\(\int \frac {a+i a \tan (c+d x)}{(e \sec (c+d x))^{7/2}} \, dx\) [192]

Optimal result

Integrand size = 26, antiderivative size = 125 \[ \int \frac {a+i a \tan (c+d x)}{(e \sec (c+d x))^{7/2}} \, dx=-\frac {2 i a}{7 d (e \sec (c+d x))^{7/2}}+\frac {10 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{21 d e^4}+\frac {2 a \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}+\frac {10 a \sin (c+d x)}{21 d e^3 \sqrt {e \sec (c+d x)}} \] Output:

-2/7*I*a/d/(e*sec(d*x+c))^(7/2)+10/21*a*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/e^4+2/7*a*sin(d*x+c)/d/e/(e*s 
ec(d*x+c))^(5/2)+10/21*a*sin(d*x+c)/d/e^3/(e*sec(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 1.08 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.97 \[ \int \frac {a+i a \tan (c+d x)}{(e \sec (c+d x))^{7/2}} \, dx=\frac {a \sqrt {e \sec (c+d x)} (\cos (c+d x)+i \sin (c+d x)) \left (-14 i \cos (c+d x)+2 i \cos (3 (c+d x))+20 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (c+d x)-i \sin (c+d x))+5 \sin (c+d x)+5 \sin (3 (c+d x))\right )}{42 d e^4} \] Input:

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

Output:

(a*Sqrt[e*Sec[c + d*x]]*(Cos[c + d*x] + I*Sin[c + d*x])*((-14*I)*Cos[c + d 
*x] + (2*I)*Cos[3*(c + d*x)] + 20*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2 
, 2]*(Cos[c + d*x] - I*Sin[c + d*x]) + 5*Sin[c + d*x] + 5*Sin[3*(c + d*x)] 
))/(42*d*e^4)
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 3967, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3967

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4256

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4256

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

(((-2*I)/7)*a)/(d*(e*Sec[c + d*x])^(7/2)) + a*((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*Sin[c + d*x])/(3*d*e*Sqrt[e*Sec[c + 
d*x]])))/(7*e^2))
 

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 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 = 10.94 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.93

method	result	size
default	\(-\frac {2 a \left (\sin \left (d x +c \right ) \left (-3 \cos \left (d x +c \right )^{2}-5\right )+i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (5+5 \sec \left (d x +c \right )\right )+3 i \cos \left (d x +c \right )^{3}\right )}{21 d \sqrt {e \sec \left (d x +c \right )}\, e^{3}}\)	\(116\)
parts	\(-\frac {2 a \left (\sin \left (d x +c \right ) \left (-3 \cos \left (d x +c \right )^{2}-5\right )+i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (5+5 \sec \left (d x +c \right )\right )\right )}{21 d \sqrt {e \sec \left (d x +c \right )}\, e^{3}}-\frac {2 i a}{7 d \left (e \sec \left (d x +c \right )\right )^{\frac {7}{2}}}\)	\(123\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.94 \[ \int \frac {a+i a \tan (c+d x)}{(e \sec (c+d x))^{7/2}} \, dx=\frac {{\left (-40 i \, \sqrt {2} a \sqrt {e} e^{\left (2 i \, d x + 2 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \sqrt {2} {\left (-3 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 19 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 9 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i \, a\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{84 \, d e^{4}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

I*a*(Integral(-I/(e*sec(c + d*x))**(7/2), x) + Integral(tan(c + d*x)/(e*se 
c(c + d*x))**(7/2), x))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {a+i a \tan (c+d x)}{(e \sec (c+d x))^{7/2}} \, dx=\frac {\sqrt {e}\, a \left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{4}}d x +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )^{4}}d x \right ) i \right )}{e^{4}} \] Input:

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

Output:

(sqrt(e)*a*(int(sqrt(sec(c + d*x))/sec(c + d*x)**4,x) + int((sqrt(sec(c + 
d*x))*tan(c + d*x))/sec(c + d*x)**4,x)*i))/e**4