\(\int \frac {1}{(1+x) \sqrt {x+\sqrt {1+x^2}}} \, dx\) [1404]

Optimal result

Integrand size = 21, antiderivative size = 100 \[ \int \frac {1}{(1+x) \sqrt {x+\sqrt {1+x^2}}} \, dx=\frac {2}{\sqrt {x+\sqrt {1+x^2}}}+2 \sqrt {-1+\sqrt {2}} \arctan \left (\frac {\sqrt {x+\sqrt {1+x^2}}}{\sqrt {1+\sqrt {2}}}\right )-2 \sqrt {1+\sqrt {2}} \text {arctanh}\left (\frac {\sqrt {x+\sqrt {1+x^2}}}{\sqrt {-1+\sqrt {2}}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2144, 1642, 842, 840, 1180, 213, 209} \[ \int \frac {1}{(1+x) \sqrt {x+\sqrt {1+x^2}}} \, dx=\frac {2 \arctan \left (\sqrt {\sqrt {2}-1} \sqrt {\sqrt {x^2+1}+x}\right )}{\sqrt {1+\sqrt {2}}}-\frac {2 \text {arctanh}\left (\sqrt {1+\sqrt {2}} \sqrt {\sqrt {x^2+1}+x}\right )}{\sqrt {\sqrt {2}-1}}+\frac {2}{\sqrt {\sqrt {x^2+1}+x}} \]

[In]

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

[Out]

2/Sqrt[x + Sqrt[1 + x^2]] + (2*ArcTan[Sqrt[-1 + Sqrt[2]]*Sqrt[x + Sqrt[1 + x^2]]])/Sqrt[1 + Sqrt[2]] - (2*ArcT
anh[Sqrt[1 + Sqrt[2]]*Sqrt[x + Sqrt[1 + x^2]]])/Sqrt[-1 + Sqrt[2]]

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 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 840

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 842

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e
*f - d*g)*((d + e*x)^(m + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[(d +
e*x)^(m + 1)*(Simp[c*d*f - f*b*e + a*e*g - c*(e*f - d*g)*x, x]/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c,
d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && FractionQ[m] && LtQ[m, -1]

Rule 1180

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 1642

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2144

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]

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(1+x) \sqrt {x+\sqrt {1+x^2}}} \, dx=\frac {2}{\sqrt {x+\sqrt {1+x^2}}}+2 \sqrt {-1+\sqrt {2}} \arctan \left (\sqrt {-1+\sqrt {2}} \sqrt {x+\sqrt {1+x^2}}\right )-2 \sqrt {1+\sqrt {2}} \text {arctanh}\left (\sqrt {1+\sqrt {2}} \sqrt {x+\sqrt {1+x^2}}\right ) \]

[In]

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

[Out]

2/Sqrt[x + Sqrt[1 + x^2]] + 2*Sqrt[-1 + Sqrt[2]]*ArcTan[Sqrt[-1 + Sqrt[2]]*Sqrt[x + Sqrt[1 + x^2]]] - 2*Sqrt[1
+ Sqrt[2]]*ArcTanh[Sqrt[1 + Sqrt[2]]*Sqrt[x + Sqrt[1 + x^2]]]

Maple [F]

\[\int \frac {1}{\left (1+x \right ) \sqrt {x +\sqrt {x^{2}+1}}}d x\]

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (72) = 144\).

Time = 0.27 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.83 \[ \int \frac {1}{(1+x) \sqrt {x+\sqrt {1+x^2}}} \, dx=-2 \, \sqrt {x + \sqrt {x^{2} + 1}} {\left (x - \sqrt {x^{2} + 1}\right )} - \sqrt {\sqrt {2} + 1} \log \left (2 \, \sqrt {\sqrt {2} + 1} {\left (\sqrt {2} - 1\right )} + 2 \, \sqrt {x + \sqrt {x^{2} + 1}}\right ) + \sqrt {\sqrt {2} + 1} \log \left (-2 \, \sqrt {\sqrt {2} + 1} {\left (\sqrt {2} - 1\right )} + 2 \, \sqrt {x + \sqrt {x^{2} + 1}}\right ) + \frac {1}{2} \, \sqrt {-4 \, \sqrt {2} + 4} \log \left ({\left (\sqrt {2} + 1\right )} \sqrt {-4 \, \sqrt {2} + 4} + 2 \, \sqrt {x + \sqrt {x^{2} + 1}}\right ) - \frac {1}{2} \, \sqrt {-4 \, \sqrt {2} + 4} \log \left (-{\left (\sqrt {2} + 1\right )} \sqrt {-4 \, \sqrt {2} + 4} + 2 \, \sqrt {x + \sqrt {x^{2} + 1}}\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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