Integrand size = 26, antiderivative size = 58 \[ \int \frac {(-\sec (e+f x))^n}{\sqrt {a-a \sec (e+f x)}} \, dx=\frac {\operatorname {AppellF1}\left (\frac {1}{2},1-n,1,\frac {3}{2},1+\sec (e+f x),\frac {1}{2} (1+\sec (e+f x))\right ) \tan (e+f x)}{f \sqrt {a-a \sec (e+f x)}} \]
\[ \int \frac {(-\sec (e+f x))^n}{\sqrt {a-a \sec (e+f x)}} \, dx=\int \frac {(-\sec (e+f x))^n}{\sqrt {a-a \sec (e+f x)}} \, dx \]
Time = 0.42 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3042, 4315, 3042, 4312, 148, 333}
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.
\(\displaystyle \int \frac {(-\sec (e+f x))^n}{\sqrt {a-a \sec (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (-\csc \left (e+f x+\frac {\pi }{2}\right )\right )^n}{\sqrt {a-a \csc \left (e+f x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4315 |
\(\displaystyle \frac {\sqrt {1-\sec (e+f x)} \int \frac {(-\sec (e+f x))^n}{\sqrt {1-\sec (e+f x)}}dx}{\sqrt {a-a \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {1-\sec (e+f x)} \int \frac {\left (-\csc \left (e+f x+\frac {\pi }{2}\right )\right )^n}{\sqrt {1-\csc \left (e+f x+\frac {\pi }{2}\right )}}dx}{\sqrt {a-a \sec (e+f x)}}\) |
\(\Big \downarrow \) 4312 |
\(\displaystyle \frac {\tan (e+f x) \int \frac {(-\sec (e+f x))^{n-1}}{(1-\sec (e+f x)) \sqrt {\sec (e+f x)+1}}d(\sec (e+f x)+1)}{f \sqrt {\sec (e+f x)+1} \sqrt {a-a \sec (e+f x)}}\) |
\(\Big \downarrow \) 148 |
\(\displaystyle \frac {2 \tan (e+f x) \int \frac {(-\sec (e+f x))^{n-1}}{1-\sec (e+f x)}d\sqrt {\sec (e+f x)+1}}{f \sqrt {\sec (e+f x)+1} \sqrt {a-a \sec (e+f x)}}\) |
\(\Big \downarrow \) 333 |
\(\displaystyle \frac {\tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1-n,1,\frac {3}{2},\sec (e+f x)+1,\frac {1}{2} (\sec (e+f x)+1)\right )}{f \sqrt {a-a \sec (e+f x)}}\) |
(AppellF1[1/2, 1 - n, 1, 3/2, 1 + Sec[e + f*x], (1 + Sec[e + f*x])/2]*Tan[ e + f*x])/(f*Sqrt[a - a*Sec[e + f*x]])
3.4.23.3.1 Defintions of rubi rules used
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))^(p_.), x_] :> With[{k = Denominator[m]}, Simp[k/b Subst[Int[x^(k*(m + 1) - 1)*(c + d*(x^k/b))^n*(e + f*(x^k/b))^p, x], x, (b*x)^(1/k)], x]] /; FreeQ[{b, c, d, e, f, n, p}, x] && FractionQ[m] && IntegerQ[p]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-(a*(d/b))^n)*(Cot[e + f*x]/(a^(n - 2)*f*Sqrt [a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]])) Subst[Int[(a - x)^(n - 1) *((2*a - x)^(m - 1/2)/Sqrt[x]), x], x, a - b*Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m] && GtQ[a, 0] & & !IntegerQ[n] && GtQ[a*(d/b), 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[a^IntPart[m]*((a + b*Csc[e + f*x])^FracPart[m ]/(1 + (b/a)*Csc[e + f*x])^FracPart[m]) Int[(1 + (b/a)*Csc[e + f*x])^m*(d *Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^ 2, 0] && !IntegerQ[m] && !GtQ[a, 0]
\[\int \frac {\left (-\sec \left (f x +e \right )\right )^{n}}{\sqrt {a -a \sec \left (f x +e \right )}}d x\]
\[ \int \frac {(-\sec (e+f x))^n}{\sqrt {a-a \sec (e+f x)}} \, dx=\int { \frac {\left (-\sec \left (f x + e\right )\right )^{n}}{\sqrt {-a \sec \left (f x + e\right ) + a}} \,d x } \]
\[ \int \frac {(-\sec (e+f x))^n}{\sqrt {a-a \sec (e+f x)}} \, dx=\int \frac {\left (- \sec {\left (e + f x \right )}\right )^{n}}{\sqrt {- a \left (\sec {\left (e + f x \right )} - 1\right )}}\, dx \]
\[ \int \frac {(-\sec (e+f x))^n}{\sqrt {a-a \sec (e+f x)}} \, dx=\int { \frac {\left (-\sec \left (f x + e\right )\right )^{n}}{\sqrt {-a \sec \left (f x + e\right ) + a}} \,d x } \]
\[ \int \frac {(-\sec (e+f x))^n}{\sqrt {a-a \sec (e+f x)}} \, dx=\int { \frac {\left (-\sec \left (f x + e\right )\right )^{n}}{\sqrt {-a \sec \left (f x + e\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {(-\sec (e+f x))^n}{\sqrt {a-a \sec (e+f x)}} \, dx=\int \frac {{\left (-\frac {1}{\cos \left (e+f\,x\right )}\right )}^n}{\sqrt {a-\frac {a}{\cos \left (e+f\,x\right )}}} \,d x \]