Integrand size = 25, antiderivative size = 98 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{(a-b)^{3/2} f}+\frac {a^2}{(a-b) b^2 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\sqrt {a+b \tan ^2(e+f x)}}{b^2 f} \]
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Time = 0.21 (sec) , antiderivative size = 98, 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} \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {a^2}{b^2 f (a-b) \sqrt {a+b \tan ^2(e+f x)}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f (a-b)^{3/2}}+\frac {\sqrt {a+b \tan ^2(e+f x)}}{b^2 f} \]
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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 )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {x^2}{(1+x) (a+b x)^{3/2}} \, 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)^{3/2}}+\frac {1}{b \sqrt {a+b x}}+\frac {1}{(a-b) (1+x) \sqrt {a+b x}}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f} \\ & = \frac {a^2}{(a-b) b^2 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\sqrt {a+b \tan ^2(e+f x)}}{b^2 f}+\frac {\text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{2 (a-b) f} \\ & = \frac {a^2}{(a-b) b^2 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\sqrt {a+b \tan ^2(e+f x)}}{b^2 f}+\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) b f} \\ & = -\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{(a-b)^{3/2} f}+\frac {a^2}{(a-b) b^2 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\sqrt {a+b \tan ^2(e+f x)}}{b^2 f} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.42 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.86 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan ^2(e+f x)}{a-b}\right )+(a-b) \left (2 a+b+b \tan ^2(e+f x)\right )}{(a-b) b^2 f \sqrt {a+b \tan ^2(e+f x)}} \]
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Time = 0.07 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.33
method | result | size |
derivativedivides | \(\frac {\frac {\tan \left (f x +e \right )^{2}}{b \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {2 a}{b^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {1}{b \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {1}{\left (a -b \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\left (a -b \right ) \sqrt {-a +b}}}{f}\) | \(130\) |
default | \(\frac {\frac {\tan \left (f x +e \right )^{2}}{b \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {2 a}{b^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {1}{b \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {1}{\left (a -b \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}+\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\left (a -b \right ) \sqrt {-a +b}}}{f}\) | \(130\) |
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Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (88) = 176\).
Time = 0.33 (sec) , antiderivative size = 432, normalized size of antiderivative = 4.41 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\left [-\frac {{\left (b^{3} \tan \left (f x + e\right )^{2} + a b^{2}\right )} \sqrt {a - b} \log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} + 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) - 4 \, {\left (2 \, a^{3} - 3 \, a^{2} b + a b^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{4 \, {\left ({\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4}\right )} f\right )}}, \frac {{\left (b^{3} \tan \left (f x + e\right )^{2} + a b^{2}\right )} \sqrt {-a + b} \arctan \left (\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) + 2 \, {\left (2 \, a^{3} - 3 \, a^{2} b + a b^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{2 \, {\left ({\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4}\right )} f\right )}}\right ] \]
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\[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\tan ^{5}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
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\[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\tan \left (f x + e\right )^{5}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
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Timed out. \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \]
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Time = 14.11 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.14 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{b^2\,f}+\frac {a^2}{b^2\,f\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\left (a-b\right )}+\frac {\mathrm {atan}\left (\frac {a\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,1{}\mathrm {i}-b\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,1{}\mathrm {i}}{{\left (a-b\right )}^{3/2}}\right )\,1{}\mathrm {i}}{f\,{\left (a-b\right )}^{3/2}} \]
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