Integrand size = 23, antiderivative size = 61 \[ \int \frac {\sec ^n(e+f x)}{\sqrt {a+a \sec (e+f x)}} \, dx=\frac {\operatorname {AppellF1}\left (\frac {1}{2},1-n,1,\frac {3}{2},1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right ) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)}} \]
Leaf count is larger than twice the leaf count of optimal. \(2964\) vs. \(2(61)=122\).
Time = 6.21 (sec) , antiderivative size = 2964, normalized size of antiderivative = 48.59 \[ \int \frac {\sec ^n(e+f x)}{\sqrt {a+a \sec (e+f x)}} \, dx=\text {Result too large to show} \]
(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 + (-1 + 2*n)/2)*(C os[(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)*AppellF 1[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*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)*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[(...
Time = 0.39 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, 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.
\(\displaystyle \int \frac {\sec ^n(e+f x)}{\sqrt {a \sec (e+f x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )^n}{\sqrt {a \csc \left (e+f x+\frac {\pi }{2}\right )+a}}dx\) |
\(\Big \downarrow \) 4315 |
\(\displaystyle \frac {\sqrt {\sec (e+f x)+1} \int \frac {\sec ^n(e+f x)}{\sqrt {\sec (e+f x)+1}}dx}{\sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sec (e+f x)+1} \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )^n}{\sqrt {\csc \left (e+f x+\frac {\pi }{2}\right )+1}}dx}{\sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 4312 |
\(\displaystyle \frac {\tan (e+f x) \int \frac {\sec ^{n-1}(e+f x)}{\sqrt {1-\sec (e+f x)} (\sec (e+f x)+1)}d(1-\sec (e+f x))}{f \sqrt {1-\sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 148 |
\(\displaystyle \frac {2 \tan (e+f x) \int \frac {\sec ^{n-1}(e+f x)}{\sec (e+f x)+1}d\sqrt {1-\sec (e+f x)}}{f \sqrt {1-\sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
\(\Big \downarrow \) 333 |
\(\displaystyle \frac {\tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1-n,1,\frac {3}{2},1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right )}{f \sqrt {a \sec (e+f x)+a}}\) |
(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]])
3.4.10.3.1 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]
\[\int \frac {\sec \left (f x +e \right )^{n}}{\sqrt {a +a \sec \left (f x +e \right )}}d x\]
\[ \int \frac {\sec ^n(e+f x)}{\sqrt {a+a \sec (e+f x)}} \, dx=\int { \frac {\sec \left (f x + e\right )^{n}}{\sqrt {a \sec \left (f x + e\right ) + a}} \,d x } \]
\[ \int \frac {\sec ^n(e+f x)}{\sqrt {a+a \sec (e+f x)}} \, dx=\int \frac {\sec ^{n}{\left (e + f x \right )}}{\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )}}\, dx \]
\[ \int \frac {\sec ^n(e+f x)}{\sqrt {a+a \sec (e+f x)}} \, dx=\int { \frac {\sec \left (f x + e\right )^{n}}{\sqrt {a \sec \left (f x + e\right ) + a}} \,d x } \]
\[ \int \frac {\sec ^n(e+f x)}{\sqrt {a+a \sec (e+f x)}} \, dx=\int { \frac {\sec \left (f x + e\right )^{n}}{\sqrt {a \sec \left (f x + e\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {\sec ^n(e+f x)}{\sqrt {a+a \sec (e+f x)}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (e+f\,x\right )}\right )}^n}{\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}} \,d x \]