\(\int \frac {\sqrt {x+\sqrt {1+x}}}{1+\sqrt {1+x}} \, dx\) [1354]

Optimal result

Integrand size = 25, antiderivative size = 97 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{1+\sqrt {1+x}} \, dx=-\frac {3}{2} \sqrt {x+\sqrt {1+x}}+\sqrt {1+x} \sqrt {x+\sqrt {1+x}}-4 \arctan \left (1+\sqrt {1+x}-\sqrt {x+\sqrt {1+x}}\right )+\frac {1}{4} \log \left (1+2 \sqrt {1+x}-2 \sqrt {x+\sqrt {1+x}}\right ) \]

[Out]

-3/2*(x+(1+x)^(1/2))^(1/2)+(1+x)^(1/2)*(x+(1+x)^(1/2))^(1/2)-4*arctan(1+(1+x)^(1/2)-(x+(1+x)^(1/2))^(1/2))+1/4
*ln(1+2*(1+x)^(1/2)-2*(x+(1+x)^(1/2))^(1/2))

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {828, 857, 635, 212, 738, 210} \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{1+\sqrt {1+x}} \, dx=-2 \arctan \left (\frac {\sqrt {x+1}+3}{2 \sqrt {x+\sqrt {x+1}}}\right )-\frac {1}{4} \text {arctanh}\left (\frac {2 \sqrt {x+1}+1}{2 \sqrt {x+\sqrt {x+1}}}\right )-\frac {1}{2} \sqrt {x+\sqrt {x+1}} \left (3-2 \sqrt {x+1}\right ) \]

[In]

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

[Out]

-1/2*((3 - 2*Sqrt[1 + x])*Sqrt[x + Sqrt[1 + x]]) - 2*ArcTan[(3 + Sqrt[1 + x])/(2*Sqrt[x + Sqrt[1 + x]])] - Arc
Tanh[(1 + 2*Sqrt[1 + x])/(2*Sqrt[x + Sqrt[1 + x]])]/4

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[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 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 738

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

Rule 828

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*((a + b*x + c*x^
2)^p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], 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] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 857

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

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {x \sqrt {-1+x+x^2}}{1+x} \, dx,x,\sqrt {1+x}\right ) \\ & = -\frac {1}{2} \left (3-2 \sqrt {1+x}\right ) \sqrt {x+\sqrt {1+x}}-\frac {1}{2} \text {Subst}\left (\int \frac {-\frac {7}{2}+\frac {x}{2}}{(1+x) \sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right ) \\ & = -\frac {1}{2} \left (3-2 \sqrt {1+x}\right ) \sqrt {x+\sqrt {1+x}}-\frac {1}{4} \text {Subst}\left (\int \frac {1}{\sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right )+2 \text {Subst}\left (\int \frac {1}{(1+x) \sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right ) \\ & = -\frac {1}{2} \left (3-2 \sqrt {1+x}\right ) \sqrt {x+\sqrt {1+x}}-\frac {1}{2} \text {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {1+2 \sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}}\right )-4 \text {Subst}\left (\int \frac {1}{-4-x^2} \, dx,x,\frac {-3-\sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}}\right ) \\ & = -\frac {1}{2} \left (3-2 \sqrt {1+x}\right ) \sqrt {x+\sqrt {1+x}}-2 \arctan \left (\frac {3+\sqrt {1+x}}{2 \sqrt {x+\sqrt {1+x}}}\right )-\frac {1}{4} \text {arctanh}\left (\frac {1+2 \sqrt {1+x}}{2 \sqrt {x+\sqrt {1+x}}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{1+\sqrt {1+x}} \, dx=\frac {1}{2} \sqrt {x+\sqrt {1+x}} \left (-3+2 \sqrt {1+x}\right )-4 \arctan \left (1+\sqrt {1+x}-\sqrt {x+\sqrt {1+x}}\right )+\frac {1}{4} \log \left (-1-2 \sqrt {1+x}+2 \sqrt {x+\sqrt {1+x}}\right ) \]

