\(\int \frac {(-1+x^2) \sqrt {1+\sqrt {1+x}}}{1+x^2} \, dx\) [1432]

Optimal result

Integrand size = 26, antiderivative size = 101 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+\sqrt {1+x}}}{1+x^2} \, dx=\frac {4}{15} \sqrt {1+x} \sqrt {1+\sqrt {1+x}}+\frac {4}{15} (1+3 x) \sqrt {1+\sqrt {1+x}}-\text {RootSum}\left [1+4 \text {$\#$1}^4-4 \text {$\#$1}^6+\text {$\#$1}^8\&,\frac {\log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right )}{-2 \text {$\#$1}+\text {$\#$1}^3}\&\right ] \]

[Out]

Unintegrable

Rubi [F]

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

[In]

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

[Out]

(-4*(1 + Sqrt[1 + x])^(3/2))/3 + (4*(1 + Sqrt[1 + x])^(5/2))/5 + 8*Defer[Subst][Defer[Int][x^2/(1 + 4*x^4 - 4*
x^6 + x^8), x], x, Sqrt[1 + Sqrt[1 + x]]] - 8*Defer[Subst][Defer[Int][x^4/(1 + 4*x^4 - 4*x^6 + x^8), x], x, Sq
rt[1 + Sqrt[1 + x]]]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {x^3 \sqrt {1+x} \left (-2+x^2\right )}{2-2 x^2+x^4} \, dx,x,\sqrt {1+x}\right ) \\ & = 4 \text {Subst}\left (\int \frac {x^2 \left (-1+x^2\right )^3 \left (-1-2 x^2+x^4\right )}{1+4 x^4-4 x^6+x^8} \, dx,x,\sqrt {1+\sqrt {1+x}}\right ) \\ & = 4 \text {Subst}\left (\int \left (-x^2+x^4-\frac {2 x^2 \left (-1+x^2\right )}{1+4 x^4-4 x^6+x^8}\right ) \, dx,x,\sqrt {1+\sqrt {1+x}}\right ) \\ & = -\frac {4}{3} \left (1+\sqrt {1+x}\right )^{3/2}+\frac {4}{5} \left (1+\sqrt {1+x}\right )^{5/2}-8 \text {Subst}\left (\int \frac {x^2 \left (-1+x^2\right )}{1+4 x^4-4 x^6+x^8} \, dx,x,\sqrt {1+\sqrt {1+x}}\right ) \\ & = -\frac {4}{3} \left (1+\sqrt {1+x}\right )^{3/2}+\frac {4}{5} \left (1+\sqrt {1+x}\right )^{5/2}-8 \text {Subst}\left (\int \left (-\frac {x^2}{1+4 x^4-4 x^6+x^8}+\frac {x^4}{1+4 x^4-4 x^6+x^8}\right ) \, dx,x,\sqrt {1+\sqrt {1+x}}\right ) \\ & = -\frac {4}{3} \left (1+\sqrt {1+x}\right )^{3/2}+\frac {4}{5} \left (1+\sqrt {1+x}\right )^{5/2}+8 \text {Subst}\left (\int \frac {x^2}{1+4 x^4-4 x^6+x^8} \, dx,x,\sqrt {1+\sqrt {1+x}}\right )-8 \text {Subst}\left (\int \frac {x^4}{1+4 x^4-4 x^6+x^8} \, dx,x,\sqrt {1+\sqrt {1+x}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.85 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+\sqrt {1+x}}}{1+x^2} \, dx=\frac {4}{15} \sqrt {1+\sqrt {1+x}} \left (-2+\sqrt {1+x}+3 (1+x)\right )-\text {RootSum}\left [1+4 \text {$\#$1}^4-4 \text {$\#$1}^6+\text {$\#$1}^8\&,\frac {\log \left (\sqrt {1+\sqrt {1+x}}-\text {$\#$1}\right )}{-2 \text {$\#$1}+\text {$\#$1}^3}\&\right ] \]

[In]

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

[Out]

(4*Sqrt[1 + Sqrt[1 + x]]*(-2 + Sqrt[1 + x] + 3*(1 + x)))/15 - RootSum[1 + 4*#1^4 - 4*#1^6 + #1^8 & , Log[Sqrt[
1 + Sqrt[1 + x]] - #1]/(-2*#1 + #1^3) & ]

Maple [N/A] (verified)

Time = 0.06 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\frac {4 \left (1+\sqrt {1+x}\right )^{\frac {5}{2}}}{5}-\frac {4 \left (1+\sqrt {1+x}\right )^{\frac {3}{2}}}{3}-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-4 \textit {\_Z}^{6}+4 \textit {\_Z}^{4}+1\right )}{\sum }\frac {\left (\textit {\_R}^{4}-\textit {\_R}^{2}\right ) \ln \left (\sqrt {1+\sqrt {1+x}}-\textit {\_R} \right )}{\textit {\_R}^{7}-3 \textit {\_R}^{5}+2 \textit {\_R}^{3}}\right )\)	\(85\)
default	\(\frac {4 \left (1+\sqrt {1+x}\right )^{\frac {5}{2}}}{5}-\frac {4 \left (1+\sqrt {1+x}\right )^{\frac {3}{2}}}{3}-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-4 \textit {\_Z}^{6}+4 \textit {\_Z}^{4}+1\right )}{\sum }\frac {\left (\textit {\_R}^{4}-\textit {\_R}^{2}\right ) \ln \left (\sqrt {1+\sqrt {1+x}}-\textit {\_R} \right )}{\textit {\_R}^{7}-3 \textit {\_R}^{5}+2 \textit {\_R}^{3}}\right )\)	\(85\)

[In]

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

[Out]

