Integrand size = 24, antiderivative size = 107 \[ \int \frac {\sec ^{10}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {8 i (a-i a \tan (c+d x))^5}{5 a^6 d}-\frac {2 i (a-i a \tan (c+d x))^6}{a^7 d}+\frac {6 i (a-i a \tan (c+d x))^7}{7 a^8 d}-\frac {i (a-i a \tan (c+d x))^8}{8 a^9 d} \] Output:
8/5*I*(a-I*a*tan(d*x+c))^5/a^6/d-2*I*(a-I*a*tan(d*x+c))^6/a^7/d+6/7*I*(a-I *a*tan(d*x+c))^7/a^8/d-1/8*I*(a-I*a*tan(d*x+c))^8/a^9/d
Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.52 \[ \int \frac {\sec ^{10}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {(i+\tan (c+d x))^5 \left (93+185 i \tan (c+d x)-135 \tan ^2(c+d x)-35 i \tan ^3(c+d x)\right )}{280 a d} \] Input:
Integrate[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x]),x]
Output:
((I + Tan[c + d*x])^5*(93 + (185*I)*Tan[c + d*x] - 135*Tan[c + d*x]^2 - (3 5*I)*Tan[c + d*x]^3))/(280*a*d)
Time = 0.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.86, 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.
\(\displaystyle \int \frac {\sec ^{10}(c+d x)}{a+i a \tan (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^{10}}{a+i a \tan (c+d x)}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i \int (a-i a \tan (c+d x))^4 (i \tan (c+d x) a+a)^3d(i a \tan (c+d x))}{a^9 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {i \int \left (-(a-i a \tan (c+d x))^7+6 a (a-i a \tan (c+d x))^6-12 a^2 (a-i a \tan (c+d x))^5+8 a^3 (a-i a \tan (c+d x))^4\right )d(i a \tan (c+d x))}{a^9 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i \left (-\frac {8}{5} a^3 (a-i a \tan (c+d x))^5+2 a^2 (a-i a \tan (c+d x))^6+\frac {1}{8} (a-i a \tan (c+d x))^8-\frac {6}{7} a (a-i a \tan (c+d x))^7\right )}{a^9 d}\) |
Input:
Int[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x]),x]
Output:
((-I)*((-8*a^3*(a - I*a*Tan[c + d*x])^5)/5 + 2*a^2*(a - I*a*Tan[c + d*x])^ 6 - (6*a*(a - I*a*Tan[c + d*x])^7)/7 + (a - I*a*Tan[c + d*x])^8/8))/(a^9*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 = 0.49 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.54
method | result | size |
risch | \(\frac {32 i \left (56 \,{\mathrm e}^{6 i \left (d x +c \right )}+28 \,{\mathrm e}^{4 i \left (d x +c \right )}+8 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{35 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{8}}\) | \(58\) |
derivativedivides | \(-\frac {-\tan \left (d x +c \right )+\frac {i \tan \left (d x +c \right )^{8}}{8}-\frac {\tan \left (d x +c \right )^{7}}{7}+\frac {i \tan \left (d x +c \right )^{6}}{2}-\frac {3 \tan \left (d x +c \right )^{5}}{5}+\frac {3 i \tan \left (d x +c \right )^{4}}{4}-\tan \left (d x +c \right )^{3}+\frac {i \tan \left (d x +c \right )^{2}}{2}}{a d}\) | \(92\) |
default | \(-\frac {-\tan \left (d x +c \right )+\frac {i \tan \left (d x +c \right )^{8}}{8}-\frac {\tan \left (d x +c \right )^{7}}{7}+\frac {i \tan \left (d x +c \right )^{6}}{2}-\frac {3 \tan \left (d x +c \right )^{5}}{5}+\frac {3 i \tan \left (d x +c \right )^{4}}{4}-\tan \left (d x +c \right )^{3}+\frac {i \tan \left (d x +c \right )^{2}}{2}}{a d}\) | \(92\) |
Input:
int(sec(d*x+c)^10/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
32/35*I*(56*exp(6*I*(d*x+c))+28*exp(4*I*(d*x+c))+8*exp(2*I*(d*x+c))+1)/d/a /(exp(2*I*(d*x+c))+1)^8
Time = 0.09 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.36 \[ \int \frac {\sec ^{10}(c+d x)}{a+i a \tan (c+d x)} \, dx=-\frac {32 \, {\left (-56 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 28 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 8 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{35 \, {\left (a d e^{\left (16 i \, d x + 16 i \, c\right )} + 8 \, a d e^{\left (14 i \, d x + 14 i \, c\right )} + 28 \, a d e^{\left (12 i \, d x + 12 i \, c\right )} + 56 \, a d e^{\left (10 i \, d x + 10 i \, c\right )} + 70 \, a d e^{\left (8 i \, d x + 8 i \, c\right )} + 56 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 28 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \] Input:
integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c)),x, algorithm="fricas")
Output:
-32/35*(-56*I*e^(6*I*d*x + 6*I*c) - 28*I*e^(4*I*d*x + 4*I*c) - 8*I*e^(2*I* d*x + 2*I*c) - I)/(a*d*e^(16*I*d*x + 16*I*c) + 8*a*d*e^(14*I*d*x + 14*I*c) + 28*a*d*e^(12*I*d*x + 12*I*c) + 56*a*d*e^(10*I*d*x + 10*I*c) + 70*a*d*e^ (8*I*d*x + 8*I*c) + 56*a*d*e^(6*I*d*x + 6*I*c) + 28*a*d*e^(4*I*d*x + 4*I*c ) + 8*a*d*e^(2*I*d*x + 2*I*c) + a*d)
\[ \int \frac {\sec ^{10}(c+d x)}{a+i a \tan (c+d x)} \, dx=- \frac {i \int \frac {\sec ^{10}{\left (c + d x \right )}}{\tan {\left (c + d x \right )} - i}\, dx}{a} \] Input:
integrate(sec(d*x+c)**10/(a+I*a*tan(d*x+c)),x)
Output:
-I*Integral(sec(c + d*x)**10/(tan(c + d*x) - I), x)/a
Time = 0.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.81 \[ \int \frac {\sec ^{10}(c+d x)}{a+i a \tan (c+d x)} \, dx=-\frac {35 i \, \tan \left (d x + c\right )^{8} - 40 \, \tan \left (d x + c\right )^{7} + 140 i \, \tan \left (d x + c\right )^{6} - 168 \, \tan \left (d x + c\right )^{5} + 210 i \, \tan \left (d x + c\right )^{4} - 280 \, \tan \left (d x + c\right )^{3} + 140 i \, \tan \left (d x + c\right )^{2} - 280 \, \tan \left (d x + c\right )}{280 \, a d} \] Input:
integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c)),x, algorithm="maxima")
Output:
-1/280*(35*I*tan(d*x + c)^8 - 40*tan(d*x + c)^7 + 140*I*tan(d*x + c)^6 - 1 68*tan(d*x + c)^5 + 210*I*tan(d*x + c)^4 - 280*tan(d*x + c)^3 + 140*I*tan( d*x + c)^2 - 280*tan(d*x + c))/(a*d)
Time = 0.17 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.81 \[ \int \frac {\sec ^{10}(c+d x)}{a+i a \tan (c+d x)} \, dx=-\frac {35 i \, \tan \left (d x + c\right )^{8} - 40 \, \tan \left (d x + c\right )^{7} + 140 i \, \tan \left (d x + c\right )^{6} - 168 \, \tan \left (d x + c\right )^{5} + 210 i \, \tan \left (d x + c\right )^{4} - 280 \, \tan \left (d x + c\right )^{3} + 140 i \, \tan \left (d x + c\right )^{2} - 280 \, \tan \left (d x + c\right )}{280 \, a d} \] Input:
integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c)),x, algorithm="giac")
Output:
-1/280*(35*I*tan(d*x + c)^8 - 40*tan(d*x + c)^7 + 140*I*tan(d*x + c)^6 - 1 68*tan(d*x + c)^5 + 210*I*tan(d*x + c)^4 - 280*tan(d*x + c)^3 + 140*I*tan( d*x + c)^2 - 280*tan(d*x + c))/(a*d)
Time = 0.60 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.86 \[ \int \frac {\sec ^{10}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {{\cos \left (c+d\,x\right )}^8\,35{}\mathrm {i}+128\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^7+64\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^5+48\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^3+40\,\sin \left (c+d\,x\right )\,\cos \left (c+d\,x\right )-35{}\mathrm {i}}{280\,a\,d\,{\cos \left (c+d\,x\right )}^8} \] Input:
int(1/(cos(c + d*x)^10*(a + a*tan(c + d*x)*1i)),x)
Output:
(40*cos(c + d*x)*sin(c + d*x) + 48*cos(c + d*x)^3*sin(c + d*x) + 64*cos(c + d*x)^5*sin(c + d*x) + 128*cos(c + d*x)^7*sin(c + d*x) + cos(c + d*x)^8*3 5i - 35i)/(280*a*d*cos(c + d*x)^8)
\[ \int \frac {\sec ^{10}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {\int \frac {\sec \left (d x +c \right )^{10}}{\tan \left (d x +c \right ) i +1}d x}{a} \] Input:
int(sec(d*x+c)^10/(a+I*a*tan(d*x+c)),x)
Output:
int(sec(c + d*x)**10/(tan(c + d*x)*i + 1),x)/a