\(\int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [369]

Optimal result

Integrand size = 26, antiderivative size = 147 \[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {256 i a^4 \sec ^{13}(c+d x)}{20995 d (a+i a \tan (c+d x))^{13/2}}+\frac {64 i a^3 \sec ^{13}(c+d x)}{1615 d (a+i a \tan (c+d x))^{11/2}}+\frac {24 i a^2 \sec ^{13}(c+d x)}{323 d (a+i a \tan (c+d x))^{9/2}}+\frac {2 i a \sec ^{13}(c+d x)}{19 d (a+i a \tan (c+d x))^{7/2}} \] Output:

256/20995*I*a^4*sec(d*x+c)^13/d/(a+I*a*tan(d*x+c))^(13/2)+64/1615*I*a^3*se 
c(d*x+c)^13/d/(a+I*a*tan(d*x+c))^(11/2)+24/323*I*a^2*sec(d*x+c)^13/d/(a+I* 
a*tan(d*x+c))^(9/2)+2/19*I*a*sec(d*x+c)^13/d/(a+I*a*tan(d*x+c))^(7/2)
 

Mathematica [A] (verified)

Time = 1.90 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.76 \[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sec ^{12}(c+d x) (798 \cos (c+d x)+1631 \cos (3 (c+d x))+13 i (38 \sin (c+d x)+123 \sin (3 (c+d x)))) (-2 i \cos (4 (c+d x))-2 \sin (4 (c+d x)))}{20995 a^2 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \] Input:

Integrate[Sec[c + d*x]^13/(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

(Sec[c + d*x]^12*(798*Cos[c + d*x] + 1631*Cos[3*(c + d*x)] + (13*I)*(38*Si 
n[c + d*x] + 123*Sin[3*(c + d*x)]))*((-2*I)*Cos[4*(c + d*x)] - 2*Sin[4*(c 
+ d*x)]))/(20995*a^2*d*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 3975, 3042, 3975, 3042, 3975, 3042, 3974}

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.

\(\displaystyle \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (c+d x)^{13}}{(a+i a \tan (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 3975

\(\displaystyle \frac {12}{19} a \int \frac {\sec ^{13}(c+d x)}{(i \tan (c+d x) a+a)^{7/2}}dx+\frac {2 i a \sec ^{13}(c+d x)}{19 d (a+i a \tan (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {12}{19} a \int \frac {\sec (c+d x)^{13}}{(i \tan (c+d x) a+a)^{7/2}}dx+\frac {2 i a \sec ^{13}(c+d x)}{19 d (a+i a \tan (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3975

\(\displaystyle \frac {12}{19} a \left (\frac {8}{17} a \int \frac {\sec ^{13}(c+d x)}{(i \tan (c+d x) a+a)^{9/2}}dx+\frac {2 i a \sec ^{13}(c+d x)}{17 d (a+i a \tan (c+d x))^{9/2}}\right )+\frac {2 i a \sec ^{13}(c+d x)}{19 d (a+i a \tan (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {12}{19} a \left (\frac {8}{17} a \int \frac {\sec (c+d x)^{13}}{(i \tan (c+d x) a+a)^{9/2}}dx+\frac {2 i a \sec ^{13}(c+d x)}{17 d (a+i a \tan (c+d x))^{9/2}}\right )+\frac {2 i a \sec ^{13}(c+d x)}{19 d (a+i a \tan (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3975

\(\displaystyle \frac {12}{19} a \left (\frac {8}{17} a \left (\frac {4}{15} a \int \frac {\sec ^{13}(c+d x)}{(i \tan (c+d x) a+a)^{11/2}}dx+\frac {2 i a \sec ^{13}(c+d x)}{15 d (a+i a \tan (c+d x))^{11/2}}\right )+\frac {2 i a \sec ^{13}(c+d x)}{17 d (a+i a \tan (c+d x))^{9/2}}\right )+\frac {2 i a \sec ^{13}(c+d x)}{19 d (a+i a \tan (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {12}{19} a \left (\frac {8}{17} a \left (\frac {4}{15} a \int \frac {\sec (c+d x)^{13}}{(i \tan (c+d x) a+a)^{11/2}}dx+\frac {2 i a \sec ^{13}(c+d x)}{15 d (a+i a \tan (c+d x))^{11/2}}\right )+\frac {2 i a \sec ^{13}(c+d x)}{17 d (a+i a \tan (c+d x))^{9/2}}\right )+\frac {2 i a \sec ^{13}(c+d x)}{19 d (a+i a \tan (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3974

