3.14.55 \(\int \frac {x+\sqrt {1+x^2}}{1+\sqrt {x+\sqrt {1+x^2}}} \, dx\)

Optimal. Leaf size=97 \[ \sqrt {x^2+1} \left (\frac {1}{3} \sqrt {\sqrt {x^2+1}+x}-\frac {1}{2}\right )+\frac {1}{3} (x+3) \sqrt {\sqrt {x^2+1}+x}+\frac {1}{2} \log \left (\sqrt {x^2+1}+x\right )-2 \log \left (\sqrt {\sqrt {x^2+1}+x}+1\right )-\frac {x}{2} \]

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Rubi [A]  time = 0.86, antiderivative size = 91, normalized size of antiderivative = 0.94, number of steps used = 41, number of rules used = 22, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.710, Rules used = {6742, 195, 215, 2117, 14, 2119, 448, 2122, 270, 266, 50, 63, 207, 2120, 462, 459, 329, 298, 203, 206, 466, 461} \begin {gather*} \frac {1}{3} \left (\sqrt {x^2+1}+x\right )^{3/2}+\sqrt {\sqrt {x^2+1}+x}-\frac {\sqrt {x^2+1}}{2}+\frac {1}{2} \tanh ^{-1}\left (\sqrt {x^2+1}\right )-2 \tanh ^{-1}\left (\sqrt {\sqrt {x^2+1}+x}\right )-\frac {x}{2}-\frac {\log (x)}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*x - Sqrt[1 + x^2]/2 + Sqrt[x + Sqrt[1 + x^2]] + (x + Sqrt[1 + x^2])^(3/2)/3 + ArcTanh[Sqrt[1 + x^2]]/2 -
2*ArcTanh[Sqrt[x + Sqrt[1 + x^2]]] - Log[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 50

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 63

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 195

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 203

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

Rule 206

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

Rule 207

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

Rule 215

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 266

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 270

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 298

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 329

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 448

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

Rule 459

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 461

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 462

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 466

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 2117

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 2119

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

Rule 2120

Int[(x_)^(p_.)*((g_) + (i_.)*(x_)^2)^(m_.)*((e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_.), x_Symbol] :>
Dist[(1*(i/c)^m)/(2^(2*m + p + 1)*e^(p + 1)*f^(2*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 2122

Int[((g_) + (i_.)*(x_)^2)^(m_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Dis
t[(1*(i/c)^m)/(2^(2*m + 1)*e*f^(2*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 6742

