\(\int \sec ^6(e+f x) (a+b (c \tan (e+f x))^n)^p \, dx\) [493]

Optimal result

Integrand size = 25, antiderivative size = 247 \[ \int \sec ^6(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b (c \tan (e+f x))^n}{a}\right ) \tan (e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \left (\frac {a+b (c \tan (e+f x))^n}{a}\right )^{-p}}{f}+\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {3}{n},-p,\frac {3+n}{n},-\frac {b (c \tan (e+f x))^n}{a}\right ) \tan ^3(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \left (\frac {a+b (c \tan (e+f x))^n}{a}\right )^{-p}}{3 f}+\frac {\operatorname {Hypergeometric2F1}\left (\frac {5}{n},-p,\frac {5+n}{n},-\frac {b (c \tan (e+f x))^n}{a}\right ) \tan ^5(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \left (\frac {a+b (c \tan (e+f x))^n}{a}\right )^{-p}}{5 f} \] Output:

hypergeom([-p, 1/n],[1+1/n],-b*(c*tan(f*x+e))^n/a)*tan(f*x+e)*(a+b*(c*tan( 
f*x+e))^n)^p/f/(((a+b*(c*tan(f*x+e))^n)/a)^p)+2/3*hypergeom([-p, 3/n],[(3+ 
n)/n],-b*(c*tan(f*x+e))^n/a)*tan(f*x+e)^3*(a+b*(c*tan(f*x+e))^n)^p/f/(((a+ 
b*(c*tan(f*x+e))^n)/a)^p)+1/5*hypergeom([-p, 5/n],[(5+n)/n],-b*(c*tan(f*x+ 
e))^n/a)*tan(f*x+e)^5*(a+b*(c*tan(f*x+e))^n)^p/f/(((a+b*(c*tan(f*x+e))^n)/ 
a)^p)
 

Mathematica [A] (verified)

Time = 0.93 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.67 \[ \int \sec ^6(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\frac {\tan (e+f x) \left (15 \operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b (c \tan (e+f x))^n}{a}\right )+10 \operatorname {Hypergeometric2F1}\left (\frac {3}{n},-p,\frac {3+n}{n},-\frac {b (c \tan (e+f x))^n}{a}\right ) \tan ^2(e+f x)+3 \operatorname {Hypergeometric2F1}\left (\frac {5}{n},-p,\frac {5+n}{n},-\frac {b (c \tan (e+f x))^n}{a}\right ) \tan ^4(e+f x)\right ) \left (a+b (c \tan (e+f x))^n\right )^p \left (1+\frac {b (c \tan (e+f x))^n}{a}\right )^{-p}}{15 f} \] Input:

Integrate[Sec[e + f*x]^6*(a + b*(c*Tan[e + f*x])^n)^p,x]
 

Output:

(Tan[e + f*x]*(15*Hypergeometric2F1[n^(-1), -p, 1 + n^(-1), -((b*(c*Tan[e 
+ f*x])^n)/a)] + 10*Hypergeometric2F1[3/n, -p, (3 + n)/n, -((b*(c*Tan[e + 
f*x])^n)/a)]*Tan[e + f*x]^2 + 3*Hypergeometric2F1[5/n, -p, (5 + n)/n, -((b 
*(c*Tan[e + f*x])^n)/a)]*Tan[e + f*x]^4)*(a + b*(c*Tan[e + f*x])^n)^p)/(15 
*f*(1 + (b*(c*Tan[e + f*x])^n)/a)^p)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 4158, 2432, 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.

\(\displaystyle \int \sec ^6(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sec (e+f x)^6 \left (a+b (c \tan (e+f x))^n\right )^pdx\)

\(\Big \downarrow \) 4158

\(\displaystyle \frac {\int \left (\tan ^2(e+f x) c^2+c^2\right )^2 \left (b (c \tan (e+f x))^n+a\right )^pd(c \tan (e+f x))}{c^5 f}\)

\(\Big \downarrow \) 2432

\(\displaystyle \frac {\int \left (c^4 \left (b (c \tan (e+f x))^n+a\right )^p+c^4 \tan ^4(e+f x) \left (b (c \tan (e+f x))^n+a\right )^p+2 c^4 \tan ^2(e+f x) \left (b (c \tan (e+f x))^n+a\right )^p\right )d(c \tan (e+f x))}{c^5 f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{5} c^5 \tan ^5(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \left (\frac {b (c \tan (e+f x))^n}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{n},-p,\frac {n+5}{n},-\frac {b (c \tan (e+f x))^n}{a}\right )+\frac {2}{3} c^5 \tan ^3(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \left (\frac {b (c \tan (e+f x))^n}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{n},-p,\frac {n+3}{n},-\frac {b (c \tan (e+f x))^n}{a}\right )+c^5 \tan (e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \left (\frac {b (c \tan (e+f x))^n}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b (c \tan (e+f x))^n}{a}\right )}{c^5 f}\)

