Integrand size = 26, antiderivative size = 88 \[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx=\frac {\operatorname {AppellF1}(1+n,1-m,1,2+n,-i \tan (e+f x),i \tan (e+f x)) (1+i \tan (e+f x))^{-m} (d \tan (e+f x))^{1+n} (a+i a \tan (e+f x))^m}{d f (1+n)} \]
AppellF1(1+n,1-m,1,2+n,-I*tan(f*x+e),I*tan(f*x+e))*(d*tan(f*x+e))^(1+n)*(a +I*a*tan(f*x+e))^m/d/f/(1+n)/((1+I*tan(f*x+e))^m)
\[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx=\int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx \]
Time = 0.29 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3042, 4047, 25, 27, 152, 150}
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.
\(\displaystyle \int (a+i a \tan (e+f x))^m (d \tan (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+i a \tan (e+f x))^m (d \tan (e+f x))^ndx\) |
\(\Big \downarrow \) 4047 |
\(\displaystyle \frac {i a^2 \int -\frac {(d \tan (e+f x))^n (i \tan (e+f x) a+a)^{m-1}}{a (a-i a \tan (e+f x))}d(i a \tan (e+f x))}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {i a^2 \int \frac {(d \tan (e+f x))^n (i \tan (e+f x) a+a)^{m-1}}{a (a-i a \tan (e+f x))}d(i a \tan (e+f x))}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {i a \int \frac {(d \tan (e+f x))^n (i \tan (e+f x) a+a)^{m-1}}{a-i a \tan (e+f x)}d(i a \tan (e+f x))}{f}\) |
\(\Big \downarrow \) 152 |
\(\displaystyle -\frac {i (1+i \tan (e+f x))^{-m} (a+i a \tan (e+f x))^m \int \frac {(i \tan (e+f x)+1)^{m-1} (d \tan (e+f x))^n}{a-i a \tan (e+f x)}d(i a \tan (e+f x))}{f}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {(1+i \tan (e+f x))^{-m} (a+i a \tan (e+f x))^m (d \tan (e+f x))^{n+1} \operatorname {AppellF1}(n+1,1-m,1,n+2,-i \tan (e+f x),i \tan (e+f x))}{d f (n+1)}\) |
(AppellF1[1 + n, 1 - m, 1, 2 + n, (-I)*Tan[e + f*x], I*Tan[e + f*x]]*(d*Ta n[e + f*x])^(1 + n)*(a + I*a*Tan[e + f*x])^m)/(d*f*(1 + n)*(1 + I*Tan[e + f*x])^m)
3.4.24.3.1 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_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 , (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !GtQ[c, 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]
\[\int \left (d \tan \left (f x +e \right )\right )^{n} \left (a +i a \tan \left (f x +e \right )\right )^{m}d x\]
\[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx=\int { {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \]
integral((2*a*e^(2*I*f*x + 2*I*e)/(e^(2*I*f*x + 2*I*e) + 1))^m*((-I*d*e^(2 *I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))^n, x)
\[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx=\int \left (d \tan {\left (e + f x \right )}\right )^{n} \left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{m}\, dx \]
\[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx=\int { {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \]
\[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx=\int { {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \]
Timed out. \[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx=\int {\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^m \,d x \]