\(\int \frac {\sec ^9(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [340]

Optimal result

Integrand size = 26, antiderivative size = 147 \[ \int \frac {\sec ^9(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {256 i a^4 \sec ^9(c+d x)}{6435 d (a+i a \tan (c+d x))^{9/2}}+\frac {64 i a^3 \sec ^9(c+d x)}{715 d (a+i a \tan (c+d x))^{7/2}}+\frac {8 i a^2 \sec ^9(c+d x)}{65 d (a+i a \tan (c+d x))^{5/2}}+\frac {2 i a \sec ^9(c+d x)}{15 d (a+i a \tan (c+d x))^{3/2}} \] Output:

256/6435*I*a^4*sec(d*x+c)^9/d/(a+I*a*tan(d*x+c))^(9/2)+64/715*I*a^3*sec(d* 
x+c)^9/d/(a+I*a*tan(d*x+c))^(7/2)+8/65*I*a^2*sec(d*x+c)^9/d/(a+I*a*tan(d*x 
+c))^(5/2)+2/15*I*a*sec(d*x+c)^9/d/(a+I*a*tan(d*x+c))^(3/2)
 

Mathematica [A] (verified)

Time = 1.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.65 \[ \int \frac {\sec ^9(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 \sec ^8(c+d x) (510 \cos (c+d x)+731 \cos (3 (c+d x))+3 i (90 \sin (c+d x)+233 \sin (3 (c+d x)))) (i \cos (4 (c+d x))+\sin (4 (c+d x)))}{6435 d \sqrt {a+i a \tan (c+d x)}} \] Input:

Integrate[Sec[c + d*x]^9/Sqrt[a + I*a*Tan[c + d*x]],x]
 

Output:

(2*Sec[c + d*x]^8*(510*Cos[c + d*x] + 731*Cos[3*(c + d*x)] + (3*I)*(90*Sin 
[c + d*x] + 233*Sin[3*(c + d*x)]))*(I*Cos[4*(c + d*x)] + Sin[4*(c + d*x)]) 
)/(6435*d*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.

\(\displaystyle \int \frac {\sec ^9(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (c+d x)^9}{\sqrt {a+i a \tan (c+d x)}}dx\)

\(\Big \downarrow \) 3975

\(\displaystyle \frac {4}{5} a \int \frac {\sec ^9(c+d x)}{(i \tan (c+d x) a+a)^{3/2}}dx+\frac {2 i a \sec ^9(c+d x)}{15 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4}{5} a \int \frac {\sec (c+d x)^9}{(i \tan (c+d x) a+a)^{3/2}}dx+\frac {2 i a \sec ^9(c+d x)}{15 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3975

\(\displaystyle \frac {4}{5} a \left (\frac {8}{13} a \int \frac {\sec ^9(c+d x)}{(i \tan (c+d x) a+a)^{5/2}}dx+\frac {2 i a \sec ^9(c+d x)}{13 d (a+i a \tan (c+d x))^{5/2}}\right )+\frac {2 i a \sec ^9(c+d x)}{15 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4}{5} a \left (\frac {8}{13} a \int \frac {\sec (c+d x)^9}{(i \tan (c+d x) a+a)^{5/2}}dx+\frac {2 i a \sec ^9(c+d x)}{13 d (a+i a \tan (c+d x))^{5/2}}\right )+\frac {2 i a \sec ^9(c+d x)}{15 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3975

\(\displaystyle \frac {4}{5} a \left (\frac {8}{13} a \left (\frac {4}{11} a \int \frac {\sec ^9(c+d x)}{(i \tan (c+d x) a+a)^{7/2}}dx+\frac {2 i a \sec ^9(c+d x)}{11 d (a+i a \tan (c+d x))^{7/2}}\right )+\frac {2 i a \sec ^9(c+d x)}{13 d (a+i a \tan (c+d x))^{5/2}}\right )+\frac {2 i a \sec ^9(c+d x)}{15 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4}{5} a \left (\frac {8}{13} a \left (\frac {4}{11} a \int \frac {\sec (c+d x)^9}{(i \tan (c+d x) a+a)^{7/2}}dx+\frac {2 i a \sec ^9(c+d x)}{11 d (a+i a \tan (c+d x))^{7/2}}\right )+\frac {2 i a \sec ^9(c+d x)}{13 d (a+i a \tan (c+d x))^{5/2}}\right )+\frac {2 i a \sec ^9(c+d x)}{15 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3974

