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]
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]
[Out]
Rule 210
Rule 212
Rule 738
Rule 1047
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*}
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]
[Out]
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]
[Out]
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]
[Out]
\[ \int \frac {1+2 \sqrt {1+x}}{x \sqrt {1+x} \sqrt {x+\sqrt {1+x}}} \, dx=\int \frac {2 \sqrt {x + 1} + 1}{x \sqrt {x + 1} \sqrt {x + \sqrt {x + 1}}}\, dx \]
[In]
[Out]
\[ \int \frac {1+2 \sqrt {1+x}}{x \sqrt {1+x} \sqrt {x+\sqrt {1+x}}} \, dx=\int { \frac {2 \, \sqrt {x + 1} + 1}{\sqrt {x + \sqrt {x + 1}} \sqrt {x + 1} x} \,d x } \]
[In]
[Out]
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]
[Out]
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]
[Out]