Integrand size = 36, antiderivative size = 247 \[ \int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (3 B+2 i A n) (a+i a \tan (c+d x))^n}{3 d \sqrt {\tan (c+d x)}}-\frac {2 (A-i B) \operatorname {AppellF1}\left (\frac {1}{2},1-n,1,\frac {3}{2},-i \tan (c+d x),i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d}-\frac {2 (1-2 n) (3 i B-2 A n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},-i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{3 d} \]
-2/3*(3*B+2*I*A*n)*(a+I*a*tan(d*x+c))^n/d/tan(d*x+c)^(1/2)-2*(A-I*B)*Appel lF1(1/2,1-n,1,3/2,-I*tan(d*x+c),I*tan(d*x+c))*tan(d*x+c)^(1/2)*(a+I*a*tan( d*x+c))^n/d/((1+I*tan(d*x+c))^n)-2/3*(1-2*n)*(3*I*B-2*A*n)*hypergeom([1/2, 1-n],[3/2],-I*tan(d*x+c))*tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^n/d/((1+I*t an(d*x+c))^n)-2/3*A*(a+I*a*tan(d*x+c))^n/d/tan(d*x+c)^(3/2)
\[ \int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx \]
Time = 1.40 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.13, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.528, Rules used = {3042, 4081, 27, 3042, 4081, 27, 3042, 4084, 3042, 4047, 25, 27, 148, 27, 334, 333, 4082, 76, 74}
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.
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan (c+d x)^{5/2}}dx\) |
| \(\Big \downarrow \) 4081 |
| \(\displaystyle \frac {2 \int \frac {(i \tan (c+d x) a+a)^n (a (3 B+2 i A n)-a A (3-2 n) \tan (c+d x))}{2 \tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(i \tan (c+d x) a+a)^n (a (3 B+2 i A n)-a A (3-2 n) \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(i \tan (c+d x) a+a)^n (a (3 B+2 i A n)-a A (3-2 n) \tan (c+d x))}{\tan (c+d x)^{3/2}}dx}{3 a}-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4081 |
| \(\displaystyle \frac {\frac {2 \int \frac {(i \tan (c+d x) a+a)^n \left (a^2 \left (6 i B n-A \left (4 n^2-2 n+3\right )\right )-a^2 (1-2 n) (3 B+2 i A n) \tan (c+d x)\right )}{2 \sqrt {\tan (c+d x)}}dx}{a}-\frac {2 a (3 B+2 i A n) (a+i a \tan (c+d x))^n}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {(i \tan (c+d x) a+a)^n \left (a^2 \left (6 i B n-A \left (4 n^2-2 n+3\right )\right )-a^2 (1-2 n) (3 B+2 i A n) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {2 a (3 B+2 i A n) (a+i a \tan (c+d x))^n}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {(i \tan (c+d x) a+a)^n \left (a^2 \left (6 i B n-A \left (4 n^2-2 n+3\right )\right )-a^2 (1-2 n) (3 B+2 i A n) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {2 a (3 B+2 i A n) (a+i a \tan (c+d x))^n}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4084 |
| \(\displaystyle \frac {\frac {-3 a^2 (A-i B) \int \frac {(i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx-a (1-2 n) (-2 A n+3 i B) \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {2 a (3 B+2 i A n) (a+i a \tan (c+d x))^n}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {-3 a^2 (A-i B) \int \frac {(i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx-a (1-2 n) (-2 A n+3 i B) \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {2 a (3 B+2 i A n) (a+i a \tan (c+d x))^n}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4047 |
| \(\displaystyle \frac {\frac {-\frac {3 i a^4 (A-i B) \int -\frac {(i \tan (c+d x) a+a)^{n-1}}{a \sqrt {\tan (c+d x)} (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d}-a (1-2 n) (-2 A n+3 i B) \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {2 a (3 B+2 i A n) (a+i a \tan (c+d x))^n}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {3 i a^4 (A-i B) \int \frac {(i \tan (c+d x) a+a)^{n-1}}{a \sqrt {\tan (c+d x)} (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d}-a (1-2 n) (-2 A n+3 i B) \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {2 a (3 B+2 i A n) (a+i a \tan (c+d x))^n}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3 i a^3 (A-i B) \int \frac {(i \tan (c+d x) a+a)^{n-1}}{\sqrt {\tan (c+d x)} (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d}-a (1-2 n) (-2 A n+3 i B) \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {2 a (3 B+2 i A n) (a+i a \tan (c+d x))^n}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 148 |
| \(\displaystyle \frac {\frac {-\frac {6 a^4 (A-i B) \int \frac {\left (a-i a^3 \tan ^2(c+d x)\right )^{n-1}}{a \left (i a^2 \tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}-a (1-2 n) (-2 A n+3 i B) \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {2 a (3 B+2 i A n) (a+i a \tan (c+d x))^n}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {-\frac {6 a^3 (A-i B) \int \frac {\left (a-i a^3 \tan ^2(c+d x)\right )^{n-1}}{i a^2 \tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}-a (1-2 n) (-2 A n+3 i B) \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {2 a (3 B+2 i A n) (a+i a \tan (c+d x))^n}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 334 |
| \(\displaystyle \frac {\frac {-\frac {6 a^2 (A-i B) \left (a-i a^3 \tan ^2(c+d x)\right )^n \left (1-i a^2 \tan ^2(c+d x)\right )^{-n} \int \frac {\left (1-i a^2 \tan ^2(c+d x)\right )^{n-1}}{i a^2 \tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}-a (1-2 n) (-2 A n+3 i B) \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {2 a (3 B+2 i A n) (a+i a \tan (c+d x))^n}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 333 |