\(\displaystyle \frac {4}{5} a \left (\frac {8}{13} a \left (\frac {8 i a^2 \sec ^9(c+d x)}{99 d (a+i a \tan (c+d x))^{9/2}}+\frac {2 i a \sec ^9(c+d x)}{11 d (a+i a \tan (c+d x))^{7/2}}\right )+\frac {2 i a \sec ^9(c+d x)}{13 d (a+i a \tan (c+d x))^{5/2}}\right )+\frac {2 i a \sec ^9(c+d x)}{15 d (a+i a \tan (c+d x))^{3/2}}\)

Input:

Int[Sec[c + d*x]^9/Sqrt[a + I*a*Tan[c + d*x]],x]
 

Output:

(((2*I)/15)*a*Sec[c + d*x]^9)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (4*a*(((( 
2*I)/13)*a*Sec[c + d*x]^9)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (8*a*((((8*I 
)/99)*a^2*Sec[c + d*x]^9)/(d*(a + I*a*Tan[c + d*x])^(9/2)) + (((2*I)/11)*a 
*Sec[c + d*x]^9)/(d*(a + I*a*Tan[c + d*x])^(7/2))))/13))/5
 

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 = 6.71 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.76

method	result	size
default	\(\frac {\frac {2 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{7} \left (1024 \cos \left (d x +c \right )^{6}+640 \cos \left (d x +c \right )^{4}+504 \cos \left (d x +c \right )^{2}+429\right )}{6435}+\frac {2 i \left (1024 \sec \left (d x +c \right )+128 \sec \left (d x +c \right )^{3}+56 \sec \left (d x +c \right )^{5}+33 \sec \left (d x +c \right )^{7}\right )}{6435}}{d \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\)	\(111\)

Input:

int(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2/6435/d/(a*(1+I*tan(d*x+c)))^(1/2)*(tan(d*x+c)*sec(d*x+c)^7*(1024*cos(d*x 
+c)^6+640*cos(d*x+c)^4+504*cos(d*x+c)^2+429)+I*(1024*sec(d*x+c)+128*sec(d* 
x+c)^3+56*sec(d*x+c)^5+33*sec(d*x+c)^7))
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.04 \[ \int \frac {\sec ^9(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {256 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-715 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 390 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 120 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i\right )}}{6435 \, {\left (a d e^{\left (14 i \, d x + 14 i \, c\right )} + 7 \, a d e^{\left (12 i \, d x + 12 i \, c\right )} + 21 \, a d e^{\left (10 i \, d x + 10 i \, c\right )} + 35 \, a d e^{\left (8 i \, d x + 8 i \, c\right )} + 35 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 21 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 7 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \] Input:

integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

-256/6435*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-715*I*e^(6*I*d*x + 6 
*I*c) - 390*I*e^(4*I*d*x + 4*I*c) - 120*I*e^(2*I*d*x + 2*I*c) - 16*I)/(a*d 
*e^(14*I*d*x + 14*I*c) + 7*a*d*e^(12*I*d*x + 12*I*c) + 21*a*d*e^(10*I*d*x 
+ 10*I*c) + 35*a*d*e^(8*I*d*x + 8*I*c) + 35*a*d*e^(6*I*d*x + 6*I*c) + 21*a 
*d*e^(4*I*d*x + 4*I*c) + 7*a*d*e^(2*I*d*x + 2*I*c) + a*d)
 

Sympy [F]

\[ \int \frac {\sec ^9(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\sec ^{9}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \] Input:

integrate(sec(d*x+c)**9/(a+I*a*tan(d*x+c))**(1/2),x)
 

Output:

Integral(sec(c + d*x)**9/sqrt(I*a*(tan(c + d*x) - I)), x)
 

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. 608 vs. \(2 (115) = 230\).

