\(\int \sec ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) [300]

Optimal result

Integrand size = 26, antiderivative size = 147 \[ \int \sec ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {256 i a^4 \sec ^5(c+d x)}{1155 d (a+i a \tan (c+d x))^{5/2}}+\frac {64 i a^3 \sec ^5(c+d x)}{231 d (a+i a \tan (c+d x))^{3/2}}+\frac {8 i a^2 \sec ^5(c+d x)}{33 d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i a \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{11 d} \] Output:

256/1155*I*a^4*sec(d*x+c)^5/d/(a+I*a*tan(d*x+c))^(5/2)+64/231*I*a^3*sec(d* 
x+c)^5/d/(a+I*a*tan(d*x+c))^(3/2)+8/33*I*a^2*sec(d*x+c)^5/d/(a+I*a*tan(d*x 
+c))^(1/2)+2/11*I*a*sec(d*x+c)^5*(a+I*a*tan(d*x+c))^(1/2)/d
 

Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.74 \[ \int \sec ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {2 a \sec ^4(c+d x) (\cos (d x)-i \sin (d x)) (i \cos (3 c+2 d x)+\sin (3 c+2 d x)) (39+494 \cos (2 (c+d x))+215 i \sec (c+d x) \sin (3 (c+d x))+110 i \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{1155 d} \] Input:

Integrate[Sec[c + d*x]^5*(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

(2*a*Sec[c + d*x]^4*(Cos[d*x] - I*Sin[d*x])*(I*Cos[3*c + 2*d*x] + Sin[3*c 
+ 2*d*x])*(39 + 494*Cos[2*(c + d*x)] + (215*I)*Sec[c + d*x]*Sin[3*(c + d*x 
)] + (110*I)*Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(1155*d)
 

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3975

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3975

\(\displaystyle \frac {12}{11} a \left (\frac {8}{9} a \int \frac {\sec ^5(c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx+\frac {2 i a \sec ^5(c+d x)}{9 d \sqrt {a+i a \tan (c+d x)}}\right )+\frac {2 i a \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{11 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {12}{11} a \left (\frac {8}{9} a \int \frac {\sec (c+d x)^5}{\sqrt {i \tan (c+d x) a+a}}dx+\frac {2 i a \sec ^5(c+d x)}{9 d \sqrt {a+i a \tan (c+d x)}}\right )+\frac {2 i a \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{11 d}\)

\(\Big \downarrow \) 3975

\(\displaystyle \frac {12}{11} a \left (\frac {8}{9} a \left (\frac {4}{7} a \int \frac {\sec ^5(c+d x)}{(i \tan (c+d x) a+a)^{3/2}}dx+\frac {2 i a \sec ^5(c+d x)}{7 d (a+i a \tan (c+d x))^{3/2}}\right )+\frac {2 i a \sec ^5(c+d x)}{9 d \sqrt {a+i a \tan (c+d x)}}\right )+\frac {2 i a \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{11 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {12}{11} a \left (\frac {8}{9} a \left (\frac {4}{7} a \int \frac {\sec (c+d x)^5}{(i \tan (c+d x) a+a)^{3/2}}dx+\frac {2 i a \sec ^5(c+d x)}{7 d (a+i a \tan (c+d x))^{3/2}}\right )+\frac {2 i a \sec ^5(c+d x)}{9 d \sqrt {a+i a \tan (c+d x)}}\right )+\frac {2 i a \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{11 d}\)

\(\Big \downarrow \) 3974

\(\displaystyle \frac {12}{11} a \left (\frac {8}{9} a \left (\frac {8 i a^2 \sec ^5(c+d x)}{35 d (a+i a \tan (c+d x))^{5/2}}+\frac {2 i a \sec ^5(c+d x)}{7 d (a+i a \tan (c+d x))^{3/2}}\right )+\frac {2 i a \sec ^5(c+d x)}{9 d \sqrt {a+i a \tan (c+d x)}}\right )+\frac {2 i a \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{11 d}\)

Input:

Int[Sec[c + d*x]^5*(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

(((2*I)/11)*a*Sec[c + d*x]^5*Sqrt[a + I*a*Tan[c + d*x]])/d + (12*a*((((2*I 
)/9)*a*Sec[c + d*x]^5)/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (8*a*((((8*I)/35)* 
a^2*Sec[c + d*x]^5)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((2*I)/7)*a*Sec[c 
+ d*x]^5)/(d*(a + I*a*Tan[c + d*x])^(3/2))))/9))/11
 

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 = 2.87 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.68

method	result	size
default	\(\frac {\left (\frac {2 i \left (105 \sec \left (d x +c \right )^{5}-20 \sec \left (d x +c \right )^{3}+512 \cos \left (d x +c \right )-64 \sec \left (d x +c \right )\right )}{1155}+\frac {8 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{33}+\frac {128 \sec \left (d x +c \right ) \tan \left (d x +c \right )}{385}+\frac {1024 \sin \left (d x +c \right )}{1155}\right ) a \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}{d}\)	\(100\)

