Integrand size = 25, antiderivative size = 295 \[ \int \frac {(e \tan (c+d x))^{7/2}}{a+a \sec (c+d x)} \, dx=-\frac {e^{7/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {e^{7/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{7/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}+\frac {e^{7/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {e^4 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 a d \sqrt {e \tan (c+d x)}}-\frac {2 e^3 (3-\sec (c+d x)) \sqrt {e \tan (c+d x)}}{3 a d} \]
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Time = 0.45 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3973, 3966, 3969, 3557, 335, 217, 1179, 642, 1176, 631, 210, 2694, 2653, 2720} \[ \int \frac {(e \tan (c+d x))^{7/2}}{a+a \sec (c+d x)} \, dx=-\frac {e^{7/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {e^{7/2} \arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a d}-\frac {e^{7/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}+\frac {e^{7/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}-\frac {e^4 \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{3 a d \sqrt {e \tan (c+d x)}}-\frac {2 e^3 (3-\sec (c+d x)) \sqrt {e \tan (c+d x)}}{3 a d} \]
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Rule 210
Rule 217
Rule 335
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 2653
Rule 2694
Rule 2720
Rule 3557
Rule 3966
Rule 3969
Rule 3973
Rubi steps \begin{align*} \text {integral}& = \frac {e^2 \int (-a+a \sec (c+d x)) (e \tan (c+d x))^{3/2} \, dx}{a^2} \\ & = -\frac {2 e^3 (3-\sec (c+d x)) \sqrt {e \tan (c+d x)}}{3 a d}-\frac {\left (2 e^4\right ) \int \frac {-\frac {3 a}{2}+\frac {1}{2} a \sec (c+d x)}{\sqrt {e \tan (c+d x)}} \, dx}{3 a^2} \\ & = -\frac {2 e^3 (3-\sec (c+d x)) \sqrt {e \tan (c+d x)}}{3 a d}-\frac {e^4 \int \frac {\sec (c+d x)}{\sqrt {e \tan (c+d x)}} \, dx}{3 a}+\frac {e^4 \int \frac {1}{\sqrt {e \tan (c+d x)}} \, dx}{a} \\ & = -\frac {2 e^3 (3-\sec (c+d x)) \sqrt {e \tan (c+d x)}}{3 a d}+\frac {e^5 \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (e^2+x^2\right )} \, dx,x,e \tan (c+d x)\right )}{a d}-\frac {\left (e^4 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)}} \, dx}{3 a \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}} \\ & = -\frac {2 e^3 (3-\sec (c+d x)) \sqrt {e \tan (c+d x)}}{3 a d}+\frac {\left (2 e^5\right ) \text {Subst}\left (\int \frac {1}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}-\frac {\left (e^4 \sec (c+d x) \sqrt {\sin (2 c+2 d x)}\right ) \int \frac {1}{\sqrt {\sin (2 c+2 d x)}} \, dx}{3 a \sqrt {e \tan (c+d x)}} \\ & = -\frac {e^4 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 a d \sqrt {e \tan (c+d x)}}-\frac {2 e^3 (3-\sec (c+d x)) \sqrt {e \tan (c+d x)}}{3 a d}+\frac {e^4 \text {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}+\frac {e^4 \text {Subst}\left (\int \frac {e+x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d} \\ & = -\frac {e^4 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 a d \sqrt {e \tan (c+d x)}}-\frac {2 e^3 (3-\sec (c+d x)) \sqrt {e \tan (c+d x)}}{3 a d}-\frac {e^{7/2} \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}+2 x}{-e-\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {e^{7/2} \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}-2 x}{-e+\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}+\frac {e^4 \text {Subst}\left (\int \frac {1}{e-\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 a d}+\frac {e^4 \text {Subst}\left (\int \frac {1}{e+\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 a d} \\ & = -\frac {e^{7/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}+\frac {e^{7/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {e^4 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 a d \sqrt {e \tan (c+d x)}}-\frac {2 e^3 (3-\sec (c+d x)) \sqrt {e \tan (c+d x)}}{3 a d}+\frac {e^{7/2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{7/2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d} \\ & = -\frac {e^{7/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {e^{7/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{7/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}+\frac {e^{7/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {e^4 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 a d \sqrt {e \tan (c+d x)}}-\frac {2 e^3 (3-\sec (c+d x)) \sqrt {e \tan (c+d x)}}{3 a d} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 22.33 (sec) , antiderivative size = 271, normalized size of antiderivative = 0.92 \[ \int \frac {(e \tan (c+d x))^{7/2}}{a+a \sec (c+d x)} \, dx=\frac {e^3 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (1+\sqrt {\sec ^2(c+d x)}\right ) \left (-2 \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )+2 \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )-\sqrt {2} \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )+\sqrt {2} \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-8 \sqrt {\tan (c+d x)}+8 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\tan ^2(c+d x)\right ) \sqrt {\tan (c+d x)}-8 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\tan ^2(c+d x)\right ) \sqrt {\tan (c+d x)}\right ) \sqrt {e \tan (c+d x)}}{2 a d (1+\sec (c+d x))^2 \sqrt {\tan (c+d x)}} \]
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Result contains complex when optimal does not.
Time = 4.03 (sec) , antiderivative size = 976, normalized size of antiderivative = 3.31
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Timed out. \[ \int \frac {(e \tan (c+d x))^{7/2}}{a+a \sec (c+d x)} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {(e \tan (c+d x))^{7/2}}{a+a \sec (c+d x)} \, dx=\text {Timed out} \]
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\[ \int \frac {(e \tan (c+d x))^{7/2}}{a+a \sec (c+d x)} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {7}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \]
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\[ \int \frac {(e \tan (c+d x))^{7/2}}{a+a \sec (c+d x)} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {7}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {(e \tan (c+d x))^{7/2}}{a+a \sec (c+d x)} \, dx=\int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{7/2}}{a\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \]
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