Integrand size = 24, antiderivative size = 104 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {64 i a^3 \sec (c+d x)}{15 d \sqrt {a+i a \tan (c+d x)}}+\frac {16 i a^2 \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d} \] Output:
64/15*I*a^3*sec(d*x+c)/d/(a+I*a*tan(d*x+c))^(1/2)+16/15*I*a^2*sec(d*x+c)*( a+I*a*tan(d*x+c))^(1/2)/d+2/5*I*a*sec(d*x+c)*(a+I*a*tan(d*x+c))^(3/2)/d
Time = 0.53 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {2 a^2 \sec ^2(c+d x) (i \cos (c-d x)+\sin (c-d x)) (20+23 \cos (2 (c+d x))+7 i \sin (2 (c+d x))) \sqrt {a+i a \tan (c+d x)}}{15 d (\cos (d x)+i \sin (d x))^2} \] Input:
Integrate[Sec[c + d*x]*(a + I*a*Tan[c + d*x])^(5/2),x]
Output:
(2*a^2*Sec[c + d*x]^2*(I*Cos[c - d*x] + Sin[c - d*x])*(20 + 23*Cos[2*(c + d*x)] + (7*I)*Sin[2*(c + d*x)])*Sqrt[a + I*a*Tan[c + d*x]])/(15*d*(Cos[d*x ] + I*Sin[d*x])^2)
Time = 0.45 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {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.
\(\displaystyle \int \sec (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x) (a+i a \tan (c+d x))^{5/2}dx\) |
\(\Big \downarrow \) 3975 |
\(\displaystyle \frac {8}{5} a \int \sec (c+d x) (i \tan (c+d x) a+a)^{3/2}dx+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8}{5} a \int \sec (c+d x) (i \tan (c+d x) a+a)^{3/2}dx+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3975 |
\(\displaystyle \frac {8}{5} a \left (\frac {4}{3} a \int \sec (c+d x) \sqrt {i \tan (c+d x) a+a}dx+\frac {2 i a \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8}{5} a \left (\frac {4}{3} a \int \sec (c+d x) \sqrt {i \tan (c+d x) a+a}dx+\frac {2 i a \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\) |
\(\Big \downarrow \) 3974 |
\(\displaystyle \frac {8}{5} a \left (\frac {8 i a^2 \sec (c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i a \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )+\frac {2 i a \sec (c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\) |
Input:
Int[Sec[c + d*x]*(a + I*a*Tan[c + d*x])^(5/2),x]
Output:
(((2*I)/5)*a*Sec[c + d*x]*(a + I*a*Tan[c + d*x])^(3/2))/d + (8*a*((((8*I)/ 3)*a^2*Sec[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((2*I)/3)*a*Sec[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d))/5
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]
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]
Time = 2.59 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.63
method | result | size |
default | \(\frac {\left (\frac {2 i \left (32 \cos \left (d x +c \right )+11 \sec \left (d x +c \right )\right )}{15}-\frac {2 \sec \left (d x +c \right ) \tan \left (d x +c \right )}{5}+\frac {64 \sin \left (d x +c \right )}{15}\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2}}{d}\) | \(66\) |
Input:
int(sec(d*x+c)*(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
1/d*(2/15*I*(32*cos(d*x+c)+11*sec(d*x+c))-2/5*sec(d*x+c)*tan(d*x+c)+64/15* sin(d*x+c))*(a*(1+I*tan(d*x+c)))^(1/2)*a^2
Time = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.80 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {8 \, \sqrt {2} {\left (-15 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 20 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 8 i \, a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:
integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
Output:
-8/15*sqrt(2)*(-15*I*a^2*e^(4*I*d*x + 4*I*c) - 20*I*a^2*e^(2*I*d*x + 2*I*c ) - 8*I*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(4*I*d*x + 4*I*c) + 2* d*e^(2*I*d*x + 2*I*c) + d)
\[ \int \sec (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}} \sec {\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))**(5/2),x)
Output:
Integral((I*a*(tan(c + d*x) - I))**(5/2)*sec(c + d*x), x)
\[ \int \sec (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right ) \,d x } \] Input:
integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate((I*a*tan(d*x + c) + a)^(5/2)*sec(d*x + c), x)
Exception generated. \[ \int \sec (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(sec(d*x+c)*(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
Time = 3.41 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.01 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {8\,a^2\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,20{}\mathrm {i}+{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,15{}\mathrm {i}+8{}\mathrm {i}\right )}{15\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2} \] Input:
int((a + a*tan(c + d*x)*1i)^(5/2)/cos(c + d*x),x)
Output:
(8*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)*(exp(c*2i + d*x*2i)*20i + exp(c*4i + d*x*4i)*15 i + 8i))/(15*d*(exp(c*2i + d*x*2i) + 1)^2)
\[ \int \sec (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {\sqrt {a}\, a^{2} \left (-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right ) i -\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right ) \tan \left (d x +c \right )^{2}d x \right ) d +5 \left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right ) \tan \left (d x +c \right )d x \right ) d i \right )}{d} \] Input:
int(sec(d*x+c)*(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)*i - int(sqrt(tan (c + d*x)*i + 1)*sec(c + d*x)*tan(c + d*x)**2,x)*d + 5*int(sqrt(tan(c + d* x)*i + 1)*sec(c + d*x)*tan(c + d*x),x)*d*i))/d