Input:

int(sec(d*x+c)^5*(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/d*(2/1155*I*(105*sec(d*x+c)^5-20*sec(d*x+c)^3+512*cos(d*x+c)-64*sec(d*x+ 
c))+8/33*tan(d*x+c)*sec(d*x+c)^3+128/385*sec(d*x+c)*tan(d*x+c)+1024/1155*s 
in(d*x+c))*a*(a*(1+I*tan(d*x+c)))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.85 \[ \int \sec ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {64 \, \sqrt {2} {\left (-231 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 198 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 88 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i \, a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{1155 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:

integrate(sec(d*x+c)^5*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

-64/1155*sqrt(2)*(-231*I*a*e^(6*I*d*x + 6*I*c) - 198*I*a*e^(4*I*d*x + 4*I* 
c) - 88*I*a*e^(2*I*d*x + 2*I*c) - 16*I*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1) 
)/(d*e^(10*I*d*x + 10*I*c) + 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6 
*I*c) + 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) + d)
 

Sympy [F]

\[ \int \sec ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \sec ^{5}{\left (c + d x \right )}\, dx \] Input:

integrate(sec(d*x+c)**5*(a+I*a*tan(d*x+c))**(3/2),x)
 

Output:

Integral((I*a*(tan(c + d*x) - I))**(3/2)*sec(c + d*x)**5, 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. 994 vs. \(2 (115) = 230\).

Time = 9.30 (sec) , antiderivative size = 994, normalized size of antiderivative = 6.76 \[ \int \sec ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)^5*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

64/1155*(231*I*sqrt(2)*a*cos(6*d*x + 6*c) + 198*I*sqrt(2)*a*cos(4*d*x + 4* 
c) + 88*I*sqrt(2)*a*cos(2*d*x + 2*c) - 231*sqrt(2)*a*sin(6*d*x + 6*c) - 19 
8*sqrt(2)*a*sin(4*d*x + 4*c) - 88*sqrt(2)*a*sin(2*d*x + 2*c) + 16*I*sqrt(2 
)*a)*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1 
/4)*sqrt(a)/(((4*cos(2*d*x + 2*c)^3 + (4*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 
 2*c)^2 + 4*I*sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^ 
2 + 2*cos(2*d*x + 2*c) + 1)*cos(8*d*x + 8*c) + 4*(cos(2*d*x + 2*c)^2 + sin 
(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c) + 6*(cos(2*d*x 
+ 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 
 9*cos(2*d*x + 2*c)^2 + (I*cos(2*d*x + 2*c)^2 + I*sin(2*d*x + 2*c)^2 + 2*I 
*cos(2*d*x + 2*c) + I)*sin(8*d*x + 8*c) + 4*(I*cos(2*d*x + 2*c)^2 + I*sin( 
2*d*x + 2*c)^2 + 2*I*cos(2*d*x + 2*c) + I)*sin(6*d*x + 6*c) + 6*(I*cos(2*d 
*x + 2*c)^2 + I*sin(2*d*x + 2*c)^2 + 2*I*cos(2*d*x + 2*c) + I)*sin(4*d*x + 
 4*c) + 4*(I*cos(2*d*x + 2*c)^2 + 2*I*cos(2*d*x + 2*c) + I)*sin(2*d*x + 2* 
c) + 6*cos(2*d*x + 2*c) + 1)*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 
 2*c) + 1)) + (4*I*cos(2*d*x + 2*c)^3 + (4*I*cos(2*d*x + 2*c) + I)*sin(2*d 
*x + 2*c)^2 - 4*sin(2*d*x + 2*c)^3 + (I*cos(2*d*x + 2*c)^2 + I*sin(2*d*x + 
 2*c)^2 + 2*I*cos(2*d*x + 2*c) + I)*cos(8*d*x + 8*c) + 4*(I*cos(2*d*x + 2* 
c)^2 + I*sin(2*d*x + 2*c)^2 + 2*I*cos(2*d*x + 2*c) + I)*cos(6*d*x + 6*c) + 
 6*(I*cos(2*d*x + 2*c)^2 + I*sin(2*d*x + 2*c)^2 + 2*I*cos(2*d*x + 2*c) ...
 

Giac [F(-2)]

Exception generated. \[ \int \sec ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(sec(d*x+c)^5*(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 = 4.66 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.99 \[ \int \sec ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {a\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,64{}\mathrm {i}}{5\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {a\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,192{}\mathrm {i}}{7\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {a\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,64{}\mathrm {i}}{3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}-\frac {a\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,64{}\mathrm {i}}{11\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5} \] Input:

int((a + a*tan(c + d*x)*1i)^(3/2)/cos(c + d*x)^5,x)
 

Output:

(a*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)*64i)/(5*d*(exp(c*2i + d*x*2i) + 1)^2) - (a*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)*192i)/(7*d*(exp(c*2i + d*x*2i) + 1)^3) + (a*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 
)*64i)/(3*d*(exp(c*2i + d*x*2i) + 1)^4) - (a*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)*64i)/(11* 
d*(exp(c*2i + d*x*2i) + 1)^5)
 

Reduce [F]

\[ \int \sec ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {2 \sqrt {a}\, a i \left (-\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{5}+6 \left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{5} \tan \left (d x +c \right )d x \right ) d \right )}{d} \] Input:

int(sec(d*x+c)^5*(a+I*a*tan(d*x+c))^(3/2),x)
 

Output:

(2*sqrt(a)*a*i*( - sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**5 + 6*int(sqrt(t 
an(c + d*x)*i + 1)*sec(c + d*x)**5*tan(c + d*x),x)*d))/d