Integrand size = 21, antiderivative size = 424 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx=-\frac {3 \operatorname {AppellF1}\left (-\frac {1}{2},\frac {5}{2},-n,\frac {1}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \cot (c+d x) \sqrt {1+\sec (c+d x)} (a+b \sec (c+d x))^n \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-n}}{2 \sqrt {2} d}-\frac {\operatorname {AppellF1}\left (-\frac {3}{2},\frac {5}{2},-n,-\frac {1}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \cot ^3(c+d x) (1+\sec (c+d x))^{3/2} (a+b \sec (c+d x))^n \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-n}}{6 \sqrt {2} d}+\frac {\operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^n \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-n} \tan (c+d x)}{\sqrt {2} d \sqrt {1+\sec (c+d x)}}+\frac {\operatorname {AppellF1}\left (\frac {1}{2},\frac {5}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^n \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-n} \tan (c+d x)}{2 \sqrt {2} d \sqrt {1+\sec (c+d x)}} \]
-1/12*AppellF1(-3/2,-n,5/2,-1/2,b*(1-sec(d*x+c))/(a+b),1/2-1/2*sec(d*x+c)) *cot(d*x+c)^3*(1+sec(d*x+c))^(3/2)*(a+b*sec(d*x+c))^n/d/(((a+b*sec(d*x+c)) /(a+b))^n)*2^(1/2)-3/4*AppellF1(-1/2,-n,5/2,1/2,b*(1-sec(d*x+c))/(a+b),1/2 -1/2*sec(d*x+c))*cot(d*x+c)*(a+b*sec(d*x+c))^n*(1+sec(d*x+c))^(1/2)/d/(((a +b*sec(d*x+c))/(a+b))^n)*2^(1/2)+1/2*AppellF1(1/2,-n,3/2,3/2,b*(1-sec(d*x+ c))/(a+b),1/2-1/2*sec(d*x+c))*(a+b*sec(d*x+c))^n*tan(d*x+c)/d/(((a+b*sec(d *x+c))/(a+b))^n)*2^(1/2)/(1+sec(d*x+c))^(1/2)+1/4*AppellF1(1/2,-n,5/2,3/2, b*(1-sec(d*x+c))/(a+b),1/2-1/2*sec(d*x+c))*(a+b*sec(d*x+c))^n*tan(d*x+c)/d /(((a+b*sec(d*x+c))/(a+b))^n)*2^(1/2)/(1+sec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(5928\) vs. \(2(424)=848\).
Time = 23.44 (sec) , antiderivative size = 5928, normalized size of antiderivative = 13.98 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx=\text {Result too large to show} \]
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.
\(\displaystyle \int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^n}{\cos \left (c+d x-\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 4365 |
\(\displaystyle \int \csc ^4(c+d x) (a+b \sec (c+d x))^ndx\) |
3.3.76.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Unintegrable[(g*Cos[e + f*x])^p*(a + b*Csc[e + f* x])^m, x] /; FreeQ[{a, b, e, f, g, m, p}, x]
\[\int \csc \left (d x +c \right )^{4} \left (a +b \sec \left (d x +c \right )\right )^{n}d x\]
\[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4} \,d x } \]
Timed out. \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx=\text {Timed out} \]
\[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4} \,d x } \]
\[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4} \,d x } \]
Timed out. \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^n}{{\sin \left (c+d\,x\right )}^4} \,d x \]