\(\int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx\) [242]

Optimal result

Integrand size = 28, antiderivative size = 150 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=\frac {10 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{33 a^2 d e^2}+\frac {2 e \sin (c+d x)}{11 a^2 d (e \sec (c+d x))^{5/2}}+\frac {10 \sin (c+d x)}{33 a^2 d e \sqrt {e \sec (c+d x)}}+\frac {4 i e^2}{11 d (e \sec (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )} \] Output:

10/33*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^2+2/11*e*sin(d*x+c)/a^2/d/(e*sec(d*x+c))^(5/2)+10/33*sin( 
d*x+c)/a^2/d/e/(e*sec(d*x+c))^(1/2)+4/11*I*e^2/d/(e*sec(d*x+c))^(7/2)/(a^2 
+I*a^2*tan(d*x+c))
 

Mathematica [A] (verified)

Time = 1.46 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.89 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=-\frac {\sec ^4(c+d x) \left (28 i+24 i \cos (2 (c+d x))-4 i \cos (4 (c+d x))+40 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))-6 \sin (2 (c+d x))+7 \sin (4 (c+d x))\right )}{132 a^2 d (e \sec (c+d x))^{3/2} (-i+\tan (c+d x))^2} \] Input:

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

Output:

-1/132*(Sec[c + d*x]^4*(28*I + (24*I)*Cos[2*(c + d*x)] - (4*I)*Cos[4*(c + 
d*x)] + 40*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[2*(c + d*x)] 
+ I*Sin[2*(c + d*x)]) - 6*Sin[2*(c + d*x)] + 7*Sin[4*(c + d*x)]))/(a^2*d*( 
e*Sec[c + d*x])^(3/2)*(-I + Tan[c + d*x])^2)
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 3981, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3981

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4256

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4256

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

(7*e^2*((2*Sin[c + d*x])/(7*d*e*(e*Sec[c + d*x])^(5/2)) + (5*((2*Sqrt[Cos[ 
c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(3*d*e^2) + (2*S 
in[c + d*x])/(3*d*e*Sqrt[e*Sec[c + d*x]])))/(7*e^2)))/(11*a^2) + (((4*I)/1 
1)*e^2)/(d*(e*Sec[c + d*x])^(7/2)*(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 = 4.24 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.85

method	result	size
default	\(\frac {\frac {2 \sin \left (d x +c \right ) \left (6 \cos \left (d x +c \right )^{4}+3 \cos \left (d x +c \right )^{2}+5\right )}{33}+\frac {2 i \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-5-5 \sec \left (d x +c \right )\right )}{33}+\frac {4 i \cos \left (d x +c \right )^{5}}{11}}{e \sqrt {e \sec \left (d x +c \right )}\, a^{2} d}\)	\(128\)

Input:

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

Output:

2/33/a^2/d/(e*sec(d*x+c))^(1/2)*(sin(d*x+c)*(6*cos(d*x+c)^4+3*cos(d*x+c)^2 
+5)+I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF 
(I*(csc(d*x+c)-cot(d*x+c)),I)*(-5-5*sec(d*x+c))+6*I*cos(d*x+c)^5)/e
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=\frac {{\left (\sqrt {2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-11 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 30 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 56 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 18 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} - 80 i \, \sqrt {2} \sqrt {e} e^{\left (6 i \, d x + 6 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{264 \, a^{2} d e^{2}} \] Input:

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

Output:

1/264*(sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-11*I*e^(8*I*d*x + 8*I*c 
) + 30*I*e^(6*I*d*x + 6*I*c) + 56*I*e^(4*I*d*x + 4*I*c) + 18*I*e^(2*I*d*x 
+ 2*I*c) + 3*I)*e^(1/2*I*d*x + 1/2*I*c) - 80*I*sqrt(2)*sqrt(e)*e^(6*I*d*x 
+ 6*I*c)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)))*e^(-6*I*d*x - 6*I*c) 
/(a^2*d*e^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(1/(e*sec(d*x+c))^(3/2)/(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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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