Integrand size = 25, antiderivative size = 115 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{(a-b)^{5/2} f}+\frac {a^2}{3 (a-b) b^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (a-2 b)}{(a-b)^2 b^2 f \sqrt {a+b \tan ^2(e+f x)}} \]
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Time = 0.24 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3751, 457, 89, 65, 214} \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {a^2}{3 b^2 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f (a-b)^{5/2}}-\frac {a (a-2 b)}{b^2 f (a-b)^2 \sqrt {a+b \tan ^2(e+f x)}} \]
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Rule 65
Rule 89
Rule 214
Rule 457
Rule 3751
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^5}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {x^2}{(1+x) (a+b x)^{5/2}} \, dx,x,\tan ^2(e+f x)\right )}{2 f} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {a^2}{(a-b) b (a+b x)^{5/2}}+\frac {a (a-2 b)}{(a-b)^2 b (a+b x)^{3/2}}+\frac {1}{(a-b)^2 (1+x) \sqrt {a+b x}}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f} \\ & = \frac {a^2}{3 (a-b) b^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (a-2 b)}{(a-b)^2 b^2 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{2 (a-b)^2 f} \\ & = \frac {a^2}{3 (a-b) b^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (a-2 b)}{(a-b)^2 b^2 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\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 )}{(a-b)^2 b f} \\ & = -\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{(a-b)^{5/2} f}+\frac {a^2}{3 (a-b) b^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (a-2 b)}{(a-b)^2 b^2 f \sqrt {a+b \tan ^2(e+f x)}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.45 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.79 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {b^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \tan ^2(e+f x)}{a-b}\right )-(a-b) \left (2 a-b+3 b \tan ^2(e+f x)\right )}{3 (a-b) b^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]
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Time = 0.08 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.35
method | result | size |
derivativedivides | \(\frac {-\frac {\tan \left (f x +e \right )^{2}}{b \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {2 a}{3 b^{2} \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {1}{3 b \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {1}{\left (a -b \right )^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {1}{3 \left (a -b \right ) \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\left (a -b \right )^{2} \sqrt {-a +b}}}{f}\) | \(155\) |
default | \(\frac {-\frac {\tan \left (f x +e \right )^{2}}{b \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {2 a}{3 b^{2} \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {1}{3 b \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {1}{\left (a -b \right )^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {1}{3 \left (a -b \right ) \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\left (a -b \right )^{2} \sqrt {-a +b}}}{f}\) | \(155\) |
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Leaf count of result is larger than twice the leaf count of optimal. 270 vs. \(2 (103) = 206\).
Time = 0.35 (sec) , antiderivative size = 608, normalized size of antiderivative = 5.29 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (b^{4} \tan \left (f x + e\right )^{4} + 2 \, a b^{3} \tan \left (f x + e\right )^{2} + a^{2} b^{2}\right )} \sqrt {a - b} \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 (2 \, a^{4} - 7 \, a^{3} b + 5 \, a^{2} b^{2} + 3 \, {\left (a^{3} b - 3 \, a^{2} b^{2} + 2 \, a b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{12 \, {\left ({\left (a^{3} b^{4} - 3 \, a^{2} b^{5} + 3 \, a b^{6} - b^{7}\right )} f \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b^{3} - 3 \, a^{3} b^{4} + 3 \, a^{2} b^{5} - a b^{6}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - a^{2} b^{5}\right )} f\right )}}, \frac {3 \, {\left (b^{4} \tan \left (f x + e\right )^{4} + 2 \, a b^{3} \tan \left (f x + e\right )^{2} + a^{2} b^{2}\right )} \sqrt {-a + b} \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 (2 \, a^{4} - 7 \, a^{3} b + 5 \, a^{2} b^{2} + 3 \, {\left (a^{3} b - 3 \, a^{2} b^{2} + 2 \, a b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{6 \, {\left ({\left (a^{3} b^{4} - 3 \, a^{2} b^{5} + 3 \, a b^{6} - b^{7}\right )} f \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b^{3} - 3 \, a^{3} b^{4} + 3 \, a^{2} b^{5} - a b^{6}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - a^{2} b^{5}\right )} f\right )}}\right ] \]
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\[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\tan ^{5}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
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Exception generated. \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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Timed out. \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \]
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Time = 15.78 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.29 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {\frac {a^2}{3\,\left (a-b\right )}+\frac {\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )\,\left (2\,a\,b-a^2\right )}{{\left (a-b\right )}^2}}{b^2\,f\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}}+\frac {\mathrm {atan}\left (\frac {a^2\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,1{}\mathrm {i}+b^2\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,1{}\mathrm {i}-a\,b\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,2{}\mathrm {i}}{{\left (a-b\right )}^{5/2}}\right )\,1{}\mathrm {i}}{f\,{\left (a-b\right )}^{5/2}} \]
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