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

Rubi steps

\begin {align*} \int \frac {x+\sqrt {1+x^2}}{1+\sqrt {x+\sqrt {1+x^2}}} \, dx &=\int \left (\frac {x}{1+\sqrt {x+\sqrt {1+x^2}}}+\frac {\sqrt {1+x^2}}{1+\sqrt {x+\sqrt {1+x^2}}}\right ) \, dx\\ &=\int \frac {x}{1+\sqrt {x+\sqrt {1+x^2}}} \, dx+\int \frac {\sqrt {1+x^2}}{1+\sqrt {x+\sqrt {1+x^2}}} \, dx\\ &=\int \left (-\frac {1}{2}+\frac {x}{2}-\frac {\sqrt {1+x^2}}{2}+\frac {1}{2} \sqrt {x+\sqrt {1+x^2}}-\frac {1}{2} x \sqrt {x+\sqrt {1+x^2}}+\frac {1}{2} \sqrt {1+x^2} \sqrt {x+\sqrt {1+x^2}}\right ) \, dx+\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 {x}{2}+\frac {x^2}{4}-\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 {x+\sqrt {1+x^2}} \, dx-\frac {1}{2} \int x \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\\ &=-\frac {x}{2}+\frac {x^2}{4}-\frac {1}{8} \operatorname {Subst}\left (\int \frac {\left (-1+x^2\right ) \left (1+x^2\right )}{x^{5/2}} \, dx,x,x+\sqrt {1+x^2}\right )+\frac {1}{8} \operatorname {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} \operatorname {Subst}\left (\int \frac {\sqrt {1+x}}{x} \, dx,x,x^2\right )+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1+x^2}{x^{3/2}} \, dx,x,x+\sqrt {1+x^2}\right )+\frac {1}{4} \operatorname {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\\ &=-\frac {x}{2}-\frac {\sqrt {1+x^2}}{2}+\frac {1}{2 \sqrt {x+\sqrt {1+x^2}}}-\frac {\log (x)}{2}-\frac {1}{8} \operatorname {Subst}\left (\int \left (-\frac {1}{x^{5/2}}+x^{3/2}\right ) \, dx,x,x+\sqrt {1+x^2}\right )+\frac {1}{4} \operatorname {Subst}\left (\int \left (\frac {1}{x^{3/2}}+\sqrt {x}\right ) \, dx,x,x+\sqrt {1+x^2}\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+x}} \, dx,x,x^2\right )+\frac {1}{4} \operatorname {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} \operatorname {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}-\frac {\sqrt {1+x^2}}{2}-\frac {1}{12 \left (x+\sqrt {1+x^2}\right )^{3/2}}+\frac {1}{3} \left (x+\sqrt {1+x^2}\right )^{3/2}-\frac {1}{20} \left (x+\sqrt {1+x^2}\right )^{5/2}-\frac {\log (x)}{2}+\frac {1}{4} \operatorname {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} \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+x^2}\right )+\operatorname {Subst}\left (\int \frac {\sqrt {x}}{-1+x^2} \, dx,x,x+\sqrt {1+x^2}\right )\\ &=-\frac {x}{2}-\frac {\sqrt {1+x^2}}{2}+\sqrt {x+\sqrt {1+x^2}}+\frac {1}{3} \left (x+\sqrt {1+x^2}\right )^{3/2}+\frac {1}{2} \tanh ^{-1}\left (\sqrt {1+x^2}\right )-\frac {\log (x)}{2}+2 \operatorname {Subst}\left (\int \frac {x^2}{-1+x^4} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )+\operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )-\operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )\\ &=-\frac {x}{2}-\frac {\sqrt {1+x^2}}{2}+\sqrt {x+\sqrt {1+x^2}}+\frac {1}{3} \left (x+\sqrt {1+x^2}\right )^{3/2}-\tan ^{-1}\left (\sqrt {x+\sqrt {1+x^2}}\right )+\frac {1}{2} \tanh ^{-1}\left (\sqrt {1+x^2}\right )-\tanh ^{-1}\left (\sqrt {x+\sqrt {1+x^2}}\right )-\frac {\log (x)}{2}-\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )+\operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )\\ &=-\frac {x}{2}-\frac {\sqrt {1+x^2}}{2}+\sqrt {x+\sqrt {1+x^2}}+\frac {1}{3} \left (x+\sqrt {1+x^2}\right )^{3/2}+\frac {1}{2} \tanh ^{-1}\left (\sqrt {1+x^2}\right )-2 \tanh ^{-1}\left (\sqrt {x+\sqrt {1+x^2}}\right )-\frac {\log (x)}{2}\\ \end {align*}

