\(\int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx\) [233]

Optimal result

Integrand size = 28, antiderivative size = 183 \[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx=-\frac {22 e^8 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^2 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {22 e^7 \sqrt {e \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{45 a^2 d}+\frac {22 e^3 (e \sec (c+d x))^{9/2} \sin (c+d x)}{63 a^2 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )} \] Output:

-22/15*e^8*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d/cos(d*x+c)^(1/2)/(e 
*sec(d*x+c))^(1/2)+22/15*e^7*(e*sec(d*x+c))^(1/2)*sin(d*x+c)/a^2/d+22/45*e 
^5*(e*sec(d*x+c))^(5/2)*sin(d*x+c)/a^2/d+22/63*e^3*(e*sec(d*x+c))^(9/2)*si 
n(d*x+c)/a^2/d-4/7*I*e^2*(e*sec(d*x+c))^(11/2)/d/(a^2+I*a^2*tan(d*x+c))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 2.75 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.65 \[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx=\frac {(e \sec (c+d x))^{15/2} (\cos (d x)+i \sin (d x))^2 \left (\frac {22 i \sqrt {2} e^{3 i c-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}+\frac {1}{56} \csc (c) \sec ^{\frac {9}{2}}(c+d x) (\cos (2 c)+i \sin (2 c)) (1260 \cos (d x)+1050 \cos (2 c+d x)+1078 \cos (2 c+3 d x)+77 \cos (4 c+3 d x)+231 \cos (4 c+5 d x)+720 i \sin (d x)-720 i \sin (2 c+d x))\right )}{45 d \sec ^{\frac {11}{2}}(c+d x) (a+i a \tan (c+d x))^2} \] Input:

Integrate[(e*Sec[c + d*x])^(15/2)/(a + I*a*Tan[c + d*x])^2,x]
 

Output:

((e*Sec[c + d*x])^(15/2)*(Cos[d*x] + I*Sin[d*x])^2*(((22*I)*Sqrt[2]*E^((3* 
I)*c - I*d*x)*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^( 
(2*I)*(c + d*x))]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + 
E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/(-1 
+ E^((2*I)*c)) + (Csc[c]*Sec[c + d*x]^(9/2)*(Cos[2*c] + I*Sin[2*c])*(1260* 
Cos[d*x] + 1050*Cos[2*c + d*x] + 1078*Cos[2*c + 3*d*x] + 77*Cos[4*c + 3*d* 
x] + 231*Cos[4*c + 5*d*x] + (720*I)*Sin[d*x] - (720*I)*Sin[2*c + d*x]))/56 
))/(45*d*Sec[c + d*x]^(11/2)*(a + I*a*Tan[c + d*x])^2)
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3981, 3042, 4255, 3042, 4255, 3042, 4255, 3042, 4258, 3042, 3119}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2}dx\)

\(\Big \downarrow \) 3981

\(\displaystyle \frac {11 e^2 \int (e \sec (c+d x))^{11/2}dx}{7 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11 e^2 \int \left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{11/2}dx}{7 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {11 e^2 \left (\frac {7}{9} e^2 \int (e \sec (c+d x))^{7/2}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{9/2}}{9 d}\right )}{7 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11 e^2 \left (\frac {7}{9} e^2 \int \left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{9/2}}{9 d}\right )}{7 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {11 e^2 \left (\frac {7}{9} e^2 \left (\frac {3}{5} e^2 \int (e \sec (c+d x))^{3/2}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{9/2}}{9 d}\right )}{7 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11 e^2 \left (\frac {7}{9} e^2 \left (\frac {3}{5} e^2 \int \left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{9/2}}{9 d}\right )}{7 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {11 e^2 \left (\frac {7}{9} e^2 \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-e^2 \int \frac {1}{\sqrt {e \sec (c+d x)}}dx\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{9/2}}{9 d}\right )}{7 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11 e^2 \left (\frac {7}{9} e^2 \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-e^2 \int \frac {1}{\sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{9/2}}{9 d}\right )}{7 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {11 e^2 \left (\frac {7}{9} e^2 \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {e^2 \int \sqrt {\cos (c+d x)}dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{9/2}}{9 d}\right )}{7 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11 e^2 \left (\frac {7}{9} e^2 \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {e^2 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{9/2}}{9 d}\right )}{7 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {11 e^2 \left (\frac {7}{9} e^2 \left (\frac {3}{5} e^2 \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d}\right )+\frac {2 e \sin (c+d x) (e \sec (c+d x))^{9/2}}{9 d}\right )}{7 a^2}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}\)

Input:

Int[(e*Sec[c + d*x])^(15/2)/(a + I*a*Tan[c + d*x])^2,x]
 

Output:

