\(\int (a+b (c \sec (e+f x))^n)^p \tan ^5(e+f x) \, dx\) [463]

Optimal result

Integrand size = 25, antiderivative size = 229 \[ \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^5(e+f x) \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {a+b (c \sec (e+f x))^n}{a}\right ) \left (a+b (c \sec (e+f x))^n\right )^{1+p}}{a f n (1+p)}-\frac {\operatorname {Hypergeometric2F1}\left (\frac {2}{n},-p,\frac {2+n}{n},-\frac {b (c \sec (e+f x))^n}{a}\right ) \sec ^2(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (\frac {a+b (c \sec (e+f x))^n}{a}\right )^{-p}}{f}+\frac {\operatorname {Hypergeometric2F1}\left (\frac {4}{n},-p,\frac {4+n}{n},-\frac {b (c \sec (e+f x))^n}{a}\right ) \sec ^4(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (\frac {a+b (c \sec (e+f x))^n}{a}\right )^{-p}}{4 f} \] Output:

-hypergeom([1, p+1],[2+p],(a+b*(c*sec(f*x+e))^n)/a)*(a+b*(c*sec(f*x+e))^n) 
^(p+1)/a/f/n/(p+1)-hypergeom([-p, 2/n],[(2+n)/n],-b*(c*sec(f*x+e))^n/a)*se 
c(f*x+e)^2*(a+b*(c*sec(f*x+e))^n)^p/f/(((a+b*(c*sec(f*x+e))^n)/a)^p)+1/4*h 
ypergeom([-p, 4/n],[(4+n)/n],-b*(c*sec(f*x+e))^n/a)*sec(f*x+e)^4*(a+b*(c*s 
ec(f*x+e))^n)^p/f/(((a+b*(c*sec(f*x+e))^n)/a)^p)
 

Mathematica [A] (warning: unable to verify)

Time = 8.59 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.07 \[ \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^5(e+f x) \, dx=\frac {\left (a+b (c \sec (e+f x))^n\right )^p \left (1+\frac {b \left (c \sqrt {\sec ^2(e+f x)}\right )^n}{a}\right )^{-p} \left (-4 a n (1+p) \operatorname {Hypergeometric2F1}\left (\frac {2}{n},-p,\frac {2+n}{n},-\frac {b \left (c \sqrt {\sec ^2(e+f x)}\right )^n}{a}\right ) \sec ^2(e+f x)+a n (1+p) \operatorname {Hypergeometric2F1}\left (\frac {4}{n},-p,\frac {4+n}{n},-\frac {b \left (c \sqrt {\sec ^2(e+f x)}\right )^n}{a}\right ) \sec ^4(e+f x)-4 \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \left (c \sqrt {\sec ^2(e+f x)}\right )^n}{a}\right ) \left (a+b \left (c \sqrt {\sec ^2(e+f x)}\right )^n\right ) \left (1+\frac {b \left (c \sqrt {\sec ^2(e+f x)}\right )^n}{a}\right )^p\right )}{4 a f n (1+p)} \] Input:

Integrate[(a + b*(c*Sec[e + f*x])^n)^p*Tan[e + f*x]^5,x]
 

Output:

((a + b*(c*Sec[e + f*x])^n)^p*(-4*a*n*(1 + p)*Hypergeometric2F1[2/n, -p, ( 
2 + n)/n, -((b*(c*Sqrt[Sec[e + f*x]^2])^n)/a)]*Sec[e + f*x]^2 + a*n*(1 + p 
)*Hypergeometric2F1[4/n, -p, (4 + n)/n, -((b*(c*Sqrt[Sec[e + f*x]^2])^n)/a 
)]*Sec[e + f*x]^4 - 4*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*(c*Sqrt[Se 
c[e + f*x]^2])^n)/a]*(a + b*(c*Sqrt[Sec[e + f*x]^2])^n)*(1 + (b*(c*Sqrt[Se 
c[e + f*x]^2])^n)/a)^p))/(4*a*f*n*(1 + p)*(1 + (b*(c*Sqrt[Sec[e + f*x]^2]) 
^n)/a)^p)
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 4627, 7293, 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.

\(\displaystyle \int \tan ^5(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (e+f x)^5 \left (a+b (c \sec (e+f x))^n\right )^pdx\)

\(\Big \downarrow \) 4627

\(\displaystyle \frac {\int \cos (e+f x) \left (1-\sec ^2(e+f x)\right )^2 \left (b (c \sec (e+f x))^n+a\right )^pd\sec (e+f x)}{f}\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{4} \sec ^4(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (\frac {b (c \sec (e+f x))^n}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {4}{n},-p,\frac {n+4}{n},-\frac {b (c \sec (e+f x))^n}{a}\right )-\sec ^2(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (\frac {b (c \sec (e+f x))^n}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {2}{n},-p,\frac {n+2}{n},-\frac {b (c \sec (e+f x))^n}{a}\right )-\frac {\left (a+b (c \sec (e+f x))^n\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b (c \sec (e+f x))^n}{a}+1\right )}{a n (p+1)}}{f}\)

Input:

Int[(a + b*(c*Sec[e + f*x])^n)^p*Tan[e + f*x]^5,x]
 

Output:

(-((Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*(c*Sec[e + f*x])^n)/a]*(a + 
b*(c*Sec[e + f*x])^n)^(1 + p))/(a*n*(1 + p))) - (Hypergeometric2F1[2/n, -p 
, (2 + n)/n, -((b*(c*Sec[e + f*x])^n)/a)]*Sec[e + f*x]^2*(a + b*(c*Sec[e + 
 f*x])^n)^p)/(1 + (b*(c*Sec[e + f*x])^n)/a)^p + (Hypergeometric2F1[4/n, -p 
, (4 + n)/n, -((b*(c*Sec[e + f*x])^n)/a)]*Sec[e + f*x]^4*(a + b*(c*Sec[e + 
 f*x])^n)^p)/(4*(1 + (b*(c*Sec[e + f*x])^n)/a)^p))/f
 

Defintions of rubi rules used

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 4627

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + ( 
f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Si 
mp[1/f   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p/x), x] 
, x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[( 
m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4] || IGtQ[p, 0] || Integers 
Q[2*n, p])
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [F]

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

Input:

int((a+b*(c*sec(f*x+e))^n)^p*tan(f*x+e)^5,x)
 

Output:

int((a+b*(c*sec(f*x+e))^n)^p*tan(f*x+e)^5,x)
 

Fricas [F]

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

integrate((a+b*(c*sec(f*x+e))^n)^p*tan(f*x+e)^5,x, algorithm="fricas")
 

Output:

integral(((c*sec(f*x + e))^n*b + a)^p*tan(f*x + e)^5, x)
 

Sympy [F(-1)]

Timed out. \[ \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^5(e+f x) \, dx=\text {Timed out} \] Input:

integrate((a+b*(c*sec(f*x+e))**n)**p*tan(f*x+e)**5,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((a+b*(c*sec(f*x+e))^n)^p*tan(f*x+e)^5,x, algorithm="maxima")
 

Output:

integrate(((c*sec(f*x + e))^n*b + a)^p*tan(f*x + e)^5, x)
 

Giac [F]

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

integrate((a+b*(c*sec(f*x+e))^n)^p*tan(f*x+e)^5,x, algorithm="giac")
 

Output:

integrate(((c*sec(f*x + e))^n*b + a)^p*tan(f*x + e)^5, x)
 

Mupad [F(-1)]

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

int(tan(e + f*x)^5*(a + b*(c/cos(e + f*x))^n)^p,x)
 

Output:

int(tan(e + f*x)^5*(a + b*(c/cos(e + f*x))^n)^p, x)
 

Reduce [F]

\[ \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^5(e+f x) \, dx =\text {Too large to display} \] Input:

int((a+b*(c*sec(f*x+e))^n)^p*tan(f*x+e)^5,x)
 

Output:

((c**n*sec(e + f*x)**n*b + a)**p*tan(e + f*x)**4*n**2*p**2 + 2*(c**n*sec(e 
 + f*x)**n*b + a)**p*tan(e + f*x)**4*n*p - 4*(c**n*sec(e + f*x)**n*b + a)* 
*p*tan(e + f*x)**2*n*p + 8*(c**n*sec(e + f*x)**n*b + a)**p + int(((c**n*se 
c(e + f*x)**n*b + a)**p*tan(e + f*x)**5)/(c**n*sec(e + f*x)**n*b*n**2*p**2 
 + 6*c**n*sec(e + f*x)**n*b*n*p + 8*c**n*sec(e + f*x)**n*b + a*n**2*p**2 + 
 6*a*n*p + 8*a),x)*a*f*n**5*p**5 + 8*int(((c**n*sec(e + f*x)**n*b + a)**p* 
tan(e + f*x)**5)/(c**n*sec(e + f*x)**n*b*n**2*p**2 + 6*c**n*sec(e + f*x)** 
n*b*n*p + 8*c**n*sec(e + f*x)**n*b + a*n**2*p**2 + 6*a*n*p + 8*a),x)*a*f*n 
**4*p**4 + 20*int(((c**n*sec(e + f*x)**n*b + a)**p*tan(e + f*x)**5)/(c**n* 
sec(e + f*x)**n*b*n**2*p**2 + 6*c**n*sec(e + f*x)**n*b*n*p + 8*c**n*sec(e 
+ f*x)**n*b + a*n**2*p**2 + 6*a*n*p + 8*a),x)*a*f*n**3*p**3 + 16*int(((c** 
n*sec(e + f*x)**n*b + a)**p*tan(e + f*x)**5)/(c**n*sec(e + f*x)**n*b*n**2* 
p**2 + 6*c**n*sec(e + f*x)**n*b*n*p + 8*c**n*sec(e + f*x)**n*b + a*n**2*p* 
*2 + 6*a*n*p + 8*a),x)*a*f*n**2*p**2 - 4*int(((c**n*sec(e + f*x)**n*b + a) 
**p*tan(e + f*x)**3)/(c**n*sec(e + f*x)**n*b*n**2*p**2 + 6*c**n*sec(e + f* 
x)**n*b*n*p + 8*c**n*sec(e + f*x)**n*b + a*n**2*p**2 + 6*a*n*p + 8*a),x)*a 
*f*n**4*p**4 - 24*int(((c**n*sec(e + f*x)**n*b + a)**p*tan(e + f*x)**3)/(c 
**n*sec(e + f*x)**n*b*n**2*p**2 + 6*c**n*sec(e + f*x)**n*b*n*p + 8*c**n*se 
c(e + f*x)**n*b + a*n**2*p**2 + 6*a*n*p + 8*a),x)*a*f*n**3*p**3 - 32*int(( 
(c**n*sec(e + f*x)**n*b + a)**p*tan(e + f*x)**3)/(c**n*sec(e + f*x)**n*...