\(\displaystyle \frac {12}{19} a \left (\frac {8}{17} a \left (\frac {8 i a^2 \sec ^{13}(c+d x)}{195 d (a+i a \tan (c+d x))^{13/2}}+\frac {2 i a \sec ^{13}(c+d x)}{15 d (a+i a \tan (c+d x))^{11/2}}\right )+\frac {2 i a \sec ^{13}(c+d x)}{17 d (a+i a \tan (c+d x))^{9/2}}\right )+\frac {2 i a \sec ^{13}(c+d x)}{19 d (a+i a \tan (c+d x))^{7/2}}\)

Input:

Int[Sec[c + d*x]^13/(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

(((2*I)/19)*a*Sec[c + d*x]^13)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + (12*a*(( 
((2*I)/17)*a*Sec[c + d*x]^13)/(d*(a + I*a*Tan[c + d*x])^(9/2)) + (8*a*(((( 
8*I)/195)*a^2*Sec[c + d*x]^13)/(d*(a + I*a*Tan[c + d*x])^(13/2)) + (((2*I) 
/15)*a*Sec[c + d*x]^13)/(d*(a + I*a*Tan[c + d*x])^(11/2))))/17))/19
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3974

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

rule 3975

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

Maple [A] (verified)

Time = 25.75 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.99

method	result	size
default	\(\frac {\frac {2 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{9} \left (1024 \cos \left (d x +c \right )^{6}+1152 \cos \left (d x +c \right )^{4}+1144 \cos \left (d x +c \right )^{2}+1105\right )}{20995}+\frac {2 i \left (1024 \sec \left (d x +c \right )^{3}+640 \sec \left (d x +c \right )^{5}+440 \sec \left (d x +c \right )^{7}+325 \sec \left (d x +c \right )^{9}\right )}{20995}}{d \left (2 i \sin \left (d x +c \right ) \cos \left (d x +c \right )+2 \cos \left (d x +c \right )^{2}-1\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2}}\)	\(145\)

Input:

int(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2/20995/d/(2*I*sin(d*x+c)*cos(d*x+c)+2*cos(d*x+c)^2-1)/(a*(1+I*tan(d*x+c)) 
)^(1/2)/a^2*(tan(d*x+c)*sec(d*x+c)^9*(1024*cos(d*x+c)^6+1152*cos(d*x+c)^4+ 
1144*cos(d*x+c)^2+1105)+I*(1024*sec(d*x+c)^3+640*sec(d*x+c)^5+440*sec(d*x+ 
c)^7+325*sec(d*x+c)^9))
 

Fricas [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.35 \[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {1024 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-1615 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 646 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 152 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i\right )}}{20995 \, {\left (a^{3} d e^{\left (18 i \, d x + 18 i \, c\right )} + 9 \, a^{3} d e^{\left (16 i \, d x + 16 i \, c\right )} + 36 \, a^{3} d e^{\left (14 i \, d x + 14 i \, c\right )} + 84 \, a^{3} d e^{\left (12 i \, d x + 12 i \, c\right )} + 126 \, a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 126 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 84 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 36 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 9 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \] Input:

integrate(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

-1024/20995*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-1615*I*e^(6*I*d*x 
+ 6*I*c) - 646*I*e^(4*I*d*x + 4*I*c) - 152*I*e^(2*I*d*x + 2*I*c) - 16*I)/( 
a^3*d*e^(18*I*d*x + 18*I*c) + 9*a^3*d*e^(16*I*d*x + 16*I*c) + 36*a^3*d*e^( 
14*I*d*x + 14*I*c) + 84*a^3*d*e^(12*I*d*x + 12*I*c) + 126*a^3*d*e^(10*I*d* 
x + 10*I*c) + 126*a^3*d*e^(8*I*d*x + 8*I*c) + 84*a^3*d*e^(6*I*d*x + 6*I*c) 
 + 36*a^3*d*e^(4*I*d*x + 4*I*c) + 9*a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)**13/(a+I*a*tan(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 902 vs. \(2 (115) = 230\).