(11*e^2*((2*e*(e*Sec[c + d*x])^(9/2)*Sin[c + d*x])/(9*d) + (7*e^2*((2*e*(e 
*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5*d) + (3*e^2*((-2*e^2*EllipticE[(c + 
d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (2*e*Sqrt[e*Sec[ 
c + d*x]]*Sin[c + d*x])/d))/5))/9))/(7*a^2) - (((4*I)/7)*e^2*(e*Sec[c + d* 
x])^(11/2))/(d*(a^2 + I*a^2*Tan[c + d*x]))
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3981

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)])^(n_), x_Symbol] :> Simp[2*d^2*(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + 
 f*x])^(n + 1)/(b*f*(m + 2*n))), x] - Simp[d^2*((m - 2)/(b^2*(m + 2*n))) 
Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[ 
{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] 
 && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (IntegersQ[n, m + 
1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]
 

rule 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

Maple [A] (verified)

Time = 5.67 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.41

method	result	size
default	\(-\frac {2 e^{7} \left (\tan \left (d x +c \right ) \sec \left (d x +c \right )^{3} \left (-231 \cos \left (d x +c \right )^{4}-77 \cos \left (d x +c \right )^{3}-77 \cos \left (d x +c \right )^{2}+35 \cos \left (d x +c \right )+35\right )+90 i \sec \left (d x +c \right )^{2}+90 i \sec \left (d x +c \right )^{3}+231 i \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+231 i \left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sqrt {e \sec \left (d x +c \right )}}{315 a^{2} d \left (\cos \left (d x +c \right )+1\right )}\)	\(258\)

Input:

int((e*sec(d*x+c))^(15/2)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

-2/315*e^7/a^2/d*(tan(d*x+c)*sec(d*x+c)^3*(-231*cos(d*x+c)^4-77*cos(d*x+c) 
^3-77*cos(d*x+c)^2+35*cos(d*x+c)+35)+90*I*sec(d*x+c)^2+90*I*sec(d*x+c)^3+2 
31*I*(cos(d*x+c)^2+2*cos(d*x+c)+1)*(1/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(c 
sc(d*x+c)-cot(d*x+c)),I)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+231*I*(-cos(d*x 
+c)^2-2*cos(d*x+c)-1)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot 
(d*x+c)),I)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(e*sec(d*x+c))^(1/2)/(cos(d 
*x+c)+1)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.40 \[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (231 i \, e^{7} e^{\left (9 i \, d x + 9 i \, c\right )} + 1078 i \, e^{7} e^{\left (7 i \, d x + 7 i \, c\right )} + 1980 i \, e^{7} e^{\left (5 i \, d x + 5 i \, c\right )} + 1770 i \, e^{7} e^{\left (3 i \, d x + 3 i \, c\right )} + 77 i \, e^{7} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 231 \, \sqrt {2} {\left (i \, e^{7} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 i \, e^{7} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 i \, e^{7} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, e^{7} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e^{7}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{315 \, {\left (a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \] Input:

integrate((e*sec(d*x+c))^(15/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas" 
)
 

Output:

-2/315*(sqrt(2)*(231*I*e^7*e^(9*I*d*x + 9*I*c) + 1078*I*e^7*e^(7*I*d*x + 7 
*I*c) + 1980*I*e^7*e^(5*I*d*x + 5*I*c) + 1770*I*e^7*e^(3*I*d*x + 3*I*c) + 
77*I*e^7*e^(I*d*x + I*c))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 
 1/2*I*c) + 231*sqrt(2)*(I*e^7*e^(8*I*d*x + 8*I*c) + 4*I*e^7*e^(6*I*d*x + 
6*I*c) + 6*I*e^7*e^(4*I*d*x + 4*I*c) + 4*I*e^7*e^(2*I*d*x + 2*I*c) + I*e^7 
)*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c 
))))/(a^2*d*e^(8*I*d*x + 8*I*c) + 4*a^2*d*e^(6*I*d*x + 6*I*c) + 6*a^2*d*e^ 
(4*I*d*x + 4*I*c) + 4*a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate((e*sec(d*x+c))**(15/2)/(a+I*a*tan(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((e*sec(d*x+c))^(15/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima" 
)
 

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F]

\[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {15}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((e*sec(d*x+c))^(15/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")
 

Output:

integrate((e*sec(d*x + c))^(15/2)/(I*a*tan(d*x + c) + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{15/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \] Input:

int((e/cos(c + d*x))^(15/2)/(a + a*tan(c + d*x)*1i)^2,x)
 

Output:

int((e/cos(c + d*x))^(15/2)/(a + a*tan(c + d*x)*1i)^2, x)
 

Reduce [F]

\[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{7}}{\tan \left (d x +c \right )^{2}-2 \tan \left (d x +c \right ) i -1}d x \right ) e^{7}}{a^{2}} \] Input:

int((e*sec(d*x+c))^(15/2)/(a+I*a*tan(d*x+c))^2,x)
 

Output:

( - sqrt(e)*int((sqrt(sec(c + d*x))*sec(c + d*x)**7)/(tan(c + d*x)**2 - 2* 
tan(c + d*x)*i - 1),x)*e**7)/a**2