\(\int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [354]

Optimal result

Integrand size = 26, antiderivative size = 147 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {256 i a^4 \sec ^{11}(c+d x)}{12155 d (a+i a \tan (c+d x))^{11/2}}+\frac {64 i a^3 \sec ^{11}(c+d x)}{1105 d (a+i a \tan (c+d x))^{9/2}}+\frac {8 i a^2 \sec ^{11}(c+d x)}{85 d (a+i a \tan (c+d x))^{7/2}}+\frac {2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}} \] Output:

256/12155*I*a^4*sec(d*x+c)^11/d/(a+I*a*tan(d*x+c))^(11/2)+64/1105*I*a^3*se 
c(d*x+c)^11/d/(a+I*a*tan(d*x+c))^(9/2)+8/85*I*a^2*sec(d*x+c)^11/d/(a+I*a*t 
an(d*x+c))^(7/2)+2/17*I*a*sec(d*x+c)^11/d/(a+I*a*tan(d*x+c))^(5/2)
 

Mathematica [A] (verified)

Time = 1.84 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.73 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 \sec ^9(c+d x) (i \cos (4 (c+d x))+\sin (4 (c+d x))) (475 i-2242 i \cos (2 (c+d x))+1089 \sec (c+d x) \sin (3 (c+d x))+374 \tan (c+d x))}{12155 a d (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \] Input:

Integrate[Sec[c + d*x]^11/(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

(2*Sec[c + d*x]^9*(I*Cos[4*(c + d*x)] + Sin[4*(c + d*x)])*(475*I - (2242*I 
)*Cos[2*(c + d*x)] + 1089*Sec[c + d*x]*Sin[3*(c + d*x)] + 374*Tan[c + d*x] 
))/(12155*a*d*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.75 (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.

\(\displaystyle \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (c+d x)^{11}}{(a+i a \tan (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 3975

\(\displaystyle \frac {12}{17} a \int \frac {\sec ^{11}(c+d x)}{(i \tan (c+d x) a+a)^{5/2}}dx+\frac {2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {12}{17} a \int \frac {\sec (c+d x)^{11}}{(i \tan (c+d x) a+a)^{5/2}}dx+\frac {2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3975

\(\displaystyle \frac {12}{17} a \left (\frac {8}{15} a \int \frac {\sec ^{11}(c+d x)}{(i \tan (c+d x) a+a)^{7/2}}dx+\frac {2 i a \sec ^{11}(c+d x)}{15 d (a+i a \tan (c+d x))^{7/2}}\right )+\frac {2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {12}{17} a \left (\frac {8}{15} a \int \frac {\sec (c+d x)^{11}}{(i \tan (c+d x) a+a)^{7/2}}dx+\frac {2 i a \sec ^{11}(c+d x)}{15 d (a+i a \tan (c+d x))^{7/2}}\right )+\frac {2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3975

\(\displaystyle \frac {12}{17} a \left (\frac {8}{15} a \left (\frac {4}{13} a \int \frac {\sec ^{11}(c+d x)}{(i \tan (c+d x) a+a)^{9/2}}dx+\frac {2 i a \sec ^{11}(c+d x)}{13 d (a+i a \tan (c+d x))^{9/2}}\right )+\frac {2 i a \sec ^{11}(c+d x)}{15 d (a+i a \tan (c+d x))^{7/2}}\right )+\frac {2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {12}{17} a \left (\frac {8}{15} a \left (\frac {4}{13} a \int \frac {\sec (c+d x)^{11}}{(i \tan (c+d x) a+a)^{9/2}}dx+\frac {2 i a \sec ^{11}(c+d x)}{13 d (a+i a \tan (c+d x))^{9/2}}\right )+\frac {2 i a \sec ^{11}(c+d x)}{15 d (a+i a \tan (c+d x))^{7/2}}\right )+\frac {2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3974

