\(\int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))} \, dx\) [230]

Optimal result

Integrand size = 28, antiderivative size = 114 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))} \, dx=\frac {10 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{21 a d e^2}+\frac {10 \sin (c+d x)}{21 a d e \sqrt {e \sec (c+d x)}}+\frac {2 i}{7 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))} \] Output:

10/21*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/d/e^2+10/21*sin(d*x+c)/a/d/e/(e*sec(d*x+c))^(1/2)+2/7*I/d/(e*se 
c(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))
 

Mathematica [A] (verified)

Time = 1.42 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.10 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))} \, dx=-\frac {\sec ^3(c+d x) \left (-14 \cos (c+d x)+2 \cos (3 (c+d x))+20 i \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (c+d x)+i \sin (c+d x))+5 i \sin (c+d x)+5 i \sin (3 (c+d x))\right )}{42 a d (e \sec (c+d x))^{3/2} (-i+\tan (c+d x))} \] Input:

Integrate[1/((e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])),x]
 

Output:

-1/42*(Sec[c + d*x]^3*(-14*Cos[c + d*x] + 2*Cos[3*(c + d*x)] + (20*I)*Sqrt 
[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[c + d*x] + I*Sin[c + d*x]) + 
 (5*I)*Sin[c + d*x] + (5*I)*Sin[3*(c + d*x)]))/(a*d*(e*Sec[c + d*x])^(3/2) 
*(-I + Tan[c + d*x]))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3983, 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.

\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x)) (e \sec (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x)) (e \sec (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {5 \int \frac {1}{(e \sec (c+d x))^{3/2}}dx}{7 a}+\frac {2 i}{7 d (a+i a \tan (c+d x)) (e \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4256

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

Input:

Int[1/((e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])),x]
 

Output:

(5*((2*Sqrt[Cos[c + d*x]]*EllipticF[(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*a) + ((2*I) 
/7)/(d*(e*Sec[c + d*x])^(3/2)*(a + I*a*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 3983

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

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 = 3.77 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.04

method	result	size
default	\(\frac {\frac {2 \sin \left (d x +c \right ) \left (3 \cos \left (d x +c \right )^{2}+5\right )}{21}+\frac {2 i \cos \left (d x +c \right )^{3}}{7}+\frac {2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \left (5+5 \sec \left (d x +c \right )\right )}{21}}{e \sqrt {e \sec \left (d x +c \right )}\, a d}\)	\(118\)

Input:

int(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

2/21/a/d/(e*sec(d*x+c))^(1/2)*(sin(d*x+c)*(3*cos(d*x+c)^2+5)+3*I*cos(d*x+c 
)^3+I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF 
(I*(cot(d*x+c)-csc(d*x+c)),I)*(5+5*sec(d*x+c)))/e
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.01 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))} \, dx=\frac {{\left (\sqrt {2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-7 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 9 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 19 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} - 40 i \, \sqrt {2} \sqrt {e} e^{\left (4 i \, d x + 4 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{84 \, a d e^{2}} \] Input:

integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c)),x, algorithm="fricas")
 

Output:

1/84*(sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-7*I*e^(6*I*d*x + 6*I*c) 
+ 9*I*e^(4*I*d*x + 4*I*c) + 19*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(1/2*I*d*x + 
 1/2*I*c) - 40*I*sqrt(2)*sqrt(e)*e^(4*I*d*x + 4*I*c)*weierstrassPInverse(- 
4, 0, e^(I*d*x + I*c)))*e^(-4*I*d*x - 4*I*c)/(a*d*e^2)
 

Sympy [F]

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

integrate(1/(e*sec(d*x+c))**(3/2)/(a+I*a*tan(d*x+c)),x)
 

Output:

-I*Integral(1/((e*sec(c + d*x))**(3/2)*tan(c + d*x) - I*(e*sec(c + d*x))** 
(3/2)), x)/a
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c)),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c)),x, algorithm="giac")
 

Output:

integrate(1/((e*sec(d*x + c))^(3/2)*(I*a*tan(d*x + c) + a)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))} \, dx=\int \frac {1}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \] Input:

int(1/((e/cos(c + d*x))^(3/2)*(a + a*tan(c + d*x)*1i)),x)
 

Output:

int(1/((e/cos(c + d*x))^(3/2)*(a + a*tan(c + d*x)*1i)), x)
 

Reduce [F]

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

int(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c)),x)
 

Output:

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