {\_R}^{2}+1\right ) \ln \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-\textit {\_R} \right )}{\textit {\_R}^{15}-7 \textit {\_R}^{13}+18 \textit {\_R}^{11}-20 \textit {\_R}^{9}+7 \textit {\_R}^{7}+3 \textit {\_R}^{5}-2 \textit {\_R}^{3}}\right )\)	\(133\)

input

int((x^2-1)/(1+x)^(1/2)/(x^2+1)/(1+(1+(1+x)^(1/2))^(1/2))^(1/2),x,method=_ 
RETURNVERBOSE)
 

output

8/3*(1+(1+(1+x)^(1/2))^(1/2))^(3/2)-8*(1+(1+(1+x)^(1/2))^(1/2))^(1/2)+sum( 
(-_R^2+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))
 

3.23.57.5 Fricas [F(-1)]

Timed out. \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\text {Timed out} \]

input

integrate((x^2-1)/(1+x)^(1/2)/(x^2+1)/(1+(1+(1+x)^(1/2))^(1/2))^(1/2),x, a 
lgorithm="fricas")
 

output

Timed out
 

3.23.57.6 Sympy [F(-1)]

Timed out. \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\text {Timed out} \]

input

integrate((x**2-1)/(1+x)**(1/2)/(x**2+1)/(1+(1+(1+x)**(1/2))**(1/2))**(1/2 
),x)
 

output

Timed out
 

3.23.57.7 Maxima [N/A]

Not integrable

Time = 0.60 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.19 \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int { \frac {x^{2} - 1}{{\left (x^{2} + 1\right )} \sqrt {x + 1} \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}} \,d x } \]

input

integrate((x^2-1)/(1+x)^(1/2)/(x^2+1)/(1+(1+(1+x)^(1/2))^(1/2))^(1/2),x, a 
lgorithm="maxima")
 

output

integrate((x^2 - 1)/((x^2 + 1)*sqrt(x + 1)*sqrt(sqrt(sqrt(x + 1) + 1) + 1) 
), x)
 

3.23.57.8 Giac [N/A]

Not integrable

Time = 1.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.19 \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int { \frac {x^{2} - 1}{{\left (x^{2} + 1\right )} \sqrt {x + 1} \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}} \,d x } \]

input

integrate((x^2-1)/(1+x)^(1/2)/(x^2+1)/(1+(1+(1+x)^(1/2))^(1/2))^(1/2),x, a 
lgorithm="giac")
 

output

integrate((x^2 - 1)/((x^2 + 1)*sqrt(x + 1)*sqrt(sqrt(sqrt(x + 1) + 1) + 1) 
), x)
 

3.23.57.9 Mupad [N/A]

Not integrable

Time = 0.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.19 \[ \int \frac {-1+x^2}{\sqrt {1+x} \left (1+x^2\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx=\int \frac {x^2-1}{\left (x^2+1\right )\,\sqrt {\sqrt {\sqrt {x+1}+1}+1}\,\sqrt {x+1}} \,d x \]

input

int((x^2 - 1)/((x^2 + 1)*(((x + 1)^(1/2) + 1)^(1/2) + 1)^(1/2)*(x + 1)^(1/ 
2)),x)
 

output

int((x^2 - 1)/((x^2 + 1)*(((x + 1)^(1/2) + 1)^(1/2) + 1)^(1/2)*(x + 1)^(1/ 
2)), x)