\(\int \frac {\sqrt {1+\sqrt {x+\sqrt {1+x^2}}}}{\sqrt {1+x^2}} \, dx\) [618]

Optimal result

Integrand size = 31, antiderivative size = 48 \[ \int \frac {\sqrt {1+\sqrt {x+\sqrt {1+x^2}}}}{\sqrt {1+x^2}} \, dx=4 \sqrt {1+\sqrt {x+\sqrt {1+x^2}}}-4 \text {arctanh}\left (\sqrt {1+\sqrt {x+\sqrt {1+x^2}}}\right ) \]

[Out]

4*(1+(x+(x^2+1)^(1/2))^(1/2))^(1/2)-4*arctanh((1+(x+(x^2+1)^(1/2))^(1/2))^(1/2))

Rubi [F]

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

[In]

Int[Sqrt[1 + Sqrt[x + Sqrt[1 + x^2]]]/Sqrt[1 + x^2],x]

[Out]

Defer[Int][Sqrt[1 + Sqrt[x + Sqrt[1 + x^2]]]/Sqrt[1 + x^2], x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\sqrt {1+\sqrt {x+\sqrt {1+x^2}}}}{\sqrt {1+x^2}} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {1+\sqrt {x+\sqrt {1+x^2}}}}{\sqrt {1+x^2}} \, dx=4 \sqrt {1+\sqrt {x+\sqrt {1+x^2}}}-4 \text {arctanh}\left (\sqrt {1+\sqrt {x+\sqrt {1+x^2}}}\right ) \]

[In]

Integrate[Sqrt[1 + Sqrt[x + Sqrt[1 + x^2]]]/Sqrt[1 + x^2],x]

[Out]

4*Sqrt[1 + Sqrt[x + Sqrt[1 + x^2]]] - 4*ArcTanh[Sqrt[1 + Sqrt[x + Sqrt[1 + x^2]]]]

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.23

method	result	size
derivativedivides	\(4 \sqrt {1+\sqrt {x +\sqrt {x^{2}+1}}}+2 \ln \left (\sqrt {1+\sqrt {x +\sqrt {x^{2}+1}}}-1\right )-2 \ln \left (\sqrt {1+\sqrt {x +\sqrt {x^{2}+1}}}+1\right )\)	\(59\)
default	\(4 \sqrt {1+\sqrt {x +\sqrt {x^{2}+1}}}+2 \ln \left (\sqrt {1+\sqrt {x +\sqrt {x^{2}+1}}}-1\right )-2 \ln \left (\sqrt {1+\sqrt {x +\sqrt {x^{2}+1}}}+1\right )\)	\(59\)

[In]

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

[Out]

4*(1+(x+(x^2+1)^(1/2))^(1/2))^(1/2)+2*ln((1+(x+(x^2+1)^(1/2))^(1/2))^(1/2)-1)-2*ln((1+(x+(x^2+1)^(1/2))^(1/2))
^(1/2)+1)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt {1+\sqrt {x+\sqrt {1+x^2}}}}{\sqrt {1+x^2}} \, dx=4 \, \sqrt {\sqrt {x + \sqrt {x^{2} + 1}} + 1} - 2 \, \log \left (\sqrt {\sqrt {x + \sqrt {x^{2} + 1}} + 1} + 1\right ) + 2 \, \log \left (\sqrt {\sqrt {x + \sqrt {x^{2} + 1}} + 1} - 1\right ) \]

[In]

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

[Out]

4*sqrt(sqrt(x + sqrt(x^2 + 1)) + 1) - 2*log(sqrt(sqrt(x + sqrt(x^2 + 1)) + 1) + 1) + 2*log(sqrt(sqrt(x + sqrt(
x^2 + 1)) + 1) - 1)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {1+\sqrt {x+\sqrt {1+x^2}}}}{\sqrt {1+x^2}} \, dx=\int \frac {\sqrt {\sqrt {x+\sqrt {x^2+1}}+1}}{\sqrt {x^2+1}} \,d x \]

[In]

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

[Out]

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