Integrand size = 33, antiderivative size = 161 \[ \int \frac {\sqrt {1+x} \left (-1+x^4\right ) \sqrt {1+\sqrt {1+x}}}{1+x^4} \, dx=\frac {4}{105} (11+3 x) \sqrt {1+\sqrt {1+x}}+\frac {4}{105} \sqrt {1+x} (11+15 x) \sqrt {1+\sqrt {1+x}}-\frac {1}{2} \text {RootSum}\left [1+16 \text {$\#$1}^8-32 \text {$\#$1}^{10}+24 \text {$\#$1}^{12}-8 \text {$\#$1}^{14}+\text {$\#$1}^{16}\&,\frac {-\log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right )+\log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^2}{-8 \text {$\#$1}^5+12 \text {$\#$1}^7-6 \text {$\#$1}^9+\text {$\#$1}^{11}}\&\right ] \]
Time = 0.00 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {1+x} \left (-1+x^4\right ) \sqrt {1+\sqrt {1+x}}}{1+x^4} \, dx=\frac {4}{105} \sqrt {1+\sqrt {1+x}} \left (8-4 \sqrt {1+x}+3 (1+x)+15 (1+x)^{3/2}\right )-\frac {1}{2} \text {RootSum}\left [1+16 \text {$\#$1}^8-32 \text {$\#$1}^{10}+24 \text {$\#$1}^{12}-8 \text {$\#$1}^{14}+\text {$\#$1}^{16}\&,\frac {-\log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right )+\log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^2}{-8 \text {$\#$1}^5+12 \text {$\#$1}^7-6 \text {$\#$1}^9+\text {$\#$1}^{11}}\&\right ] \]
(4*Sqrt[1 + Sqrt[1 + x]]*(8 - 4*Sqrt[1 + x] + 3*(1 + x) + 15*(1 + x)^(3/2) ))/105 - RootSum[1 + 16*#1^8 - 32*#1^10 + 24*#1^12 - 8*#1^14 + #1^16 & , ( -Log[Sqrt[1 + Sqrt[1 + x]] - #1] + Log[Sqrt[1 + Sqrt[1 + x]] - #1]*#1^2)/( -8*#1^5 + 12*#1^7 - 6*#1^9 + #1^11) & ]/2
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.
\(\displaystyle \int \frac {\sqrt {x+1} \left (x^4-1\right ) \sqrt {\sqrt {x+1}+1}}{x^4+1} \, dx\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 2 \int -\frac {(x+1) \left (1-x^4\right ) \sqrt {\sqrt {x+1}+1}}{x^4+1}d\sqrt {x+1}\) |
\(\Big \downarrow \) 9 |
\(\displaystyle 2 \int -\frac {(x+1)^2 \sqrt {\sqrt {x+1}+1} \left (-(x+1)^3+4 (x+1)^2-6 (x+1)+4\right )}{x^4+1}d\sqrt {x+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {(x+1)^2 \sqrt {\sqrt {x+1}+1} \left (-(x+1)^3+4 (x+1)^2-6 (x+1)+4\right )}{x^4+1}d\sqrt {x+1}\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle -4 \int \frac {x^4 (x+1) \left (-x^6+4 x^4-6 x^2+4\right )}{(1-x)^4 (x+1)^4+1}d\sqrt {\sqrt {x+1}+1}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -4 \int \left (-(x+1)^3+2 (x+1)^2+\frac {2 \left ((x+1)^2-2 (x+1)+1\right ) (x+1)}{(1-x)^4 (x+1)^4+1}-x-1\right )d\sqrt {\sqrt {x+1}+1}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \left (2 \int \frac {x+1}{(x+1)^8-8 (x+1)^7+24 (x+1)^6-32 (x+1)^5+16 (x+1)^4+1}d\sqrt {\sqrt {x+1}+1}-4 \int \frac {(x+1)^2}{(x+1)^8-8 (x+1)^7+24 (x+1)^6-32 (x+1)^5+16 (x+1)^4+1}d\sqrt {\sqrt {x+1}+1}+2 \int \frac {(x+1)^3}{(x+1)^8-8 (x+1)^7+24 (x+1)^6-32 (x+1)^5+16 (x+1)^4+1}d\sqrt {\sqrt {x+1}+1}-\frac {1}{7} (x+1)^{7/2}+\frac {2}{5} (x+1)^{5/2}-\frac {1}{3} (x+1)^{3/2}\right )\) |
