\(\int \tan (c+d x) (a+b \tan (c+d x))^n \, dx\) [710]

Optimal result

Integrand size = 19, antiderivative size = 127 \[ \int \tan (c+d x) (a+b \tan (c+d x))^n \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a-i b) d (1+n)}-\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a+i b) d (1+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.92 \[ \int \tan (c+d x) (a+b \tan (c+d x))^n \, dx=-\frac {\left ((a+i b) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right )+(a-i b) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right )\right ) (a+b \tan (c+d x))^{1+n}}{2 (a-i b) (a+i b) d (1+n)} \] Input:

Integrate[Tan[c + d*x]*(a + b*Tan[c + d*x])^n,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 4022, 3042, 4020, 25, 78}

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.

\(\displaystyle \int \tan (c+d x) (a+b \tan (c+d x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x) (a+b \tan (c+d x))^ndx\)

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 78

\(\displaystyle -\frac {(a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (a-i b)}-\frac {(a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (a+i b)}\)

Input:

Int[Tan[c + d*x]*(a + b*Tan[c + d*x])^n,x]
 

Output:

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

Defintions of rubi rules used

rule 25

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

rule 78

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b 
*c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m 
 + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] 
 &&  !IntegerQ[m] && IntegerQ[n]
 

rule 3042

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

rule 4020

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f)   Subst[Int[(a + (b/d)*x)^m/(d^2 + 
c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ 
b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
 

rule 4022

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2   Int[(a + b*Tan[e + f*x])^m*( 
1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2   Int[(a + b*Tan[e + f*x])^m 
*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c 
 - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]
 

Maple [F]

\[\int \tan \left (d x +c \right ) \left (a +b \tan \left (d x +c \right )\right )^{n}d x\]

Input:

int(tan(d*x+c)*(a+b*tan(d*x+c))^n,x)
 

Output:

int(tan(d*x+c)*(a+b*tan(d*x+c))^n,x)
 

Fricas [F]

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

integrate(tan(d*x+c)*(a+b*tan(d*x+c))^n,x, algorithm="fricas")
 

Output:

integral((b*tan(d*x + c) + a)^n*tan(d*x + c), x)
 

Sympy [F]

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

integrate(tan(d*x+c)*(a+b*tan(d*x+c))**n,x)
 

Output:

Integral((a + b*tan(c + d*x))**n*tan(c + d*x), x)
 

Maxima [F]

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

integrate(tan(d*x+c)*(a+b*tan(d*x+c))^n,x, algorithm="maxima")
 

Output:

integrate((b*tan(d*x + c) + a)^n*tan(d*x + c), x)
 

Giac [F]

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

integrate(tan(d*x+c)*(a+b*tan(d*x+c))^n,x, algorithm="giac")
 

Output:

integrate((b*tan(d*x + c) + a)^n*tan(d*x + c), x)
 

Mupad [F(-1)]

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

int(tan(c + d*x)*(a + b*tan(c + d*x))^n,x)
 

Output:

int(tan(c + d*x)*(a + b*tan(c + d*x))^n, x)
 

Reduce [F]

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

int(tan(d*x+c)*(a+b*tan(d*x+c))^n,x)
 

Output:

int((tan(c + d*x)*b + a)**n*tan(c + d*x),x)