\(\int \tan ^m(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\) [218]

Optimal result

Integrand size = 34, antiderivative size = 167 \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\frac {(A-i B) \operatorname {AppellF1}(1+m,1-n,1,2+m,-i \tan (c+d x),i \tan (c+d x)) (1+i \tan (c+d x))^{-n} \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^n}{d (1+m)}+\frac {i B \operatorname {Hypergeometric2F1}(1+m,1-n,2+m,-i \tan (c+d x)) (1+i \tan (c+d x))^{-n} \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^n}{d (1+m)} \] Output:

(A-I*B)*AppellF1(1+m,1-n,1,2+m,-I*tan(d*x+c),I*tan(d*x+c))*tan(d*x+c)^(1+m 
)*(a+I*a*tan(d*x+c))^n/d/(1+m)/((1+I*tan(d*x+c))^n)+I*B*hypergeom([1+m, 1- 
n],[2+m],-I*tan(d*x+c))*tan(d*x+c)^(1+m)*(a+I*a*tan(d*x+c))^n/d/(1+m)/((1+ 
I*tan(d*x+c))^n)
 

Mathematica [F]

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

Integrate[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]),x]
 

Output:

Integrate[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]), x]
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.324, Rules used = {3042, 4084, 3042, 4047, 25, 27, 152, 150, 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.

\(\displaystyle \int \tan ^m(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x)^m (a+i a \tan (c+d x))^n (A+B \tan (c+d x))dx\)

\(\Big \downarrow \) 4084

\(\displaystyle (A-i B) \int \tan ^m(c+d x) (i \tan (c+d x) a+a)^ndx+\frac {i B \int \tan ^m(c+d x) (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^ndx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle (A-i B) \int \tan (c+d x)^m (i \tan (c+d x) a+a)^ndx+\frac {i B \int \tan (c+d x)^m (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^ndx}{a}\)

\(\Big \downarrow \) 4047

\(\displaystyle \frac {i a^2 (A-i B) \int -\frac {\tan ^m(c+d x) (i \tan (c+d x) a+a)^{n-1}}{a (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d}+\frac {i B \int \tan (c+d x)^m (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^ndx}{a}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {i B \int \tan (c+d x)^m (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^ndx}{a}-\frac {i a^2 (A-i B) \int \frac {\tan ^m(c+d x) (i \tan (c+d x) a+a)^{n-1}}{a (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {i B \int \tan (c+d x)^m (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^ndx}{a}-\frac {i a (A-i B) \int \frac {\tan ^m(c+d x) (i \tan (c+d x) a+a)^{n-1}}{a-i a \tan (c+d x)}d(i a \tan (c+d x))}{d}\)

\(\Big \downarrow \) 152

\(\displaystyle \frac {i B \int \tan (c+d x)^m (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^ndx}{a}-\frac {i (A-i B) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \int \frac {(i \tan (c+d x)+1)^{n-1} \tan ^m(c+d x)}{a-i a \tan (c+d x)}d(i a \tan (c+d x))}{d}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {i B \int \tan (c+d x)^m (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^ndx}{a}+\frac {(A-i B) \tan ^{m+1}(c+d x) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \operatorname {AppellF1}(m+1,1-n,1,m+2,-i \tan (c+d x),i \tan (c+d x))}{d (m+1)}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {i a B \int \tan ^m(c+d x) (i \tan (c+d x) a+a)^{n-1}d\tan (c+d x)}{d}+\frac {(A-i B) \tan ^{m+1}(c+d x) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \operatorname {AppellF1}(m+1,1-n,1,m+2,-i \tan (c+d x),i \tan (c+d x))}{d (m+1)}\)

\(\Big \downarrow \) 76

\(\displaystyle \frac {i B (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \int (i \tan (c+d x)+1)^{n-1} \tan ^m(c+d x)d\tan (c+d x)}{d}+\frac {(A-i B) \tan ^{m+1}(c+d x) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \operatorname {AppellF1}(m+1,1-n,1,m+2,-i \tan (c+d x),i \tan (c+d x))}{d (m+1)}\)

\(\Big \downarrow \) 74

\(\displaystyle \frac {(A-i B) \tan ^{m+1}(c+d x) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \operatorname {AppellF1}(m+1,1-n,1,m+2,-i \tan (c+d x),i \tan (c+d x))}{d (m+1)}+\frac {i B \tan ^{m+1}(c+d x) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \operatorname {Hypergeometric2F1}(m+1,1-n,m+2,-i \tan (c+d x))}{d (m+1)}\)

Input:

Int[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]),x]
 

Output:

((A - I*B)*AppellF1[1 + m, 1 - n, 1, 2 + m, (-I)*Tan[c + d*x], I*Tan[c + d 
*x]]*Tan[c + d*x]^(1 + m)*(a + I*a*Tan[c + d*x])^n)/(d*(1 + m)*(1 + I*Tan[ 
c + d*x])^n) + (I*B*Hypergeometric2F1[1 + m, 1 - n, 2 + m, (-I)*Tan[c + d* 
x]]*Tan[c + d*x]^(1 + m)*(a + I*a*Tan[c + d*x])^n)/(d*(1 + m)*(1 + I*Tan[c 
 + d*x])^n)
 

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 74

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])))
 

rule 76

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])
 

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]
 

rule 4082

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]
 

rule 4084

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]
 

Maple [F]

\[\int \tan \left (d x +c \right )^{m} \left (a +i a \tan \left (d x +c \right )\right )^{n} \left (A +B \tan \left (d x +c \right )\right )d x\]

Input:

int(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)
 

Output:

int(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)
 

Fricas [F]

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

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm= 
"fricas")
 

Output:

integral(((A - I*B)*e^(2*I*d*x + 2*I*c) + A + I*B)*(2*a*e^(2*I*d*x + 2*I*c 
)/(e^(2*I*d*x + 2*I*c) + 1))^n*((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 
 2*I*c) + 1))^m/(e^(2*I*d*x + 2*I*c) + 1), x)
 

Sympy [F]

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

integrate(tan(d*x+c)**m*(a+I*a*tan(d*x+c))**n*(A+B*tan(d*x+c)),x)
 

Output:

Integral((I*a*(tan(c + d*x) - I))**n*(A + B*tan(c + d*x))*tan(c + d*x)**m, 
 x)
 

Maxima [F]

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

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm= 
"maxima")
 

Output:

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^n*tan(d*x + c)^m, x)
 

Giac [F]

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

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm= 
"giac")
 

Output:

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^n*tan(d*x + c)^m, x)
 

Mupad [F(-1)]

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

int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^n,x)
                                                                                    
                                                                                    
 

Output:

int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^n, x)
 

Reduce [F]

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

int(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)
 

Output:

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