Time = 0.37 (sec) , antiderivative size = 902, normalized size of antiderivative = 6.14 \[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

-2/20995*(-2429*I*sqrt(a) - 8850*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 
 5122*I*sqrt(a)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 45190*sqrt(a)*sin(d* 
x + c)^3/(cos(d*x + c) + 1)^3 - 12924*I*sqrt(a)*sin(d*x + c)^4/(cos(d*x + 
c) + 1)^4 - 152478*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 40470*I*s 
qrt(a)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 397594*sqrt(a)*sin(d*x + c)^7 
/(cos(d*x + c) + 1)^7 - 50065*I*sqrt(a)*sin(d*x + c)^8/(cos(d*x + c) + 1)^ 
8 - 722228*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 19380*I*sqrt(a)*s 
in(d*x + c)^10/(cos(d*x + c) + 1)^10 - 936700*sqrt(a)*sin(d*x + c)^11/(cos 
(d*x + c) + 1)^11 - 936700*sqrt(a)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 + 
 19380*I*sqrt(a)*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 - 722228*sqrt(a)*si 
n(d*x + c)^15/(cos(d*x + c) + 1)^15 + 50065*I*sqrt(a)*sin(d*x + c)^16/(cos 
(d*x + c) + 1)^16 - 397594*sqrt(a)*sin(d*x + c)^17/(cos(d*x + c) + 1)^17 + 
 40470*I*sqrt(a)*sin(d*x + c)^18/(cos(d*x + c) + 1)^18 - 152478*sqrt(a)*si 
n(d*x + c)^19/(cos(d*x + c) + 1)^19 + 12924*I*sqrt(a)*sin(d*x + c)^20/(cos 
(d*x + c) + 1)^20 - 45190*sqrt(a)*sin(d*x + c)^21/(cos(d*x + c) + 1)^21 + 
5122*I*sqrt(a)*sin(d*x + c)^22/(cos(d*x + c) + 1)^22 - 8850*sqrt(a)*sin(d* 
x + c)^23/(cos(d*x + c) + 1)^23 + 2429*I*sqrt(a)*sin(d*x + c)^24/(cos(d*x 
+ c) + 1)^24)*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(sin(d*x + c)/(c 
os(d*x + c) + 1) - 1)^(5/2)/((a^3 - 12*a^3*sin(d*x + c)^2/(cos(d*x + c) + 
1)^2 + 66*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 220*a^3*sin(d*x + c...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [B] (verification not implemented)

Time = 9.03 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.05 \[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,1024{}\mathrm {i}}{13\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6}-\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,1024{}\mathrm {i}}{5\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^7}+\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,3072{}\mathrm {i}}{17\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^8}-\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,1024{}\mathrm {i}}{19\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^9} \] Input:

int(1/(cos(c + d*x)^13*(a + a*tan(c + d*x)*1i)^(5/2)),x)
 

Output:

(exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + 
 d*x*2i) + 1))^(1/2)*1024i)/(13*a^3*d*(exp(c*2i + d*x*2i) + 1)^6) - (exp(- 
 c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2 
i) + 1))^(1/2)*1024i)/(5*a^3*d*(exp(c*2i + d*x*2i) + 1)^7) + (exp(- c*1i - 
 d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1) 
)^(1/2)*3072i)/(17*a^3*d*(exp(c*2i + d*x*2i) + 1)^8) - (exp(- c*1i - d*x*1 
i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2 
)*1024i)/(19*a^3*d*(exp(c*2i + d*x*2i) + 1)^9)
 

Reduce [F]

\[ \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\int \frac {\sec \left (d x +c \right )^{13}}{\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right ) i +1}}d x}{\sqrt {a}\, a^{2}} \] Input:

int(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^(5/2),x)
 

Output:

( - int(sec(c + d*x)**13/(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 2*sqr 
t(tan(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + d*x)*i + 1)),x))/(sqrt 
(a)*a**2)