\(\int \frac {1+\sqrt {1+x^2}}{1+\sqrt {x+\sqrt {1+x^2}}} \, dx\) [2132]

Optimal result

Integrand size = 31, antiderivative size = 154 \[ \int \frac {1+\sqrt {1+x^2}}{1+\sqrt {x+\sqrt {1+x^2}}} \, dx=\frac {104-36 x+214 x^2-24 x^3+\left (40+48 x+72 x^2+16 x^3\right ) \sqrt {x+\sqrt {1+x^2}}+\sqrt {1+x^2} \left (-24+214 x-24 x^2+\left (40+72 x+16 x^2\right ) \sqrt {x+\sqrt {1+x^2}}\right )}{48 x \sqrt {1+x^2}+24 \left (1+2 x^2\right )}+\frac {5}{4} \log \left (x+\sqrt {1+x^2}\right )-4 \log \left (1+\sqrt {x+\sqrt {1+x^2}}\right ) \]

[Out]

(104-36*x+214*x^2-24*x^3+(16*x^3+72*x^2+48*x+40)*(x+(x^2+1)^(1/2))^(1/2)+(x^2+1)^(1/2)*(-24+214*x-24*x^2+(16*x
^2+72*x+40)*(x+(x^2+1)^(1/2))^(1/2)))/(48*x*(x^2+1)^(1/2)+48*x^2+24)+5/4*ln(x+(x^2+1)^(1/2))-4*ln(1+(x+(x^2+1)
^(1/2))^(1/2))

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.35, number of steps used = 32, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.710, Rules used = {6874, 2142, 1835, 1634, 201, 221, 272, 52, 65, 213, 14, 2147, 276, 2145, 473, 470, 335, 304, 209, 212, 477, 472} \[ \int \frac {1+\sqrt {1+x^2}}{1+\sqrt {x+\sqrt {1+x^2}}} \, dx=\frac {\text {arcsinh}(x)}{4}+\frac {1}{2} \text {arctanh}\left (\sqrt {x^2+1}\right )-2 \text {arctanh}\left (\sqrt {\sqrt {x^2+1}+x}\right )-\frac {x^2}{4}+\frac {1}{4} \sqrt {x^2+1} x+\frac {1}{6} \left (\sqrt {x^2+1}+x\right )^{3/2}-\frac {\sqrt {x^2+1}}{2}+\frac {3}{2} \sqrt {\sqrt {x^2+1}+x}+\frac {3}{2 \sqrt {\sqrt {x^2+1}+x}}-\frac {1}{2 \left (\sqrt {x^2+1}+x\right )}+\frac {1}{6 \left (\sqrt {x^2+1}+x\right )^{3/2}}+\frac {1}{2} \log \left (\sqrt {x^2+1}+x\right )-2 \log \left (\sqrt {\sqrt {x^2+1}+x}+1\right )-\frac {\log (x)}{2} \]

[In]

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

[Out]

-1/4*x^2 - Sqrt[1 + x^2]/2 + (x*Sqrt[1 + x^2])/4 + 1/(6*(x + Sqrt[1 + x^2])^(3/2)) - 1/(2*(x + Sqrt[1 + x^2]))
+ 3/(2*Sqrt[x + Sqrt[1 + x^2]]) + (3*Sqrt[x + Sqrt[1 + x^2]])/2 + (x + Sqrt[1 + x^2])^(3/2)/6 + ArcSinh[x]/4
+ ArcTanh[Sqrt[1 + x^2]]/2 - 2*ArcTanh[Sqrt[x + Sqrt[1 + x^2]]] - Log[x]/2 + Log[x + Sqrt[1 + x^2]]/2 - 2*Log[
1 + Sqrt[x + Sqrt[1 + x^2]]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 272

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

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 472

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[(e*x)^m*((a + b*x^n)^p/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 473

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[c^2*(e*x)^(m
+ 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rule 477

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/e^n))^p*(c + d*(x^(k*n)/e^n))^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

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]

Rule 2145

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

Rule 2147

Int[((g_) + (i_.)*(x_)^2)^(m_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Dis
t[(1/(2^(2*m + 1)*e*f^(2*m)))*(i/c)^m, Subst[Int[x^n*((d^2 + a*f^2 - 2*d*x + x^2)^(2*m + 1)/(-d + x)^(2*(m + 1
))), x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /; FreeQ[{a, c, d, e, f, g, i, n}, x] && EqQ[e^2 - c*f^2, 0] && E
qQ[c*g - a*i, 0] && IntegerQ[2*m] && (IntegerQ[m] || GtQ[i/c, 0])