\(\displaystyle \frac {12}{17} a \left (\frac {8}{15} a \left (\frac {8 i a^2 \sec ^{11}(c+d x)}{143 d (a+i a \tan (c+d x))^{11/2}}+\frac {2 i a \sec ^{11}(c+d x)}{13 d (a+i a \tan (c+d x))^{9/2}}\right )+\frac {2 i a \sec ^{11}(c+d x)}{15 d (a+i a \tan (c+d x))^{7/2}}\right )+\frac {2 i a \sec ^{11}(c+d x)}{17 d (a+i a \tan (c+d x))^{5/2}}\)

Input:

Int[Sec[c + d*x]^11/(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

(((2*I)/17)*a*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (12*a*(( 
((2*I)/15)*a*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + (8*a*(((( 
8*I)/143)*a^2*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(11/2)) + (((2*I) 
/13)*a*Sec[c + d*x]^11)/(d*(a + I*a*Tan[c + d*x])^(9/2))))/15))/17
 

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 = 9.92 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {\frac {2 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{8} \left (1024 \cos \left (d x +c \right )^{6}+896 \cos \left (d x +c \right )^{4}+792 \cos \left (d x +c \right )^{2}+715\right )}{12155}+\frac {2 i \left (1024 \sec \left (d x +c \right )^{2}+384 \sec \left (d x +c \right )^{4}+216 \sec \left (d x +c \right )^{6}+143 \sec \left (d x +c \right )^{8}\right )}{12155}}{d \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a}\)	\(134\)

Input:

int(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2/12155/d/(I*sin(d*x+c)+cos(d*x+c))/(a*(1+I*tan(d*x+c)))^(1/2)/a*(tan(d*x+ 
c)*sec(d*x+c)^8*(1024*cos(d*x+c)^6+896*cos(d*x+c)^4+792*cos(d*x+c)^2+715)+ 
I*(1024*sec(d*x+c)^2+384*sec(d*x+c)^4+216*sec(d*x+c)^6+143*sec(d*x+c)^8))
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.25 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {512 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-1105 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 510 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 136 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i\right )}}{12155 \, {\left (a^{2} d e^{\left (16 i \, d x + 16 i \, c\right )} + 8 \, a^{2} d e^{\left (14 i \, d x + 14 i \, c\right )} + 28 \, a^{2} d e^{\left (12 i \, d x + 12 i \, c\right )} + 56 \, a^{2} d e^{\left (10 i \, d x + 10 i \, c\right )} + 70 \, a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 56 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 28 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \] Input:

integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

-512/12155*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-1105*I*e^(6*I*d*x + 
 6*I*c) - 510*I*e^(4*I*d*x + 4*I*c) - 136*I*e^(2*I*d*x + 2*I*c) - 16*I)/(a 
^2*d*e^(16*I*d*x + 16*I*c) + 8*a^2*d*e^(14*I*d*x + 14*I*c) + 28*a^2*d*e^(1 
2*I*d*x + 12*I*c) + 56*a^2*d*e^(10*I*d*x + 10*I*c) + 70*a^2*d*e^(8*I*d*x + 
 8*I*c) + 56*a^2*d*e^(6*I*d*x + 6*I*c) + 28*a^2*d*e^(4*I*d*x + 4*I*c) + 8* 
a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)**11/(a+I*a*tan(d*x+c))**(3/2),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. 764 vs. \(2 (115) = 230\).

Time = 0.30 (sec) , antiderivative size = 764, normalized size of antiderivative = 5.20 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

