Integrand size = 25, antiderivative size = 117 \[ \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=-\frac {\sqrt {a-b} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}+\frac {\sqrt {a+b \tan ^2(e+f x)}}{f}-\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f} \]
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Time = 0.18 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3751, 457, 90, 52, 65, 214} \[ \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=-\frac {\sqrt {a-b} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}-\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac {\sqrt {a+b \tan ^2(e+f x)}}{f} \]
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Rule 52
Rule 65
Rule 90
Rule 214
Rule 457
Rule 3751
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^5 \sqrt {a+b x^2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {x^2 \sqrt {a+b x}}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f} \\ & = \frac {\text {Subst}\left (\int \left (\frac {(-a-b) \sqrt {a+b x}}{b}+\frac {\sqrt {a+b x}}{1+x}+\frac {(a+b x)^{3/2}}{b}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f} \\ & = -\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac {\text {Subst}\left (\int \frac {\sqrt {a+b x}}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f} \\ & = \frac {\sqrt {a+b \tan ^2(e+f x)}}{f}-\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac {(a-b) \text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{2 f} \\ & = \frac {\sqrt {a+b \tan ^2(e+f x)}}{f}-\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac {(a-b) \text {Subst}\left (\int \frac {1}{1-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \tan ^2(e+f x)}\right )}{b f} \\ & = -\frac {\sqrt {a-b} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}+\frac {\sqrt {a+b \tan ^2(e+f x)}}{f}-\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f} \\ \end{align*}
Time = 1.47 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.93 \[ \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {-15 \sqrt {a-b} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )+\frac {\sqrt {a+b \tan ^2(e+f x)} \left (-2 a^2-5 a b+15 b^2+(a-5 b) b \tan ^2(e+f x)+3 b^2 \tan ^4(e+f x)\right )}{b^2}}{15 f} \]
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Time = 0.09 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.34
method | result | size |
derivativedivides | \(\frac {\frac {\tan \left (f x +e \right )^{2} \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{15 b^{2}}-\frac {\left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{3 b}+b \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{b}-\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}\right )+\frac {a \arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}}{f}\) | \(157\) |
default | \(\frac {\frac {\tan \left (f x +e \right )^{2} \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{15 b^{2}}-\frac {\left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{3 b}+b \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{b}-\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}\right )+\frac {a \arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}}{f}\) | \(157\) |
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Time = 0.32 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.74 \[ \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\left [\frac {15 \, \sqrt {a - b} b^{2} \log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, {\left (3 \, b^{2} \tan \left (f x + e\right )^{4} + {\left (a b - 5 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2} - 5 \, a b + 15 \, b^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{60 \, b^{2} f}, \frac {15 \, \sqrt {-a + b} b^{2} \arctan \left (\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) + 2 \, {\left (3 \, b^{2} \tan \left (f x + e\right )^{4} + {\left (a b - 5 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2} - 5 \, a b + 15 \, b^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{30 \, b^{2} f}\right ] \]
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\[ \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int \sqrt {a + b \tan ^{2}{\left (e + f x \right )}} \tan ^{5}{\left (e + f x \right )}\, dx \]
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\[ \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int { \sqrt {b \tan \left (f x + e\right )^{2} + a} \tan \left (f x + e\right )^{5} \,d x } \]
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Timed out. \[ \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\text {Timed out} \]
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Time = 19.81 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.34 \[ \int \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{5/2}}{5\,b^2\,f}-\left (\frac {2\,a}{3\,b^2\,f}-\frac {a-b}{3\,b^2\,f}\right )\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}-\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\left (\left (\frac {2\,a}{b^2\,f}-\frac {a-b}{b^2\,f}\right )\,\left (a-b\right )-\frac {a^2}{b^2\,f}\right )+\frac {\mathrm {atan}\left (\frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,1{}\mathrm {i}}{\sqrt {a-b}}\right )\,\sqrt {a-b}\,1{}\mathrm {i}}{f} \]
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