Integrand size = 26, antiderivative size = 73 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {8 i a^2 \sec ^7(c+d x)}{63 d (a+i a \tan (c+d x))^{7/2}}+\frac {2 i a \sec ^7(c+d x)}{9 d (a+i a \tan (c+d x))^{5/2}} \] Output:
8/63*I*a^2*sec(d*x+c)^7/d/(a+I*a*tan(d*x+c))^(7/2)+2/9*I*a*sec(d*x+c)^7/d/ (a+I*a*tan(d*x+c))^(5/2)
Time = 0.95 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.10 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 \sec ^5(c+d x) (i \cos (2 (c+d x))+\sin (2 (c+d x))) (-11 i+7 \tan (c+d x))}{63 a d (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \] Input:
Integrate[Sec[c + d*x]^7/(a + I*a*Tan[c + d*x])^(3/2),x]
Output:
(2*Sec[c + d*x]^5*(I*Cos[2*(c + d*x)] + Sin[2*(c + d*x)])*(-11*I + 7*Tan[c + d*x]))/(63*a*d*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])
Time = 0.40 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {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.
\(\displaystyle \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^7}{(a+i a \tan (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 3975 |
\(\displaystyle \frac {4}{9} a \int \frac {\sec ^7(c+d x)}{(i \tan (c+d x) a+a)^{5/2}}dx+\frac {2 i a \sec ^7(c+d x)}{9 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4}{9} a \int \frac {\sec (c+d x)^7}{(i \tan (c+d x) a+a)^{5/2}}dx+\frac {2 i a \sec ^7(c+d x)}{9 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3974 |
\(\displaystyle \frac {8 i a^2 \sec ^7(c+d x)}{63 d (a+i a \tan (c+d x))^{7/2}}+\frac {2 i a \sec ^7(c+d x)}{9 d (a+i a \tan (c+d x))^{5/2}}\) |
Input:
Int[Sec[c + d*x]^7/(a + I*a*Tan[c + d*x])^(3/2),x]
Output:
(((8*I)/63)*a^2*Sec[c + d*x]^7)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + (((2*I) /9)*a*Sec[c + d*x]^7)/(d*(a + I*a*Tan[c + d*x])^(5/2))
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 = 4.79 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.18
method | result | size |
default | \(\frac {\frac {32 i \sec \left (d x +c \right )}{63}+\frac {32 \sec \left (d x +c \right ) \tan \left (d x +c \right )}{63}+\frac {4 i \sec \left (d x +c \right )^{3}}{63}+\frac {20 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{63}-\frac {2 i \sec \left (d x +c \right )^{5}}{9}}{d \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a}\) | \(86\) |
Input:
int(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
2/63/d/(a*(1+I*tan(d*x+c)))^(1/2)/a*(16*I*sec(d*x+c)+16*sec(d*x+c)*tan(d*x +c)+2*I*sec(d*x+c)^3+10*tan(d*x+c)*sec(d*x+c)^3-7*I*sec(d*x+c)^5)
Time = 0.13 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.40 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {32 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-9 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i\right )}}{63 \, {\left (a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \] Input:
integrate(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
Output:
-32/63*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-9*I*e^(2*I*d*x + 2*I*c) - 2*I)/(a^2*d*e^(8*I*d*x + 8*I*c) + 4*a^2*d*e^(6*I*d*x + 6*I*c) + 6*a^2*d *e^(4*I*d*x + 4*I*c) + 4*a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)
\[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {\sec ^{7}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(sec(d*x+c)**7/(a+I*a*tan(d*x+c))**(3/2),x)
Output:
Integral(sec(c + d*x)**7/(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. 488 vs. \(2 (57) = 114\).
Time = 0.22 (sec) , antiderivative size = 488, normalized size of antiderivative = 6.68 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:
integrate(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
Output:
-2/63*(-11*I*sqrt(a) - 30*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 12*I*s qrt(a)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 86*sqrt(a)*sin(d*x + c)^3/(co s(d*x + c) + 1)^3 + 9*I*sqrt(a)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 108* sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 108*sqrt(a)*sin(d*x + c)^7/( cos(d*x + c) + 1)^7 - 9*I*sqrt(a)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 86 *sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 12*I*sqrt(a)*sin(d*x + c)^1 0/(cos(d*x + c) + 1)^10 - 30*sqrt(a)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 11*I*sqrt(a)*sin(d*x + c)^12/(cos(d*x + c) + 1)^12)*(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 - 6*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*a^2*sin(d*x + c)^4/(cos( d*x + c) + 1)^4 - 20*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 15*a^2*sin( d*x + c)^8/(cos(d*x + c) + 1)^8 - 6*a^2*sin(d*x + c)^10/(cos(d*x + c) + 1) ^10 + a^2*sin(d*x + c)^12/(cos(d*x + c) + 1)^12)*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. \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(sec(d*x+c)^7/(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 = 3.69 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.25 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {32\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,9{}\mathrm {i}+2{}\mathrm {i}\right )\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}}{63\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4} \] Input:
int(1/(cos(c + d*x)^7*(a + a*tan(c + d*x)*1i)^(3/2)),x)
Output:
(32*exp(- c*1i - d*x*1i)*(exp(c*2i + d*x*2i)*9i + 2i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2))/(63*a^2*d*(exp(c*2i + d*x*2i) + 1)^4)
\[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{7} i -11 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{7} \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) \tan \left (d x +c \right )^{2} d -11 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{7} \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) d +9 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{7} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) \tan \left (d x +c \right )^{2} d i +9 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )^{7} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) d i \right )}{a^{2} d \left (\tan \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)^7/(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)**7*i - 11*int((sqrt(t an(c + d*x)*i + 1)*sec(c + d*x)**7*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 - 11*int((sqrt(t an(c + d*x)*i + 1)*sec(c + d*x)**7*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 + 9*int((sqrt(tan(c + d*x)*i + 1 )*sec(c + d*x)**7*tan(c + d*x))/(tan(c + d*x)**3*i + tan(c + d*x)**2 + tan (c + d*x)*i + 1),x)*tan(c + d*x)**2*d*i + 9*int((sqrt(tan(c + d*x)*i + 1)* sec(c + d*x)**7*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))