Integrand size = 26, antiderivative size = 110 \[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {64 i a^3 \sec ^9(c+d x)}{1287 d (a+i a \tan (c+d x))^{9/2}}+\frac {16 i a^2 \sec ^9(c+d x)}{143 d (a+i a \tan (c+d x))^{7/2}}+\frac {2 i a \sec ^9(c+d x)}{13 d (a+i a \tan (c+d x))^{5/2}} \] Output:
64/1287*I*a^3*sec(d*x+c)^9/d/(a+I*a*tan(d*x+c))^(9/2)+16/143*I*a^2*sec(d*x +c)^9/d/(a+I*a*tan(d*x+c))^(7/2)+2/13*I*a*sec(d*x+c)^9/d/(a+I*a*tan(d*x+c) )^(5/2)
Time = 1.40 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.84 \[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 \sec ^8(c+d x) (52+151 \cos (2 (c+d x))+135 i \sin (2 (c+d x))) (\cos (3 (c+d x))-i \sin (3 (c+d x)))}{1287 a d (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \] Input:
Integrate[Sec[c + d*x]^9/(a + I*a*Tan[c + d*x])^(3/2),x]
Output:
(2*Sec[c + d*x]^8*(52 + 151*Cos[2*(c + d*x)] + (135*I)*Sin[2*(c + d*x)])*( Cos[3*(c + d*x)] - I*Sin[3*(c + d*x)]))/(1287*a*d*(-I + Tan[c + d*x])*Sqrt [a + I*a*Tan[c + d*x]])
Time = 0.53 (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.
\(\displaystyle \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^9}{(a+i a \tan (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 3975 |
\(\displaystyle \frac {8}{13} a \int \frac {\sec ^9(c+d x)}{(i \tan (c+d x) a+a)^{5/2}}dx+\frac {2 i a \sec ^9(c+d x)}{13 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8}{13} a \int \frac {\sec (c+d x)^9}{(i \tan (c+d x) a+a)^{5/2}}dx+\frac {2 i a \sec ^9(c+d x)}{13 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3975 |
\(\displaystyle \frac {8}{13} a \left (\frac {4}{11} a \int \frac {\sec ^9(c+d x)}{(i \tan (c+d x) a+a)^{7/2}}dx+\frac {2 i a \sec ^9(c+d x)}{11 d (a+i a \tan (c+d x))^{7/2}}\right )+\frac {2 i a \sec ^9(c+d x)}{13 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8}{13} a \left (\frac {4}{11} a \int \frac {\sec (c+d x)^9}{(i \tan (c+d x) a+a)^{7/2}}dx+\frac {2 i a \sec ^9(c+d x)}{11 d (a+i a \tan (c+d x))^{7/2}}\right )+\frac {2 i a \sec ^9(c+d x)}{13 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3974 |
\(\displaystyle \frac {8}{13} a \left (\frac {8 i a^2 \sec ^9(c+d x)}{99 d (a+i a \tan (c+d x))^{9/2}}+\frac {2 i a \sec ^9(c+d x)}{11 d (a+i a \tan (c+d x))^{7/2}}\right )+\frac {2 i a \sec ^9(c+d x)}{13 d (a+i a \tan (c+d x))^{5/2}}\) |
Input:
Int[Sec[c + d*x]^9/(a + I*a*Tan[c + d*x])^(3/2),x]
Output:
(((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
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 = 5.38 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.04
method | result | size |
default | \(\frac {\frac {2 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{6} \left (128 \cos \left (d x +c \right )^{4}+112 \cos \left (d x +c \right )^{2}+99\right )}{1287}+\frac {2 i \left (128 \sec \left (d x +c \right )^{2}+48 \sec \left (d x +c \right )^{4}+27 \sec \left (d x +c \right )^{6}\right )}{1287}}{d \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a}\) | \(114\) |
Input:
int(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
2/1287/d/(I*sin(d*x+c)+cos(d*x+c))/(a*(1+I*tan(d*x+c)))^(1/2)/a*(tan(d*x+c )*sec(d*x+c)^6*(128*cos(d*x+c)^4+112*cos(d*x+c)^2+99)+I*(128*sec(d*x+c)^2+ 48*sec(d*x+c)^4+27*sec(d*x+c)^6))
Time = 0.12 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.30 \[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {128 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-143 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 52 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 8 i\right )}}{1287 \, {\left (a^{2} d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{2} d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \] Input:
integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
Output:
-128/1287*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-143*I*e^(4*I*d*x + 4 *I*c) - 52*I*e^(2*I*d*x + 2*I*c) - 8*I)/(a^2*d*e^(12*I*d*x + 12*I*c) + 6*a ^2*d*e^(10*I*d*x + 10*I*c) + 15*a^2*d*e^(8*I*d*x + 8*I*c) + 20*a^2*d*e^(6* I*d*x + 6*I*c) + 15*a^2*d*e^(4*I*d*x + 4*I*c) + 6*a^2*d*e^(2*I*d*x + 2*I*c ) + a^2*d)
\[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {\sec ^{9}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(sec(d*x+c)**9/(a+I*a*tan(d*x+c))**(3/2),x)
Output:
Integral(sec(c + d*x)**9/(I*a*(tan(c + d*x) - I))**(3/2), x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 626 vs. \(2 (86) = 172\).