| \(\displaystyle \frac {\frac {-a (1-2 n) (-2 A n+3 i B) \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx-\frac {6 i a^3 (A-i B) \tan (c+d x) \left (a-i a^3 \tan ^2(c+d x)\right )^n \left (1-i a^2 \tan ^2(c+d x)\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},1,1-n,\frac {3}{2},-i a^2 \tan ^2(c+d x),i a^2 \tan ^2(c+d x)\right )}{d}}{a}-\frac {2 a (3 B+2 i A n) (a+i a \tan (c+d x))^n}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4082 |
| \(\displaystyle \frac {\frac {-\frac {a^3 (1-2 n) (-2 A n+3 i B) \int \frac {(i \tan (c+d x) a+a)^{n-1}}{\sqrt {\tan (c+d x)}}d\tan (c+d x)}{d}-\frac {6 i a^3 (A-i B) \tan (c+d x) \left (a-i a^3 \tan ^2(c+d x)\right )^n \left (1-i a^2 \tan ^2(c+d x)\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},1,1-n,\frac {3}{2},-i a^2 \tan ^2(c+d x),i a^2 \tan ^2(c+d x)\right )}{d}}{a}-\frac {2 a (3 B+2 i A n) (a+i a \tan (c+d x))^n}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 76 |
| \(\displaystyle \frac {\frac {-\frac {a^2 (1-2 n) (-2 A n+3 i B) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \int \frac {(i \tan (c+d x)+1)^{n-1}}{\sqrt {\tan (c+d x)}}d\tan (c+d x)}{d}-\frac {6 i a^3 (A-i B) \tan (c+d x) \left (1-i a^2 \tan ^2(c+d x)\right )^{-n} \left (a-i a^3 \tan ^2(c+d x)\right )^n \operatorname {AppellF1}\left (\frac {1}{2},1,1-n,\frac {3}{2},-i a^2 \tan ^2(c+d x),i a^2 \tan ^2(c+d x)\right )}{d}}{a}-\frac {2 a (3 B+2 i A n) (a+i a \tan (c+d x))^n}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 74 |
| \(\displaystyle \frac {\frac {-\frac {2 a^2 (1-2 n) (-2 A n+3 i B) \sqrt {\tan (c+d x)} (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},-i \tan (c+d x)\right )}{d}-\frac {6 i a^3 (A-i B) \tan (c+d x) \left (1-i a^2 \tan ^2(c+d x)\right )^{-n} \left (a-i a^3 \tan ^2(c+d x)\right )^n \operatorname {AppellF1}\left (\frac {1}{2},1,1-n,\frac {3}{2},-i a^2 \tan ^2(c+d x),i a^2 \tan ^2(c+d x)\right )}{d}}{a}-\frac {2 a (3 B+2 i A n) (a+i a \tan (c+d x))^n}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
(-2*A*(a + I*a*Tan[c + d*x])^n)/(3*d*Tan[c + d*x]^(3/2)) + ((-2*a*(3*B + ( 2*I)*A*n)*(a + I*a*Tan[c + d*x])^n)/(d*Sqrt[Tan[c + d*x]]) + ((-2*a^2*(1 - 2*n)*((3*I)*B - 2*A*n)*Hypergeometric2F1[1/2, 1 - n, 3/2, (-I)*Tan[c + d* x]]*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^n)/(d*(1 + I*Tan[c + d*x])^n ) - ((6*I)*a^3*(A - I*B)*AppellF1[1/2, 1, 1 - n, 3/2, (-I)*a^2*Tan[c + d*x ]^2, I*a^2*Tan[c + d*x]^2]*Tan[c + d*x]*(a - I*a^3*Tan[c + d*x]^2)^n)/(d*( 1 - I*a^2*Tan[c + d*x]^2)^n))/a)/(3*a)
3.3.31.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x )^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] && !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^IntPart [n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) Int[(b*x)^m*(1 + d* (x/c))^n, x], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && !Integer Q[n] && !GtQ[c, 0] && !GtQ[-d/(b*c), 0] && ((RationalQ[m] && !(EqQ[n, -2 ^(-1)] && EqQ[c^2 - d^2, 0])) || !RationalQ[n])
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))^(p_.), x_] :> With[{k = Denominator[m]}, Simp[k/b Subst[Int[x^(k*(m + 1) - 1)*(c + d*(x^k/b))^n*(e + f*(x^k/b))^p, x], x, (b*x)^(1/k)], x]] /; FreeQ[{b, c, d, e, f, n, p}, x] && FractionQ[m] && IntegerQ[p]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[ (1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && !(IntegerQ[p] || GtQ[a, 0])
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(b/f) Subst[Int[(a + x)^(m - 1)*(( c + (d/b)*x)^n/(b^2 + a*x)), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d ^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(A*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(a*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c* m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Fr eeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(B/f) Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[ a^2 + b^2, 0] && EqQ[A*b + a*B, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(A*b + a*B)/b Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x], x] - Simp[B/b Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[ e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
\[\int \frac {\left (a +i a \tan \left (d x +c \right )\right )^{n} \left (A +B \tan \left (d x +c \right )\right )}{\tan \left (d x +c \right )^{\frac {5}{2}}}d x\]
\[ \int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\tan \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
integral(((-I*A - B)*e^(6*I*d*x + 6*I*c) + (-3*I*A - B)*e^(4*I*d*x + 4*I*c ) + (-3*I*A + B)*e^(2*I*d*x + 2*I*c) - I*A + B)*(2*a*e^(2*I*d*x + 2*I*c)/( e^(2*I*d*x + 2*I*c) + 1))^n*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))/(e^(6*I*d*x + 6*I*c) - 3*e^(4*I*d*x + 4*I*c) + 3*e^(2*I*d*x + 2*I*c) - 1), x)
\[ \int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n} \left (A + B \tan {\left (c + d x \right )}\right )}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\tan \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\tan \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \]