Integrand size = 30, antiderivative size = 205 \[ \int (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^n \, dx=\frac {i (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^n}{d (3-n)}+\frac {3 i (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^{1+n}}{a d \left (3-4 n+n^2\right )}-\frac {6 i (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^{2+n}}{a^2 d (3-n) \left (1-n^2\right )}+\frac {6 i (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^{3+n}}{a^3 d \left (9-10 n^2+n^4\right )} \]
I*(e*sec(d*x+c))^(-3-n)*(a+I*a*tan(d*x+c))^n/d/(3-n)+3*I*(e*sec(d*x+c))^(- 3-n)*(a+I*a*tan(d*x+c))^(1+n)/a/d/(n^2-4*n+3)-6*I*(e*sec(d*x+c))^(-3-n)*(a +I*a*tan(d*x+c))^(2+n)/a^2/d/(3-n)/(-n^2+1)+6*I*(e*sec(d*x+c))^(-3-n)*(a+I *a*tan(d*x+c))^(3+n)/a^3/d/(n^4-10*n^2+9)
Time = 1.89 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.58 \[ \int (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^n \, dx=\frac {(e \sec (c+d x))^{-n} \left (-3 i n \left (-9+n^2\right ) \cos (c+d x)-i n \left (-1+n^2\right ) \cos (3 (c+d x))-6 \left (-5+n^2+\left (-1+n^2\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right ) (a+i a \tan (c+d x))^n}{4 d e^3 (-3+n) (-1+n) (1+n) (3+n)} \]
(((-3*I)*n*(-9 + n^2)*Cos[c + d*x] - I*n*(-1 + n^2)*Cos[3*(c + d*x)] - 6*( -5 + n^2 + (-1 + n^2)*Cos[2*(c + d*x)])*Sin[c + d*x])*(a + I*a*Tan[c + d*x ])^n)/(4*d*e^3*(-3 + n)*(-1 + n)*(1 + n)*(3 + n)*(e*Sec[c + d*x])^n)
Time = 0.89 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3042, 3985, 3042, 3985, 3042, 3985, 3042, 3969}
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.
| \(\displaystyle \int (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n-3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n-3}dx\) |
| \(\Big \downarrow \) 3985 |
| \(\displaystyle \frac {3 \int (e \sec (c+d x))^{-n-3} (i \tan (c+d x) a+a)^{n+1}dx}{a (3-n)}+\frac {i (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n-3}}{d (3-n)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \int (e \sec (c+d x))^{-n-3} (i \tan (c+d x) a+a)^{n+1}dx}{a (3-n)}+\frac {i (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n-3}}{d (3-n)}\) |
| \(\Big \downarrow \) 3985 |
| \(\displaystyle \frac {3 \left (\frac {2 \int (e \sec (c+d x))^{-n-3} (i \tan (c+d x) a+a)^{n+2}dx}{a (1-n)}+\frac {i (a+i a \tan (c+d x))^{n+1} (e \sec (c+d x))^{-n-3}}{d (1-n)}\right )}{a (3-n)}+\frac {i (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n-3}}{d (3-n)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \left (\frac {2 \int (e \sec (c+d x))^{-n-3} (i \tan (c+d x) a+a)^{n+2}dx}{a (1-n)}+\frac {i (a+i a \tan (c+d x))^{n+1} (e \sec (c+d x))^{-n-3}}{d (1-n)}\right )}{a (3-n)}+\frac {i (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n-3}}{d (3-n)}\) |
| \(\Big \downarrow \) 3985 |
| \(\displaystyle \frac {3 \left (\frac {2 \left (-\frac {\int (e \sec (c+d x))^{-n-3} (i \tan (c+d x) a+a)^{n+3}dx}{a (n+1)}-\frac {i (a+i a \tan (c+d x))^{n+2} (e \sec (c+d x))^{-n-3}}{d (n+1)}\right )}{a (1-n)}+\frac {i (a+i a \tan (c+d x))^{n+1} (e \sec (c+d x))^{-n-3}}{d (1-n)}\right )}{a (3-n)}+\frac {i (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n-3}}{d (3-n)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \left (\frac {2 \left (-\frac {\int (e \sec (c+d x))^{-n-3} (i \tan (c+d x) a+a)^{n+3}dx}{a (n+1)}-\frac {i (a+i a \tan (c+d x))^{n+2} (e \sec (c+d x))^{-n-3}}{d (n+1)}\right )}{a (1-n)}+\frac {i (a+i a \tan (c+d x))^{n+1} (e \sec (c+d x))^{-n-3}}{d (1-n)}\right )}{a (3-n)}+\frac {i (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n-3}}{d (3-n)}\) |
