Integrand size = 24, antiderivative size = 110 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x \sqrt {1+x}} \, dx=-2 \arctan \left (\frac {\sqrt {1+x}-\sqrt {x+\sqrt {1+x}}}{2+\sqrt {1+x}-\sqrt {x+\sqrt {1+x}}}\right )-2 \text {arctanh}\left (1-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}\right )-2 \log \left (1+2 \sqrt {1+x}-2 \sqrt {x+\sqrt {1+x}}\right ) \]
-2*arctan(((1+x)^(1/2)-(x+(1+x)^(1/2))^(1/2))/(2+(1+x)^(1/2)-(x+(1+x)^(1/2 ))^(1/2)))+2*arctanh(-1+(1+x)^(1/2)-(x+(1+x)^(1/2))^(1/2))-2*ln(1+2*(1+x)^ (1/2)-2*(x+(1+x)^(1/2))^(1/2))
Time = 0.10 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x \sqrt {1+x}} \, dx=-2 \arctan \left (1+\sqrt {1+x}-\sqrt {x+\sqrt {1+x}}\right )-2 \text {arctanh}\left (1-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}\right )-2 \log \left (-1-2 \sqrt {1+x}+2 \sqrt {x+\sqrt {1+x}}\right ) \]
-2*ArcTan[1 + Sqrt[1 + x] - Sqrt[x + Sqrt[1 + x]]] - 2*ArcTanh[1 - Sqrt[1 + x] + Sqrt[x + Sqrt[1 + x]]] - 2*Log[-1 - 2*Sqrt[1 + x] + 2*Sqrt[x + Sqrt [1 + x]]]
Time = 0.50 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.87, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {7267, 25, 1321, 1092, 219, 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.
\(\displaystyle \int \frac {\sqrt {x+\sqrt {x+1}}}{x \sqrt {x+1}} \, dx\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 2 \int \frac {\sqrt {x+\sqrt {x+1}}}{x}d\sqrt {x+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int -\frac {\sqrt {x+\sqrt {x+1}}}{x}d\sqrt {x+1}\) |
\(\Big \downarrow \) 1321 |
\(\displaystyle 2 \left (\int \frac {1}{\sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}-\int -\frac {\sqrt {x+1}}{x \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}\right )\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle 2 \left (2 \int \frac {1}{3-x}d\frac {2 \sqrt {x+1}+1}{\sqrt {x+\sqrt {x+1}}}-\int -\frac {\sqrt {x+1}}{x \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 2 \left (\text {arctanh}\left (\frac {2 \sqrt {x+1}+1}{2 \sqrt {x+\sqrt {x+1}}}\right )-\int -\frac {\sqrt {x+1}}{x \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}\right )\) |
\(\Big \downarrow \) 1366 |
\(\displaystyle 2 \left (-\frac {1}{2} \int \frac {1}{\left (1-\sqrt {x+1}\right ) \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}-\frac {1}{2} \int -\frac {1}{\left (\sqrt {x+1}+1\right ) \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}+\text {arctanh}\left (\frac {2 \sqrt {x+1}+1}{2 \sqrt {x+\sqrt {x+1}}}\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle 2 \left (-\frac {1}{2} \int \frac {1}{\left (1-\sqrt {x+1}\right ) \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}+\frac {1}{2} \int \frac {1}{\left (\sqrt {x+1}+1\right ) \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}+\text {arctanh}\left (\frac {2 \sqrt {x+1}+1}{2 \sqrt {x+\sqrt {x+1}}}\right )\right )\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle 2 \left (-\int \frac {1}{-x-5}d\left (-\frac {\sqrt {x+1}+3}{\sqrt {x+\sqrt {x+1}}}\right )+\int \frac {1}{3-x}d\frac {1-3 \sqrt {x+1}}{\sqrt {x+\sqrt {x+1}}}+\text {arctanh}\left (\frac {2 \sqrt {x+1}+1}{2 \sqrt {x+\sqrt {x+1}}}\right )\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle 2 \left (\int \frac {1}{3-x}d\frac {1-3 \sqrt {x+1}}{\sqrt {x+\sqrt {x+1}}}-\frac {1}{2} \arctan \left (\frac {\sqrt {x+1}+3}{2 \sqrt {x+\sqrt {x+1}}}\right )+\text {arctanh}\left (\frac {2 \sqrt {x+1}+1}{2 \sqrt {x+\sqrt {x+1}}}\right )\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 2 \left (-\frac {1}{2} \arctan \left (\frac {\sqrt {x+1}+3}{2 \sqrt {x+\sqrt {x+1}}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {1-3 \sqrt {x+1}}{2 \sqrt {x+\sqrt {x+1}}}\right )+\text {arctanh}\left (\frac {2 \sqrt {x+1}+1}{2 \sqrt {x+\sqrt {x+1}}}\right )\right )\) |
