Integrand size = 25, antiderivative size = 119 \[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=-\frac {15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}-\frac {3 \left (2 a^2-3 b^2\right ) d \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(e+f x)\right ) \sin (e+f x)}{8 f (d \sec (e+f x))^{4/3} \sqrt {\sin ^2(e+f x)}}+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}} \]
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Time = 0.18 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3589, 3567, 3857, 2722} \[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=-\frac {3 d \left (2 a^2-3 b^2\right ) \sin (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(e+f x)\right )}{8 f \sqrt {\sin ^2(e+f x)} (d \sec (e+f x))^{4/3}}-\frac {15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}} \]
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Rule 2722
Rule 3567
Rule 3589
Rule 3857
Rubi steps \begin{align*} \text {integral}& = \frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac {3}{2} \int \frac {\frac {2 a^2}{3}-b^2+\frac {5}{3} a b \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx \\ & = -\frac {15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac {1}{2} \left (2 a^2-3 b^2\right ) \int \frac {1}{\sqrt [3]{d \sec (e+f x)}} \, dx \\ & = -\frac {15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac {1}{2} \left (\left (2 a^2-3 b^2\right ) \left (\frac {\cos (e+f x)}{d}\right )^{2/3} (d \sec (e+f x))^{2/3}\right ) \int \sqrt [3]{\frac {\cos (e+f x)}{d}} \, dx \\ & = -\frac {15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}-\frac {3 \left (2 a^2-3 b^2\right ) \cos ^2(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(e+f x)\right ) (d \sec (e+f x))^{2/3} \sin (e+f x)}{8 d f \sqrt {\sin ^2(e+f x)}}+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}} \\ \end{align*}
Time = 1.01 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.90 \[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=-\frac {3 \left (b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{6},\frac {5}{6},\sec ^2(e+f x)\right ) \tan (e+f x)+a \left (-a \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\sec ^2(e+f x)\right ) \tan (e+f x)+2 b \sqrt {-\tan ^2(e+f x)}\right )\right )}{f \sqrt [3]{d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}} \]
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\[\int \frac {\left (a +b \tan \left (f x +e \right )\right )^{2}}{\left (d \sec \left (f x +e \right )\right )^{\frac {1}{3}}}d x\]
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\[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \]
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\[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{2}}{\sqrt [3]{d \sec {\left (e + f x \right )}}}\, dx \]
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\[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \]
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\[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \]
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Timed out. \[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{1/3}} \,d x \]
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