Time = 0.26 (sec) , antiderivative size = 608, normalized size of antiderivative = 4.14 \[ \int \frac {\sec ^9(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx =\text {Too large to display} \] Input:

integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

-2/6435*(-1241*I*sqrt(a) - 5194*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) + 
6090*I*sqrt(a)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 2490*sqrt(a)*sin(d*x 
+ c)^3/(cos(d*x + c) + 1)^3 - 14430*I*sqrt(a)*sin(d*x + c)^4/(cos(d*x + c) 
 + 1)^4 - 33618*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 13442*I*sqrt 
(a)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 18590*sqrt(a)*sin(d*x + c)^7/(co 
s(d*x + c) + 1)^7 - 18590*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 13 
442*I*sqrt(a)*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 33618*sqrt(a)*sin(d* 
x + c)^11/(cos(d*x + c) + 1)^11 + 14430*I*sqrt(a)*sin(d*x + c)^12/(cos(d*x 
 + c) + 1)^12 + 2490*sqrt(a)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - 6090* 
I*sqrt(a)*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 - 5194*sqrt(a)*sin(d*x + c 
)^15/(cos(d*x + c) + 1)^15 + 1241*I*sqrt(a)*sin(d*x + c)^16/(cos(d*x + c) 
+ 1)^16)*sqrt(sin(d*x + c)/(cos(d*x + c) + 1) + 1)*sqrt(sin(d*x + c)/(cos( 
d*x + c) + 1) - 1)/((a - 8*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 28*a*si 
n(d*x + c)^4/(cos(d*x + c) + 1)^4 - 56*a*sin(d*x + c)^6/(cos(d*x + c) + 1) 
^6 + 70*a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 56*a*sin(d*x + c)^10/(cos( 
d*x + c) + 1)^10 + 28*a*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 8*a*sin(d* 
x + c)^14/(cos(d*x + c) + 1)^14 + a*sin(d*x + c)^16/(cos(d*x + c) + 1)^16) 
*d*sqrt(-2*I*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^2/(cos(d*x + c 
) + 1)^2 - 1))
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sec ^9(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^(1/2),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 = 6.32 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.05 \[ \int \frac {\sec ^9(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,256{}\mathrm {i}}{9\,a\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}-\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,768{}\mathrm {i}}{11\,a\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5}+\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,768{}\mathrm {i}}{13\,a\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6}-\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,256{}\mathrm {i}}{15\,a\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^7} \] Input:

int(1/(cos(c + d*x)^9*(a + a*tan(c + d*x)*1i)^(1/2)),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)*256i)/(9*a*d*(exp(c*2i + d*x*2i) + 1)^4) - (exp(- c*1 
i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 
 1))^(1/2)*768i)/(11*a*d*(exp(c*2i + d*x*2i) + 1)^5) + (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 
)*768i)/(13*a*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)*256i)/( 
15*a*d*(exp(c*2i + d*x*2i) + 1)^7)
 

Reduce [F]

\[ \int \frac {\sec ^9(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 \sqrt {a}\, i \left (-\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{9}+7 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{9} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{2}+1}d x \right ) \tan \left (d x +c \right )^{2} d +7 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{9} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{2}+1}d x \right ) d \right )}{a d \left (\tan \left (d x +c \right )^{2}+1\right )} \] Input:

int(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^(1/2),x)
 

Output:

(2*sqrt(a)*i*( - sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**9 + 7*int((sqrt(ta 
n(c + d*x)*i + 1)*sec(c + d*x)**9*tan(c + d*x))/(tan(c + d*x)**2 + 1),x)*t 
an(c + d*x)**2*d + 7*int((sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**9*tan(c + 
 d*x))/(tan(c + d*x)**2 + 1),x)*d))/(a*d*(tan(c + d*x)**2 + 1))