Integrand size = 23, antiderivative size = 73 \[ \int \frac {(d \sec (e+f x))^n}{(1+\sec (e+f x))^{3/2}} \, dx=-\frac {\operatorname {AppellF1}\left (n,\frac {1}{2},2,1+n,\sec (e+f x),-\sec (e+f x)\right ) (d \sec (e+f x))^n \tan (e+f x)}{f n \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}} \]
-AppellF1(n,2,1/2,1+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)/(1+sec(f*x+e))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(3003\) vs. \(2(73)=146\).
Time = 6.18 (sec) , antiderivative size = 3003, normalized size of antiderivative = 41.14 \[ \int \frac {(d \sec (e+f x))^n}{(1+\sec (e+f x))^{3/2}} \, dx=\text {Result too large to show} \]
(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 - n + (-3 + 2*n)/2)*(d*Sec[ e + f*x])^n*(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, 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)*((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, T an[(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)*Appe llF1[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*Ap pellF1[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, -Ta n[(e + f*x)/2]^2] + (-3 + 2*n)*AppellF1[3/2, -1/2 + n, 1 - n, 5/2, Tan[...
Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {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.
\(\displaystyle \int \frac {(d \sec (e+f x))^n}{(\sec (e+f x)+1)^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^n}{\left (\csc \left (e+f x+\frac {\pi }{2}\right )+1\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4314 |
\(\displaystyle -\frac {d \tan (e+f x) \int \frac {(d \sec (e+f x))^{n-1}}{\sqrt {1-\sec (e+f x)} (\sec (e+f x)+1)^2}d\sec (e+f x)}{f \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle -\frac {\tan (e+f x) \operatorname {AppellF1}\left (n,\frac {1}{2},2,n+1,\sec (e+f x),-\sec (e+f x)\right ) (d \sec (e+f x))^n}{f n \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}\) |
-((AppellF1[n, 1/2, 2, 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[1 + Sec[e + f*x]]))
3.4.6.3.1 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]
\[\int \frac {\left (d \sec \left (f x +e \right )\right )^{n}}{\left (\sec \left (f x +e \right )+1\right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(d \sec (e+f x))^n}{(1+\sec (e+f x))^{3/2}} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{n}}{{\left (\sec \left (f x + e\right ) + 1\right )}^{\frac {3}{2}}} \,d x } \]
integral((d*sec(f*x + e))^n*sqrt(sec(f*x + e) + 1)/(sec(f*x + e)^2 + 2*sec (f*x + e) + 1), x)
\[ \int \frac {(d \sec (e+f x))^n}{(1+\sec (e+f x))^{3/2}} \, dx=\int \frac {\left (d \sec {\left (e + f x \right )}\right )^{n}}{\left (\sec {\left (e + f x \right )} + 1\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(d \sec (e+f x))^n}{(1+\sec (e+f x))^{3/2}} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{n}}{{\left (\sec \left (f x + e\right ) + 1\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(d \sec (e+f x))^n}{(1+\sec (e+f x))^{3/2}} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{n}}{{\left (\sec \left (f x + e\right ) + 1\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(d \sec (e+f x))^n}{(1+\sec (e+f x))^{3/2}} \, dx=\int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^n}{{\left (\frac {1}{\cos \left (e+f\,x\right )}+1\right )}^{3/2}} \,d x \]