Integrand size = 21, antiderivative size = 62 \[ \int \frac {\sec ^n(e+f x)}{(1+\sec (e+f x))^{3/2}} \, dx=\frac {\operatorname {AppellF1}\left (\frac {1}{2},1-n,2,\frac {3}{2},1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right ) \tan (e+f x)}{2 f \sqrt {1+\sec (e+f x)}} \]
1/2*AppellF1(1/2,1-n,2,3/2,1-sec(f*x+e),1/2-1/2*sec(f*x+e))*tan(f*x+e)/f/( 1+sec(f*x+e))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(2990\) vs. \(2(62)=124\).
Time = 16.38 (sec) , antiderivative size = 2990, normalized size of antiderivative = 48.23 \[ \int \frac {\sec ^n(e+f x)}{(1+\sec (e+f x))^{3/2}} \, dx=\text {Result too large to show} \]
(6*AppellF1[1/2, -3/2 + 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 + (-3 + 2*n)/2)*(Cos[(e + f *x)/2]^2*Sec[e + f*x])^(3/2 + n)*Tan[(e + f*x)/2]*(-1 + Tan[(e + f*x)/2]^2 )^2)/(f*(1 + Sec[e + f*x])^(3/2)*(3*AppellF1[1/2, -3/2 + n, 1 - n, 3/2, Ta n[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (2*(-1 + n)*AppellF1[3/2, -3/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (-3 + 2*n)*Appel lF1[3/2, -1/2 + n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*T an[(e + f*x)/2]^2)*((12*AppellF1[1/2, -3/2 + 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])^(3/2 + n)*Tan[(e + f*x)/2]^2*(-1 + Tan[(e + f *x)/2]^2))/(3*AppellF1[1/2, -3/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan [(e + f*x)/2]^2] + (2*(-1 + n)*AppellF1[3/2, -3/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (-3 + 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*AppellF1[1/2, -3/2 + 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])^(3/2 + n)*(-1 + Tan[(e + f*x)/2]^2)^2)/(3*AppellF1[1/2, -3/2 + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (2*(-1 + n)*AppellF1 [3/2, -3/2 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (-3 + 2*n)*AppellF1[3/2, -1/2 + n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e...
Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {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.
\(\displaystyle \int \frac {\sec ^n(e+f x)}{(\sec (e+f x)+1)^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )^n}{\left (\csc \left (e+f x+\frac {\pi }{2}\right )+1\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4312 |
\(\displaystyle \frac {\tan (e+f x) \int \frac {\sec ^{n-1}(e+f x)}{\sqrt {1-\sec (e+f x)} (\sec (e+f x)+1)^2}d(1-\sec (e+f x))}{f \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}\) |
\(\Big \downarrow \) 148 |
\(\displaystyle \frac {2 \tan (e+f x) \int \frac {\sec ^{n-1}(e+f x)}{(\sec (e+f x)+1)^2}d\sqrt {1-\sec (e+f x)}}{f \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}\) |
\(\Big \downarrow \) 333 |
\(\displaystyle \frac {\tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1-n,2,\frac {3}{2},1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right )}{2 f \sqrt {\sec (e+f x)+1}}\) |
(AppellF1[1/2, 1 - n, 2, 3/2, 1 - Sec[e + f*x], (1 - Sec[e + f*x])/2]*Tan[ e + f*x])/(2*f*Sqrt[1 + Sec[e + f*x]])
3.3.98.3.1 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]
\[\int \frac {\sec \left (f x +e \right )^{n}}{\left (\sec \left (f x +e \right )+1\right )^{\frac {3}{2}}}d x\]
\[ \int \frac {\sec ^n(e+f x)}{(1+\sec (e+f x))^{3/2}} \, dx=\int { \frac {\sec \left (f x + e\right )^{n}}{{\left (\sec \left (f x + e\right ) + 1\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\sec ^n(e+f x)}{(1+\sec (e+f x))^{3/2}} \, dx=\int \frac {\sec ^{n}{\left (e + f x \right )}}{\left (\sec {\left (e + f x \right )} + 1\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\sec ^n(e+f x)}{(1+\sec (e+f x))^{3/2}} \, dx=\int { \frac {\sec \left (f x + e\right )^{n}}{{\left (\sec \left (f x + e\right ) + 1\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\sec ^n(e+f x)}{(1+\sec (e+f x))^{3/2}} \, dx=\int { \frac {\sec \left (f x + e\right )^{n}}{{\left (\sec \left (f x + e\right ) + 1\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^n(e+f x)}{(1+\sec (e+f x))^{3/2}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (e+f\,x\right )}\right )}^n}{{\left (\frac {1}{\cos \left (e+f\,x\right )}+1\right )}^{3/2}} \,d x \]