\(\int \frac {\sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}} \, dx\) [926]

Optimal result

Integrand size = 21, antiderivative size = 70 \[ \int \frac {\sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}} \, dx=-\frac {3}{2} \sqrt {x+\sqrt {1+x}}+\sqrt {1+x} \sqrt {x+\sqrt {1+x}}-\frac {7}{4} \log \left (1+2 \sqrt {1+x}-2 \sqrt {x+\sqrt {1+x}}\right ) \]

[Out]

-3/2*(x+(1+x)^(1/2))^(1/2)+(1+x)^(1/2)*(x+(1+x)^(1/2))^(1/2)-7/4*ln(1+2*(1+x)^(1/2)-2*(x+(1+x)^(1/2))^(1/2))

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {756, 654, 635, 212} \[ \int \frac {\sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}} \, dx=\frac {7}{4} \text {arctanh}\left (\frac {2 \sqrt {x+1}+1}{2 \sqrt {x+\sqrt {x+1}}}\right )+\sqrt {x+1} \sqrt {x+\sqrt {x+1}}-\frac {3}{2} \sqrt {x+\sqrt {x+1}} \]

[In]

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

[Out]

(-3*Sqrt[x + Sqrt[1 + x]])/2 + Sqrt[1 + x]*Sqrt[x + Sqrt[1 + x]] + (7*ArcTanh[(1 + 2*Sqrt[1 + x])/(2*Sqrt[x +
Sqrt[1 + x]])])/4

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 654

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

Rule 756

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}} \, dx=\frac {1}{2} \sqrt {x+\sqrt {1+x}} \left (-3+2 \sqrt {1+x}\right )-\frac {7}{4} \log \left (-1-2 \sqrt {1+x}+2 \sqrt {x+\sqrt {1+x}}\right ) \]

[In]

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

[Out]

(Sqrt[x + Sqrt[1 + x]]*(-3 + 2*Sqrt[1 + x]))/2 - (7*Log[-1 - 2*Sqrt[1 + x] + 2*Sqrt[x + Sqrt[1 + x]]])/4

Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.67

method	result	size
derivativedivides	\(\sqrt {1+x}\, \sqrt {x +\sqrt {1+x}}-\frac {3 \sqrt {x +\sqrt {1+x}}}{2}+\frac {7 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{4}\)	\(47\)
default	\(\sqrt {1+x}\, \sqrt {x +\sqrt {1+x}}-\frac {3 \sqrt {x +\sqrt {1+x}}}{2}+\frac {7 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{4}\)	\(47\)

[In]

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

[Out]

(1+x)^(1/2)*(x+(1+x)^(1/2))^(1/2)-3/2*(x+(1+x)^(1/2))^(1/2)+7/4*ln(1/2+(1+x)^(1/2)+(x+(1+x)^(1/2))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.51 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.80 \[ \int \frac {\sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}} \, dx=\frac {1}{2} \, \sqrt {x + \sqrt {x + 1}} {\left (2 \, \sqrt {x + 1} - 3\right )} + \frac {7}{8} \, \log \left (4 \, \sqrt {x + \sqrt {x + 1}} {\left (2 \, \sqrt {x + 1} + 1\right )} + 8 \, x + 8 \, \sqrt {x + 1} + 5\right ) \]

[In]

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

[Out]

1/2*sqrt(x + sqrt(x + 1))*(2*sqrt(x + 1) - 3) + 7/8*log(4*sqrt(x + sqrt(x + 1))*(2*sqrt(x + 1) + 1) + 8*x + 8*
sqrt(x + 1) + 5)

Sympy [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}} \, dx=2 \sqrt {x + \sqrt {x + 1}} \left (\frac {\sqrt {x + 1}}{2} - \frac {3}{4}\right ) + \frac {7 \log {\left (2 \sqrt {x + 1} + 2 \sqrt {x + \sqrt {x + 1}} + 1 \right )}}{4} \]

[In]

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

[Out]

2*sqrt(x + sqrt(x + 1))*(sqrt(x + 1)/2 - 3/4) + 7*log(2*sqrt(x + 1) + 2*sqrt(x + sqrt(x + 1)) + 1)/4

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.63 \[ \int \frac {\sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}} \, dx=\frac {1}{2} \, \sqrt {x + \sqrt {x + 1}} {\left (2 \, \sqrt {x + 1} - 3\right )} - \frac {7}{4} \, \log \left (-2 \, \sqrt {x + \sqrt {x + 1}} + 2 \, \sqrt {x + 1} + 1\right ) \]

[In]

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

[Out]

1/2*sqrt(x + sqrt(x + 1))*(2*sqrt(x + 1) - 3) - 7/4*log(-2*sqrt(x + sqrt(x + 1)) + 2*sqrt(x + 1) + 1)

Mupad [F(-1)]

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

[In]

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

[Out]

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