Optimal. Leaf size=247 \[ -\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{f}+\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{f}+\frac {\log \left (\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f}-\frac {\log \left (\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f} \]
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Rubi [A] time = 0.19, antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3485, 700, 1127, 1161, 618, 204, 1164, 628} \[ -\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{f}+\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{f}+\frac {\log \left (\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f}-\frac {\log \left (\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f} \]
Antiderivative was successfully verified.
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Rule 204
Rule 618
Rule 628
Rule 700
Rule 1127
Rule 1161
Rule 1164
Rule 3485
Rubi steps
\begin {align*} \int \sqrt {1+\tan (e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sqrt {1+x}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {2 \operatorname {Subst}\left (\int \frac {x^2}{2-2 x^2+x^4} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{f}\\ &=-\frac {\operatorname {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 {\operatorname {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 {\operatorname {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 {\operatorname {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 {\operatorname {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 {\operatorname {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 {\operatorname {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 {\operatorname {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 {\tan ^{-1}\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 {\tan ^{-1}\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*}
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Mathematica [C] time = 0.03, size = 67, normalized size = 0.27 \[ -\frac {i \left (\sqrt {1-i} \tanh ^{-1}\left (\frac {\sqrt {\tan (e+f x)+1}}{\sqrt {1-i}}\right )-\sqrt {1+i} \tanh ^{-1}\left (\frac {\sqrt {\tan (e+f x)+1}}{\sqrt {1+i}}\right )\right )}{f} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.49, size = 767, normalized size = 3.11 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.24, size = 266, normalized size = 1.08 \[ \frac {{\left (f^{2} \sqrt {\sqrt {2} + 1} + f \sqrt {\sqrt {2} - 1} {\left | f \right |}\right )} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{2 \, f^{3}} + \frac {{\left (f^{2} \sqrt {\sqrt {2} + 1} + f \sqrt {\sqrt {2} - 1} {\left | f \right |}\right )} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} - 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{2 \, f^{3}} + \frac {{\left (f^{2} \sqrt {\sqrt {2} - 1} - f \sqrt {\sqrt {2} + 1} {\left | f \right |}\right )} \log \left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{4 \, f^{3}} - \frac {{\left (f^{2} \sqrt {\sqrt {2} - 1} - f \sqrt {\sqrt {2} + 1} {\left | f \right |}\right )} \log \left (-2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{4 \, f^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.15, size = 288, normalized size = 1.17 \[ \frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (1+\sqrt {2}-\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}+\frac {\arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (1+\sqrt {2}-\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}-\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (1+\sqrt {2}+\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}+\frac {\arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (1+\sqrt {2}+\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.18, size = 73, normalized size = 0.30 \[ 2\,\mathrm {atanh}\left (4\,f^3\,{\left (\frac {-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}\right )}^{3/2}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}}+2\,\mathrm {atanh}\left (4\,f^3\,{\left (\frac {-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}\right )}^{3/2}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\tan {\left (e + f x \right )} + 1}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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