Integrand size = 25, antiderivative size = 836 \[ \int \frac {1}{(a+b \sec (c+d x)) (e \tan (c+d x))^{5/2}} \, dx=\frac {a \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}-\frac {b^2 \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}-\frac {a \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}+\frac {b^2 \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}+\frac {a \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}-\frac {b^2 \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}-\frac {a \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}+\frac {b^2 \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}-\frac {2 (a-b \sec (c+d x))}{3 \left (a^2-b^2\right ) d e (e \tan (c+d x))^{3/2}}-\frac {2 \sqrt {2} b^3 \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right ),-1\right ) \sqrt {\sin (c+d x)}}{a \left (a^2-b^2\right )^{3/2} d e^2 \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {2 \sqrt {2} b^3 \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right ),-1\right ) \sqrt {\sin (c+d x)}}{a \left (a^2-b^2\right )^{3/2} d e^2 \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {b \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 \left (a^2-b^2\right ) d e^2 \sqrt {e \tan (c+d x)}} \]
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Time = 1.23 (sec) , antiderivative size = 836, normalized size of antiderivative = 1.00, number of steps used = 36, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {3978, 3967, 3969, 3557, 335, 217, 1179, 642, 1176, 631, 210, 2694, 2653, 2720, 3977, 2812, 2808, 2986, 1227, 551} \[ \int \frac {1}{(a+b \sec (c+d x)) (e \tan (c+d x))^{5/2}} \, dx=-\frac {2 \sqrt {2} \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right ) \sqrt {\sin (c+d x)} b^3}{a \left (a^2-b^2\right )^{3/2} d e^2 \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {2 \sqrt {2} \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right ) \sqrt {\sin (c+d x)} b^3}{a \left (a^2-b^2\right )^{3/2} d e^2 \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right ) b^2}{\sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}+\frac {\arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right ) b^2}{\sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}-\frac {\log \left (\sqrt {e} \tan (c+d x)+\sqrt {e}-\sqrt {2} \sqrt {e \tan (c+d x)}\right ) b^2}{2 \sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}+\frac {\log \left (\sqrt {e} \tan (c+d x)+\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}\right ) b^2}{2 \sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}+\frac {\operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)} b}{3 \left (a^2-b^2\right ) d e^2 \sqrt {e \tan (c+d x)}}+\frac {a \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}-\frac {a \arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}+\frac {a \log \left (\sqrt {e} \tan (c+d x)+\sqrt {e}-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}-\frac {a \log \left (\sqrt {e} \tan (c+d x)+\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}-\frac {2 (a-b \sec (c+d x))}{3 \left (a^2-b^2\right ) d e (e \tan (c+d x))^{3/2}} \]
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Rule 210
Rule 217
Rule 335
Rule 551
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 1227
Rule 2653
Rule 2694
Rule 2720
Rule 2808
Rule 2812
Rule 2986
Rule 3557
Rule 3967
Rule 3969
Rule 3977
Rule 3978
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {a-b \sec (c+d x)}{(e \tan (c+d x))^{5/2}} \, dx}{a^2-b^2}+\frac {b^2 \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \tan (c+d x)}} \, dx}{\left (a^2-b^2\right ) e^2} \\ & = -\frac {2 (a-b \sec (c+d x))}{3 \left (a^2-b^2\right ) d e (e \tan (c+d x))^{3/2}}+\frac {2 \int \frac {-\frac {3 a}{2}+\frac {1}{2} b \sec (c+d x)}{\sqrt {e \tan (c+d x)}} \, dx}{3 \left (a^2-b^2\right ) e^2}+\frac {b^2 \int \frac {1}{\sqrt {e \tan (c+d x)}} \, dx}{a \left (a^2-b^2\right ) e^2}-\frac {b^3 \int \frac {1}{(b+a \cos (c+d x)) \sqrt {e \tan (c+d x)}} \, dx}{a \left (a^2-b^2\right ) e^2} \\ & = -\frac {2 (a-b \sec (c+d x))}{3 \left (a^2-b^2\right ) d e (e \tan (c+d x))^{3/2}}-\frac {a \int \frac {1}{\sqrt {e \tan (c+d x)}} \, dx}{\left (a^2-b^2\right ) e^2}+\frac {b \int \frac {\sec (c+d x)}{\sqrt {e \tan (c+d x)}} \, dx}{3 \left (a^2-b^2\right ) e^2}+\frac {b^2 \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (e^2+x^2\right )} \, dx,x,e \tan (c+d x)\right )}{a \left (a^2-b^2\right ) d e}-\frac {b^3 \int \frac {\sqrt {e \cot (c+d x)}}{b+a \cos (c+d x)} \, dx}{a \left (a^2-b^2\right ) e^2 \sqrt {e \cot (c+d x)} \sqrt {e \tan (c+d x)}} \\ & = -\frac {2 (a-b \sec (c+d x))}{3 \left (a^2-b^2\right ) d e (e \tan (c+d x))^{3/2}}-\frac {a \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (e^2+x^2\right )} \, dx,x,e \tan (c+d x)\right )}{\left (a^2-b^2\right ) d e}+\frac {\left (2 b^2\right ) \text {Subst}\left (\int \frac {1}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a \left (a^2-b^2\right ) d e}-\frac {\left (b^3 \sqrt {\sin (c+d x)}\right ) \int \frac {\sqrt {-\cos (c+d x)}}{(b+a \cos (c+d x)) \sqrt {\sin (c+d x)}} \, dx}{a \left (a^2-b^2\right ) e^2 \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {\left (b \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right ) e^2 \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}} \\ & = -\frac {2 (a-b \sec (c+d x))}{3 \left (a^2-b^2\right ) d e (e \tan (c+d x))^{3/2}}+\frac {b^2 \text {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a \left (a^2-b^2\right ) d e^2}+\frac {b^2 \text {Subst}\left (\int \frac {e+x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a \left (a^2-b^2\right ) d e^2}-\frac {(2 a) \text {Subst}\left (\int \frac {1}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{\left (a^2-b^2\right ) d e}-\frac {\left (2 \sqrt {2} b^3 \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \sqrt {\sin (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\left (-a+\sqrt {a^2-b^2}+b x^2\right ) \sqrt {1-x^4}} \, dx,x,\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right )}{a \left (a^2-b^2\right ) d e^2 \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}-\frac {\left (2 \sqrt {2} b^3 \left (1+\frac {a}{\sqrt {a^2-b^2}}\right ) \sqrt {\sin (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\left (-a-\sqrt {a^2-b^2}+b x^2\right ) \sqrt {1-x^4}} \, dx,x,\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right )}{a \left (a^2-b^2\right ) d e^2 \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {\left (b \sec (c+d x) \sqrt {\sin (2 c+2 d x)}\right ) \int \frac {1}{\sqrt {\sin (2 c+2 d x)}} \, dx}{3 \left (a^2-b^2\right ) e^2 \sqrt {e \tan (c+d x)}} \\ & = -\frac {2 (a-b \sec (c+d x))}{3 \left (a^2-b^2\right ) d e (e \tan (c+d x))^{3/2}}+\frac {b \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 \left (a^2-b^2\right ) d e^2 \sqrt {e \tan (c+d x)}}-\frac {b^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 \left (a^2-b^2\right ) d e^{5/2}}-\frac {b^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 \left (a^2-b^2\right ) d e^{5/2}}-\frac {a \text {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{\left (a^2-b^2\right ) d e^2}-\frac {a \text {Subst}\left (\int \frac {e+x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{\left (a^2-b^2\right ) d e^2}+\frac {b^2 \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 \left (a^2-b^2\right ) d e^2}+\frac {b^2 \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 \left (a^2-b^2\right ) d e^2}-\frac {\left (2 \sqrt {2} b^3 \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \sqrt {\sin (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (-a+\sqrt {a^2-b^2}+b x^2\right )} \, dx,x,\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right )}{a \left (a^2-b^2\right ) d e^2 \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}-\frac {\left (2 \sqrt {2} b^3 \left (1+\frac {a}{\sqrt {a^2-b^2}}\right ) \sqrt {\sin (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (-a-\sqrt {a^2-b^2}+b x^2\right )} \, dx,x,\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right )}{a \left (a^2-b^2\right ) d e^2 \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}} \\ & = -\frac {b^2 \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}+\frac {b^2 \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}-\frac {2 (a-b \sec (c+d x))}{3 \left (a^2-b^2\right ) d e (e \tan (c+d x))^{3/2}}-\frac {2 \sqrt {2} b^3 \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right ),-1\right ) \sqrt {\sin (c+d x)}}{a \left (a^2-b^2\right )^{3/2} d e^2 \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {2 \sqrt {2} b^3 \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right ),-1\right ) \sqrt {\sin (c+d x)}}{a \left (a^2-b^2\right )^{3/2} d e^2 \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {b \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 \left (a^2-b^2\right ) d e^2 \sqrt {e \tan (c+d x)}}+\frac {a \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} \left (a^2-b^2\right ) d e^{5/2}}+\frac {a \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} \left (a^2-b^2\right ) d e^{5/2}}+\frac {b^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 \left (a^2-b^2\right ) d e^{5/2}}-\frac {b^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 \left (a^2-b^2\right ) d e^{5/2}}-\frac {a \text {Subst}\left (\int \frac {1}{e-\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 \left (a^2-b^2\right ) d e^2}-\frac {a \text {Subst}\left (\int \frac {1}{e+\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 \left (a^2-b^2\right ) d e^2} \\ & = -\frac {b^2 \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}+\frac {b^2 \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}+\frac {a \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}-\frac {b^2 \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}-\frac {a \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}+\frac {b^2 \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}-\frac {2 (a-b \sec (c+d x))}{3 \left (a^2-b^2\right ) d e (e \tan (c+d x))^{3/2}}-\frac {2 \sqrt {2} b^3 \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right ),-1\right ) \sqrt {\sin (c+d x)}}{a \left (a^2-b^2\right )^{3/2} d e^2 \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {2 \sqrt {2} b^3 \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right ),-1\right ) \sqrt {\sin (c+d x)}}{a \left (a^2-b^2\right )^{3/2} d e^2 \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {b \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 \left (a^2-b^2\right ) d e^2 \sqrt {e \tan (c+d x)}}-\frac {a \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} \left (a^2-b^2\right ) d e^{5/2}}+\frac {a \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} \left (a^2-b^2\right ) d e^{5/2}} \\ & = \frac {a \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}-\frac {b^2 \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}-\frac {a \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}+\frac {b^2 \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}+\frac {a \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}-\frac {b^2 \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}-\frac {a \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2-b^2\right ) d e^{5/2}}+\frac {b^2 \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a \left (a^2-b^2\right ) d e^{5/2}}-\frac {2 (a-b \sec (c+d x))}{3 \left (a^2-b^2\right ) d e (e \tan (c+d x))^{3/2}}-\frac {2 \sqrt {2} b^3 \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right ),-1\right ) \sqrt {\sin (c+d x)}}{a \left (a^2-b^2\right )^{3/2} d e^2 \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {2 \sqrt {2} b^3 \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right ),-1\right ) \sqrt {\sin (c+d x)}}{a \left (a^2-b^2\right )^{3/2} d e^2 \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {b \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 \left (a^2-b^2\right ) d e^2 \sqrt {e \tan (c+d x)}} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 33.62 (sec) , antiderivative size = 2169, normalized size of antiderivative = 2.59 \[ \int \frac {1}{(a+b \sec (c+d x)) (e \tan (c+d x))^{5/2}} \, dx=\text {Result too large to show} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 7580 vs. \(2 (725 ) = 1450\).
Time = 4.79 (sec) , antiderivative size = 7581, normalized size of antiderivative = 9.07
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Timed out. \[ \int \frac {1}{(a+b \sec (c+d x)) (e \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
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\[ \int \frac {1}{(a+b \sec (c+d x)) (e \tan (c+d x))^{5/2}} \, dx=\int \frac {1}{\left (e \tan {\left (c + d x \right )}\right )^{\frac {5}{2}} \left (a + b \sec {\left (c + d x \right )}\right )}\, dx \]
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\[ \int \frac {1}{(a+b \sec (c+d x)) (e \tan (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )} \left (e \tan \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
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\[ \int \frac {1}{(a+b \sec (c+d x)) (e \tan (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )} \left (e \tan \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {1}{(a+b \sec (c+d x)) (e \tan (c+d x))^{5/2}} \, dx=\int \frac {\cos \left (c+d\,x\right )}{{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}\,\left (b+a\,\cos \left (c+d\,x\right )\right )} \,d x \]
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