[In]

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

[Out]

(Sqrt[x + Sqrt[1 + x]]*(-3 + 2*Sqrt[1 + x]))/2 - 4*ArcTan[1 + Sqrt[1 + x] - Sqrt[x + Sqrt[1 + x]]] + Log[-1 -
2*Sqrt[1 + x] + 2*Sqrt[x + Sqrt[1 + x]]]/4

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.29

method	result	size
derivativedivides	\(\frac {\left (2 \sqrt {1+x}+1\right ) \sqrt {x +\sqrt {1+x}}}{2}-\frac {5 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{4}-2 \sqrt {\left (\sqrt {1+x}+1\right )^{2}-\sqrt {1+x}-2}+\ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (\sqrt {1+x}+1\right )^{2}-\sqrt {1+x}-2}\right )+2 \arctan \left (\frac {-3-\sqrt {1+x}}{2 \sqrt {\left (\sqrt {1+x}+1\right )^{2}-\sqrt {1+x}-2}}\right )\)	\(125\)
default	\(\frac {\left (2 \sqrt {1+x}+1\right ) \sqrt {x +\sqrt {1+x}}}{2}-\frac {5 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{4}-2 \sqrt {\left (\sqrt {1+x}+1\right )^{2}-\sqrt {1+x}-2}+\ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (\sqrt {1+x}+1\right )^{2}-\sqrt {1+x}-2}\right )+2 \arctan \left (\frac {-3-\sqrt {1+x}}{2 \sqrt {\left (\sqrt {1+x}+1\right )^{2}-\sqrt {1+x}-2}}\right )\)	\(125\)

[In]

int((x+(1+x)^(1/2))^(1/2)/((1+x)^(1/2)+1),x,method=_RETURNVERBOSE)

[Out]

1/2*(2*(1+x)^(1/2)+1)*(x+(1+x)^(1/2))^(1/2)-5/4*ln(1/2+(1+x)^(1/2)+(x+(1+x)^(1/2))^(1/2))-2*(((1+x)^(1/2)+1)^2
-(1+x)^(1/2)-2)^(1/2)+ln(1/2+(1+x)^(1/2)+(((1+x)^(1/2)+1)^2-(1+x)^(1/2)-2)^(1/2))+2*arctan(1/2*(-3-(1+x)^(1/2)
)/(((1+x)^(1/2)+1)^2-(1+x)^(1/2)-2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 1.36 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{1+\sqrt {1+x}} \, dx=\frac {1}{2} \, \sqrt {x + \sqrt {x + 1}} {\left (2 \, \sqrt {x + 1} - 3\right )} + 2 \, \arctan \left (\frac {2 \, \sqrt {x + \sqrt {x + 1}} {\left (\sqrt {x + 1} - 3\right )}}{x - 8}\right ) + \frac {1}{8} \, \log \left (4 \, \sqrt {x + \sqrt {x + 1}} {\left (2 \, \sqrt {x + 1} + 1\right )} - 8 \, x - 8 \, \sqrt {x + 1} - 5\right ) \]

[In]

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

[Out]

1/2*sqrt(x + sqrt(x + 1))*(2*sqrt(x + 1) - 3) + 2*arctan(2*sqrt(x + sqrt(x + 1))*(sqrt(x + 1) - 3)/(x - 8)) +
1/8*log(4*sqrt(x + sqrt(x + 1))*(2*sqrt(x + 1) + 1) - 8*x - 8*sqrt(x + 1) - 5)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{1+\sqrt {1+x}} \, dx=\frac {1}{2} \, \sqrt {x + \sqrt {x + 1}} {\left (2 \, \sqrt {x + 1} - 3\right )} + 4 \, \arctan \left (\sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} - 1\right ) + \frac {1}{4} \, \log \left (-2 \, \sqrt {x + \sqrt {x + 1}} + 2 \, \sqrt {x + 1} + 1\right ) \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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