Integrand size = 25, antiderivative size = 310 \[ \int \frac {(e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {e^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}+\frac {e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}-\frac {4 e^3}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a^2 d \sqrt {\sin (2 c+2 d x)}} \]
[Out]
Time = 0.60 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.680, Rules used = {3973, 3971, 3555, 3557, 335, 303, 1176, 631, 210, 1179, 642, 2688, 2695, 2652, 2719, 2687, 32} \[ \int \frac {(e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {e^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{5/2} \arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a^2 d}-\frac {e^{5/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a^2 d}+\frac {e^{5/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a^2 d}-\frac {4 e^3}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{a^2 d \sqrt {\sin (2 c+2 d x)}} \]
[In]
[Out]
Rule 32
Rule 210
Rule 303
Rule 335
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 2652
Rule 2687
Rule 2688
Rule 2695
Rule 2719
Rule 3555
Rule 3557
Rule 3971
Rule 3973
Rubi steps \begin{align*} \text {integral}& = \frac {e^4 \int \frac {(-a+a \sec (c+d x))^2}{(e \tan (c+d x))^{3/2}} \, dx}{a^4} \\ & = \frac {e^4 \int \left (\frac {a^2}{(e \tan (c+d x))^{3/2}}-\frac {2 a^2 \sec (c+d x)}{(e \tan (c+d x))^{3/2}}+\frac {a^2 \sec ^2(c+d x)}{(e \tan (c+d x))^{3/2}}\right ) \, dx}{a^4} \\ & = \frac {e^4 \int \frac {1}{(e \tan (c+d x))^{3/2}} \, dx}{a^2}+\frac {e^4 \int \frac {\sec ^2(c+d x)}{(e \tan (c+d x))^{3/2}} \, dx}{a^2}-\frac {\left (2 e^4\right ) \int \frac {\sec (c+d x)}{(e \tan (c+d x))^{3/2}} \, dx}{a^2} \\ & = -\frac {2 e^3}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}-\frac {e^2 \int \sqrt {e \tan (c+d x)} \, dx}{a^2}+\frac {\left (4 e^2\right ) \int \cos (c+d x) \sqrt {e \tan (c+d x)} \, dx}{a^2}+\frac {e^4 \text {Subst}\left (\int \frac {1}{(e x)^{3/2}} \, dx,x,\tan (c+d x)\right )}{a^2 d} \\ & = -\frac {4 e^3}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}-\frac {e^3 \text {Subst}\left (\int \frac {\sqrt {x}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{a^2 d}+\frac {\left (4 e^2 \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)} \, dx}{a^2 \sqrt {\sin (c+d x)}} \\ & = -\frac {4 e^3}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}-\frac {\left (2 e^3\right ) \text {Subst}\left (\int \frac {x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a^2 d}+\frac {\left (4 e^2 \cos (c+d x) \sqrt {e \tan (c+d x)}\right ) \int \sqrt {\sin (2 c+2 d x)} \, dx}{a^2 \sqrt {\sin (2 c+2 d x)}} \\ & = -\frac {4 e^3}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a^2 d \sqrt {\sin (2 c+2 d x)}}+\frac {e^3 \text {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a^2 d}-\frac {e^3 \text {Subst}\left (\int \frac {e+x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a^2 d} \\ & = -\frac {4 e^3}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a^2 d \sqrt {\sin (2 c+2 d x)}}-\frac {e^{5/2} \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^2 d}-\frac {e^{5/2} \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^2 d}-\frac {e^3 \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^2 d}-\frac {e^3 \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^2 d} \\ & = -\frac {e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}+\frac {e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}-\frac {4 e^3}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a^2 d \sqrt {\sin (2 c+2 d x)}}-\frac {e^{5/2} \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^2 d}+\frac {e^{5/2} \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^2 d} \\ & = \frac {e^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}+\frac {e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}-\frac {4 e^3}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a^2 d \sqrt {\sin (2 c+2 d x)}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.36 (sec) , antiderivative size = 812, normalized size of antiderivative = 2.62 \[ \int \frac {(e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {\cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc ^2(c+d x) \left (\frac {32 \cos \left (\frac {c}{2}\right ) \cos (d x) \sec (2 c) \sin \left (\frac {c}{2}\right )}{d}+\frac {16 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {d x}{2}\right )}{d}-\frac {16 \cos (c) \sec (2 c) \sin (d x)}{d}+\frac {16 \tan \left (\frac {c}{2}\right )}{d}\right ) (e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2}+\frac {e^{-2 i c} \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 ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (2 c) \sec ^2(c+d x) (e \tan (c+d x))^{5/2}}{d \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 ) (a+a \sec (c+d x))^2 \tan ^{\frac {5}{2}}(c+d x)}-\frac {e^{-2 i c} \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 ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (2 c) \sec ^2(c+d x) (e \tan (c+d x))^{5/2}}{d \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 ) (a+a \sec (c+d x))^2 \tan ^{\frac {5}{2}}(c+d x)}-\frac {8 e^{i (c-d x)} \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (3-3 e^{4 i (c+d x)}+e^{4 i d x} \left (1+e^{4 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 ^2(c+d x) (e \tan (c+d x))^{5/2}}{3 d \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 ) (a+a \sec (c+d x))^2 \tan ^{\frac {5}{2}}(c+d x)} \]
[In]
[Out]
Result contains complex when optimal does not.
Time = 4.49 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.84
method | result | size |
default | \(-\frac {\sqrt {2}\, e^{2} \left (i \operatorname {EllipticPi}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-i \operatorname {EllipticPi}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+\operatorname {EllipticPi}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+\operatorname {EllipticPi}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-8 \operatorname {EllipticE}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+4 \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {e \tan \left (d x +c \right )}\, \sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \sin \left (d x +c \right )}{2 a^{2} d \left (\cos \left (d x +c \right )-1\right )}\) | \(261\) |
[In]
[Out]
Timed out. \[ \int \frac {(e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int \frac {(e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\left (e \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
[In]
[Out]
\[ \int \frac {(e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
[In]
[Out]
\[ \int \frac {(e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {(e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{a^2\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]
[In]
[Out]