Integrand size = 25, antiderivative size = 162 \[ \int \sec ^4(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b (c \tan (e+f x))^n}{a}\right ) \tan (e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \left (\frac {a+b (c \tan (e+f x))^n}{a}\right )^{-p}}{f}+\frac {\operatorname {Hypergeometric2F1}\left (\frac {3}{n},-p,\frac {3+n}{n},-\frac {b (c \tan (e+f x))^n}{a}\right ) \tan ^3(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \left (\frac {a+b (c \tan (e+f x))^n}{a}\right )^{-p}}{3 f} \] 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)+1/3*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)
Time = 0.68 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.75 \[ \int \sec ^4(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\frac {\tan (e+f x) \left (3 \operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b (c \tan (e+f x))^n}{a}\right )+\operatorname {Hypergeometric2F1}\left (\frac {3}{n},-p,\frac {3+n}{n},-\frac {b (c \tan (e+f x))^n}{a}\right ) \tan ^2(e+f x)\right ) \left (a+b (c \tan (e+f x))^n\right )^p \left (1+\frac {b (c \tan (e+f x))^n}{a}\right )^{-p}}{3 f} \] Input:
Integrate[Sec[e + f*x]^4*(a + b*(c*Tan[e + f*x])^n)^p,x]
Output:
(Tan[e + f*x]*(3*Hypergeometric2F1[n^(-1), -p, 1 + n^(-1), -((b*(c*Tan[e + f*x])^n)/a)] + Hypergeometric2F1[3/n, -p, (3 + n)/n, -((b*(c*Tan[e + f*x] )^n)/a)]*Tan[e + f*x]^2)*(a + b*(c*Tan[e + f*x])^n)^p)/(3*f*(1 + (b*(c*Tan [e + f*x])^n)/a)^p)
Time = 0.35 (sec) , antiderivative size = 167, 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.
\(\displaystyle \int \sec ^4(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (e+f x)^4 \left (a+b (c \tan (e+f x))^n\right )^pdx\) |
\(\Big \downarrow \) 4158 |
\(\displaystyle \frac {\int \left (\tan ^2(e+f x) c^2+c^2\right ) \left (b (c \tan (e+f x))^n+a\right )^pd(c \tan (e+f x))}{c^3 f}\) |
\(\Big \downarrow \) 2432 |
\(\displaystyle \frac {\int \left (c^2 \left (b (c \tan (e+f x))^n+a\right )^p+c^2 \tan ^2(e+f x) \left (b (c \tan (e+f x))^n+a\right )^p\right )d(c \tan (e+f x))}{c^3 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{3} c^3 \tan ^3(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \left (\frac {b (c \tan (e+f x))^n}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{n},-p,\frac {n+3}{n},-\frac {b (c \tan (e+f x))^n}{a}\right )+c^3 \tan (e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \left (\frac {b (c \tan (e+f x))^n}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b (c \tan (e+f x))^n}{a}\right )}{c^3 f}\) |
Input:
Int[Sec[e + f*x]^4*(a + b*(c*Tan[e + f*x])^n)^p,x]
Output:
((c^3*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 + (c^3*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^3*f)
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])
\[\int \sec \left (f x +e \right )^{4} \left (a +b \left (c \tan \left (f x +e \right )\right )^{n}\right )^{p}d x\]
Input:
int(sec(f*x+e)^4*(a+b*(c*tan(f*x+e))^n)^p,x)
Output:
int(sec(f*x+e)^4*(a+b*(c*tan(f*x+e))^n)^p,x)
\[ \int \sec ^4(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\int { {\left (\left (c \tan \left (f x + e\right )\right )^{n} b + a\right )}^{p} \sec \left (f x + e\right )^{4} \,d x } \] Input:
integrate(sec(f*x+e)^4*(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)^4, x)
Timed out. \[ \int \sec ^4(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\text {Timed out} \] Input:
integrate(sec(f*x+e)**4*(a+b*(c*tan(f*x+e))**n)**p,x)
Output:
Timed out
\[ \int \sec ^4(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\int { {\left (\left (c \tan \left (f x + e\right )\right )^{n} b + a\right )}^{p} \sec \left (f x + e\right )^{4} \,d x } \] Input:
integrate(sec(f*x+e)^4*(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)^4, x)
\[ \int \sec ^4(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\int { {\left (\left (c \tan \left (f x + e\right )\right )^{n} b + a\right )}^{p} \sec \left (f x + e\right )^{4} \,d x } \] Input:
integrate(sec(f*x+e)^4*(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)^4, x)
Timed out. \[ \int \sec ^4(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\int \frac {{\left (a+b\,{\left (c\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\right )}^p}{{\cos \left (e+f\,x\right )}^4} \,d x \] Input:
int((a + b*(c*tan(e + f*x))^n)^p/cos(e + f*x)^4,x)
Output:
int((a + b*(c*tan(e + f*x))^n)^p/cos(e + f*x)^4, x)
\[ \int \sec ^4(e+f x) \left (a+b (c \tan (e+f x))^n\right )^p \, dx=\int \left (c^{n} \tan \left (f x +e \right )^{n} b +a \right )^{p} \sec \left (f x +e \right )^{4}d x \] Input:
int(sec(f*x+e)^4*(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)**4,x)