Integrand size = 25, antiderivative size = 244 \[ \int \sec ^6(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b (c \tan (e+f x))^n}{a}\right ) \tan (e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \left (1+\frac {b (c \tan (e+f x))^n}{a}\right )^{-p}}{f}+\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {3}{n},-p,\frac {3+n}{n},-\frac {b (c \tan (e+f x))^n}{a}\right ) \tan ^3(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \left (1+\frac {b (c \tan (e+f x))^n}{a}\right )^{-p}}{3 f}+\frac {\operatorname {Hypergeometric2F1}\left (\frac {5}{n},-p,\frac {5+n}{n},-\frac {b (c \tan (e+f x))^n}{a}\right ) \tan ^5(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \left (1+\frac {b (c \tan (e+f x))^n}{a}\right )^{-p}}{5 f} \]
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/((1+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/((1+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/((1+b*(c*tan(f*x+e))^n/a)^p)
Time = 1.58 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.68 \[ \int \sec ^6(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\frac {\tan (e+f x) \left (15 \operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b (c \tan (e+f x))^n}{a}\right )+10 \operatorname {Hypergeometric2F1}\left (\frac {3}{n},-p,\frac {3+n}{n},-\frac {b (c \tan (e+f x))^n}{a}\right ) \tan ^2(e+f x)+3 \operatorname {Hypergeometric2F1}\left (\frac {5}{n},-p,\frac {5+n}{n},-\frac {b (c \tan (e+f x))^n}{a}\right ) \tan ^4(e+f x)\right ) \left (a+b (c \tan (e+f x))^n\right )^p \left (1+\frac {b (c \tan (e+f x))^n}{a}\right )^{-p}}{15 f} \]
(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)
Time = 0.43 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.03, 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 |
\(\displaystyle \frac {\frac {1}{5} c^5 \tan ^5(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \left (\frac {b (c \tan (e+f x))^n}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{n},-p,\frac {n+5}{n},-\frac {b (c \tan (e+f x))^n}{a}\right )+\frac {2}{3} c^5 \tan ^3(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \left (\frac {b (c \tan (e+f x))^n}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{n},-p,\frac {n+3}{n},-\frac {b (c \tan (e+f x))^n}{a}\right )+c^5 \tan (e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \left (\frac {b (c \tan (e+f x))^n}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b (c \tan (e+f x))^n}{a}\right )}{c^5 f}\) |
((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)
3.5.91.3.1 Defintions of rubi rules used
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])
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])
\[\int \sec \left (f x +e \right )^{6} \left (a +b \left (c \tan \left (f x +e \right )\right )^{n}\right )^{p}d x\]
\[ \int \sec ^6(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\int { {\left (\left (c \tan \left (f x + e\right )\right )^{n} b + a\right )}^{p} \sec \left (f x + e\right )^{6} \,d x } \]
Timed out. \[ \int \sec ^6(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\text {Timed out} \]
\[ \int \sec ^6(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\int { {\left (\left (c \tan \left (f x + e\right )\right )^{n} b + a\right )}^{p} \sec \left (f x + e\right )^{6} \,d x } \]
Exception generated. \[ \int \sec ^6(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{1,[0,1,4,0,0]%%%}+%%%{2,[0,1,2,2,0]%%%}+%%%{1,[0,1,0,4,0]% %%} / %%%
Timed out. \[ \int \sec ^6(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\int \frac {{\left (a+b\,{\left (c\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\right )}^p}{{\cos \left (e+f\,x\right )}^6} \,d x \]