\(\int \frac {1}{(e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx\) [244]

Optimal result

Integrand size = 28, antiderivative size = 181 \[ \int \frac {1}{(e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx=\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{7 a^2 d e^4}+\frac {2 e \sin (c+d x)}{15 a^2 d (e \sec (c+d x))^{9/2}}+\frac {6 \sin (c+d x)}{35 a^2 d e (e \sec (c+d x))^{5/2}}+\frac {2 \sin (c+d x)}{7 a^2 d e^3 \sqrt {e \sec (c+d x)}}+\frac {4 i e^2}{15 d (e \sec (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )} \] Output:

2/7*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*(e*sec(d*x+c)) 
^(1/2)/a^2/d/e^4+2/15*e*sin(d*x+c)/a^2/d/(e*sec(d*x+c))^(9/2)+6/35*sin(d*x 
+c)/a^2/d/e/(e*sec(d*x+c))^(5/2)+2/7*sin(d*x+c)/a^2/d/e^3/(e*sec(d*x+c))^( 
1/2)+4/15*I*e^2/d/(e*sec(d*x+c))^(11/2)/(a^2+I*a^2*tan(d*x+c))
 

Mathematica [A] (verified)

Time = 1.83 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.83 \[ \int \frac {1}{(e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx=-\frac {(e \sec (c+d x))^{5/2} \left (296 i+228 i \cos (2 (c+d x))-72 i \cos (4 (c+d x))-4 i \cos (6 (c+d x))+480 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))-17 \sin (2 (c+d x))+128 \sin (4 (c+d x))+11 \sin (6 (c+d x))\right )}{1680 a^2 d e^6 (-i+\tan (c+d x))^2} \] Input:

Integrate[1/((e*Sec[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x])^2),x]
 

Output:

-1/1680*((e*Sec[c + d*x])^(5/2)*(296*I + (228*I)*Cos[2*(c + d*x)] - (72*I) 
*Cos[4*(c + d*x)] - (4*I)*Cos[6*(c + d*x)] + 480*Sqrt[Cos[c + d*x]]*Ellipt 
icF[(c + d*x)/2, 2]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) - 17*Sin[2*(c 
+ d*x)] + 128*Sin[4*(c + d*x)] + 11*Sin[6*(c + d*x)]))/(a^2*d*e^6*(-I + Ta 
n[c + d*x])^2)
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.09, 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, 4256, 3042, 4256, 3042, 4256, 3042, 4258, 3042, 3120}

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.

\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^2 (e \sec (c+d x))^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^2 (e \sec (c+d x))^{7/2}}dx\)

\(\Big \downarrow \) 3981

\(\displaystyle \frac {11 e^2 \int \frac {1}{(e \sec (c+d x))^{11/2}}dx}{15 a^2}+\frac {4 i e^2}{15 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{11/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11 e^2 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{11/2}}dx}{15 a^2}+\frac {4 i e^2}{15 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{11/2}}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {11 e^2 \left (\frac {9 \int \frac {1}{(e \sec (c+d x))^{7/2}}dx}{11 e^2}+\frac {2 \sin (c+d x)}{11 d e (e \sec (c+d x))^{9/2}}\right )}{15 a^2}+\frac {4 i e^2}{15 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{11/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11 e^2 \left (\frac {9 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}dx}{11 e^2}+\frac {2 \sin (c+d x)}{11 d e (e \sec (c+d x))^{9/2}}\right )}{15 a^2}+\frac {4 i e^2}{15 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{11/2}}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {11 e^2 \left (\frac {9 \left (\frac {5 \int \frac {1}{(e \sec (c+d x))^{3/2}}dx}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}+\frac {2 \sin (c+d x)}{11 d e (e \sec (c+d x))^{9/2}}\right )}{15 a^2}+\frac {4 i e^2}{15 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{11/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11 e^2 \left (\frac {9 \left (\frac {5 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}+\frac {2 \sin (c+d x)}{11 d e (e \sec (c+d x))^{9/2}}\right )}{15 a^2}+\frac {4 i e^2}{15 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{11/2}}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {11 e^2 \left (\frac {9 \left (\frac {5 \left (\frac {\int \sqrt {e \sec (c+d x)}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}+\frac {2 \sin (c+d x)}{11 d e (e \sec (c+d x))^{9/2}}\right )}{15 a^2}+\frac {4 i e^2}{15 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{11/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11 e^2 \left (\frac {9 \left (\frac {5 \left (\frac {\int \sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}+\frac {2 \sin (c+d x)}{11 d e (e \sec (c+d x))^{9/2}}\right )}{15 a^2}+\frac {4 i e^2}{15 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{11/2}}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {11 e^2 \left (\frac {9 \left (\frac {5 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}+\frac {2 \sin (c+d x)}{11 d e (e \sec (c+d x))^{9/2}}\right )}{15 a^2}+\frac {4 i e^2}{15 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{11/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11 e^2 \left (\frac {9 \left (\frac {5 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}+\frac {2 \sin (c+d x)}{11 d e (e \sec (c+d x))^{9/2}}\right )}{15 a^2}+\frac {4 i e^2}{15 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{11/2}}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {11 e^2 \left (\frac {9 \left (\frac {5 \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}+\frac {2 \sin (c+d x)}{11 d e (e \sec (c+d x))^{9/2}}\right )}{15 a^2}+\frac {4 i e^2}{15 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{11/2}}\)