| \(\Big \downarrow \) 3969 |
| \(\displaystyle \frac {i (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n-3}}{d (3-n)}+\frac {3 \left (\frac {i (a+i a \tan (c+d x))^{n+1} (e \sec (c+d x))^{-n-3}}{d (1-n)}+\frac {2 \left (\frac {i (a+i a \tan (c+d x))^{n+3} (e \sec (c+d x))^{-n-3}}{a d (n+1) (n+3)}-\frac {i (a+i a \tan (c+d x))^{n+2} (e \sec (c+d x))^{-n-3}}{d (n+1)}\right )}{a (1-n)}\right )}{a (3-n)}\) |
(I*(e*Sec[c + d*x])^(-3 - n)*(a + I*a*Tan[c + d*x])^n)/(d*(3 - n)) + (3*(( I*(e*Sec[c + d*x])^(-3 - n)*(a + I*a*Tan[c + d*x])^(1 + n))/(d*(1 - n)) + (2*(((-I)*(e*Sec[c + d*x])^(-3 - n)*(a + I*a*Tan[c + d*x])^(2 + n))/(d*(1 + n)) + (I*(e*Sec[c + d*x])^(-3 - n)*(a + I*a*Tan[c + d*x])^(3 + n))/(a*d* (1 + n)*(3 + n))))/(a*(1 - n))))/(a*(3 - n))
3.5.84.3.1 Defintions of rubi rules used
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/ (a*f*m)), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] && EqQ [Simplify[m + n], 0]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_), x_Symbol] :> Simp[a*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/ (b*f*(m + 2*n))), x] + Simp[Simplify[m + n]/(a*(m + 2*n)) Int[(d*Sec[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] && ILtQ[Simplify[m + n], 0] && NeQ[m + 2*n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 6.88 (sec) , antiderivative size = 4982, normalized size of antiderivative = 24.30
-1/8*I/(-3+n)/d/(e^n)*exp(I*(d*x+c))^n/e^3*a^n*exp(-1/2*I*(6*c-3*Pi*csgn(I *e*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))^3-csgn(I*a*exp(2*I*(d*x+c))/(exp(2 *I*(d*x+c))+1))^2*csgn(I*a)*Pi*n-2*Pi*csgn(I*exp(2*I*(d*x+c)))^2*csgn(I*ex p(I*(d*x+c)))*n+Pi*csgn(I*exp(2*I*(d*x+c)))*csgn(I*exp(I*(d*x+c)))^2*n+6*d *x-n*Pi*csgn(I*e)*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))*csgn(I*e*exp (I*(d*x+c))/(exp(2*I*(d*x+c))+1))+csgn(I/(exp(2*I*(d*x+c))+1)*exp(2*I*(d*x +c)))^3*Pi*n+csgn(I*a*exp(2*I*(d*x+c))/(exp(2*I*(d*x+c))+1))^3*Pi*n+Pi*csg n(I*exp(2*I*(d*x+c)))^3*n-csgn(I/(exp(2*I*(d*x+c))+1)*exp(2*I*(d*x+c)))^2* Pi*csgn(I/(exp(2*I*(d*x+c))+1))*n-csgn(I/(exp(2*I*(d*x+c))+1)*exp(2*I*(d*x +c)))^2*Pi*csgn(I*exp(2*I*(d*x+c)))*n-csgn(I/(exp(2*I*(d*x+c))+1)*exp(2*I* (d*x+c)))*csgn(I*a*exp(2*I*(d*x+c))/(exp(2*I*(d*x+c))+1))^2*Pi*n+3*Pi*csgn (I*e)*csgn(I*e*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))^2+3*Pi*csgn(I/(exp(2*I *(d*x+c))+1))*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))^2+3*Pi*csgn(I*ex p(I*(d*x+c)))*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))^2-n*Pi*csgn(I*ex p(I*(d*x+c))/(exp(2*I*(d*x+c))+1))^3-3*Pi*csgn(I*e)*csgn(I*exp(I*(d*x+c))/ (exp(2*I*(d*x+c))+1))*csgn(I*e*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))+3*Pi*c sgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))*csgn(I*e*exp(I*(d*x+c))/(exp(2* I*(d*x+c))+1))^2-n*Pi*csgn(I*e*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))^3+n*Pi *csgn(I*exp(I*(d*x+c)))*csgn(I*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))^2+n*Pi *csgn(I*e)*csgn(I*e*exp(I*(d*x+c))/(exp(2*I*(d*x+c))+1))^2-3*Pi*csgn(I/...
Time = 0.24 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.29 \[ \int (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^n \, dx=\frac {{\left (-i \, n^{3} - 3 i \, n^{2} + {\left (-i \, n^{3} + 3 i \, n^{2} + i \, n - 3 i\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, {\left (i \, n^{3} - i \, n^{2} - 9 i \, n + 9 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 3 \, {\left (i \, n^{3} + i \, n^{2} - 9 i \, n - 9 i\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, n + 3 i\right )} \left (\frac {2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{-n - 3} e^{\left (i \, d n x + i \, c n + n \log \left (\frac {2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right ) + n \log \left (\frac {a}{e}\right )\right )}}{d n^{4} - 10 \, d n^{2} + {\left (d n^{4} - 10 \, d n^{2} + 9 \, d\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, {\left (d n^{4} - 10 \, d n^{2} + 9 \, d\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\left (d n^{4} - 10 \, d n^{2} + 9 \, d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 9 \, d} \]
(-I*n^3 - 3*I*n^2 + (-I*n^3 + 3*I*n^2 + I*n - 3*I)*e^(6*I*d*x + 6*I*c) - 3 *(I*n^3 - I*n^2 - 9*I*n + 9*I)*e^(4*I*d*x + 4*I*c) - 3*(I*n^3 + I*n^2 - 9* I*n - 9*I)*e^(2*I*d*x + 2*I*c) + I*n + 3*I)*(2*e*e^(I*d*x + I*c)/(e^(2*I*d *x + 2*I*c) + 1))^(-n - 3)*e^(I*d*n*x + I*c*n + n*log(2*e*e^(I*d*x + I*c)/ (e^(2*I*d*x + 2*I*c) + 1)) + n*log(a/e))/(d*n^4 - 10*d*n^2 + (d*n^4 - 10*d *n^2 + 9*d)*e^(6*I*d*x + 6*I*c) + 3*(d*n^4 - 10*d*n^2 + 9*d)*e^(4*I*d*x + 4*I*c) + 3*(d*n^4 - 10*d*n^2 + 9*d)*e^(2*I*d*x + 2*I*c) + 9*d)
\[ \int (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^n \, dx=\int \left (e \sec {\left (c + d x \right )}\right )^{- n - 3} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n}\, dx \]
Time = 0.40 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.69 \[ \int (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^n \, dx=\frac {{\left (-i \, a^{n} n^{3} + 3 i \, a^{n} n^{2} + i \, a^{n} n - 3 i \, a^{n}\right )} \cos \left ({\left (d x + c\right )} {\left (n + 3\right )}\right ) - 3 \, {\left (i \, a^{n} n^{3} - i \, a^{n} n^{2} - 9 i \, a^{n} n + 9 i \, a^{n}\right )} \cos \left ({\left (d x + c\right )} {\left (n + 1\right )}\right ) - 3 \, {\left (i \, a^{n} n^{3} + i \, a^{n} n^{2} - 9 i \, a^{n} n - 9 i \, a^{n}\right )} \cos \left ({\left (d x + c\right )} {\left (n - 1\right )}\right ) + {\left (-i \, a^{n} n^{3} - 3 i \, a^{n} n^{2} + i \, a^{n} n + 3 i \, a^{n}\right )} \cos \left ({\left (d x + c\right )} {\left (n - 3\right )}\right ) + {\left (a^{n} n^{3} - 3 \, a^{n} n^{2} - a^{n} n + 3 \, a^{n}\right )} \sin \left ({\left (d x + c\right )} {\left (n + 3\right )}\right ) + 3 \, {\left (a^{n} n^{3} - a^{n} n^{2} - 9 \, a^{n} n + 9 \, a^{n}\right )} \sin \left ({\left (d x + c\right )} {\left (n + 1\right )}\right ) + 3 \, {\left (a^{n} n^{3} + a^{n} n^{2} - 9 \, a^{n} n - 9 \, a^{n}\right )} \sin \left ({\left (d x + c\right )} {\left (n - 1\right )}\right ) + {\left (a^{n} n^{3} + 3 \, a^{n} n^{2} - a^{n} n - 3 \, a^{n}\right )} \sin \left ({\left (d x + c\right )} {\left (n - 3\right )}\right )}{8 \, {\left (e^{n + 3} n^{4} - 10 \, e^{n + 3} n^{2} + 9 \, e^{n + 3}\right )} d} \]
