\(\int \frac {\sqrt {1+\sqrt {1+x^2}}}{x+\sqrt {1+x^2}} \, dx\) [1500]

Optimal result

Integrand size = 29, antiderivative size = 104 \[ \int \frac {\sqrt {1+\sqrt {1+x^2}}}{x+\sqrt {1+x^2}} \, dx=\frac {2 x \left (11+3 x^2\right )}{15 \sqrt {1+\sqrt {1+x^2}}}-\frac {2}{15} \left (1+3 x^2\right ) \sqrt {1+\sqrt {1+x^2}}+\sqrt {1+x^2} \left (\frac {8 x}{15 \sqrt {1+\sqrt {1+x^2}}}-\frac {2}{15} \sqrt {1+\sqrt {1+x^2}}\right ) \]

[Out]

2/15*x*(3*x^2+11)/(1+(x^2+1)^(1/2))^(1/2)-2/15*(3*x^2+1)*(1+(x^2+1)^(1/2))^(1/2)+(x^2+1)^(1/2)*(8/15*x/(1+(x^2
+1)^(1/2))^(1/2)-2/15*(1+(x^2+1)^(1/2))^(1/2))

Rubi [F]

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

[In]

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

[Out]

(2*(1 + Sqrt[1 + x^2])^(3/2))/3 - (2*(1 + Sqrt[1 + x^2])^(5/2))/5 + Defer[Int][Sqrt[1 + x^2]*Sqrt[1 + Sqrt[1 +
x^2]], x]

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt {1+\sqrt {1+x^2}}}{x+\sqrt {1+x^2}} \, dx=\frac {6 x^3-4 \left (1+\sqrt {1+x^2}\right )-2 x^2 \left (4+3 \sqrt {1+x^2}\right )+x \left (22+8 \sqrt {1+x^2}\right )}{15 \sqrt {1+\sqrt {1+x^2}}} \]

[In]

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

[Out]

(6*x^3 - 4*(1 + Sqrt[1 + x^2]) - 2*x^2*(4 + 3*Sqrt[1 + x^2]) + x*(22 + 8*Sqrt[1 + x^2]))/(15*Sqrt[1 + Sqrt[1 +
x^2]])

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.46 \[ \int \frac {\sqrt {1+\sqrt {1+x^2}}}{x+\sqrt {1+x^2}} \, dx=-\frac {2 \, {\left (3 \, x^{3} - x^{2} - {\left (3 \, x^{2} - x + 7\right )} \sqrt {x^{2} + 1} + x + 7\right )} \sqrt {\sqrt {x^{2} + 1} + 1}}{15 \, x} \]

[In]

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

[Out]

-2/15*(3*x^3 - x^2 - (3*x^2 - x + 7)*sqrt(x^2 + 1) + x + 7)*sqrt(sqrt(x^2 + 1) + 1)/x

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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