\(\int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx\) [324]

Optimal result

Integrand size = 26, antiderivative size = 88 \[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx=\frac {\operatorname {AppellF1}(1+n,1-m,1,2+n,-i \tan (e+f x),i \tan (e+f x)) (1+i \tan (e+f x))^{-m} (d \tan (e+f x))^{1+n} (a+i a \tan (e+f x))^m}{d f (1+n)} \] Output:

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)
 

Mathematica [F]

\[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx=\int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx \] Input:

Integrate[(d*Tan[e + f*x])^n*(a + I*a*Tan[e + f*x])^m,x]
 

Output:

Integrate[(d*Tan[e + f*x])^n*(a + I*a*Tan[e + f*x])^m, x]
 

Rubi [A] (verified)

Time = 0.30 (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

\(\displaystyle \frac {(1+i \tan (e+f x))^{-m} (a+i a \tan (e+f x))^m (d \tan (e+f x))^{n+1} \operatorname {AppellF1}(n+1,1-m,1,n+2,-i \tan (e+f x),i \tan (e+f x))}{d f (n+1)}\)

Input:

Int[(d*Tan[e + f*x])^n*(a + I*a*Tan[e + f*x])^m,x]
 

Output:

(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)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 150

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])
 

rule 152

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4047

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]
 

Maple [F]

\[\int \left (d \tan \left (f x +e \right )\right )^{n} \left (a +i a \tan \left (f x +e \right )\right )^{m}d x\]

Input:

int((d*tan(f*x+e))^n*(a+I*a*tan(f*x+e))^m,x)
 

Output:

int((d*tan(f*x+e))^n*(a+I*a*tan(f*x+e))^m,x)
 

Fricas [F]

\[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx=\int { {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \] Input:

integrate((d*tan(f*x+e))^n*(a+I*a*tan(f*x+e))^m,x, algorithm="fricas")
 

Output:

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)
 

Sympy [F]

\[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx=\int \left (d \tan {\left (e + f x \right )}\right )^{n} \left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{m}\, dx \] Input:

integrate((d*tan(f*x+e))**n*(a+I*a*tan(f*x+e))**m,x)
 

Output:

Integral((d*tan(e + f*x))**n*(I*a*(tan(e + f*x) - I))**m, x)
 

Maxima [F]

\[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx=\int { {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \] Input:

integrate((d*tan(f*x+e))^n*(a+I*a*tan(f*x+e))^m,x, algorithm="maxima")
 

Output:

integrate((I*a*tan(f*x + e) + a)^m*(d*tan(f*x + e))^n, x)
 

Giac [F]

\[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx=\int { {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \] Input:

integrate((d*tan(f*x+e))^n*(a+I*a*tan(f*x+e))^m,x, algorithm="giac")
 

Output:

integrate((I*a*tan(f*x + e) + a)^m*(d*tan(f*x + e))^n, x)
 

Mupad [F(-1)]

Timed out. \[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx=\int {\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^m \,d x \] Input:

int((d*tan(e + f*x))^n*(a + a*tan(e + f*x)*1i)^m,x)
 

Output:

int((d*tan(e + f*x))^n*(a + a*tan(e + f*x)*1i)^m, x)
 

Reduce [F]

\[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx=\frac {d^{n} i \left (-\tan \left (f x +e \right )^{n} \left (\tan \left (f x +e \right ) a i +a \right )^{m}+\left (\int \frac {\tan \left (f x +e \right )^{n} \left (\tan \left (f x +e \right ) a i +a \right )^{m}}{\tan \left (f x +e \right )}d x \right ) f n +\left (\int \tan \left (f x +e \right )^{n} \left (\tan \left (f x +e \right ) a i +a \right )^{m} \tan \left (f x +e \right )d x \right ) f m +\left (\int \tan \left (f x +e \right )^{n} \left (\tan \left (f x +e \right ) a i +a \right )^{m} \tan \left (f x +e \right )d x \right ) f n \right )}{f m} \] Input:

int((d*tan(f*x+e))^n*(a+I*a*tan(f*x+e))^m,x)
                                                                                    
                                                                                    
 

Output:

(d**n*i*( - tan(e + f*x)**n*(tan(e + f*x)*a*i + a)**m + int((tan(e + f*x)* 
*n*(tan(e + f*x)*a*i + a)**m)/tan(e + f*x),x)*f*n + int(tan(e + f*x)**n*(t 
an(e + f*x)*a*i + a)**m*tan(e + f*x),x)*f*m + int(tan(e + f*x)**n*(tan(e + 
 f*x)*a*i + a)**m*tan(e + f*x),x)*f*n))/(f*m)