\(\int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx\) [220]

Optimal result

Integrand size = 21, antiderivative size = 97 \[ \int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx=-\frac {3 (a+a \sec (c+d x))^{3+n}}{a^3 d (3+n)}-\frac {\operatorname {Hypergeometric2F1}(1,3+n,4+n,1+\sec (c+d x)) (a+a \sec (c+d x))^{3+n}}{a^3 d (3+n)}+\frac {(a+a \sec (c+d x))^{4+n}}{a^4 d (4+n)} \] Output:

-3*(a+a*sec(d*x+c))^(3+n)/a^3/d/(3+n)-hypergeom([1, 3+n],[4+n],1+sec(d*x+c 
))*(a+a*sec(d*x+c))^(3+n)/a^3/d/(3+n)+(a+a*sec(d*x+c))^(4+n)/a^4/d/(4+n)
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74 \[ \int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx=\frac {(1+\sec (c+d x))^3 (a (1+\sec (c+d x)))^n (-9-2 n-(4+n) \operatorname {Hypergeometric2F1}(1,3+n,4+n,1+\sec (c+d x))+(3+n) \sec (c+d x))}{d (3+n) (4+n)} \] Input:

Integrate[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^5,x]
 

Output:

((1 + Sec[c + d*x])^3*(a*(1 + Sec[c + d*x]))^n*(-9 - 2*n - (4 + n)*Hyperge 
ometric2F1[1, 3 + n, 4 + n, 1 + Sec[c + d*x]] + (3 + n)*Sec[c + d*x]))/(d* 
(3 + n)*(4 + n))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 25, 4368, 27, 99, 2009}

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.

\(\displaystyle \int \tan ^5(c+d x) (a \sec (c+d x)+a)^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\cot \left (c+d x+\frac {\pi }{2}\right )^5 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^ndx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^5 \left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^ndx\)

\(\Big \downarrow \) 4368

\(\displaystyle \frac {\int a^2 \cos (c+d x) (1-\sec (c+d x))^2 (\sec (c+d x) a+a)^{n+2}d\sec (c+d x)}{a^4 d}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 99

\(\displaystyle \frac {\int \left (\cos (c+d x) (\sec (c+d x) a+a)^{n+2}-3 (\sec (c+d x) a+a)^{n+2}+\frac {(\sec (c+d x) a+a)^{n+3}}{a}\right )d\sec (c+d x)}{a^2 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {(a \sec (c+d x)+a)^{n+4}}{a^2 (n+4)}-\frac {(a \sec (c+d x)+a)^{n+3} \operatorname {Hypergeometric2F1}(1,n+3,n+4,\sec (c+d x)+1)}{a (n+3)}-\frac {3 (a \sec (c+d x)+a)^{n+3}}{a (n+3)}}{a^2 d}\)

Input:

Int[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^5,x]
 

Output:

((-3*(a + a*Sec[c + d*x])^(3 + n))/(a*(3 + n)) - (Hypergeometric2F1[1, 3 + 
 n, 4 + n, 1 + Sec[c + d*x]]*(a + a*Sec[c + d*x])^(3 + n))/(a*(3 + n)) + ( 
a + a*Sec[c + d*x])^(4 + n)/(a^2*(4 + n)))/(a^2*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 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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]

\[\int \left (a +a \sec \left (d x +c \right )\right )^{n} \tan \left (d x +c \right )^{5}d x\]

Input:

int((a+a*sec(d*x+c))^n*tan(d*x+c)^5,x)
 

Output:

int((a+a*sec(d*x+c))^n*tan(d*x+c)^5,x)
 

Fricas [F]

\[ \int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5} \,d x } \] Input:

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^5,x, algorithm="fricas")
 

Output:

integral((a*sec(d*x + c) + a)^n*tan(d*x + c)^5, x)
 

Sympy [F]

\[ \int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n} \tan ^{5}{\left (c + d x \right )}\, dx \] Input:

integrate((a+a*sec(d*x+c))**n*tan(d*x+c)**5,x)
 

Output:

Integral((a*(sec(c + d*x) + 1))**n*tan(c + d*x)**5, x)
 

Maxima [F]

\[ \int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5} \,d x } \] Input:

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^5,x, algorithm="maxima")
 

Output:

integrate((a*sec(d*x + c) + a)^n*tan(d*x + c)^5, x)
 

Giac [F]

\[ \int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5} \,d x } \] Input:

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^5,x, algorithm="giac")
 

Output:

integrate((a*sec(d*x + c) + a)^n*tan(d*x + c)^5, x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^5\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \] Input:

int(tan(c + d*x)^5*(a + a/cos(c + d*x))^n,x)
 

Output:

int(tan(c + d*x)^5*(a + a/cos(c + d*x))^n, x)
 

Reduce [F]

\[ \int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx =\text {Too large to display} \] Input:

int((a+a*sec(d*x+c))^n*tan(d*x+c)^5,x)
                                                                                    
                                                                                    
 

Output:

((sec(c + d*x)*a + a)**n*tan(c + d*x)**4*n**2 + 2*(sec(c + d*x)*a + a)**n* 
tan(c + d*x)**4*n - 4*(sec(c + d*x)*a + a)**n*tan(c + d*x)**2*n + 8*(sec(c 
 + d*x)*a + a)**n + int(((sec(c + d*x)*a + a)**n*tan(c + d*x)**5)/(sec(c + 
 d*x)*n**2 + 6*sec(c + d*x)*n + 8*sec(c + d*x) + n**2 + 6*n + 8),x)*d*n**5 
 + 8*int(((sec(c + d*x)*a + a)**n*tan(c + d*x)**5)/(sec(c + d*x)*n**2 + 6* 
sec(c + d*x)*n + 8*sec(c + d*x) + n**2 + 6*n + 8),x)*d*n**4 + 20*int(((sec 
(c + d*x)*a + a)**n*tan(c + d*x)**5)/(sec(c + d*x)*n**2 + 6*sec(c + d*x)*n 
 + 8*sec(c + d*x) + n**2 + 6*n + 8),x)*d*n**3 + 16*int(((sec(c + d*x)*a + 
a)**n*tan(c + d*x)**5)/(sec(c + d*x)*n**2 + 6*sec(c + d*x)*n + 8*sec(c + d 
*x) + n**2 + 6*n + 8),x)*d*n**2 - 4*int(((sec(c + d*x)*a + a)**n*tan(c + d 
*x)**3)/(sec(c + d*x)*n**2 + 6*sec(c + d*x)*n + 8*sec(c + d*x) + n**2 + 6* 
n + 8),x)*d*n**4 - 24*int(((sec(c + d*x)*a + a)**n*tan(c + d*x)**3)/(sec(c 
 + d*x)*n**2 + 6*sec(c + d*x)*n + 8*sec(c + d*x) + n**2 + 6*n + 8),x)*d*n* 
*3 - 32*int(((sec(c + d*x)*a + a)**n*tan(c + d*x)**3)/(sec(c + d*x)*n**2 + 
 6*sec(c + d*x)*n + 8*sec(c + d*x) + n**2 + 6*n + 8),x)*d*n**2 + 8*int(((s 
ec(c + d*x)*a + a)**n*tan(c + d*x))/(sec(c + d*x)*n**2 + 6*sec(c + d*x)*n 
+ 8*sec(c + d*x) + n**2 + 6*n + 8),x)*d*n**3 + 48*int(((sec(c + d*x)*a + a 
)**n*tan(c + d*x))/(sec(c + d*x)*n**2 + 6*sec(c + d*x)*n + 8*sec(c + d*x) 
+ n**2 + 6*n + 8),x)*d*n**2 + 64*int(((sec(c + d*x)*a + a)**n*tan(c + d*x) 
)/(sec(c + d*x)*n**2 + 6*sec(c + d*x)*n + 8*sec(c + d*x) + n**2 + 6*n +...