Integrand size = 12, antiderivative size = 247 \[ \int \sqrt {1+\tan (e+f x)} \, dx=-\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{f}+\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{f}+\frac {\log \left (1+\sqrt {2}+\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f}-\frac {\log \left (1+\sqrt {2}+\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f} \]
[Out]
Time = 0.22 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3566, 714, 1141, 1175, 632, 210, 1178, 642} \[ \int \sqrt {1+\tan (e+f x)} \, dx=-\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{f}+\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \arctan \left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{f}+\frac {\log \left (\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f}-\frac {\log \left (\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f} \]
[In]
[Out]
Rule 210
Rule 632
Rule 642
Rule 714
Rule 1141
Rule 1175
Rule 1178
Rule 3566
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\sqrt {1+x}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {2 \text {Subst}\left (\int \frac {x^2}{2-2 x^2+x^4} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{f} \\ & = -\frac {\text {Subst}\left (\int \frac {\sqrt {2}-x^2}{2-2 x^2+x^4} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{f}+\frac {\text {Subst}\left (\int \frac {\sqrt {2}+x^2}{2-2 x^2+x^4} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {1}{\sqrt {2}-\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{2 f}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {2}+\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{2 f}+\frac {\text {Subst}\left (\int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 x}{-\sqrt {2}-\sqrt {2 \left (1+\sqrt {2}\right )} x-x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f}+\frac {\text {Subst}\left (\int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 x}{-\sqrt {2}+\sqrt {2 \left (1+\sqrt {2}\right )} x-x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f} \\ & = \frac {\log \left (1+\sqrt {2}+\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f}-\frac {\log \left (1+\sqrt {2}+\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f}-\frac {\text {Subst}\left (\int \frac {1}{2 \left (1-\sqrt {2}\right )-x^2} \, dx,x,-\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}\right )}{f}-\frac {\text {Subst}\left (\int \frac {1}{2 \left (1-\sqrt {2}\right )-x^2} \, dx,x,\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}\right )}{f} \\ & = -\frac {\arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{\sqrt {2 \left (-1+\sqrt {2}\right )} f}+\frac {\arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{\sqrt {2 \left (-1+\sqrt {2}\right )} f}+\frac {\log \left (1+\sqrt {2}+\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f}-\frac {\log \left (1+\sqrt {2}+\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f} \\ \end{align*}
Result contains complex when optimal does not.
Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.27 \[ \int \sqrt {1+\tan (e+f x)} \, dx=-\frac {i \left (\sqrt {1-i} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )-\sqrt {1+i} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )\right )}{f} \]
[In]
[Out]
Time = 0.99 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.17
method | result | size |
derivativedivides | \(\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (1+\sqrt {2}-\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}+\frac {\arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (1+\sqrt {2}-\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}-\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}+\frac {\arctan \left (\frac {\sqrt {2+2 \sqrt {2}}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}\) | \(288\) |
default | \(\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (1+\sqrt {2}-\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}+\frac {\arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (1+\sqrt {2}-\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}-\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}+\frac {\arctan \left (\frac {\sqrt {2+2 \sqrt {2}}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}\) | \(288\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.05 \[ \int \sqrt {1+\tan (e+f x)} \, dx=-\frac {1}{2} \, \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (f^{3} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \sqrt {-\frac {1}{f^{4}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + \frac {1}{2} \, \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-f^{3} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \sqrt {-\frac {1}{f^{4}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + \frac {1}{2} \, \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (f^{3} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \sqrt {-\frac {1}{f^{4}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - \frac {1}{2} \, \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-f^{3} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \sqrt {-\frac {1}{f^{4}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) \]
[In]
[Out]
\[ \int \sqrt {1+\tan (e+f x)} \, dx=\int \sqrt {\tan {\left (e + f x \right )} + 1}\, dx \]
[In]
[Out]
Exception generated. \[ \int \sqrt {1+\tan (e+f x)} \, dx=\text {Exception raised: RuntimeError} \]
[In]
[Out]
none
Time = 0.74 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.05 \[ \int \sqrt {1+\tan (e+f x)} \, dx=\frac {{\left (f^{2} \sqrt {\sqrt {2} + 1} + f \sqrt {\sqrt {2} - 1} {\left | f \right |}\right )} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{2 \, f^{3}} + \frac {{\left (f^{2} \sqrt {\sqrt {2} + 1} + f \sqrt {\sqrt {2} - 1} {\left | f \right |}\right )} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} - 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{2 \, f^{3}} + \frac {{\left (f^{2} \sqrt {\sqrt {2} - 1} - f \sqrt {\sqrt {2} + 1} {\left | f \right |}\right )} \log \left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{4 \, f^{3}} - \frac {{\left (f^{2} \sqrt {\sqrt {2} - 1} - f \sqrt {\sqrt {2} + 1} {\left | f \right |}\right )} \log \left (-2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{4 \, f^{3}} \]
[In]
[Out]
Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.30 \[ \int \sqrt {1+\tan (e+f x)} \, dx=2\,\mathrm {atanh}\left (4\,f^3\,{\left (\frac {-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}\right )}^{3/2}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}}+2\,\mathrm {atanh}\left (4\,f^3\,{\left (\frac {-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}\right )}^{3/2}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}} \]
[In]
[Out]