Input:

Int[Sec[e + f*x]^6*(a + b*(c*Tan[e + f*x])^n)^p,x]
 

Output:

((c^5*Hypergeometric2F1[n^(-1), -p, 1 + n^(-1), -((b*(c*Tan[e + f*x])^n)/a 
)]*Tan[e + f*x]*(a + b*(c*Tan[e + f*x])^n)^p)/(1 + (b*(c*Tan[e + f*x])^n)/ 
a)^p + (2*c^5*Hypergeometric2F1[3/n, -p, (3 + n)/n, -((b*(c*Tan[e + f*x])^ 
n)/a)]*Tan[e + f*x]^3*(a + b*(c*Tan[e + f*x])^n)^p)/(3*(1 + (b*(c*Tan[e + 
f*x])^n)/a)^p) + (c^5*Hypergeometric2F1[5/n, -p, (5 + n)/n, -((b*(c*Tan[e 
+ f*x])^n)/a)]*Tan[e + f*x]^5*(a + b*(c*Tan[e + f*x])^n)^p)/(5*(1 + (b*(c* 
Tan[e + f*x])^n)/a)^p))/(c^5*f)
 

Defintions of rubi rules used

rule 2009

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

rule 2432

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[ 
Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n, p}, x] && (PolyQ[Pq, x] || Poly 
Q[Pq, x^n])
 

rule 3042

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

rule 4158

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ 
)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim 
p[ff/(c^(m - 1)*f)   Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)^n)^ 
p, x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && I 
ntegerQ[m/2] && (IntegersQ[n, p] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] 
 || EqQ[n^2, 16])
 

Maple [F]

\[\int \sec \left (f x +e \right )^{6} \left (a +b \left (c \tan \left (f x +e \right )\right )^{n}\right )^{p}d x\]

Input:

int(sec(f*x+e)^6*(a+b*(c*tan(f*x+e))^n)^p,x)
 

Output:

int(sec(f*x+e)^6*(a+b*(c*tan(f*x+e))^n)^p,x)
 

Fricas [F]

\[ \int \sec ^6(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\int { {\left (\left (c \tan \left (f x + e\right )\right )^{n} b + a\right )}^{p} \sec \left (f x + e\right )^{6} \,d x } \] Input:

integrate(sec(f*x+e)^6*(a+b*(c*tan(f*x+e))^n)^p,x, algorithm="fricas")
 

Output:

integral(((c*tan(f*x + e))^n*b + a)^p*sec(f*x + e)^6, x)
 

Sympy [F(-1)]

Timed out. \[ \int \sec ^6(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\text {Timed out} \] Input:

integrate(sec(f*x+e)**6*(a+b*(c*tan(f*x+e))**n)**p,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sec ^6(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\int { {\left (\left (c \tan \left (f x + e\right )\right )^{n} b + a\right )}^{p} \sec \left (f x + e\right )^{6} \,d x } \] Input:

integrate(sec(f*x+e)^6*(a+b*(c*tan(f*x+e))^n)^p,x, algorithm="maxima")
 

Output:

integrate(((c*tan(f*x + e))^n*b + a)^p*sec(f*x + e)^6, x)
 

Giac [F]

\[ \int \sec ^6(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\int { {\left (\left (c \tan \left (f x + e\right )\right )^{n} b + a\right )}^{p} \sec \left (f x + e\right )^{6} \,d x } \] Input:

integrate(sec(f*x+e)^6*(a+b*(c*tan(f*x+e))^n)^p,x, algorithm="giac")
 

Output:

integrate(((c*tan(f*x + e))^n*b + a)^p*sec(f*x + e)^6, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^6(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\int \frac {{\left (a+b\,{\left (c\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\right )}^p}{{\cos \left (e+f\,x\right )}^6} \,d x \] Input:

int((a + b*(c*tan(e + f*x))^n)^p/cos(e + f*x)^6,x)
 

Output:

int((a + b*(c*tan(e + f*x))^n)^p/cos(e + f*x)^6, x)
 

Reduce [F]

\[ \int \sec ^6(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\int \left (c^{n} \tan \left (f x +e \right )^{n} b +a \right )^{p} \sec \left (f x +e \right )^{6}d x \] Input:

int(sec(f*x+e)^6*(a+b*(c*tan(f*x+e))^n)^p,x)
 

Output:

int((c**n*tan(e + f*x)**n*b + a)**p*sec(e + f*x)**6,x)