3.22.79.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
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 = 0.11 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.74
method | result | size |
derivativedivides | \(\frac {4 \left (1+\sqrt {1+x}\right )^{\frac {7}{2}}}{7}-\frac {8 \left (1+\sqrt {1+x}\right )^{\frac {5}{2}}}{5}+\frac {4 \left (1+\sqrt {1+x}\right )^{\frac {3}{2}}}{3}-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{16}-8 \textit {\_Z}^{14}+24 \textit {\_Z}^{12}-32 \textit {\_Z}^{10}+16 \textit {\_Z}^{8}+1\right )}{\sum }\frac {\left (\textit {\_R}^{6}-2 \textit {\_R}^{4}+\textit {\_R}^{2}\right ) \ln \left (\sqrt {1+\sqrt {1+x}}-\textit {\_R} \right )}{\textit {\_R}^{15}-7 \textit {\_R}^{13}+18 \textit {\_R}^{11}-20 \textit {\_R}^{9}+8 \textit {\_R}^{7}}\right )}{2}\) | \(119\) |
default | \(\frac {4 \left (1+\sqrt {1+x}\right )^{\frac {7}{2}}}{7}-\frac {8 \left (1+\sqrt {1+x}\right )^{\frac {5}{2}}}{5}+\frac {4 \left (1+\sqrt {1+x}\right )^{\frac {3}{2}}}{3}-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{16}-8 \textit {\_Z}^{14}+24 \textit {\_Z}^{12}-32 \textit {\_Z}^{10}+16 \textit {\_Z}^{8}+1\right )}{\sum }\frac {\left (\textit {\_R}^{6}-2 \textit {\_R}^{4}+\textit {\_R}^{2}\right ) \ln \left (\sqrt {1+\sqrt {1+x}}-\textit {\_R} \right )}{\textit {\_R}^{15}-7 \textit {\_R}^{13}+18 \textit {\_R}^{11}-20 \textit {\_R}^{9}+8 \textit {\_R}^{7}}\right )}{2}\) | \(119\) |
4/7*(1+(1+x)^(1/2))^(7/2)-8/5*(1+(1+x)^(1/2))^(5/2)+4/3*(1+(1+x)^(1/2))^(3 /2)-1/2*sum((_R^6-2*_R^4+_R^2)/(_R^15-7*_R^13+18*_R^11-20*_R^9+8*_R^7)*ln( (1+(1+x)^(1/2))^(1/2)-_R),_R=RootOf(_Z^16-8*_Z^14+24*_Z^12-32*_Z^10+16*_Z^ 8+1))
Timed out. \[ \int \frac {\sqrt {1+x} \left (-1+x^4\right ) \sqrt {1+\sqrt {1+x}}}{1+x^4} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sqrt {1+x} \left (-1+x^4\right ) \sqrt {1+\sqrt {1+x}}}{1+x^4} \, dx=\text {Timed out} \]
Not integrable
Time = 0.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.18 \[ \int \frac {\sqrt {1+x} \left (-1+x^4\right ) \sqrt {1+\sqrt {1+x}}}{1+x^4} \, dx=\int { \frac {{\left (x^{4} - 1\right )} \sqrt {x + 1} \sqrt {\sqrt {x + 1} + 1}}{x^{4} + 1} \,d x } \]
Not integrable
Time = 0.40 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.18 \[ \int \frac {\sqrt {1+x} \left (-1+x^4\right ) \sqrt {1+\sqrt {1+x}}}{1+x^4} \, dx=\int { \frac {{\left (x^{4} - 1\right )} \sqrt {x + 1} \sqrt {\sqrt {x + 1} + 1}}{x^{4} + 1} \,d x } \]
Not integrable
Time = 0.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.18 \[ \int \frac {\sqrt {1+x} \left (-1+x^4\right ) \sqrt {1+\sqrt {1+x}}}{1+x^4} \, dx=\int \frac {\left (x^4-1\right )\,\sqrt {\sqrt {x+1}+1}\,\sqrt {x+1}}{x^4+1} \,d x \]