\(\int \frac {(1+x) \sqrt {x+\sqrt {1+x^2}}}{-1+x} \, dx\) [1855]

Optimal result

Integrand size = 24, antiderivative size = 127 \[ \int \frac {(1+x) \sqrt {x+\sqrt {1+x^2}}}{-1+x} \, dx=\frac {2 (6+x) \sqrt {1+x^2}+2 \left (-1+6 x+x^2\right )}{3 \sqrt {x+\sqrt {1+x^2}}}-4 \sqrt {-1+\sqrt {2}} \arctan \left (\frac {\sqrt {x+\sqrt {1+x^2}}}{\sqrt {-1+\sqrt {2}}}\right )-4 \sqrt {1+\sqrt {2}} \text {arctanh}\left (\frac {\sqrt {x+\sqrt {1+x^2}}}{\sqrt {1+\sqrt {2}}}\right ) \]

[Out]

1/3*(2*(6+x)*(x^2+1)^(1/2)+2*x^2+12*x-2)/(x+(x^2+1)^(1/2))^(1/2)-4*(2^(1/2)-1)^(1/2)*arctan((x+(x^2+1)^(1/2))^
(1/2)/(2^(1/2)-1)^(1/2))-4*(1+2^(1/2))^(1/2)*arctanh((x+(x^2+1)^(1/2))^(1/2)/(1+2^(1/2))^(1/2))

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6874, 2142, 14, 2144, 1642, 840, 1180, 213, 209} \[ \int \frac {(1+x) \sqrt {x+\sqrt {1+x^2}}}{-1+x} \, dx=-4 \sqrt {\sqrt {2}-1} \arctan \left (\sqrt {1+\sqrt {2}} \sqrt {\sqrt {x^2+1}+x}\right )-4 \sqrt {1+\sqrt {2}} \text {arctanh}\left (\sqrt {\sqrt {2}-1} \sqrt {\sqrt {x^2+1}+x}\right )+\frac {1}{3} \left (\sqrt {x^2+1}+x\right )^{3/2}+4 \sqrt {\sqrt {x^2+1}+x}-\frac {1}{\sqrt {\sqrt {x^2+1}+x}} \]

[In]

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

[Out]

