\(\int \frac {1+2 \sqrt {1+x}}{x \sqrt {1+x} \sqrt {x+\sqrt {1+x}}} \, dx\) [728]

Optimal result

Integrand size = 35, antiderivative size = 61 \[ \int \frac {1+2 \sqrt {1+x}}{x \sqrt {1+x} \sqrt {x+\sqrt {1+x}}} \, dx=-\arctan \left (\frac {3+\sqrt {1+x}}{2 \sqrt {x+\sqrt {1+x}}}\right )+3 \text {arctanh}\left (\frac {1-3 \sqrt {1+x}}{2 \sqrt {x+\sqrt {1+x}}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {1047, 738, 212, 210} \[ \int \frac {1+2 \sqrt {1+x}}{x \sqrt {1+x} \sqrt {x+\sqrt {1+x}}} \, dx=3 \text {arctanh}\left (\frac {1-3 \sqrt {x+1}}{2 \sqrt {x+\sqrt {x+1}}}\right )-\arctan \left (\frac {\sqrt {x+1}+3}{2 \sqrt {x+\sqrt {x+1}}}\right ) \]

[In]

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

[Out]

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

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 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 1047

Int[((g_.) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[(-a)*c, 2]}, Dist[h/2 + c*(g/(2*q)), Int[1/((-q + c*x)*Sqrt[d + e*x + f*x^2]), x], x] + Dist[h/2 - c*(g/
(2*q)), Int[1/((q + c*x)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, c, d, e, f, g, h}, x] && NeQ[e^2 - 4*d*f
, 0] && PosQ[(-a)*c]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {1+2 x}{\left (-1+x^2\right ) \sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right ) \\ & = 3 \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right )+\text {Subst}\left (\int \frac {1}{(1+x) \sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right ) \\ & = -\left (2 \text {Subst}\left (\int \frac {1}{-4-x^2} \, dx,x,\frac {-3-\sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}}\right )\right )-6 \text {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {-1+3 \sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}}\right ) \\ & = -\tan ^{-1}\left (\frac {3+\sqrt {1+x}}{2 \sqrt {x+\sqrt {1+x}}}\right )+3 \tanh ^{-1}\left (\frac {1-3 \sqrt {1+x}}{2 \sqrt {x+\sqrt {1+x}}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.90 \[ \int \frac {1+2 \sqrt {1+x}}{x \sqrt {1+x} \sqrt {x+\sqrt {1+x}}} \, dx=-2 \arctan \left (1+\sqrt {1+x}-\sqrt {x+\sqrt {1+x}}\right )-6 \text {arctanh}\left (1-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}\right ) \]

[In]

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

[Out]

-2*ArcTan[1 + Sqrt[1 + x] - Sqrt[x + Sqrt[1 + x]]] - 6*ArcTanh[1 - Sqrt[1 + x] + Sqrt[x + Sqrt[1 + x]]]

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.11

method	result	size
derivativedivides	\(\arctan \left (\frac {-3-\sqrt {x +1}}{2 \sqrt {\left (1+\sqrt {x +1}\right )^{2}-\sqrt {x +1}-2}}\right )-3 \,\operatorname {arctanh}\left (\frac {-1+3 \sqrt {x +1}}{2 \sqrt {\left (\sqrt {x +1}-1\right )^{2}+3 \sqrt {x +1}-2}}\right )\)	\(68\)
default	\(\arctan \left (\frac {-3-\sqrt {x +1}}{2 \sqrt {\left (1+\sqrt {x +1}\right )^{2}-\sqrt {x +1}-2}}\right )-3 \,\operatorname {arctanh}\left (\frac {-1+3 \sqrt {x +1}}{2 \sqrt {\left (\sqrt {x +1}-1\right )^{2}+3 \sqrt {x +1}-2}}\right )\)	\(68\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 1.61 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.02 \[ \int \frac {1+2 \sqrt {1+x}}{x \sqrt {1+x} \sqrt {x+\sqrt {1+x}}} \, dx=\arctan \left (\frac {2 \, \sqrt {x + \sqrt {x + 1}} {\left (\sqrt {x + 1} - 3\right )}}{x - 8}\right ) + 3 \, \log \left (\frac {2 \, \sqrt {x + \sqrt {x + 1}} {\left (\sqrt {x + 1} + 1\right )} - 3 \, x - 2 \, \sqrt {x + 1} - 2}{x}\right ) \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {1+2 \sqrt {1+x}}{x \sqrt {1+x} \sqrt {x+\sqrt {1+x}}} \, dx=\int \frac {2 \sqrt {x + 1} + 1}{x \sqrt {x + 1} \sqrt {x + \sqrt {x + 1}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1+2 \sqrt {1+x}}{x \sqrt {1+x} \sqrt {x+\sqrt {1+x}}} \, dx=\int { \frac {2 \, \sqrt {x + 1} + 1}{\sqrt {x + \sqrt {x + 1}} \sqrt {x + 1} x} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.63 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.07 \[ \int \frac {1+2 \sqrt {1+x}}{x \sqrt {1+x} \sqrt {x+\sqrt {1+x}}} \, dx=2 \, \arctan \left (\sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} - 1\right ) - 3 \, \log \left ({\left | \sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} + 2 \right |}\right ) + 3 \, \log \left ({\left | \sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} \right |}\right ) \]

[In]

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

[Out]

2*arctan(sqrt(x + sqrt(x + 1)) - sqrt(x + 1) - 1) - 3*log(abs(sqrt(x + sqrt(x + 1)) - sqrt(x + 1) + 2)) + 3*lo
g(abs(sqrt(x + sqrt(x + 1)) - sqrt(x + 1)))

Mupad [F(-1)]

Timed out. \[ \int \frac {1+2 \sqrt {1+x}}{x \sqrt {1+x} \sqrt {x+\sqrt {1+x}}} \, dx=\int \frac {2\,\sqrt {x+1}+1}{x\,\sqrt {x+\sqrt {x+1}}\,\sqrt {x+1}} \,d x \]

[In]

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

[Out]

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