\(\int \sec ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [313]

Optimal result

Integrand size = 26, antiderivative size = 147 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {256 i a^4 \sec ^3(c+d x)}{315 d (a+i a \tan (c+d x))^{3/2}}+\frac {64 i a^3 \sec ^3(c+d x)}{105 d \sqrt {a+i a \tan (c+d x)}}+\frac {8 i a^2 \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{21 d}+\frac {2 i a \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d} \] Output:

256/315*I*a^4*sec(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(3/2)+64/105*I*a^3*sec(d*x 
+c)^3/d/(a+I*a*tan(d*x+c))^(1/2)+8/21*I*a^2*sec(d*x+c)^3*(a+I*a*tan(d*x+c) 
)^(1/2)/d+2/9*I*a*sec(d*x+c)^3*(a+I*a*tan(d*x+c))^(3/2)/d
 

Mathematica [A] (verified)

Time = 1.10 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.70 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {2 a^2 \sec ^3(c+d x) (i \cos (2 c)+\sin (2 c)) (77+242 \cos (2 (c+d x))+89 i \sec (c+d x) \sin (3 (c+d x))+54 i \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{315 d (\cos (d x)+i \sin (d x))^2} \] Input:

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

Output:

(2*a^2*Sec[c + d*x]^3*(I*Cos[2*c] + Sin[2*c])*(77 + 242*Cos[2*(c + d*x)] + 
 (89*I)*Sec[c + d*x]*Sin[3*(c + d*x)] + (54*I)*Tan[c + d*x])*Sqrt[a + I*a* 
Tan[c + d*x]])/(315*d*(Cos[d*x] + I*Sin[d*x])^2)
 

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3975

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3975

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3975

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3974

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

Input:

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

Output:

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

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 = 21.63 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.63

method	result	size
default	\(\frac {\left (\frac {2 i \left (95 \sec \left (d x +c \right )^{3}-32 \sec \left (d x +c \right )+256 \cos \left (d x +c \right )\right )}{315}-\frac {2 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{9}+\frac {64 \sec \left (d x +c \right ) \tan \left (d x +c \right )}{105}+\frac {512 \sin \left (d x +c \right )}{315}\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2}}{d}\)	\(92\)

Input:

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

Output:

1/d*(2/315*I*(95*sec(d*x+c)^3-32*sec(d*x+c)+256*cos(d*x+c))-2/9*tan(d*x+c) 
*sec(d*x+c)^3+64/105*sec(d*x+c)*tan(d*x+c)+512/315*sin(d*x+c))*(a*(1+I*tan 
(d*x+c)))^(1/2)*a^2
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.82 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {32 \, \sqrt {2} {\left (-105 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 126 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 72 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i \, a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{315 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:

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

Output:

-32/315*sqrt(2)*(-105*I*a^2*e^(6*I*d*x + 6*I*c) - 126*I*a^2*e^(4*I*d*x + 4 
*I*c) - 72*I*a^2*e^(2*I*d*x + 2*I*c) - 16*I*a^2)*sqrt(a/(e^(2*I*d*x + 2*I* 
c) + 1))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x 
 + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F(-2)]

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

integrate(sec(d*x+c)^3*(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 = 5.13 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.05 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {a^2\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,32{}\mathrm {i}}{3\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {a^2\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,96{}\mathrm {i}}{5\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {a^2\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,96{}\mathrm {i}}{7\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}-\frac {a^2\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,32{}\mathrm {i}}{9\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4} \] Input:

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

Output:

(a^2*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)*32i)/(3*d*(exp(c*2i + d*x*2i) + 1)) - (a^2*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)*96i)/(5*d*(exp(c*2i + d*x*2i) + 1)^2) + (a^2*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)*96i)/(7*d*(exp(c*2i + d*x*2i) + 1)^3) - (a^2*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)*32i) 
/(9*d*(exp(c*2i + d*x*2i) + 1)^4)
 

Reduce [F]

\[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {\sqrt {a}\, a^{2} \left (-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{3} i -\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{3} \tan \left (d x +c \right )^{2}d x \right ) d +9 \left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{3} \tan \left (d x +c \right )d x \right ) d i \right )}{d} \] Input:

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

Output:

(sqrt(a)*a**2*( - 2*sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**3*i - int(sqrt( 
tan(c + d*x)*i + 1)*sec(c + d*x)**3*tan(c + d*x)**2,x)*d + 9*int(sqrt(tan( 
c + d*x)*i + 1)*sec(c + d*x)**3*tan(c + d*x),x)*d*i))/d