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Mathematica [C]  time = 1.41, size = 304, normalized size = 3.13 \begin {gather*} \frac {1}{12} \left (-\frac {4 \sqrt {x^2+1} \left (\sqrt {x^2+1}+x\right )^{9/2} \left (\left (4 x^2+4 \sqrt {x^2+1} x+2\right ) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};\left (x+\sqrt {x^2+1}\right )^2\right )-x^2-\sqrt {x^2+1} x-2\right )}{16 x^6+28 x^4+13 x^2+5 \sqrt {x^2+1} x+16 \sqrt {x^2+1} x^5+20 \sqrt {x^2+1} x^3+1}-4 \left (\sqrt {x^2+1}-2 x\right ) \sqrt {\sqrt {x^2+1}+x}-6 \sqrt {x^2+1}+6 \log \left (\sqrt {x^2+1}+1\right )+\frac {12 \sqrt {x^2+1} \left (\sqrt {x^2+1}+x\right ) \left (\sqrt {\sqrt {x^2+1}+x}-\tan ^{-1}\left (\sqrt {\sqrt {x^2+1}+x}\right )-\tanh ^{-1}\left (\sqrt {\sqrt {x^2+1}+x}\right )\right )}{x^2+\sqrt {x^2+1} x+1}-6 x-12 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*x - 6*Sqrt[1 + x^2] - 4*(-2*x + Sqrt[1 + x^2])*Sqrt[x + Sqrt[1 + x^2]] + (12*Sqrt[1 + x^2]*(x + Sqrt[1 + x
^2])*(Sqrt[x + Sqrt[1 + x^2]] - ArcTan[Sqrt[x + Sqrt[1 + x^2]]] - ArcTanh[Sqrt[x + Sqrt[1 + x^2]]]))/(1 + x^2
+ x*Sqrt[1 + x^2]) - (4*Sqrt[1 + x^2]*(x + Sqrt[1 + x^2])^(9/2)*(-2 - x^2 - x*Sqrt[1 + x^2] + (2 + 4*x^2 + 4*x
*Sqrt[1 + x^2])*Hypergeometric2F1[3/4, 1, 7/4, (x + Sqrt[1 + x^2])^2]))/(1 + 13*x^2 + 28*x^4 + 16*x^6 + 5*x*Sq
rt[1 + x^2] + 20*x^3*Sqrt[1 + x^2] + 16*x^5*Sqrt[1 + x^2]) - 12*Log[x] + 6*Log[1 + Sqrt[1 + x^2]])/12

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IntegrateAlgebraic [A]  time = 0.15, size = 97, normalized size = 1.00 \begin {gather*} -\frac {x}{2}+\frac {1}{3} (3+x) \sqrt {x+\sqrt {1+x^2}}+\sqrt {1+x^2} \left (-\frac {1}{2}+\frac {1}{3} \sqrt {x+\sqrt {1+x^2}}\right )+\frac {1}{2} \log \left (x+\sqrt {1+x^2}\right )-2 \log \left (1+\sqrt {x+\sqrt {1+x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*x + ((3 + x)*Sqrt[x + Sqrt[1 + x^2]])/3 + Sqrt[1 + x^2]*(-1/2 + Sqrt[x + Sqrt[1 + x^2]]/3) + Log[x + Sqrt
[1 + x^2]]/2 - 2*Log[1 + Sqrt[x + Sqrt[1 + x^2]]]

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fricas [A]  time = 0.45, size = 64, normalized size = 0.66 \begin {gather*} \frac {1}{3} \, {\left (x + \sqrt {x^{2} + 1} + 3\right )} \sqrt {x + \sqrt {x^{2} + 1}} - \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(x + sqrt(x^2 + 1) + 3)*sqrt(x + sqrt(x^2 + 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)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x + \sqrt {x^{2} + 1}}{\sqrt {x + \sqrt {x^{2} + 1}} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x +\sqrt {x^{2}+1}}{1+\sqrt {x +\sqrt {x^{2}+1}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{4} \, x^{2} + \frac {1}{2} \, \int \sqrt {x^{2} + 1}\,{d x} - \int \frac {2 \, x^{2} + \sqrt {x^{2} + 1} {\left (2 \, x - 1\right )} - x + 1}{2 \, {\left (x + \sqrt {x^{2} + 1} + 2 \, \sqrt {x + \sqrt {x^{2} + 1}} + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^2 + 1/2*integrate(sqrt(x^2 + 1), x) - integrate(1/2*(2*x^2 + sqrt(x^2 + 1)*(2*x - 1) - x + 1)/(x + sqrt(
x^2 + 1) + 2*sqrt(x + sqrt(x^2 + 1)) + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x+\sqrt {x^2+1}}{\sqrt {x+\sqrt {x^2+1}}+1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x + \sqrt {x^{2} + 1}}{\sqrt {x + \sqrt {x^{2} + 1}} + 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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