4/5*(1+(1+x)^(1/2))^(5/2)-4/3*(1+(1+x)^(1/2))^(3/2)-sum((_R^4-_R^2)/(_R^7-3*_R^5+2*_R^3)*ln((1+(1+x)^(1/2))^(1
/2)-_R),_R=RootOf(_Z^8-4*_Z^6+4*_Z^4+1))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 0.30 (sec) , antiderivative size = 302, normalized size of antiderivative = 2.99 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+\sqrt {1+x}}}{1+x^2} \, dx=\frac {4}{15} \, {\left (3 \, x + \sqrt {x + 1} + 1\right )} \sqrt {\sqrt {x + 1} + 1} + \frac {1}{2} \, \sqrt {2 \, \sqrt {4 i + 4} - 4} \log \left (i \, \sqrt {2 \, \sqrt {4 i + 4} - 4} + 2 \, \sqrt {\sqrt {x + 1} + 1}\right ) - \frac {1}{2} \, \sqrt {2 \, \sqrt {4 i + 4} - 4} \log \left (-i \, \sqrt {2 \, \sqrt {4 i + 4} - 4} + 2 \, \sqrt {\sqrt {x + 1} + 1}\right ) + \frac {1}{2} \, \sqrt {-2 \, \sqrt {4 i + 4} - 4} \log \left (i \, \sqrt {-2 \, \sqrt {4 i + 4} - 4} + 2 \, \sqrt {\sqrt {x + 1} + 1}\right ) - \frac {1}{2} \, \sqrt {-2 \, \sqrt {4 i + 4} - 4} \log \left (-i \, \sqrt {-2 \, \sqrt {4 i + 4} - 4} + 2 \, \sqrt {\sqrt {x + 1} + 1}\right ) - \frac {1}{2} \, \sqrt {2 \, \sqrt {-4 i + 4} - 4} \log \left (i \, \sqrt {2 \, \sqrt {-4 i + 4} - 4} + 2 \, \sqrt {\sqrt {x + 1} + 1}\right ) + \frac {1}{2} \, \sqrt {2 \, \sqrt {-4 i + 4} - 4} \log \left (-i \, \sqrt {2 \, \sqrt {-4 i + 4} - 4} + 2 \, \sqrt {\sqrt {x + 1} + 1}\right ) - \frac {1}{2} \, \sqrt {-2 \, \sqrt {-4 i + 4} - 4} \log \left (i \, \sqrt {-2 \, \sqrt {-4 i + 4} - 4} + 2 \, \sqrt {\sqrt {x + 1} + 1}\right ) + \frac {1}{2} \, \sqrt {-2 \, \sqrt {-4 i + 4} - 4} \log \left (-i \, \sqrt {-2 \, \sqrt {-4 i + 4} - 4} + 2 \, \sqrt {\sqrt {x + 1} + 1}\right ) \]

[In]

integrate((x^2-1)*(1+(1+x)^(1/2))^(1/2)/(x^2+1),x, algorithm="fricas")

[Out]

4/15*(3*x + sqrt(x + 1) + 1)*sqrt(sqrt(x + 1) + 1) + 1/2*sqrt(2*sqrt(4*I + 4) - 4)*log(I*sqrt(2*sqrt(4*I + 4)
- 4) + 2*sqrt(sqrt(x + 1) + 1)) - 1/2*sqrt(2*sqrt(4*I + 4) - 4)*log(-I*sqrt(2*sqrt(4*I + 4) - 4) + 2*sqrt(sqrt
(x + 1) + 1)) + 1/2*sqrt(-2*sqrt(4*I + 4) - 4)*log(I*sqrt(-2*sqrt(4*I + 4) - 4) + 2*sqrt(sqrt(x + 1) + 1)) - 1
/2*sqrt(-2*sqrt(4*I + 4) - 4)*log(-I*sqrt(-2*sqrt(4*I + 4) - 4) + 2*sqrt(sqrt(x + 1) + 1)) - 1/2*sqrt(2*sqrt(-
4*I + 4) - 4)*log(I*sqrt(2*sqrt(-4*I + 4) - 4) + 2*sqrt(sqrt(x + 1) + 1)) + 1/2*sqrt(2*sqrt(-4*I + 4) - 4)*log
(-I*sqrt(2*sqrt(-4*I + 4) - 4) + 2*sqrt(sqrt(x + 1) + 1)) - 1/2*sqrt(-2*sqrt(-4*I + 4) - 4)*log(I*sqrt(-2*sqrt
(-4*I + 4) - 4) + 2*sqrt(sqrt(x + 1) + 1)) + 1/2*sqrt(-2*sqrt(-4*I + 4) - 4)*log(-I*sqrt(-2*sqrt(-4*I + 4) - 4
) + 2*sqrt(sqrt(x + 1) + 1))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [N/A]

Not integrable

Time = 0.34 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.24 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+\sqrt {1+x}}}{1+x^2} \, dx=\int { \frac {{\left (x^{2} - 1\right )} \sqrt {\sqrt {x + 1} + 1}}{x^{2} + 1} \,d x } \]

[In]

integrate((x^2-1)*(1+(1+x)^(1/2))^(1/2)/(x^2+1),x, algorithm="maxima")

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+\sqrt {1+x}}}{1+x^2} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((x^2-1)*(1+(1+x)^(1/2))^(1/2)/(x^2+1),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Invalid _EXT in replace_ext Error: Bad Argument ValueInvalid _EXT in replace_ext Error: Bad Argument ValueD
one

Mupad [N/A]

Not integrable

Time = 6.77 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.24 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+\sqrt {1+x}}}{1+x^2} \, dx=\int \frac {\left (x^2-1\right )\,\sqrt {\sqrt {x+1}+1}}{x^2+1} \,d x \]

[In]

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

[Out]

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