-2/12155*(-1767*I*sqrt(a) - 6854*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) + 
 2088*I*sqrt(a)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 16438*sqrt(a)*sin(d* 
x + c)^3/(cos(d*x + c) + 1)^3 - 5661*I*sqrt(a)*sin(d*x + c)^4/(cos(d*x + c 
) + 1)^4 - 56984*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 13328*I*sqr 
t(a)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 129336*sqrt(a)*sin(d*x + c)^7/( 
cos(d*x + c) + 1)^7 + 7514*I*sqrt(a)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 
 156468*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 156468*sqrt(a)*sin(d 
*x + c)^11/(cos(d*x + c) + 1)^11 - 7514*I*sqrt(a)*sin(d*x + c)^12/(cos(d*x 
 + c) + 1)^12 - 129336*sqrt(a)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 + 133 
28*I*sqrt(a)*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 - 56984*sqrt(a)*sin(d*x 
 + c)^15/(cos(d*x + c) + 1)^15 + 5661*I*sqrt(a)*sin(d*x + c)^16/(cos(d*x + 
 c) + 1)^16 - 16438*sqrt(a)*sin(d*x + c)^17/(cos(d*x + c) + 1)^17 - 2088*I 
*sqrt(a)*sin(d*x + c)^18/(cos(d*x + c) + 1)^18 - 6854*sqrt(a)*sin(d*x + c) 
^19/(cos(d*x + c) + 1)^19 + 1767*I*sqrt(a)*sin(d*x + c)^20/(cos(d*x + c) + 
 1)^20)*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(3/2)*(sin(d*x + c)/(cos(d*x 
 + c) + 1) - 1)^(3/2)/((a^2 - 10*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 
 45*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 120*a^2*sin(d*x + c)^6/(cos( 
d*x + c) + 1)^6 + 210*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 252*a^2*si 
n(d*x + c)^10/(cos(d*x + c) + 1)^10 + 210*a^2*sin(d*x + c)^12/(cos(d*x + c 
) + 1)^12 - 120*a^2*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 + 45*a^2*sin(...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(3/2),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 = 7.53 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.05 \[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,512{}\mathrm {i}}{11\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5}-\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,1536{}\mathrm {i}}{13\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6}+\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,512{}\mathrm {i}}{5\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^7}-\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,512{}\mathrm {i}}{17\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^8} \] Input:

int(1/(cos(c + d*x)^11*(a + a*tan(c + d*x)*1i)^(3/2)),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)*512i)/(11*a^2*d*(exp(c*2i + d*x*2i) + 1)^5) - (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)*1536i)/(13*a^2*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*2i) + 1) 
)^(1/2)*512i)/(5*a^2*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)* 
512i)/(17*a^2*d*(exp(c*2i + d*x*2i) + 1)^8)
 

Reduce [F]

\[ \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{11} i -19 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{11} \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) \tan \left (d x +c \right )^{2} d -19 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{11} \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) d +17 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{11} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) \tan \left (d x +c \right )^{2} d i +17 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{11} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) d i \right )}{a^{2} d \left (\tan \left (d x +c \right )^{2}+1\right )} \] Input:

int(sec(d*x+c)^11/(a+I*a*tan(d*x+c))^(3/2),x)
 

Output:

(sqrt(a)*( - 2*sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**11*i - 19*int((sqrt( 
tan(c + d*x)*i + 1)*sec(c + d*x)**11*tan(c + d*x)**2)/(tan(c + d*x)**3*i + 
 tan(c + d*x)**2 + tan(c + d*x)*i + 1),x)*tan(c + d*x)**2*d - 19*int((sqrt 
(tan(c + d*x)*i + 1)*sec(c + d*x)**11*tan(c + d*x)**2)/(tan(c + d*x)**3*i 
+ tan(c + d*x)**2 + tan(c + d*x)*i + 1),x)*d + 17*int((sqrt(tan(c + d*x)*i 
 + 1)*sec(c + d*x)**11*tan(c + d*x))/(tan(c + d*x)**3*i + tan(c + d*x)**2 
+ tan(c + d*x)*i + 1),x)*tan(c + d*x)**2*d*i + 17*int((sqrt(tan(c + d*x)*i 
 + 1)*sec(c + d*x)**11*tan(c + d*x))/(tan(c + d*x)**3*i + tan(c + d*x)**2 
+ tan(c + d*x)*i + 1),x)*d*i))/(a**2*d*(tan(c + d*x)**2 + 1))