\(\int \tan ^3(c+d x) (a+b \tan (c+d x))^n \, dx\) [708]

Optimal result

Integrand size = 21, antiderivative size = 192 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^n \, dx=-\frac {a (a+b \tan (c+d x))^{1+n}}{b^2 d (1+n) (2+n)}+\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a-i b) d (1+n)}+\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a+i b) d (1+n)}+\frac {\tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b d (2+n)} \] Output:

-a*(a+b*tan(d*x+c))^(1+n)/b^2/d/(1+n)/(2+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)+1/2*hyperg 
eom([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)+tan(d*x+c)*(a+b*tan(d*x+c))^(1+n)/b/d/(2+n)
 

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.70 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {(a+b \tan (c+d x))^{1+n} \left (\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right )}{(a-i b) (1+n)}+\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right )}{(a+i b) (1+n)}+\frac {2 \left (-\frac {a}{1+n}+b \tan (c+d x)\right )}{b^2 (2+n)}\right )}{2 d} \] Input:

Integrate[Tan[c + d*x]^3*(a + b*Tan[c + d*x])^n,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4049, 25, 3042, 4113, 27, 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.

\(\displaystyle \int \tan ^3(c+d x) (a+b \tan (c+d x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x)^3 (a+b \tan (c+d x))^ndx\)

\(\Big \downarrow \) 4049

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4113

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 78

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

Input:

Int[Tan[c + d*x]^3*(a + b*Tan[c + d*x])^n,x]
 

Output:

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

rule 4049

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c 
+ d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) 
 Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n 
- 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ 
e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], 
x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 
, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I 
ntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) 
)
 

rule 4113

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + 
 b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si 
mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && 
NeQ[A*b^2 - a*b*B + a^2*C, 0] &&  !LeQ[m, -1]
 

Maple [F]

\[\int \tan \left (d x +c \right )^{3} \left (a +b \tan \left (d x +c \right )\right )^{n}d x\]

Input:

int(tan(d*x+c)^3*(a+b*tan(d*x+c))^n,x)
 

Output:

int(tan(d*x+c)^3*(a+b*tan(d*x+c))^n,x)
 

Fricas [F]

\[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{3} \,d x } \] Input:

integrate(tan(d*x+c)^3*(a+b*tan(d*x+c))^n,x, algorithm="fricas")
 

Output:

integral((b*tan(d*x + c) + a)^n*tan(d*x + c)^3, x)
 

Sympy [F]

\[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{n} \tan ^{3}{\left (c + d x \right )}\, dx \] Input:

integrate(tan(d*x+c)**3*(a+b*tan(d*x+c))**n,x)
 

Output:

Integral((a + b*tan(c + d*x))**n*tan(c + d*x)**3, x)
 

Maxima [F]

\[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{3} \,d x } \] Input:

integrate(tan(d*x+c)^3*(a+b*tan(d*x+c))^n,x, algorithm="maxima")
 

Output:

integrate((b*tan(d*x + c) + a)^n*tan(d*x + c)^3, x)
                                                                                    
                                                                                    
 

Giac [F]

\[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{3} \,d x } \] Input:

integrate(tan(d*x+c)^3*(a+b*tan(d*x+c))^n,x, algorithm="giac")
 

Output:

integrate((b*tan(d*x + c) + a)^n*tan(d*x + c)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^3\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n \,d x \] Input:

int(tan(c + d*x)^3*(a + b*tan(c + d*x))^n,x)
 

Output:

int(tan(c + d*x)^3*(a + b*tan(c + d*x))^n, x)
 

Reduce [F]

\[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\int \left (a +\tan \left (d x +c \right ) b \right )^{n} \tan \left (d x +c \right )^{3}d x \] Input:

int(tan(d*x+c)^3*(a+b*tan(d*x+c))^n,x)
 

Output:

int((tan(c + d*x)*b + a)**n*tan(c + d*x)**3,x)