Optimal. Leaf size=311 \[ -\frac {\sqrt {1+\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{2 f}+\frac {\sqrt {1+\sqrt {2}} \tan ^{-1}\left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{2 f}+\frac {2 \sqrt {\tan (e+f x)+1} \tan ^2(e+f x)}{5 f}-\frac {8 \sqrt {\tan (e+f x)+1} \tan (e+f x)}{15 f}-\frac {14 \sqrt {\tan (e+f x)+1}}{15 f}-\frac {\log \left (\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{4 \sqrt {1+\sqrt {2}} f}+\frac {\log \left (\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{4 \sqrt {1+\sqrt {2}} f} \]
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Rubi [A] time = 0.36, antiderivative size = 311, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3566, 3647, 3631, 3485, 708, 1094, 634, 618, 204, 628} \[ \frac {2 \sqrt {\tan (e+f x)+1} \tan ^2(e+f x)}{5 f}-\frac {\sqrt {1+\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{2 f}+\frac {\sqrt {1+\sqrt {2}} \tan ^{-1}\left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{2 f}-\frac {8 \sqrt {\tan (e+f x)+1} \tan (e+f x)}{15 f}-\frac {14 \sqrt {\tan (e+f x)+1}}{15 f}-\frac {\log \left (\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{4 \sqrt {1+\sqrt {2}} f}+\frac {\log \left (\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{4 \sqrt {1+\sqrt {2}} f} \]
Antiderivative was successfully verified.
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Rule 204
Rule 618
Rule 628
Rule 634
Rule 708
Rule 1094
Rule 3485
Rule 3566
Rule 3631
Rule 3647
Rubi steps
\begin {align*} \int \frac {\tan ^4(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx &=\frac {2 \tan ^2(e+f x) \sqrt {1+\tan (e+f x)}}{5 f}+\frac {2}{5} \int \frac {\tan (e+f x) \left (-2-\frac {5}{2} \tan (e+f x)-2 \tan ^2(e+f x)\right )}{\sqrt {1+\tan (e+f x)}} \, dx\\ &=-\frac {8 \tan (e+f x) \sqrt {1+\tan (e+f x)}}{15 f}+\frac {2 \tan ^2(e+f x) \sqrt {1+\tan (e+f x)}}{5 f}+\frac {4}{15} \int \frac {2-\frac {7}{4} \tan ^2(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx\\ &=-\frac {14 \sqrt {1+\tan (e+f x)}}{15 f}-\frac {8 \tan (e+f x) \sqrt {1+\tan (e+f x)}}{15 f}+\frac {2 \tan ^2(e+f x) \sqrt {1+\tan (e+f x)}}{5 f}+\int \frac {1}{\sqrt {1+\tan (e+f x)}} \, dx\\ &=-\frac {14 \sqrt {1+\tan (e+f x)}}{15 f}-\frac {8 \tan (e+f x) \sqrt {1+\tan (e+f x)}}{15 f}+\frac {2 \tan ^2(e+f x) \sqrt {1+\tan (e+f x)}}{5 f}+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {14 \sqrt {1+\tan (e+f x)}}{15 f}-\frac {8 \tan (e+f x) \sqrt {1+\tan (e+f x)}}{15 f}+\frac {2 \tan ^2(e+f x) \sqrt {1+\tan (e+f x)}}{5 f}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{2-2 x^2+x^4} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{f}\\ &=-\frac {14 \sqrt {1+\tan (e+f x)}}{15 f}-\frac {8 \tan (e+f x) \sqrt {1+\tan (e+f x)}}{15 f}+\frac {2 \tan ^2(e+f x) \sqrt {1+\tan (e+f x)}}{5 f}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}-x}{\sqrt {2}-\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {1+\sqrt {2}} f}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}+x}{\sqrt {2}+\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {1+\sqrt {2}} f}\\ &=-\frac {14 \sqrt {1+\tan (e+f x)}}{15 f}-\frac {8 \tan (e+f x) \sqrt {1+\tan (e+f x)}}{15 f}+\frac {2 \tan ^2(e+f x) \sqrt {1+\tan (e+f x)}}{5 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 \sqrt {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 \sqrt {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 )}{4 \sqrt {1+\sqrt {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 )}{4 \sqrt {1+\sqrt {2}} 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 )}{4 \sqrt {1+\sqrt {2}} 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 )}{4 \sqrt {1+\sqrt {2}} f}-\frac {14 \sqrt {1+\tan (e+f x)}}{15 f}-\frac {8 \tan (e+f x) \sqrt {1+\tan (e+f x)}}{15 f}+\frac {2 \tan ^2(e+f x) \sqrt {1+\tan (e+f x)}}{5 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 )}{\sqrt {2} 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 )}{\sqrt {2} 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 )}{2 \sqrt {-1+\sqrt {2}} 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 )}{2 \sqrt {-1+\sqrt {2}} 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 )}{4 \sqrt {1+\sqrt {2}} 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 )}{4 \sqrt {1+\sqrt {2}} f}-\frac {14 \sqrt {1+\tan (e+f x)}}{15 f}-\frac {8 \tan (e+f x) \sqrt {1+\tan (e+f x)}}{15 f}+\frac {2 \tan ^2(e+f x) \sqrt {1+\tan (e+f x)}}{5 f}\\ \end {align*}
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Mathematica [C] time = 0.83, size = 112, normalized size = 0.36 \[ \frac {15 (1-i)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {\tan (e+f x)+1}}{\sqrt {1-i}}\right )+15 (1+i)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {\tan (e+f x)+1}}{\sqrt {1+i}}\right )-4 \sqrt {\tan (e+f x)+1} \sec ^2(e+f x) (2 \sin (2 (e+f x))+5 \cos (2 (e+f x))+2)}{30 f} \]
Antiderivative was successfully verified.
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fricas [B] time = 1.00, size = 847, normalized size = 2.72 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.18, size = 238, normalized size = 0.77 \[ \frac {\sqrt {\sqrt {2} + 1} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{2 \, f} + \frac {\sqrt {\sqrt {2} + 1} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} - 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{2 \, f} + \frac {\sqrt {\sqrt {2} - 1} \log \left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{4 \, f} - \frac {\sqrt {\sqrt {2} - 1} \log \left (-2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{4 \, f} + \frac {2 \, {\left (3 \, f^{4} {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {5}{2}} - 10 \, f^{4} {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {3}{2}}\right )}}{15 \, f^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.23, size = 326, normalized size = 1.05 \[ \frac {2 \left (1+\tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}-\frac {4 \left (1+\tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3 f}-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (1+\sqrt {2}-\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}+\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (1+\sqrt {2}-\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{8 f}+\frac {\arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{2 f \sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (1+\sqrt {2}+\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}-\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (1+\sqrt {2}+\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{8 f}+\frac {\arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{2 f \sqrt {-2+2 \sqrt {2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (f x + e\right )^{4}}{\sqrt {\tan \left (f x + e\right ) + 1}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.55, size = 101, normalized size = 0.32 \[ \frac {2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{5/2}}{5\,f}-\frac {4\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{3/2}}{3\,f}+\mathrm {atan}\left (2\,f\,\sqrt {\frac {-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (2\,f\,\sqrt {\frac {-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \]
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 \[ \int \frac {\tan ^{4}{\left (e + f x \right )}}{\sqrt {\tan {\left (e + f x \right )} + 1}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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