Integrand size = 26, antiderivative size = 110 \[ \int \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {64 i a^3 \sec ^5(c+d x)}{315 d (a+i a \tan (c+d x))^{5/2}}+\frac {16 i a^2 \sec ^5(c+d x)}{63 d (a+i a \tan (c+d x))^{3/2}}+\frac {2 i a \sec ^5(c+d x)}{9 d \sqrt {a+i a \tan (c+d x)}} \] Output:
64/315*I*a^3*sec(d*x+c)^5/d/(a+I*a*tan(d*x+c))^(5/2)+16/63*I*a^2*sec(d*x+c )^5/d/(a+I*a*tan(d*x+c))^(3/2)+2/9*I*a*sec(d*x+c)^5/d/(a+I*a*tan(d*x+c))^( 1/2)
Time = 0.60 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.70 \[ \int \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {2 \sec ^4(c+d x) (36+71 \cos (2 (c+d x))+55 i \sin (2 (c+d x))) (i \cos (3 (c+d x))+\sin (3 (c+d x))) \sqrt {a+i a \tan (c+d x)}}{315 d} \] Input:
Integrate[Sec[c + d*x]^5*Sqrt[a + I*a*Tan[c + d*x]],x]
Output:
(2*Sec[c + d*x]^4*(36 + 71*Cos[2*(c + d*x)] + (55*I)*Sin[2*(c + d*x)])*(I* Cos[3*(c + d*x)] + Sin[3*(c + d*x)])*Sqrt[a + I*a*Tan[c + d*x]])/(315*d)
Time = 0.52 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, 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.
\(\displaystyle \int \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x)^5 \sqrt {a+i a \tan (c+d x)}dx\) |
\(\Big \downarrow \) 3975 |
\(\displaystyle \frac {8}{9} a \int \frac {\sec ^5(c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx+\frac {2 i a \sec ^5(c+d x)}{9 d \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8}{9} a \int \frac {\sec (c+d x)^5}{\sqrt {i \tan (c+d x) a+a}}dx+\frac {2 i a \sec ^5(c+d x)}{9 d \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 3975 |
\(\displaystyle \frac {8}{9} a \left (\frac {4}{7} a \int \frac {\sec ^5(c+d x)}{(i \tan (c+d x) a+a)^{3/2}}dx+\frac {2 i a \sec ^5(c+d x)}{7 d (a+i a \tan (c+d x))^{3/2}}\right )+\frac {2 i a \sec ^5(c+d x)}{9 d \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8}{9} a \left (\frac {4}{7} a \int \frac {\sec (c+d x)^5}{(i \tan (c+d x) a+a)^{3/2}}dx+\frac {2 i a \sec ^5(c+d x)}{7 d (a+i a \tan (c+d x))^{3/2}}\right )+\frac {2 i a \sec ^5(c+d x)}{9 d \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 3974 |
\(\displaystyle \frac {8}{9} a \left (\frac {8 i a^2 \sec ^5(c+d x)}{35 d (a+i a \tan (c+d x))^{5/2}}+\frac {2 i a \sec ^5(c+d x)}{7 d (a+i a \tan (c+d x))^{3/2}}\right )+\frac {2 i a \sec ^5(c+d x)}{9 d \sqrt {a+i a \tan (c+d x)}}\) |
Input:
Int[Sec[c + d*x]^5*Sqrt[a + I*a*Tan[c + d*x]],x]
Output:
(((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
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 = 2.81 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.81
method | result | size |
default | \(\frac {\left (\frac {2 i \left (-5 \sec \left (d x +c \right )^{3}+128 \cos \left (d x +c \right )-16 \sec \left (d x +c \right )\right )}{315}+\frac {2 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{9}+\frac {32 \sec \left (d x +c \right ) \tan \left (d x +c \right )}{105}+\frac {256 \sin \left (d x +c \right )}{315}\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}{d}\) | \(89\) |
Input:
int(sec(d*x+c)^5*(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/d*(2/315*I*(-5*sec(d*x+c)^3+128*cos(d*x+c)-16*sec(d*x+c))+2/9*tan(d*x+c) *sec(d*x+c)^3+32/105*sec(d*x+c)*tan(d*x+c)+256/315*sin(d*x+c))*(a*(1+I*tan (d*x+c)))^(1/2)
Time = 0.13 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.88 \[ \int \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {32 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-63 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 36 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 8 i\right )}}{315 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:
integrate(sec(d*x+c)^5*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")
Output:
-32/315*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-63*I*e^(4*I*d*x + 4*I* c) - 36*I*e^(2*I*d*x + 2*I*c) - 8*I)/(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)
\[ \int \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\int \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \sec ^{5}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**5*(a+I*a*tan(d*x+c))**(1/2),x)
Output:
Integral(sqrt(I*a*(tan(c + d*x) - I))*sec(c + d*x)**5, x)
Timed out. \[ \int \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)^5*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")
Output:
Timed out
Exception generated. \[ \int \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(sec(d*x+c)^5*(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
Time = 3.32 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.93 \[ \int \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {32\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,36{}\mathrm {i}+{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,63{}\mathrm {i}+8{}\mathrm {i}\right )}{315\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4} \] Input:
int((a + a*tan(c + d*x)*1i)^(1/2)/cos(c + d*x)^5,x)
Output:
(32*exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2 i + d*x*2i) + 1))^(1/2)*(exp(c*2i + d*x*2i)*36i + exp(c*4i + d*x*4i)*63i + 8i))/(315*d*(exp(c*2i + d*x*2i) + 1)^4)
\[ \int \sec ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {\sqrt {a}\, i \left (-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{5}+11 \left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{5} \tan \left (d x +c \right )d x \right ) d \right )}{d} \] Input:
int(sec(d*x+c)^5*(a+I*a*tan(d*x+c))^(1/2),x)
Output:
(sqrt(a)*i*( - 2*sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**5 + 11*int(sqrt(ta n(c + d*x)*i + 1)*sec(c + d*x)**5*tan(c + d*x),x)*d))/d