Integrand size = 35, antiderivative size = 61 \[ \int \frac {1+2 \sqrt {1+x}}{x \sqrt {1+x} \sqrt {x+\sqrt {1+x}}} \, dx=-\arctan \left (\frac {3+\sqrt {1+x}}{2 \sqrt {x+\sqrt {1+x}}}\right )+3 \text {arctanh}\left (\frac {1-3 \sqrt {1+x}}{2 \sqrt {x+\sqrt {1+x}}}\right ) \]
-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))
Time = 0.12 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.90 \[ \int \frac {1+2 \sqrt {1+x}}{x \sqrt {1+x} \sqrt {x+\sqrt {1+x}}} \, dx=-2 \arctan \left (1+\sqrt {1+x}-\sqrt {x+\sqrt {1+x}}\right )-6 \text {arctanh}\left (1-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}\right ) \]
-2*ArcTan[1 + Sqrt[1 + x] - Sqrt[x + Sqrt[1 + x]]] - 6*ArcTanh[1 - Sqrt[1 + x] + Sqrt[x + Sqrt[1 + x]]]
Time = 0.64 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {7267, 25, 1366, 25, 1154, 217, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {2 \sqrt {x+1}+1}{x \sqrt {x+1} \sqrt {x+\sqrt {x+1}}} \, dx\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 2 \int \frac {2 \sqrt {x+1}+1}{x \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int -\frac {2 \sqrt {x+1}+1}{x \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}\) |
\(\Big \downarrow \) 1366 |
\(\displaystyle 2 \left (-\frac {3}{2} \int \frac {1}{\left (1-\sqrt {x+1}\right ) \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}-\frac {1}{2} \int -\frac {1}{\left (\sqrt {x+1}+1\right ) \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle 2 \left (\frac {1}{2} \int \frac {1}{\left (\sqrt {x+1}+1\right ) \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}-\frac {3}{2} \int \frac {1}{\left (1-\sqrt {x+1}\right ) \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}\right )\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle 2 \left (3 \int \frac {1}{3-x}d\frac {1-3 \sqrt {x+1}}{\sqrt {x+\sqrt {x+1}}}-\int \frac {1}{-x-5}d\left (-\frac {\sqrt {x+1}+3}{\sqrt {x+\sqrt {x+1}}}\right )\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle 2 \left (3 \int \frac {1}{3-x}d\frac {1-3 \sqrt {x+1}}{\sqrt {x+\sqrt {x+1}}}-\frac {1}{2} \arctan \left (\frac {\sqrt {x+1}+3}{2 \sqrt {x+\sqrt {x+1}}}\right )\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 2 \left (\frac {3}{2} \text {arctanh}\left (\frac {1-3 \sqrt {x+1}}{2 \sqrt {x+\sqrt {x+1}}}\right )-\frac {1}{2} \arctan \left (\frac {\sqrt {x+1}+3}{2 \sqrt {x+\sqrt {x+1}}}\right )\right )\) |
2*(-1/2*ArcTan[(3 + Sqrt[1 + x])/(2*Sqrt[x + Sqrt[1 + x]])] + (3*ArcTanh[( 1 - 3*Sqrt[1 + x])/(2*Sqrt[x + Sqrt[1 + x]])])/2)
3.8.28.3.1 Defintions of rubi rules used
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])
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-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]
Int[((g_.) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + ( f_.)*(x_)^2]), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Simp[(h/2 + c*(g/(2*q ))) Int[1/((-q + c*x)*Sqrt[d + e*x + f*x^2]), x], x] + Simp[(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]
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
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 | \(\arctan \left (\frac {-3-\sqrt {x +1}}{2 \sqrt {\left (1+\sqrt {x +1}\right )^{2}-\sqrt {x +1}-2}}\right )-3 \,\operatorname {arctanh}\left (\frac {-1+3 \sqrt {x +1}}{2 \sqrt {\left (\sqrt {x +1}-1\right )^{2}+3 \sqrt {x +1}-2}}\right )\) | \(68\) |
arctan(1/2*(-3-(x+1)^(1/2))/((1+(x+1)^(1/2))^2-(x+1)^(1/2)-2)^(1/2))-3*arc tanh(1/2*(-1+3*(x+1)^(1/2))/(((x+1)^(1/2)-1)^2+3*(x+1)^(1/2)-2)^(1/2))
Time = 1.61 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.02 \[ \int \frac {1+2 \sqrt {1+x}}{x \sqrt {1+x} \sqrt {x+\sqrt {1+x}}} \, dx=\arctan \left (\frac {2 \, \sqrt {x + \sqrt {x + 1}} {\left (\sqrt {x + 1} - 3\right )}}{x - 8}\right ) + 3 \, \log \left (\frac {2 \, \sqrt {x + \sqrt {x + 1}} {\left (\sqrt {x + 1} + 1\right )} - 3 \, x - 2 \, \sqrt {x + 1} - 2}{x}\right ) \]
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)
\[ \int \frac {1+2 \sqrt {1+x}}{x \sqrt {1+x} \sqrt {x+\sqrt {1+x}}} \, dx=\int \frac {2 \sqrt {x + 1} + 1}{x \sqrt {x + 1} \sqrt {x + \sqrt {x + 1}}}\, dx \]
\[ \int \frac {1+2 \sqrt {1+x}}{x \sqrt {1+x} \sqrt {x+\sqrt {1+x}}} \, dx=\int { \frac {2 \, \sqrt {x + 1} + 1}{\sqrt {x + \sqrt {x + 1}} \sqrt {x + 1} x} \,d x } \]
Time = 0.63 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.07 \[ \int \frac {1+2 \sqrt {1+x}}{x \sqrt {1+x} \sqrt {x+\sqrt {1+x}}} \, dx=2 \, \arctan \left (\sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} - 1\right ) - 3 \, \log \left ({\left | \sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} + 2 \right |}\right ) + 3 \, \log \left ({\left | \sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} \right |}\right ) \]
2*arctan(sqrt(x + sqrt(x + 1)) - sqrt(x + 1) - 1) - 3*log(abs(sqrt(x + sqr t(x + 1)) - sqrt(x + 1) + 2)) + 3*log(abs(sqrt(x + sqrt(x + 1)) - sqrt(x + 1)))
Timed out. \[ \int \frac {1+2 \sqrt {1+x}}{x \sqrt {1+x} \sqrt {x+\sqrt {1+x}}} \, dx=\int \frac {2\,\sqrt {x+1}+1}{x\,\sqrt {x+\sqrt {x+1}}\,\sqrt {x+1}} \,d x \]