\(\int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^2} \, dx\) [239]

Optimal result

Integrand size = 28, antiderivative size = 90 \[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^2} \, dx=\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {4 i e^2}{5 d \sqrt {e \sec (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )} \] Output:

2/5*e^2*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d/cos(d*x+c)^(1/2)/(e*se 
c(d*x+c))^(1/2)+4/5*I*e^2/d/(e*sec(d*x+c))^(1/2)/(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 = 1.22 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.13 \[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^2} \, dx=\frac {i e e^{-3 i (c+d x)} \left (1+e^{2 i (c+d x)}+2 e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )\right ) \sqrt {e \sec (c+d x)}}{5 a^2 d} \] Input:

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

Output:

((I/5)*e*(1 + E^((2*I)*(c + d*x)) + 2*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I 
)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))])*Sqr 
t[e*Sec[c + d*x]])/(a^2*d*E^((3*I)*(c + d*x)))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3042, 3981, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3981

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

Input:

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

Output:

(2*e^2*EllipticE[(c + d*x)/2, 2])/(5*a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c 
 + d*x]]) + (((4*I)/5)*e^2)/(d*Sqrt[e*Sec[c + d*x]]*(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 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 [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 427 vs. \(2 (80 ) = 160\).

Time = 3.41 (sec) , antiderivative size = 428, normalized size of antiderivative = 4.76

method	result	size
default	\(-\frac {2 \left (i \left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right )+\sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \left (-\cos \left (d x +c \right )^{3}-2 \cos \left (d x +c \right )^{2}-\cos \left (d x +c \right )\right )+i \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right )+\sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \left (\cos \left (d x +c \right )^{3}+2 \cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )\right )-i \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+\cos \left (d x +c \right ) \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right )\right ) \sqrt {e \sec \left (d x +c \right )}\, e}{5 a^{2} d \left (-\cos \left (d x +c \right ) \sin \left (d x +c \right )-\sin \left (d x +c \right )+i \cos \left (d x +c \right )^{2}+i \cos \left (d x +c \right )\right )}\)	\(428\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.20 \[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^2} \, dx=\frac {{\left (2 i \, \sqrt {2} e^{\frac {3}{2}} e^{\left (3 i \, d x + 3 i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \sqrt {2} {\left (2 i \, e e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, e e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{5 \, a^{2} d} \] Input:

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

Output:

1/5*(2*I*sqrt(2)*e^(3/2)*e^(3*I*d*x + 3*I*c)*weierstrassZeta(-4, 0, weiers 
trassPInverse(-4, 0, e^(I*d*x + I*c))) + sqrt(2)*(2*I*e*e^(4*I*d*x + 4*I*c 
) + 3*I*e*e^(2*I*d*x + 2*I*c) + I*e)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^( 
1/2*I*d*x + 1/2*I*c))*e^(-3*I*d*x - 3*I*c)/(a^2*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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