Integrand size = 25, antiderivative size = 75 \[ \int \frac {(d \sec (e+f x))^n}{\sqrt {a+a \sec (e+f x)}} \, dx=-\frac {\operatorname {AppellF1}\left (n,\frac {1}{2},1,1+n,\sec (e+f x),-\sec (e+f x)\right ) (d \sec (e+f x))^n \tan (e+f x)}{f n \sqrt {1-\sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \]
-AppellF1(n,1,1/2,1+n,-sec(f*x+e),sec(f*x+e))*(d*sec(f*x+e))^n*tan(f*x+e)/ f/n/(1-sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(2977\) vs. \(2(75)=150\).
Time = 6.23 (sec) , antiderivative size = 2977, normalized size of antiderivative = 39.69 \[ \int \frac {(d \sec (e+f x))^n}{\sqrt {a+a \sec (e+f x)}} \, dx=\text {Result too large to show} \]
(3*Sqrt[2]*AppellF1[1/2, -1/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Sec[(e + f*x)/2]^2)^n*Sec[e + f*x]^(-1/2 - n + (-1 + 2*n)/2 )*(d*Sec[e + f*x])^n*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^n*Sqrt[1 + Sec[e + f*x]]*Tan[(e + f*x)/2])/(f*Sqrt[a*(1 + Sec[e + f*x])]*(3*AppellF1[1/2, -1/ 2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (2*(-1 + n)* AppellF1[3/2, -1/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^ 2] + (-1 + 2*n)*AppellF1[3/2, 1/2 + n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Ta n[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)*((3*AppellF1[1/2, -1/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[e + f*x]*(Sec[(e + f*x)/ 2]^2)^(1 + n)*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^n*Sqrt[1 + Sec[e + f*x]])/ (Sqrt[2]*(3*AppellF1[1/2, -1/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[( e + f*x)/2]^2] + (2*(-1 + n)*AppellF1[3/2, -1/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (-1 + 2*n)*AppellF1[3/2, 1/2 + n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)) - (3* Sqrt[2]*AppellF1[1/2, -1/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Sec[(e + f*x)/2]^2)^n*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^n*Sqrt [1 + Sec[e + f*x]]*Sin[e + f*x]*Tan[(e + f*x)/2])/(3*AppellF1[1/2, -1/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (2*(-1 + n)*Appe llF1[3/2, -1/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (-1 + 2*n)*AppellF1[3/2, 1/2 + n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan...
Time = 0.37 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4315, 3042, 4314, 150}
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.
\(\displaystyle \int \frac {(d \sec (e+f x))^n}{\sqrt {a \sec (e+f x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^n}{\sqrt {a \csc \left (e+f x+\frac {\pi }{2}\right )+a}}dx\) |
\(\Big \downarrow \) 4315 |
\(\displaystyle \frac {\sqrt {\sec (e+f x)+1} \int \frac {(d \sec (e+f x))^n}{\sqrt {\sec (e+f x)+1}}dx}{\sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sec (e+f x)+1} \int \frac {\left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^n}{\sqrt {\csc \left (e+f x+\frac {\pi }{2}\right )+1}}dx}{\sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 4314 |
\(\displaystyle -\frac {d \tan (e+f x) \int \frac {(d \sec (e+f x))^{n-1}}{\sqrt {1-\sec (e+f x)} (\sec (e+f x)+1)}d\sec (e+f x)}{f \sqrt {1-\sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle -\frac {\tan (e+f x) \operatorname {AppellF1}\left (n,\frac {1}{2},1,n+1,\sec (e+f x),-\sec (e+f x)\right ) (d \sec (e+f x))^n}{f n \sqrt {1-\sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
-((AppellF1[n, 1/2, 1, 1 + n, Sec[e + f*x], -Sec[e + f*x]]*(d*Sec[e + f*x] )^n*Tan[e + f*x])/(f*n*Sqrt[1 - Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]))
3.4.18.3.1 Defintions of rubi rules used
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 , (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[a^2*d*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x ]]*Sqrt[a - b*Csc[e + f*x]])) Subst[Int[(d*x)^(n - 1)*((a + b*x)^(m - 1/2 )/Sqrt[a - b*x]), x], x, 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]
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]
\[\int \frac {\left (d \sec \left (f x +e \right )\right )^{n}}{\sqrt {a +a \sec \left (f x +e \right )}}d x\]
\[ \int \frac {(d \sec (e+f x))^n}{\sqrt {a+a \sec (e+f x)}} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{n}}{\sqrt {a \sec \left (f x + e\right ) + a}} \,d x } \]
\[ \int \frac {(d \sec (e+f x))^n}{\sqrt {a+a \sec (e+f x)}} \, dx=\int \frac {\left (d \sec {\left (e + f x \right )}\right )^{n}}{\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )}}\, dx \]
\[ \int \frac {(d \sec (e+f x))^n}{\sqrt {a+a \sec (e+f x)}} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{n}}{\sqrt {a \sec \left (f x + e\right ) + a}} \,d x } \]
\[ \int \frac {(d \sec (e+f x))^n}{\sqrt {a+a \sec (e+f x)}} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{n}}{\sqrt {a \sec \left (f x + e\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {(d \sec (e+f x))^n}{\sqrt {a+a \sec (e+f x)}} \, dx=\int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^n}{\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}} \,d x \]