Time = 0.25 (sec) , antiderivative size = 626, normalized size of antiderivative = 5.69 \[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:
integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
Output:
-2/1287*(-203*I*sqrt(a) - 678*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 2* I*sqrt(a)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1802*sqrt(a)*sin(d*x + c)^ 3/(cos(d*x + c) + 1)^3 - 26*I*sqrt(a)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 3614*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 858*I*sqrt(a)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 6578*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 6578*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 858*I*sqrt(a)*s in(d*x + c)^10/(cos(d*x + c) + 1)^10 - 3614*sqrt(a)*sin(d*x + c)^11/(cos(d *x + c) + 1)^11 + 26*I*sqrt(a)*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 180 2*sqrt(a)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 + 2*I*sqrt(a)*sin(d*x + c) ^14/(cos(d*x + c) + 1)^14 - 678*sqrt(a)*sin(d*x + c)^15/(cos(d*x + c) + 1) ^15 + 203*I*sqrt(a)*sin(d*x + c)^16/(cos(d*x + c) + 1)^16)*(sin(d*x + c)/( cos(d*x + c) + 1) + 1)^(3/2)*(sin(d*x + c)/(cos(d*x + c) + 1) - 1)^(3/2)/( (a^2 - 8*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 28*a^2*sin(d*x + c)^4/( cos(d*x + c) + 1)^4 - 56*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 70*a^2* sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 56*a^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 28*a^2*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 8*a^2*sin(d*x + c)^14/(cos(d*x + c) + 1)^14 + a^2*sin(d*x + c)^16/(cos(d*x + c) + 1)^16)*d *(-2*I*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^2/(cos(d*x + c) + 1) ^2 - 1)^(3/2))
Exception generated. \[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(sec(d*x+c)^9/(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
Time = 5.76 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.95 \[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {128\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,52{}\mathrm {i}+{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,143{}\mathrm {i}+8{}\mathrm {i}\right )}{1287\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6} \] Input:
int(1/(cos(c + d*x)^9*(a + a*tan(c + d*x)*1i)^(3/2)),x)
Output:
(128*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)*52i + exp(c*4i + d*x*4i)*143i + 8i))/(1287*a^2*d*(exp(c*2i + d*x*2i) + 1)^6)
\[ \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{9} i -15 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{9} \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) \tan \left (d x +c \right )^{2} d -15 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{9} \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) d +13 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{9} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) \tan \left (d x +c \right )^{2} d i +13 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{9} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) d i \right )}{a^{2} d \left (\tan \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^(3/2),x)
Output:
(sqrt(a)*( - 2*sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**9*i - 15*int((sqrt(t an(c + d*x)*i + 1)*sec(c + d*x)**9*tan(c + d*x)**2)/(tan(c + d*x)**3*i + t an(c + d*x)**2 + tan(c + d*x)*i + 1),x)*tan(c + d*x)**2*d - 15*int((sqrt(t an(c + d*x)*i + 1)*sec(c + d*x)**9*tan(c + d*x)**2)/(tan(c + d*x)**3*i + t an(c + d*x)**2 + tan(c + d*x)*i + 1),x)*d + 13*int((sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**9*tan(c + d*x))/(tan(c + d*x)**3*i + tan(c + d*x)**2 + ta n(c + d*x)*i + 1),x)*tan(c + d*x)**2*d*i + 13*int((sqrt(tan(c + d*x)*i + 1 )*sec(c + d*x)**9*tan(c + d*x))/(tan(c + d*x)**3*i + tan(c + d*x)**2 + tan (c + d*x)*i + 1),x)*d*i))/(a**2*d*(tan(c + d*x)**2 + 1))