2*(-1/2*ArcTan[(3 + Sqrt[1 + x])/(2*Sqrt[x + Sqrt[1 + x]])] + ArcTanh[(1 - 3*Sqrt[1 + x])/(2*Sqrt[x + Sqrt[1 + x]])]/2 + ArcTanh[(1 + 2*Sqrt[1 + x]) /(2*Sqrt[x + Sqrt[1 + x]])])
3.17.26.3.1 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/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
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[Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]/((d_) + (f_.)*(x_)^2), x_Symbol] :> Simp[c/f Int[1/Sqrt[a + b*x + c*x^2], x], x] - Simp[1/f Int[(c*d - a*f - b*f*x)/(Sqrt[a + b*x + c*x^2]*(d + f*x^2)), x], x] /; FreeQ[{a, b, c, d, f}, x] && NeQ[b^2 - 4*a*c, 0]
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 = 0.11 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.55
method | result | size |
derivativedivides | \(-\sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}+\frac {\ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}\right )}{2}+\arctan \left (\frac {-3-\sqrt {1+x}}{2 \sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}}\right )+\sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}+\frac {3 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}\right )}{2}-\operatorname {arctanh}\left (\frac {-1+3 \sqrt {1+x}}{2 \sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}}\right )\) | \(170\) |
default | \(-\sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}+\frac {\ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}\right )}{2}+\arctan \left (\frac {-3-\sqrt {1+x}}{2 \sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}}\right )+\sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}+\frac {3 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}\right )}{2}-\operatorname {arctanh}\left (\frac {-1+3 \sqrt {1+x}}{2 \sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}}\right )\) | \(170\) |
-((1+(1+x)^(1/2))^2-(1+x)^(1/2)-2)^(1/2)+1/2*ln(1/2+(1+x)^(1/2)+((1+(1+x)^ (1/2))^2-(1+x)^(1/2)-2)^(1/2))+arctan(1/2*(-3-(1+x)^(1/2))/((1+(1+x)^(1/2) )^2-(1+x)^(1/2)-2)^(1/2))+((-1+(1+x)^(1/2))^2+3*(1+x)^(1/2)-2)^(1/2)+3/2*l n(1/2+(1+x)^(1/2)+((-1+(1+x)^(1/2))^2+3*(1+x)^(1/2)-2)^(1/2))-arctanh(1/2* (-1+3*(1+x)^(1/2))/((-1+(1+x)^(1/2))^2+3*(1+x)^(1/2)-2)^(1/2))
Time = 1.67 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x \sqrt {1+x}} \, dx=\arctan \left (\frac {2 \, \sqrt {x + \sqrt {x + 1}} {\left (\sqrt {x + 1} - 3\right )}}{x - 8}\right ) + \log \left (\frac {8 \, x^{2} + 2 \, {\left ({\left (4 \, x - 1\right )} \sqrt {x + 1} - 2 \, x - 1\right )} \sqrt {x + \sqrt {x + 1}} - x + 2 \, \sqrt {x + 1} + 2}{x}\right ) \]
arctan(2*sqrt(x + sqrt(x + 1))*(sqrt(x + 1) - 3)/(x - 8)) + log((8*x^2 + 2 *((4*x - 1)*sqrt(x + 1) - 2*x - 1)*sqrt(x + sqrt(x + 1)) - x + 2*sqrt(x + 1) + 2)/x)
\[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x \sqrt {1+x}} \, dx=\int \frac {\sqrt {x + \sqrt {x + 1}}}{x \sqrt {x + 1}}\, dx \]
\[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x \sqrt {1+x}} \, dx=\int { \frac {\sqrt {x + \sqrt {x + 1}}}{\sqrt {x + 1} x} \,d x } \]
Time = 0.51 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x \sqrt {1+x}} \, dx=2 \, \arctan \left (\sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} - 1\right ) - 2 \, \log \left (-2 \, \sqrt {x + \sqrt {x + 1}} + 2 \, \sqrt {x + 1} + 1\right ) - \log \left ({\left | \sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} + 2 \right |}\right ) + \log \left ({\left | \sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} \right |}\right ) \]
2*arctan(sqrt(x + sqrt(x + 1)) - sqrt(x + 1) - 1) - 2*log(-2*sqrt(x + sqrt (x + 1)) + 2*sqrt(x + 1) + 1) - log(abs(sqrt(x + sqrt(x + 1)) - sqrt(x + 1 ) + 2)) + log(abs(sqrt(x + sqrt(x + 1)) - sqrt(x + 1)))
Timed out. \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x \sqrt {1+x}} \, dx=\int \frac {\sqrt {x+\sqrt {x+1}}}{x\,\sqrt {x+1}} \,d x \]