Input:

Int[1/((e*Sec[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x])^2),x]
 

Output:

(11*e^2*((2*Sin[c + d*x])/(11*d*e*(e*Sec[c + d*x])^(9/2)) + (9*((2*Sin[c + 
 d*x])/(7*d*e*(e*Sec[c + d*x])^(5/2)) + (5*((2*Sqrt[Cos[c + d*x]]*Elliptic 
F[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(3*d*e^2) + (2*Sin[c + d*x])/(3*d* 
e*Sqrt[e*Sec[c + d*x]])))/(7*e^2)))/(11*e^2)))/(15*a^2) + (((4*I)/15)*e^2) 
/(d*(e*Sec[c + d*x])^(11/2)*(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 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 4256

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n)   Int[(b*Csc[c 
+ d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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 = 17.70 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.76

method	result	size
default	\(\frac {\frac {2 \sin \left (d x +c \right ) \left (14 \cos \left (d x +c \right )^{6}+7 \cos \left (d x +c \right )^{4}+9 \cos \left (d x +c \right )^{2}+15\right )}{105}+\frac {4 i \cos \left (d x +c \right )^{7}}{15}+\frac {2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (15+15 \sec \left (d x +c \right )\right )}{105}}{a^{2} d \sqrt {e \sec \left (d x +c \right )}\, e^{3}}\)	\(138\)

Input:

int(1/(e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/a^2/d*(2/105*sin(d*x+c)*(14*cos(d*x+c)^6+7*cos(d*x+c)^4+9*cos(d*x+c)^2+1 
5)+4/15*I*cos(d*x+c)^7+2/105*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF 
(I*(cot(d*x+c)-csc(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*(15+15*sec(d*x+c))) 
/(e*sec(d*x+c))^(1/2)/e^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.82 \[ \int \frac {1}{(e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx=\frac {{\left (\sqrt {2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-15 i \, e^{\left (12 i \, d x + 12 i \, c\right )} - 200 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 245 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 592 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 211 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 56 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} - 960 i \, \sqrt {2} \sqrt {e} e^{\left (8 i \, d x + 8 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{3360 \, a^{2} d e^{4}} \] Input:

integrate(1/(e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas 
")
 

Output:

1/3360*(sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-15*I*e^(12*I*d*x + 12* 
I*c) - 200*I*e^(10*I*d*x + 10*I*c) + 245*I*e^(8*I*d*x + 8*I*c) + 592*I*e^( 
6*I*d*x + 6*I*c) + 211*I*e^(4*I*d*x + 4*I*c) + 56*I*e^(2*I*d*x + 2*I*c) + 
7*I)*e^(1/2*I*d*x + 1/2*I*c) - 960*I*sqrt(2)*sqrt(e)*e^(8*I*d*x + 8*I*c)*w 
eierstrassPInverse(-4, 0, e^(I*d*x + I*c)))*e^(-8*I*d*x - 8*I*c)/(a^2*d*e^ 
4)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate(1/(e*sec(d*x+c))**(7/2)/(a+I*a*tan(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/(e*sec(d*x+c))^(7/2)/(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]

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

integrate(1/(e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")
 

Output:

integrate(1/((e*sec(d*x + c))^(7/2)*(I*a*tan(d*x + c) + a)^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx=\int \frac {1}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \] Input:

int(1/((e/cos(c + d*x))^(7/2)*(a + a*tan(c + d*x)*1i)^2),x)
 

Output:

int(1/((e/cos(c + d*x))^(7/2)*(a + a*tan(c + d*x)*1i)^2), x)
 

Reduce [F]

\[ \int \frac {1}{(e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx=-\frac {\int \frac {1}{\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3} \tan \left (d x +c \right )^{2}-2 \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3} \tan \left (d x +c \right ) i -\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}}d x}{\sqrt {e}\, a^{2} e^{3}} \] Input:

int(1/(e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x)
 

Output:

( - int(1/(sqrt(sec(c + d*x))*sec(c + d*x)**3*tan(c + d*x)**2 - 2*sqrt(sec 
(c + d*x))*sec(c + d*x)**3*tan(c + d*x)*i - sqrt(sec(c + d*x))*sec(c + d*x 
)**3),x))/(sqrt(e)*a**2*e**3)