1/8*((-I*a^n*n^3 + 3*I*a^n*n^2 + I*a^n*n - 3*I*a^n)*cos((d*x + c)*(n + 3)) - 3*(I*a^n*n^3 - I*a^n*n^2 - 9*I*a^n*n + 9*I*a^n)*cos((d*x + c)*(n + 1)) - 3*(I*a^n*n^3 + I*a^n*n^2 - 9*I*a^n*n - 9*I*a^n)*cos((d*x + c)*(n - 1)) + (-I*a^n*n^3 - 3*I*a^n*n^2 + I*a^n*n + 3*I*a^n)*cos((d*x + c)*(n - 3)) + ( a^n*n^3 - 3*a^n*n^2 - a^n*n + 3*a^n)*sin((d*x + c)*(n + 3)) + 3*(a^n*n^3 - a^n*n^2 - 9*a^n*n + 9*a^n)*sin((d*x + c)*(n + 1)) + 3*(a^n*n^3 + a^n*n^2 - 9*a^n*n - 9*a^n)*sin((d*x + c)*(n - 1)) + (a^n*n^3 + 3*a^n*n^2 - a^n*n - 3*a^n)*sin((d*x + c)*(n - 3)))/((e^(n + 3)*n^4 - 10*e^(n + 3)*n^2 + 9*e^( n + 3))*d)
\[ \int (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^n \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{-n - 3} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \]
Time = 11.06 (sec) , antiderivative size = 425, normalized size of antiderivative = 2.07 \[ \int (e \sec (c+d x))^{-3-n} (a+i a \tan (c+d x))^n \, dx=-\frac {\left (2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )\,\left (2\,{\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}^2+\sin \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}-1\right )\,\left (\frac {{\left (a-\frac {a\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}\right )}^n\,\left (-n^3-3\,n^2+n+3\right )}{d\,\left (n^4\,1{}\mathrm {i}-n^2\,10{}\mathrm {i}+9{}\mathrm {i}\right )}+\frac {{\left (a-\frac {a\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}\right )}^n\,\left (-2\,{\sin \left (3\,c+3\,d\,x\right )}^2+\sin \left (6\,c+6\,d\,x\right )\,1{}\mathrm {i}+1\right )\,\left (-n^3+3\,n^2+n-3\right )}{d\,\left (n^4\,1{}\mathrm {i}-n^2\,10{}\mathrm {i}+9{}\mathrm {i}\right )}+\frac {{\left (a-\frac {a\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}\right )}^n\,\left (-2\,{\sin \left (c+d\,x\right )}^2+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+1\right )\,\left (-3\,n^3-3\,n^2+27\,n+27\right )}{d\,\left (n^4\,1{}\mathrm {i}-n^2\,10{}\mathrm {i}+9{}\mathrm {i}\right )}+\frac {{\left (a-\frac {a\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}\right )}^n\,\left (-2\,{\sin \left (2\,c+2\,d\,x\right )}^2+\sin \left (4\,c+4\,d\,x\right )\,1{}\mathrm {i}+1\right )\,\left (-3\,n^3+3\,n^2+27\,n-27\right )}{d\,\left (n^4\,1{}\mathrm {i}-n^2\,10{}\mathrm {i}+9{}\mathrm {i}\right )}\right )}{8\,{\left (-\frac {e}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}\right )}^{n+3}\,{\left ({\sin \left (c+d\,x\right )}^2-1\right )}^2} \]
-((2*sin(c/2 + (d*x)/2)^2 - 1)*(sin(3*c + 3*d*x)*1i + 2*sin((3*c)/2 + (3*d *x)/2)^2 - 1)*(((a - (a*sin(c + d*x)*1i)/(2*sin(c/2 + (d*x)/2)^2 - 1))^n*( n - 3*n^2 - n^3 + 3))/(d*(n^4*1i - n^2*10i + 9i)) + ((a - (a*sin(c + d*x)* 1i)/(2*sin(c/2 + (d*x)/2)^2 - 1))^n*(sin(6*c + 6*d*x)*1i - 2*sin(3*c + 3*d *x)^2 + 1)*(n + 3*n^2 - n^3 - 3))/(d*(n^4*1i - n^2*10i + 9i)) + ((a - (a*s in(c + d*x)*1i)/(2*sin(c/2 + (d*x)/2)^2 - 1))^n*(sin(2*c + 2*d*x)*1i - 2*s in(c + d*x)^2 + 1)*(27*n - 3*n^2 - 3*n^3 + 27))/(d*(n^4*1i - n^2*10i + 9i) ) + ((a - (a*sin(c + d*x)*1i)/(2*sin(c/2 + (d*x)/2)^2 - 1))^n*(sin(4*c + 4 *d*x)*1i - 2*sin(2*c + 2*d*x)^2 + 1)*(27*n + 3*n^2 - 3*n^3 - 27))/(d*(n^4* 1i - n^2*10i + 9i))))/(8*(-e/(2*sin(c/2 + (d*x)/2)^2 - 1))^(n + 3)*(sin(c + d*x)^2 - 1)^2)