Integrand size = 25, antiderivative size = 257 \[ \int \frac {(e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\frac {e^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {e^{3/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {e^{3/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}+\frac {e^2 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{a d \sqrt {e \tan (c+d x)}} \]
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Time = 0.35 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3973, 3969, 3557, 335, 217, 1179, 642, 1176, 631, 210, 2694, 2653, 2720} \[ \int \frac {(e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\frac {e^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{3/2} \arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a d}+\frac {e^{3/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}-\frac {e^{3/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}+\frac {e^2 \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{a d \sqrt {e \tan (c+d x)}} \]
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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 3969
Rule 3973
Rubi steps \begin{align*} \text {integral}& = \frac {e^2 \int \frac {-a+a \sec (c+d x)}{\sqrt {e \tan (c+d x)}} \, dx}{a^2} \\ & = -\frac {e^2 \int \frac {1}{\sqrt {e \tan (c+d x)}} \, dx}{a}+\frac {e^2 \int \frac {\sec (c+d x)}{\sqrt {e \tan (c+d x)}} \, dx}{a} \\ & = -\frac {e^3 \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^2 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)}} \, dx}{a \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}} \\ & = -\frac {\left (2 e^3\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^2 \sec (c+d x) \sqrt {\sin (2 c+2 d x)}\right ) \int \frac {1}{\sqrt {\sin (2 c+2 d x)}} \, dx}{a \sqrt {e \tan (c+d x)}} \\ & = \frac {e^2 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{a d \sqrt {e \tan (c+d x)}}-\frac {e^2 \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^2 \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^2 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{a d \sqrt {e \tan (c+d x)}}+\frac {e^{3/2} \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^{3/2} \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^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 d}-\frac {e^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 d} \\ & = \frac {e^{3/2} \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^{3/2} \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^2 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{a d \sqrt {e \tan (c+d x)}}-\frac {e^{3/2} \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^{3/2} \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^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {e^{3/2} \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^{3/2} \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^2 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{a d \sqrt {e \tan (c+d x)}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 15.61 (sec) , antiderivative size = 1211, normalized size of antiderivative = 4.71 \[ \int \frac {(e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\frac {\cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc (c+d x) \left (\frac {8 \cos (c) \cos (d x) \sec (2 c) \sin ^2\left (\frac {c}{2}\right )}{d}-\frac {16 \cos \left (\frac {c}{2}\right ) \sec (2 c) \sin ^3\left (\frac {c}{2}\right ) \sin (d x)}{d}\right ) (e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)}-\frac {2 e^{-i (c+d x)} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right ) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (2 c) \sec (c+d x) (e \tan (c+d x))^{3/2}}{d (a+a \sec (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}-\frac {e^{-2 i c} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (e^{4 i c} \sqrt {-1+e^{4 i (c+d x)}} \arctan \left (\sqrt {-1+e^{4 i (c+d x)}}\right )+2 \sqrt {-1+e^{2 i (c+d x)}} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right ) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (2 c) \sec (c+d x) (e \tan (c+d x))^{3/2}}{2 d \left (-1+e^{2 i (c+d x)}\right ) (a+a \sec (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}-\frac {e^{-2 i c} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (\sqrt {-1+e^{4 i (c+d x)}} \arctan \left (\sqrt {-1+e^{4 i (c+d x)}}\right )+2 e^{4 i c} \sqrt {-1+e^{2 i (c+d x)}} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right ) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (2 c) \sec (c+d x) (e \tan (c+d x))^{3/2}}{2 d \left (-1+e^{2 i (c+d x)}\right ) (a+a \sec (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {e^{-i (2 c+d x)} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (3 \left (-1+e^{4 i (c+d x)}\right )+e^{4 i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1-e^{4 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},e^{4 i (c+d x)}\right )\right ) \sec (2 c) \sec (c+d x) (e \tan (c+d x))^{3/2}}{3 d \left (-1+e^{2 i (c+d x)}\right ) (a+a \sec (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}-\frac {e^{-i d x} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (3-3 e^{4 i (c+d x)}+e^{2 i (c+2 d x)} \left (-1+e^{2 i c}\right ) \sqrt {1-e^{4 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},e^{4 i (c+d x)}\right )\right ) \sec (2 c) \sec (c+d x) (e \tan (c+d x))^{3/2}}{3 d \left (-1+e^{2 i (c+d x)}\right ) (a+a \sec (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}-\frac {4 \sqrt [4]{-1} \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ),-1\right ) \sec ^4(c+d x) (e \tan (c+d x))^{3/2}}{d (a+a \sec (c+d x)) \tan ^{\frac {3}{2}}(c+d x) \left (1+\tan ^2(c+d x)\right )^{3/2}} \]
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Result contains complex when optimal does not.
Time = 3.75 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.80
method | result | size |
default | \(\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {2}\, \left (i \operatorname {EllipticPi}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+\operatorname {EllipticPi}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-2 \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 i \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {e \tan \left (d x +c \right )}\, \sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, e \sin \left (d x +c \right )}{a d \left (\cos \left (d x +c \right )-1\right )}\) | \(205\) |
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Timed out. \[ \int \frac {(e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\text {Timed out} \]
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\[ \int \frac {(e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {\left (e \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]
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\[ \int \frac {(e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {3}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \]
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\[ \int \frac {(e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {3}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {(e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{a\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \]
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