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{1+\sqrt {x+\sqrt {1+x^2}}}+\frac {\sqrt {1+x^2}}{1+\sqrt {x+\sqrt {1+x^2}}}\right ) \, dx \\ & = \int \frac {1}{1+\sqrt {x+\sqrt {1+x^2}}} \, dx+\int \frac {\sqrt {1+x^2}}{1+\sqrt {x+\sqrt {1+x^2}}} \, dx \\ & = \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 )+\int \left (\frac {\sqrt {1+x^2}}{2}-\frac {\sqrt {1+x^2}}{2 x}-\frac {1+x^2}{2 x}-\frac {1}{2} \sqrt {1+x^2} \sqrt {x+\sqrt {1+x^2}}+\frac {\sqrt {1+x^2} \sqrt {x+\sqrt {1+x^2}}}{2 x}+\frac {\left (1+x^2\right ) \sqrt {x+\sqrt {1+x^2}}}{2 x}\right ) \, dx \\ & = \frac {1}{2} \int \sqrt {1+x^2} \, dx-\frac {1}{2} \int \frac {\sqrt {1+x^2}}{x} \, dx-\frac {1}{2} \int \frac {1+x^2}{x} \, dx-\frac {1}{2} \int \sqrt {1+x^2} \sqrt {x+\sqrt {1+x^2}} \, dx+\frac {1}{2} \int \frac {\sqrt {1+x^2} \sqrt {x+\sqrt {1+x^2}}}{x} \, dx+\frac {1}{2} \int \frac {\left (1+x^2\right ) \sqrt {x+\sqrt {1+x^2}}}{x} \, dx+\text {Subst}\left (\int \frac {1+x^4}{x^3 (1+x)} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right ) \\ & = \frac {1}{4} x \sqrt {1+x^2}-\frac {1}{8} \text {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^{5/2}} \, dx,x,x+\sqrt {1+x^2}\right )+\frac {1}{8} \text {Subst}\left (\int \frac {\left (1+x^2\right )^3}{x^{5/2} \left (-1+x^2\right )} \, dx,x,x+\sqrt {1+x^2}\right )+\frac {1}{4} \int \frac {1}{\sqrt {1+x^2}} \, dx-\frac {1}{4} \text {Subst}\left (\int \frac {\sqrt {1+x}}{x} \, dx,x,x^2\right )+\frac {1}{4} \text {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^{3/2} \left (-1+x^2\right )} \, dx,x,x+\sqrt {1+x^2}\right )-\frac {1}{2} \int \left (\frac {1}{x}+x\right ) \, dx+\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 {x^2}{4}-\frac {\sqrt {1+x^2}}{2}+\frac {1}{4} x \sqrt {1+x^2}-\frac {1}{2 \left (x+\sqrt {1+x^2}\right )}+\frac {3}{2 \sqrt {x+\sqrt {1+x^2}}}+\sqrt {x+\sqrt {1+x^2}}+\frac {\text {arcsinh}(x)}{4}-\frac {\log (x)}{2}+\frac {1}{2} \log \left (x+\sqrt {1+x^2}\right )-2 \log \left (1+\sqrt {x+\sqrt {1+x^2}}\right )-\frac {1}{8} \text {Subst}\left (\int \left (\frac {1}{x^{5/2}}+\frac {2}{\sqrt {x}}+x^{3/2}\right ) \, dx,x,x+\sqrt {1+x^2}\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{x \sqrt {1+x}} \, dx,x,x^2\right )+\frac {1}{4} \text {Subst}\left (\int \frac {\left (1+x^4\right )^3}{x^4 \left (-1+x^4\right )} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )-\frac {1}{2} \text {Subst}\left (\int \frac {\sqrt {x} \left (-\frac {3}{2}-\frac {x^2}{2}\right )}{-1+x^2} \, dx,x,x+\sqrt {1+x^2}\right ) \\ & = -\frac {x^2}{4}-\frac {\sqrt {1+x^2}}{2}+\frac {1}{4} x \sqrt {1+x^2}+\frac {1}{12 \left (x+\sqrt {1+x^2}\right )^{3/2}}-\frac {1}{2 \left (x+\sqrt {1+x^2}\right )}+\frac {3}{2 \sqrt {x+\sqrt {1+x^2}}}+\frac {1}{2} \sqrt {x+\sqrt {1+x^2}}+\frac {1}{6} \left (x+\sqrt {1+x^2}\right )^{3/2}-\frac {1}{20} \left (x+\sqrt {1+x^2}\right )^{5/2}+\frac {\text {arcsinh}(x)}{4}-\frac {\log (x)}{2}+\frac {1}{2} \log \left (x+\sqrt {1+x^2}\right )-2 \log \left (1+\sqrt {x+\sqrt {1+x^2}}\right )+\frac {1}{4} \text {Subst}\left (\int \left (4-\frac {1}{x^4}+x^4+\frac {4}{-1+x^2}-\frac {4}{1+x^2}\right ) \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+x^2}\right )+\text {Subst}\left (\int \frac {\sqrt {x}}{-1+x^2} \, dx,x,x+\sqrt {1+x^2}\right ) \\ & = -\frac {x^2}{4}-\frac {\sqrt {1+x^2}}{2}+\frac {1}{4} x \sqrt {1+x^2}+\frac {1}{6 \left (x+\sqrt {1+x^2}\right )^{3/2}}-\frac {1}{2 \left (x+\sqrt {1+x^2}\right )}+\frac {3}{2 \sqrt {x+\sqrt {1+x^2}}}+\frac {3}{2} \sqrt {x+\sqrt {1+x^2}}+\frac {1}{6} \left (x+\sqrt {1+x^2}\right )^{3/2}+\frac {\text {arcsinh}(x)}{4}+\frac {1}{2} \text {arctanh}\left (\sqrt {1+x^2}\right )-\frac {\log (x)}{2}+\frac {1}{2} \log \left (x+\sqrt {1+x^2}\right )-2 \log \left (1+\sqrt {x+\sqrt {1+x^2}}\right )+2 \text {Subst}\left (\int \frac {x^2}{-1+x^4} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )+\text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )-\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right ) \\ & = -\frac {x^2}{4}-\frac {\sqrt {1+x^2}}{2}+\frac {1}{4} x \sqrt {1+x^2}+\frac {1}{6 \left (x+\sqrt {1+x^2}\right )^{3/2}}-\frac {1}{2 \left (x+\sqrt {1+x^2}\right )}+\frac {3}{2 \sqrt {x+\sqrt {1+x^2}}}+\frac {3}{2} \sqrt {x+\sqrt {1+x^2}}+\frac {1}{6} \left (x+\sqrt {1+x^2}\right )^{3/2}+\frac {\text {arcsinh}(x)}{4}-\arctan \left (\sqrt {x+\sqrt {1+x^2}}\right )+\frac {1}{2} \text {arctanh}\left (\sqrt {1+x^2}\right )-\text {arctanh}\left (\sqrt {x+\sqrt {1+x^2}}\right )-\frac {\log (x)}{2}+\frac {1}{2} \log \left (x+\sqrt {1+x^2}\right )-2 \log \left (1+\sqrt {x+\sqrt {1+x^2}}\right )-\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )+\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right ) \\ & = -\frac {x^2}{4}-\frac {\sqrt {1+x^2}}{2}+\frac {1}{4} x \sqrt {1+x^2}+\frac {1}{6 \left (x+\sqrt {1+x^2}\right )^{3/2}}-\frac {1}{2 \left (x+\sqrt {1+x^2}\right )}+\frac {3}{2 \sqrt {x+\sqrt {1+x^2}}}+\frac {3}{2} \sqrt {x+\sqrt {1+x^2}}+\frac {1}{6} \left (x+\sqrt {1+x^2}\right )^{3/2}+\frac {\text {arcsinh}(x)}{4}+\frac {1}{2} \text {arctanh}\left (\sqrt {1+x^2}\right )-2 \text {arctanh}\left (\sqrt {x+\sqrt {1+x^2}}\right )-\frac {\log (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 = 0.23 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00 \[ \int \frac {1+\sqrt {1+x^2}}{1+\sqrt {x+\sqrt {1+x^2}}} \, dx=\frac {104-36 x+214 x^2-24 x^3+\left (40+48 x+72 x^2+16 x^3\right ) \sqrt {x+\sqrt {1+x^2}}+\sqrt {1+x^2} \left (-24+214 x-24 x^2+\left (40+72 x+16 x^2\right ) \sqrt {x+\sqrt {1+x^2}}\right )}{48 x \sqrt {1+x^2}+24 \left (1+2 x^2\right )}+\frac {5}{4} \log \left (x+\sqrt {1+x^2}\right )-4 \log \left (1+\sqrt {x+\sqrt {1+x^2}}\right ) \]