-(1/Sqrt[x + Sqrt[1 + x^2]]) + 4*Sqrt[x + Sqrt[1 + x^2]] + (x + Sqrt[1 + x^2])^(3/2)/3 - 4*Sqrt[-1 + Sqrt[2]]*
ArcTan[Sqrt[1 + Sqrt[2]]*Sqrt[x + Sqrt[1 + x^2]]] - 4*Sqrt[1 + Sqrt[2]]*ArcTanh[Sqrt[-1 + Sqrt[2]]*Sqrt[x + Sq
rt[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 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 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 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 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]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\sqrt {x+\sqrt {1+x^2}}+\frac {2 \sqrt {x+\sqrt {1+x^2}}}{-1+x}\right ) \, dx \\ & = 2 \int \frac {\sqrt {x+\sqrt {1+x^2}}}{-1+x} \, dx+\int \sqrt {x+\sqrt {1+x^2}} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1+x^2}{x^{3/2}} \, dx,x,x+\sqrt {1+x^2}\right )+2 \text {Subst}\left (\int \frac {1+x^2}{\sqrt {x} \left (-1-2 x+x^2\right )} \, dx,x,x+\sqrt {1+x^2}\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{x^{3/2}}+\sqrt {x}\right ) \, dx,x,x+\sqrt {1+x^2}\right )+2 \text {Subst}\left (\int \left (\frac {1}{\sqrt {x}}+\frac {2 (1+x)}{\sqrt {x} \left (-1-2 x+x^2\right )}\right ) \, dx,x,x+\sqrt {1+x^2}\right ) \\ & = -\frac {1}{\sqrt {x+\sqrt {1+x^2}}}+4 \sqrt {x+\sqrt {1+x^2}}+\frac {1}{3} \left (x+\sqrt {1+x^2}\right )^{3/2}+4 \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 {1}{\sqrt {x+\sqrt {1+x^2}}}+4 \sqrt {x+\sqrt {1+x^2}}+\frac {1}{3} \left (x+\sqrt {1+x^2}\right )^{3/2}+8 \text {Subst}\left (\int \frac {1+x^2}{-1-2 x^2+x^4} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right ) \\ & = -\frac {1}{\sqrt {x+\sqrt {1+x^2}}}+4 \sqrt {x+\sqrt {1+x^2}}+\frac {1}{3} \left (x+\sqrt {1+x^2}\right )^{3/2}+\left (4 \left (1-\sqrt {2}\right )\right ) \text {Subst}\left (\int \frac {1}{-1+\sqrt {2}+x^2} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right )+\left (4 \left (1+\sqrt {2}\right )\right ) \text {Subst}\left (\int \frac {1}{-1-\sqrt {2}+x^2} \, dx,x,\sqrt {x+\sqrt {1+x^2}}\right ) \\ & = -\frac {1}{\sqrt {x+\sqrt {1+x^2}}}+4 \sqrt {x+\sqrt {1+x^2}}+\frac {1}{3} \left (x+\sqrt {1+x^2}\right )^{3/2}-4 \sqrt {-1+\sqrt {2}} \arctan \left (\frac {\sqrt {x+\sqrt {1+x^2}}}{\sqrt {-1+\sqrt {2}}}\right )-4 \sqrt {1+\sqrt {2}} \text {arctanh}\left (\frac {\sqrt {x+\sqrt {1+x^2}}}{\sqrt {1+\sqrt {2}}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.97 \[ \int \frac {(1+x) \sqrt {x+\sqrt {1+x^2}}}{-1+x} \, dx=\frac {2 \left (-1+6 x+x^2+(6+x) \sqrt {1+x^2}\right )}{3 \sqrt {x+\sqrt {1+x^2}}}-4 \sqrt {-1+\sqrt {2}} \arctan \left (\sqrt {1+\sqrt {2}} \sqrt {x+\sqrt {1+x^2}}\right )-4 \sqrt {1+\sqrt {2}} \text {arctanh}\left (\sqrt {-1+\sqrt {2}} \sqrt {x+\sqrt {1+x^2}}\right ) \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.31 \[ \int \frac {(1+x) \sqrt {x+\sqrt {1+x^2}}}{-1+x} \, dx=\frac {2}{3} \, {\left (2 \, x - \sqrt {x^{2} + 1} + 6\right )} \sqrt {x + \sqrt {x^{2} + 1}} - 2 \, \sqrt {\sqrt {2} + 1} \log \left (4 \, \sqrt {x + \sqrt {x^{2} + 1}} + 4 \, \sqrt {\sqrt {2} + 1}\right ) + 2 \, \sqrt {\sqrt {2} + 1} \log \left (4 \, \sqrt {x + \sqrt {x^{2} + 1}} - 4 \, \sqrt {\sqrt {2} + 1}\right ) - \frac {1}{2} \, \sqrt {-16 \, \sqrt {2} + 16} \log \left (4 \, \sqrt {x + \sqrt {x^{2} + 1}} + \sqrt {-16 \, \sqrt {2} + 16}\right ) + \frac {1}{2} \, \sqrt {-16 \, \sqrt {2} + 16} \log \left (4 \, \sqrt {x + \sqrt {x^{2} + 1}} - \sqrt {-16 \, \sqrt {2} + 16}\right ) \]

[In]

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

[Out]

2/3*(2*x - sqrt(x^2 + 1) + 6)*sqrt(x + sqrt(x^2 + 1)) - 2*sqrt(sqrt(2) + 1)*log(4*sqrt(x + sqrt(x^2 + 1)) + 4*
sqrt(sqrt(2) + 1)) + 2*sqrt(sqrt(2) + 1)*log(4*sqrt(x + sqrt(x^2 + 1)) - 4*sqrt(sqrt(2) + 1)) - 1/2*sqrt(-16*s
qrt(2) + 16)*log(4*sqrt(x + sqrt(x^2 + 1)) + sqrt(-16*sqrt(2) + 16)) + 1/2*sqrt(-16*sqrt(2) + 16)*log(4*sqrt(x
+ sqrt(x^2 + 1)) - sqrt(-16*sqrt(2) + 16))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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