Integrand size = 34, antiderivative size = 69 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{\sqrt {c-c \sec (e+f x)}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sec (e+f x))\right ) (a+a \sec (e+f x))^m \tan (e+f x)}{f (1+2 m) \sqrt {c-c \sec (e+f x)}} \]
-hypergeom([1, 1/2+m],[3/2+m],1/2+1/2*sec(f*x+e))*(a+a*sec(f*x+e))^m*tan(f *x+e)/f/(1+2*m)/(c-c*sec(f*x+e))^(1/2)
Time = 0.37 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.97 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{\sqrt {c-c \sec (e+f x)}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sec (e+f x))\right ) (a (1+\sec (e+f x)))^m \tan (e+f x)}{(f+2 f m) \sqrt {c-c \sec (e+f x)}} \]
-((Hypergeometric2F1[1, 1/2 + m, 3/2 + m, (1 + Sec[e + f*x])/2]*(a*(1 + Se c[e + f*x]))^m*Tan[e + f*x])/((f + 2*f*m)*Sqrt[c - c*Sec[e + f*x]]))
Time = 0.32 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3042, 4449, 27, 78}
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.
\(\displaystyle \int \frac {\sec (e+f x) (a \sec (e+f x)+a)^m}{\sqrt {c-c \sec (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^m}{\sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4449 |
\(\displaystyle -\frac {a c \tan (e+f x) \int \frac {(\sec (e+f x) a+a)^{m-\frac {1}{2}}}{c (1-\sec (e+f x))}d\sec (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a \tan (e+f x) \int \frac {(\sec (e+f x) a+a)^{m-\frac {1}{2}}}{1-\sec (e+f x)}d\sec (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 78 |
\(\displaystyle -\frac {\tan (e+f x) (a \sec (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (1,m+\frac {1}{2},m+\frac {3}{2},\frac {1}{2} (\sec (e+f x)+1)\right )}{f (2 m+1) \sqrt {c-c \sec (e+f x)}}\) |
-((Hypergeometric2F1[1, 1/2 + m, 3/2 + m, (1 + Sec[e + f*x])/2]*(a + a*Sec [e + f*x])^m*Tan[e + f*x])/(f*(1 + 2*m)*Sqrt[c - c*Sec[e + f*x]]))
3.2.59.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b *c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] && !IntegerQ[m] && IntegerQ[n]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(c sc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Simp[a*c*(Cot[e + f *x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]])) Subst[Int[(a + b*x)^(m - 1/2)*(c + d*x)^(n - 1/2), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]
\[\int \frac {\sec \left (f x +e \right ) \left (a +a \sec \left (f x +e \right )\right )^{m}}{\sqrt {c -c \sec \left (f x +e \right )}}d x\]
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{\sqrt {c-c \sec (e+f x)}} \, dx=\int { \frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )}{\sqrt {-c \sec \left (f x + e\right ) + c}} \,d x } \]
integral(-sqrt(-c*sec(f*x + e) + c)*(a*sec(f*x + e) + a)^m*sec(f*x + e)/(c *sec(f*x + e) - c), x)
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{\sqrt {c-c \sec (e+f x)}} \, dx=\int \frac {\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{m} \sec {\left (e + f x \right )}}{\sqrt {- c \left (\sec {\left (e + f x \right )} - 1\right )}}\, dx \]
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{\sqrt {c-c \sec (e+f x)}} \, dx=\int { \frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )}{\sqrt {-c \sec \left (f x + e\right ) + c}} \,d x } \]
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{\sqrt {c-c \sec (e+f x)}} \, dx=\int { \frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )}{\sqrt {-c \sec \left (f x + e\right ) + c}} \,d x } \]
Timed out. \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{\sqrt {c-c \sec (e+f x)}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m}{\cos \left (e+f\,x\right )\,\sqrt {c-\frac {c}{\cos \left (e+f\,x\right )}}} \,d x \]