\(\int \frac {1}{1+\sqrt {x+\sqrt {1+x^2}}} \, dx\) [9]

Optimal result

Integrand size = 19, antiderivative size = 84 \[ \int \frac {1}{1+\sqrt {x+\sqrt {1+x^2}}} \, dx=-\frac {1}{2 \left (x+\sqrt {1+x^2}\right )}+\frac {1}{\sqrt {x+\sqrt {1+x^2}}}+\sqrt {x+\sqrt {1+x^2}}+\frac {1}{2} \log \left (x+\sqrt {1+x^2}\right )-2 \log \left (1+\sqrt {x+\sqrt {1+x^2}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2142, 1835, 1634} \[ \int \frac {1}{1+\sqrt {x+\sqrt {1+x^2}}} \, dx=\sqrt {\sqrt {x^2+1}+x}+\frac {1}{\sqrt {\sqrt {x^2+1}+x}}-\frac {1}{2 \left (\sqrt {x^2+1}+x\right )}+\frac {1}{2} \log \left (\sqrt {x^2+1}+x\right )-2 \log \left (\sqrt {\sqrt {x^2+1}+x}+1\right ) \]

[In]

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

[Out]

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

Rule 1634

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 1835

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] -
1)*SubstFor[x^n, Pq, x]*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && PolyQ[Pq, x^n] && Intege
rQ[Simplify[(m + 1)/n]]

Rule 2142

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Dist[1/(2*
e), Subst[Int[(g + h*x^n)^p*((d^2 + a*f^2 - 2*d*x + x^2)/(d - x)^2), x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /
; FreeQ[{a, c, d, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.25 \[ \int \frac {1}{1+\sqrt {x+\sqrt {1+x^2}}} \, dx=\frac {1}{2} \left (\frac {-1+5 x+2 (1+x) \sqrt {x+\sqrt {1+x^2}}+\sqrt {1+x^2} \left (5+2 \sqrt {x+\sqrt {1+x^2}}\right )}{x+\sqrt {1+x^2}}+\log \left (x+\sqrt {1+x^2}\right )-4 \log \left (1+\sqrt {x+\sqrt {1+x^2}}\right )\right ) \]

[In]

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

[Out]

((-1 + 5*x + 2*(1 + x)*Sqrt[x + Sqrt[1 + x^2]] + Sqrt[1 + x^2]*(5 + 2*Sqrt[x + Sqrt[1 + x^2]]))/(x + Sqrt[1 +
x^2]) + Log[x + Sqrt[1 + x^2]] - 4*Log[1 + Sqrt[x + Sqrt[1 + x^2]]])/2

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.79 \[ \int \frac {1}{1+\sqrt {x+\sqrt {1+x^2}}} \, dx=-\sqrt {x + \sqrt {x^{2} + 1}} {\left (x - \sqrt {x^{2} + 1} - 1\right )} + \frac {1}{2} \, x - \frac {1}{2} \, \sqrt {x^{2} + 1} - 2 \, \log \left (\sqrt {x + \sqrt {x^{2} + 1}} + 1\right ) + \log \left (\sqrt {x + \sqrt {x^{2} + 1}}\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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