\(\int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx\) [186]

Optimal result

Integrand size = 26, antiderivative size = 94 \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx=\frac {2 a e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d}+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}+\frac {2 a e (e \sec (c+d x))^{3/2} \sin (c+d x)}{3 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.61 \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx=\frac {a (e \sec (c+d x))^{5/2} \left (6 i+10 \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+5 \sin (2 (c+d x))\right )}{15 d} \] Input:

Integrate[(e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x]),x]
 

Output:

(a*(e*Sec[c + d*x])^(5/2)*(6*I + 10*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x) 
/2, 2] + 5*Sin[2*(c + d*x)]))/(15*d)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 3967, 3042, 4255, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3967

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

Input:

Int[(e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x]),x]
 

Output:

(((2*I)/5)*a*(e*Sec[c + d*x])^(5/2))/d + a*((2*e^2*Sqrt[Cos[c + d*x]]*Elli 
pticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(3*d) + (2*e*(e*Sec[c + d*x])^ 
(3/2)*Sin[c + d*x])/(3*d))
 

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 3967

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

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 = 32.66 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.09

method	result	size
default	\(\frac {a \left (\frac {2 \tan \left (d x +c \right )}{3}+\frac {2 i \sec \left (d x +c \right )^{2}}{5}-\frac {2 i \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{3}\right ) e^{2} \sqrt {e \sec \left (d x +c \right )}}{d}\)	\(102\)
parts	\(\frac {a \left (-\frac {2 i \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{3}+\frac {2 \tan \left (d x +c \right )}{3}\right ) e^{2} \sqrt {e \sec \left (d x +c \right )}}{d}+\frac {2 i a \left (e \sec \left (d x +c \right )\right )^{\frac {5}{2}}}{5 d}\)	\(109\)

Input:

int((e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.64 \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (5 i \, a e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 12 i \, a e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 5 i \, a e^{2}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 5 \, \sqrt {2} {\left (i \, a e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, a e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a e^{2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:

integrate((e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c)),x, algorithm="fricas")
 

Output:

-2/15*(sqrt(2)*(5*I*a*e^2*e^(4*I*d*x + 4*I*c) - 12*I*a*e^2*e^(2*I*d*x + 2* 
I*c) - 5*I*a*e^2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c 
) + 5*sqrt(2)*(I*a*e^2*e^(4*I*d*x + 4*I*c) + 2*I*a*e^2*e^(2*I*d*x + 2*I*c) 
 + I*a*e^2)*sqrt(e)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)))/(d*e^(4*I 
*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)
 

Sympy [F]

\[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx=i a \left (\int \left (- i \left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}\right )\, dx + \int \left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \tan {\left (c + d x \right )}\, dx\right ) \] Input:

integrate((e*sec(d*x+c))**(5/2)*(a+I*a*tan(d*x+c)),x)
 

Output:

I*a*(Integral(-I*(e*sec(c + d*x))**(5/2), x) + Integral((e*sec(c + d*x))** 
(5/2)*tan(c + d*x), x))
 

Maxima [F]

\[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )} \,d x } \] Input:

integrate((e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c)),x, algorithm="maxima")
 

Output:

integrate((e*sec(d*x + c))^(5/2)*(I*a*tan(d*x + c) + a), x)
 

Giac [F]

\[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )} \,d x } \] Input:

integrate((e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c)),x, algorithm="giac")
 

Output:

integrate((e*sec(d*x + c))^(5/2)*(I*a*tan(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx=\int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right ) \,d x \] Input:

int((e/cos(c + d*x))^(5/2)*(a + a*tan(c + d*x)*1i),x)
 

Output:

int((e/cos(c + d*x))^(5/2)*(a + a*tan(c + d*x)*1i), x)
 

Reduce [F]

\[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx=\frac {\sqrt {e}\, a \,e^{2} \left (2 \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2} i +5 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right ) d \right )}{5 d} \] Input:

int((e*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c)),x)
 

Output:

(sqrt(e)*a*e**2*(2*sqrt(sec(c + d*x))*sec(c + d*x)**2*i + 5*int(sqrt(sec(c 
 + d*x))*sec(c + d*x)**2,x)*d))/(5*d)