Integrand size = 24, antiderivative size = 80 \[ \int \frac {\sec ^8(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {i (a-i a \tan (c+d x))^4}{a^5 d}-\frac {4 i (a-i a \tan (c+d x))^5}{5 a^6 d}+\frac {i (a-i a \tan (c+d x))^6}{6 a^7 d} \] Output:
I*(a-I*a*tan(d*x+c))^4/a^5/d-4/5*I*(a-I*a*tan(d*x+c))^5/a^6/d+1/6*I*(a-I*a *tan(d*x+c))^6/a^7/d
Time = 0.14 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.58 \[ \int \frac {\sec ^8(c+d x)}{a+i a \tan (c+d x)} \, dx=-\frac {i (i+\tan (c+d x))^4 \left (-11-14 i \tan (c+d x)+5 \tan ^2(c+d x)\right )}{30 a d} \] Input:
Integrate[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x]),x]
Output:
((-1/30*I)*(I + Tan[c + d*x])^4*(-11 - (14*I)*Tan[c + d*x] + 5*Tan[c + d*x ]^2))/(a*d)
Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3968, 49, 2009}
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.
\(\displaystyle \int \frac {\sec ^8(c+d x)}{a+i a \tan (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^8}{a+i a \tan (c+d x)}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i \int (a-i a \tan (c+d x))^3 (i \tan (c+d x) a+a)^2d(i a \tan (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {i \int \left ((a-i a \tan (c+d x))^5-4 a (a-i a \tan (c+d x))^4+4 a^2 (a-i a \tan (c+d x))^3\right )d(i a \tan (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i \left (-a^2 (a-i a \tan (c+d x))^4-\frac {1}{6} (a-i a \tan (c+d x))^6+\frac {4}{5} a (a-i a \tan (c+d x))^5\right )}{a^7 d}\) |
Input:
Int[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x]),x]
Output:
((-I)*(-(a^2*(a - I*a*Tan[c + d*x])^4) + (4*a*(a - I*a*Tan[c + d*x])^5)/5 - (a - I*a*Tan[c + d*x])^6/6))/(a^7*d)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(a^(m - 2)*b*f) Subst[Int[(a - x)^(m/2 - 1)*(a + x )^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
Time = 0.46 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.59
method | result | size |
risch | \(\frac {16 i \left (15 \,{\mathrm e}^{4 i \left (d x +c \right )}+6 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{15 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}\) | \(47\) |
derivativedivides | \(-\frac {-\tan \left (d x +c \right )+\frac {i \tan \left (d x +c \right )^{6}}{6}-\frac {\tan \left (d x +c \right )^{5}}{5}+\frac {i \tan \left (d x +c \right )^{4}}{2}-\frac {2 \tan \left (d x +c \right )^{3}}{3}+\frac {i \tan \left (d x +c \right )^{2}}{2}}{a d}\) | \(71\) |
default | \(-\frac {-\tan \left (d x +c \right )+\frac {i \tan \left (d x +c \right )^{6}}{6}-\frac {\tan \left (d x +c \right )^{5}}{5}+\frac {i \tan \left (d x +c \right )^{4}}{2}-\frac {2 \tan \left (d x +c \right )^{3}}{3}+\frac {i \tan \left (d x +c \right )^{2}}{2}}{a d}\) | \(71\) |
Input:
int(sec(d*x+c)^8/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
16/15*I*(15*exp(4*I*(d*x+c))+6*exp(2*I*(d*x+c))+1)/d/a/(exp(2*I*(d*x+c))+1 )^6
Time = 0.09 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.36 \[ \int \frac {\sec ^8(c+d x)}{a+i a \tan (c+d x)} \, dx=-\frac {16 \, {\left (-15 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{15 \, {\left (a d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \] Input:
integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c)),x, algorithm="fricas")
Output:
-16/15*(-15*I*e^(4*I*d*x + 4*I*c) - 6*I*e^(2*I*d*x + 2*I*c) - I)/(a*d*e^(1 2*I*d*x + 12*I*c) + 6*a*d*e^(10*I*d*x + 10*I*c) + 15*a*d*e^(8*I*d*x + 8*I* c) + 20*a*d*e^(6*I*d*x + 6*I*c) + 15*a*d*e^(4*I*d*x + 4*I*c) + 6*a*d*e^(2* I*d*x + 2*I*c) + a*d)
\[ \int \frac {\sec ^8(c+d x)}{a+i a \tan (c+d x)} \, dx=- \frac {i \int \frac {\sec ^{8}{\left (c + d x \right )}}{\tan {\left (c + d x \right )} - i}\, dx}{a} \] Input:
integrate(sec(d*x+c)**8/(a+I*a*tan(d*x+c)),x)
Output:
-I*Integral(sec(c + d*x)**8/(tan(c + d*x) - I), x)/a
Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.84 \[ \int \frac {\sec ^8(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {-5 i \, \tan \left (d x + c\right )^{6} + 6 \, \tan \left (d x + c\right )^{5} - 15 i \, \tan \left (d x + c\right )^{4} + 20 \, \tan \left (d x + c\right )^{3} - 15 i \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right )}{30 \, a d} \] Input:
integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c)),x, algorithm="maxima")
Output:
1/30*(-5*I*tan(d*x + c)^6 + 6*tan(d*x + c)^5 - 15*I*tan(d*x + c)^4 + 20*ta n(d*x + c)^3 - 15*I*tan(d*x + c)^2 + 30*tan(d*x + c))/(a*d)
Time = 0.16 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.84 \[ \int \frac {\sec ^8(c+d x)}{a+i a \tan (c+d x)} \, dx=-\frac {5 i \, \tan \left (d x + c\right )^{6} - 6 \, \tan \left (d x + c\right )^{5} + 15 i \, \tan \left (d x + c\right )^{4} - 20 \, \tan \left (d x + c\right )^{3} + 15 i \, \tan \left (d x + c\right )^{2} - 30 \, \tan \left (d x + c\right )}{30 \, a d} \] Input:
integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c)),x, algorithm="giac")
Output:
-1/30*(5*I*tan(d*x + c)^6 - 6*tan(d*x + c)^5 + 15*I*tan(d*x + c)^4 - 20*ta n(d*x + c)^3 + 15*I*tan(d*x + c)^2 - 30*tan(d*x + c))/(a*d)
Time = 0.41 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.42 \[ \int \frac {\sec ^8(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {\sin \left (c+d\,x\right )\,\left (30\,{\cos \left (c+d\,x\right )}^5-{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )\,15{}\mathrm {i}+20\,{\cos \left (c+d\,x\right )}^3\,{\sin \left (c+d\,x\right )}^2-{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^3\,15{}\mathrm {i}+6\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^4-{\sin \left (c+d\,x\right )}^5\,5{}\mathrm {i}\right )}{30\,a\,d\,{\cos \left (c+d\,x\right )}^6} \] Input:
int(1/(cos(c + d*x)^8*(a + a*tan(c + d*x)*1i)),x)
Output:
(sin(c + d*x)*(6*cos(c + d*x)*sin(c + d*x)^4 - cos(c + d*x)^4*sin(c + d*x) *15i + 30*cos(c + d*x)^5 - sin(c + d*x)^5*5i - cos(c + d*x)^2*sin(c + d*x) ^3*15i + 20*cos(c + d*x)^3*sin(c + d*x)^2))/(30*a*d*cos(c + d*x)^6)
\[ \int \frac {\sec ^8(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {\int \frac {\sec \left (d x +c \right )^{8}}{\tan \left (d x +c \right ) i +1}d x}{a} \] Input:
int(sec(d*x+c)^8/(a+I*a*tan(d*x+c)),x)
Output:
int(sec(c + d*x)**8/(tan(c + d*x)*i + 1),x)/a