Integrand size = 25, antiderivative size = 226 \[ \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^5(e+f x) \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b (c \sec (e+f x))^n}{a}\right ) \left (a+b (c \sec (e+f x))^n\right )^{1+p}}{a f n (1+p)}-\frac {\operatorname {Hypergeometric2F1}\left (\frac {2}{n},-p,\frac {2+n}{n},-\frac {b (c \sec (e+f x))^n}{a}\right ) \sec ^2(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (1+\frac {b (c \sec (e+f x))^n}{a}\right )^{-p}}{f}+\frac {\operatorname {Hypergeometric2F1}\left (\frac {4}{n},-p,\frac {4+n}{n},-\frac {b (c \sec (e+f x))^n}{a}\right ) \sec ^4(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (1+\frac {b (c \sec (e+f x))^n}{a}\right )^{-p}}{4 f} \]
-hypergeom([1, p+1],[2+p],1+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)*sec( f*x+e)^2*(a+b*(c*sec(f*x+e))^n)^p/f/((1+b*(c*sec(f*x+e))^n/a)^p)+1/4*hyper geom([-p, 4/n],[(4+n)/n],-b*(c*sec(f*x+e))^n/a)*sec(f*x+e)^4*(a+b*(c*sec(f *x+e))^n)^p/f/((1+b*(c*sec(f*x+e))^n/a)^p)
Time = 13.76 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.08 \[ \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^5(e+f x) \, dx=\frac {\left (a+b (c \sec (e+f x))^n\right )^p \left (1+\frac {b \left (c \sqrt {\sec ^2(e+f x)}\right )^n}{a}\right )^{-p} \left (-4 a n (1+p) \operatorname {Hypergeometric2F1}\left (\frac {2}{n},-p,\frac {2+n}{n},-\frac {b \left (c \sqrt {\sec ^2(e+f x)}\right )^n}{a}\right ) \sec ^2(e+f x)+a n (1+p) \operatorname {Hypergeometric2F1}\left (\frac {4}{n},-p,\frac {4+n}{n},-\frac {b \left (c \sqrt {\sec ^2(e+f x)}\right )^n}{a}\right ) \sec ^4(e+f x)-4 \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \left (c \sqrt {\sec ^2(e+f x)}\right )^n}{a}\right ) \left (a+b \left (c \sqrt {\sec ^2(e+f x)}\right )^n\right ) \left (1+\frac {b \left (c \sqrt {\sec ^2(e+f x)}\right )^n}{a}\right )^p\right )}{4 a f n (1+p)} \]
((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)
Time = 0.74 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.98, 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 |
\(\displaystyle \frac {\frac {1}{4} \sec ^4(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (\frac {b (c \sec (e+f x))^n}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {4}{n},-p,\frac {n+4}{n},-\frac {b (c \sec (e+f x))^n}{a}\right )-\sec ^2(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (\frac {b (c \sec (e+f x))^n}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {2}{n},-p,\frac {n+2}{n},-\frac {b (c \sec (e+f x))^n}{a}\right )-\frac {\left (a+b (c \sec (e+f x))^n\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b (c \sec (e+f x))^n}{a}+1\right )}{a n (p+1)}}{f}\) |
(-((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
3.5.63.3.1 Defintions of rubi rules used
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])
\[\int \left (a +b \left (c \sec \left (f x +e \right )\right )^{n}\right )^{p} \tan \left (f x +e \right )^{5}d x\]
\[ \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^5(e+f x) \, dx=\int { {\left (\left (c \sec \left (f x + e\right )\right )^{n} b + a\right )}^{p} \tan \left (f x + e\right )^{5} \,d x } \]
Timed out. \[ \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^5(e+f x) \, dx=\text {Timed out} \]
\[ \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^5(e+f x) \, dx=\int { {\left (\left (c \sec \left (f x + e\right )\right )^{n} b + a\right )}^{p} \tan \left (f x + e\right )^{5} \,d x } \]
\[ \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^5(e+f x) \, dx=\int { {\left (\left (c \sec \left (f x + e\right )\right )^{n} b + a\right )}^{p} \tan \left (f x + e\right )^{5} \,d x } \]
Timed out. \[ \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^5(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^5\,{\left (a+b\,{\left (\frac {c}{\cos \left (e+f\,x\right )}\right )}^n\right )}^p \,d x \]