\(\int \frac {\sqrt {x+\sqrt {1+x}}}{x^2-\sqrt {1+x}} \, dx\) [2486]

Optimal result

Integrand size = 29, antiderivative size = 204 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x^2-\sqrt {1+x}} \, dx=2 \text {RootSum}\left [-1+6 \text {$\#$1}+3 \text {$\#$1}^2-2 \text {$\#$1}^3-8 \text {$\#$1}^4+16 \text {$\#$1}^5-4 \text {$\#$1}^6+\text {$\#$1}^8\&,\frac {\log \left (\sqrt {1+x}-\sqrt {x+\sqrt {1+x}}+\text {$\#$1}\right )+2 \log \left (\sqrt {1+x}-\sqrt {x+\sqrt {1+x}}+\text {$\#$1}\right ) \text {$\#$1}-2 \log \left (\sqrt {1+x}-\sqrt {x+\sqrt {1+x}}+\text {$\#$1}\right ) \text {$\#$1}^5+\log \left (\sqrt {1+x}-\sqrt {x+\sqrt {1+x}}+\text {$\#$1}\right ) \text {$\#$1}^6}{3+3 \text {$\#$1}-3 \text {$\#$1}^2-16 \text {$\#$1}^3+40 \text {$\#$1}^4-12 \text {$\#$1}^5+4 \text {$\#$1}^7}\&\right ] \]

[Out]

Unintegrable

Rubi [F]

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

[In]

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

[Out]

2*Defer[Subst][Defer[Int][(x*Sqrt[-1 + x + x^2])/(1 - x - 2*x^2 + x^4), x], x, Sqrt[1 + x]]

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.04 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x^2-\sqrt {1+x}} \, dx=2 \text {RootSum}\left [-1+6 \text {$\#$1}+3 \text {$\#$1}^2-2 \text {$\#$1}^3-8 \text {$\#$1}^4+16 \text {$\#$1}^5-4 \text {$\#$1}^6+\text {$\#$1}^8\&,\frac {\log \left (-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}-\text {$\#$1}\right )+2 \log \left (-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}-2 \log \left (-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^5+\log \left (-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^6}{3+3 \text {$\#$1}-3 \text {$\#$1}^2-16 \text {$\#$1}^3+40 \text {$\#$1}^4-12 \text {$\#$1}^5+4 \text {$\#$1}^7}\&\right ] \]

[In]

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

[Out]

2*RootSum[-1 + 6*#1 + 3*#1^2 - 2*#1^3 - 8*#1^4 + 16*#1^5 - 4*#1^6 + #1^8 & , (Log[-Sqrt[1 + x] + Sqrt[x + Sqrt
[1 + x]] - #1] + 2*Log[-Sqrt[1 + x] + Sqrt[x + Sqrt[1 + x]] - #1]*#1 - 2*Log[-Sqrt[1 + x] + Sqrt[x + Sqrt[1 +
x]] - #1]*#1^5 + Log[-Sqrt[1 + x] + Sqrt[x + Sqrt[1 + x]] - #1]*#1^6)/(3 + 3*#1 - 3*#1^2 - 16*#1^3 + 40*#1^4 -
12*#1^5 + 4*#1^7) & ]

Maple [N/A] (verified)

Time = 0.05 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.52

method	result	size
derivativedivides	\(2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-4 \textit {\_Z}^{6}+16 \textit {\_Z}^{5}-8 \textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z}^{2}+6 \textit {\_Z} -1\right )}{\sum }\frac {\left (\textit {\_R}^{6}-2 \textit {\_R}^{5}+2 \textit {\_R} +1\right ) \ln \left (\sqrt {x +\sqrt {1+x}}-\sqrt {1+x}-\textit {\_R} \right )}{4 \textit {\_R}^{7}-12 \textit {\_R}^{5}+40 \textit {\_R}^{4}-16 \textit {\_R}^{3}-3 \textit {\_R}^{2}+3 \textit {\_R} +3}\right )\)	\(107\)
default	\(2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-4 \textit {\_Z}^{6}+16 \textit {\_Z}^{5}-8 \textit {\_Z}^{4}-2 \textit {\_Z}^{3}+3 \textit {\_Z}^{2}+6 \textit {\_Z} -1\right )}{\sum }\frac {\left (\textit {\_R}^{6}-2 \textit {\_R}^{5}+2 \textit {\_R} +1\right ) \ln \left (\sqrt {x +\sqrt {1+x}}-\sqrt {1+x}-\textit {\_R} \right )}{4 \textit {\_R}^{7}-12 \textit {\_R}^{5}+40 \textit {\_R}^{4}-16 \textit {\_R}^{3}-3 \textit {\_R}^{2}+3 \textit {\_R} +3}\right )\)	\(107\)

[In]

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

[Out]

2*sum((_R^6-2*_R^5+2*_R+1)/(4*_R^7-12*_R^5+40*_R^4-16*_R^3-3*_R^2+3*_R+3)*ln((x+(1+x)^(1/2))^(1/2)-(1+x)^(1/2)
-_R),_R=RootOf(_Z^8-4*_Z^6+16*_Z^5-8*_Z^4-2*_Z^3+3*_Z^2+6*_Z-1))

Fricas [C] (verification not implemented)

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

Time = 18.15 (sec) , antiderivative size = 36926, normalized size of antiderivative = 181.01 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x^2-\sqrt {1+x}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [N/A]

Not integrable

Time = 1.59 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.11 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x^2-\sqrt {1+x}} \, dx=\int \frac {\sqrt {x + \sqrt {x + 1}}}{x^{2} - \sqrt {x + 1}}\, dx \]

[In]

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

[Out]

Integral(sqrt(x + sqrt(x + 1))/(x**2 - sqrt(x + 1)), x)

Maxima [N/A]

Not integrable

Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.12 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x^2-\sqrt {1+x}} \, dx=\int { \frac {\sqrt {x + \sqrt {x + 1}}}{x^{2} - \sqrt {x + 1}} \,d x } \]

[In]

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

[Out]

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

Giac [N/A]

Not integrable

Time = 0.42 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.12 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x^2-\sqrt {1+x}} \, dx=\int { \frac {\sqrt {x + \sqrt {x + 1}}}{x^{2} - \sqrt {x + 1}} \,d x } \]

[In]

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

[Out]

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

Mupad [N/A]

Not integrable

Time = 6.97 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.13 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x^2-\sqrt {1+x}} \, dx=-\int \frac {\sqrt {x+\sqrt {x+1}}}{\sqrt {x+1}-x^2} \,d x \]

[In]

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

[Out]

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