[In]

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

[Out]

(104 - 36*x + 214*x^2 - 24*x^3 + (40 + 48*x + 72*x^2 + 16*x^3)*Sqrt[x + Sqrt[1 + x^2]] + Sqrt[1 + x^2]*(-24 +
214*x - 24*x^2 + (40 + 72*x + 16*x^2)*Sqrt[x + Sqrt[1 + x^2]]))/(48*x*Sqrt[1 + x^2] + 24*(1 + 2*x^2)) + (5*Log
[x + Sqrt[1 + x^2]])/4 - 4*Log[1 + Sqrt[x + Sqrt[1 + x^2]]]

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.55 \[ \int \frac {1+\sqrt {1+x^2}}{1+\sqrt {x+\sqrt {1+x^2}}} \, dx=-\frac {1}{4} \, x^{2} + \frac {1}{3} \, {\left (x^{2} - \sqrt {x^{2} + 1} {\left (x - 5\right )} - 4 \, x + 5\right )} \sqrt {x + \sqrt {x^{2} + 1}} + \frac {1}{4} \, \sqrt {x^{2} + 1} {\left (x - 4\right )} + \frac {1}{2} \, x - 4 \, \log \left (\sqrt {x + \sqrt {x^{2} + 1}} + 1\right ) + \frac {5}{2} \, \log \left (\sqrt {x + \sqrt {x^{2} + 1}}\right ) \]

[In]

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

[Out]

-1/4*x^2 + 1/3*(x^2 - sqrt(x^2 + 1)*(x - 5) - 4*x + 5)*sqrt(x + sqrt(x^2 + 1)) + 1/4*sqrt(x^2 + 1)*(x - 4) + 1
/2*x - 4*log(sqrt(x + sqrt(x^2 + 1)) + 1) + 5/2*log(sqrt(x + sqrt(x^2 + 1)))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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