\(\int (a+a \sec (c+d x))^n \tan ^3(c+d x) \, dx\) [221]

Optimal result

Integrand size = 21, antiderivative size = 69 \[ \int (a+a \sec (c+d x))^n \tan ^3(c+d x) \, dx=\frac {(a+a \sec (c+d x))^{2+n}}{a^2 d (2+n)}+\frac {\operatorname {Hypergeometric2F1}(1,2+n,3+n,1+\sec (c+d x)) (a+a \sec (c+d x))^{2+n}}{a^2 d (2+n)} \] Output:

(a+a*sec(d*x+c))^(2+n)/a^2/d/(2+n)+hypergeom([1, 2+n],[3+n],1+sec(d*x+c))* 
(a+a*sec(d*x+c))^(2+n)/a^2/d/(2+n)
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.71 \[ \int (a+a \sec (c+d x))^n \tan ^3(c+d x) \, dx=\frac {(1+\operatorname {Hypergeometric2F1}(1,2+n,3+n,1+\sec (c+d x))) (1+\sec (c+d x))^2 (a (1+\sec (c+d x)))^n}{d (2+n)} \] Input:

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

Output:

((1 + Hypergeometric2F1[1, 2 + n, 3 + n, 1 + Sec[c + d*x]])*(1 + Sec[c + d 
*x])^2*(a*(1 + Sec[c + d*x]))^n)/(d*(2 + n))
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 25, 4368, 25, 27, 90, 75}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int -\cot \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^ndx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^ndx\)

\(\Big \downarrow \) 4368

\(\displaystyle \frac {\int -a \cos (c+d x) (1-\sec (c+d x)) (\sec (c+d x) a+a)^{n+1}d\sec (c+d x)}{a^2 d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int a \cos (c+d x) (1-\sec (c+d x)) (\sec (c+d x) a+a)^{n+1}d\sec (c+d x)}{a^2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \cos (c+d x) (1-\sec (c+d x)) (\sec (c+d x) a+a)^{n+1}d\sec (c+d x)}{a d}\)

\(\Big \downarrow \) 90

\(\displaystyle -\frac {\int \cos (c+d x) (\sec (c+d x) a+a)^{n+1}d\sec (c+d x)-\frac {(a \sec (c+d x)+a)^{n+2}}{a (n+2)}}{a d}\)

\(\Big \downarrow \) 75

\(\displaystyle -\frac {-\frac {(a \sec (c+d x)+a)^{n+2} \operatorname {Hypergeometric2F1}(1,n+2,n+3,\sec (c+d x)+1)}{a (n+2)}-\frac {(a \sec (c+d x)+a)^{n+2}}{a (n+2)}}{a d}\)

Input:

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

Output:

-((-((a + a*Sec[c + d*x])^(2 + n)/(a*(2 + n))) - (Hypergeometric2F1[1, 2 + 
 n, 3 + n, 1 + Sec[c + d*x]]*(a + a*Sec[c + d*x])^(2 + n))/(a*(2 + n)))/(a 
*d))
 

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 75

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x 
)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + 
 d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (IntegerQ[m] 
 || GtQ[-d/(b*c), 0])
 

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 3042

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

rule 4368

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n 
_), x_Symbol] :> Simp[-(d*b^(m - 1))^(-1)   Subst[Int[(-a + b*x)^((m - 1)/2 
)*((a + b*x)^((m - 1)/2 + n)/x), x], x, Csc[c + d*x]], x] /; FreeQ[{a, b, c 
, d, n}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[n]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

\[ \int (a+a \sec (c+d x))^n \tan ^3(c+d x) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n} \tan ^{3}{\left (c + d x \right )}\, dx \] Input:

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

Output:

Integral((a*(sec(c + d*x) + 1))**n*tan(c + d*x)**3, x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (a+a \sec (c+d x))^n \tan ^3(c+d x) \, dx=\frac {\left (\sec \left (d x +c \right ) a +a \right )^{n} \tan \left (d x +c \right )^{2} n -2 \left (\sec \left (d x +c \right ) a +a \right )^{n}+\left (\int \frac {\left (\sec \left (d x +c \right ) a +a \right )^{n} \tan \left (d x +c \right )^{3}}{\sec \left (d x +c \right ) n +2 \sec \left (d x +c \right )+n +2}d x \right ) d \,n^{3}+2 \left (\int \frac {\left (\sec \left (d x +c \right ) a +a \right )^{n} \tan \left (d x +c \right )^{3}}{\sec \left (d x +c \right ) n +2 \sec \left (d x +c \right )+n +2}d x \right ) d \,n^{2}-2 \left (\int \frac {\left (\sec \left (d x +c \right ) a +a \right )^{n} \tan \left (d x +c \right )}{\sec \left (d x +c \right ) n +2 \sec \left (d x +c \right )+n +2}d x \right ) d \,n^{2}-4 \left (\int \frac {\left (\sec \left (d x +c \right ) a +a \right )^{n} \tan \left (d x +c \right )}{\sec \left (d x +c \right ) n +2 \sec \left (d x +c \right )+n +2}d x \right ) d n}{d n \left (n +2\right )} \] Input:

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

Output:

((sec(c + d*x)*a + a)**n*tan(c + d*x)**2*n - 2*(sec(c + d*x)*a + a)**n + i 
nt(((sec(c + d*x)*a + a)**n*tan(c + d*x)**3)/(sec(c + d*x)*n + 2*sec(c + d 
*x) + n + 2),x)*d*n**3 + 2*int(((sec(c + d*x)*a + a)**n*tan(c + d*x)**3)/( 
sec(c + d*x)*n + 2*sec(c + d*x) + n + 2),x)*d*n**2 - 2*int(((sec(c + d*x)* 
a + a)**n*tan(c + d*x))/(sec(c + d*x)*n + 2*sec(c + d*x) + n + 2),x)*d*n** 
2 - 4*int(((sec(c + d*x)*a + a)**n*tan(c + d*x))/(sec(c + d*x)*n + 2*sec(c 
 + d*x) + n + 2),x)*d*n)/(d*n*(n + 2))