Integrand size = 21, antiderivative size = 424 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx=-\frac {3 \operatorname {AppellF1}\left (-\frac {1}{2},\frac {5}{2},-n,\frac {1}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \cot (c+d x) \sqrt {1+\sec (c+d x)} (a+b \sec (c+d x))^n \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-n}}{2 \sqrt {2} d}-\frac {\operatorname {AppellF1}\left (-\frac {3}{2},\frac {5}{2},-n,-\frac {1}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \cot ^3(c+d x) (1+\sec (c+d x))^{3/2} (a+b \sec (c+d x))^n \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-n}}{6 \sqrt {2} d}+\frac {\operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^n \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-n} \tan (c+d x)}{\sqrt {2} d \sqrt {1+\sec (c+d x)}}+\frac {\operatorname {AppellF1}\left (\frac {1}{2},\frac {5}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^n \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{-n} \tan (c+d x)}{2 \sqrt {2} d \sqrt {1+\sec (c+d x)}} \] Output:
-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))*c ot(d*x+c)*(1+sec(d*x+c))^(1/2)*(a+b*sec(d*x+c))^n*2^(1/2)/d/(((a+b*sec(d*x +c))/(a+b))^n)-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*2^(1/2 )/d/(((a+b*sec(d*x+c))/(a+b))^n)+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)*2^(1/2)/d/(1+s ec(d*x+c))^(1/2)/(((a+b*sec(d*x+c))/(a+b))^n)+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)*2 ^(1/2)/d/(1+sec(d*x+c))^(1/2)/(((a+b*sec(d*x+c))/(a+b))^n)
Leaf count is larger than twice the leaf count of optimal. \(5928\) vs. \(2(424)=848\).
Time = 21.86 (sec) , antiderivative size = 5928, normalized size of antiderivative = 13.98 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx=\text {Result too large to show} \] Input:
Integrate[Csc[c + d*x]^4*(a + b*Sec[c + d*x])^n,x]
Output:
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 |
\(\displaystyle \int \csc ^4(c+d x) (a+b \sec (c+d x))^ndx\) |
Input:
Int[Csc[c + d*x]^4*(a + b*Sec[c + d*x])^n,x]
Output:
$Aborted
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]
\[\int \csc \left (d x +c \right )^{4} \left (a +b \sec \left (d x +c \right )\right )^{n}d x\]
Input:
int(csc(d*x+c)^4*(a+b*sec(d*x+c))^n,x)
Output:
int(csc(d*x+c)^4*(a+b*sec(d*x+c))^n,x)
\[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4} \,d x } \] Input:
integrate(csc(d*x+c)^4*(a+b*sec(d*x+c))^n,x, algorithm="fricas")
Output:
integral((b*sec(d*x + c) + a)^n*csc(d*x + c)^4, x)
Timed out. \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx=\text {Timed out} \] Input:
integrate(csc(d*x+c)**4*(a+b*sec(d*x+c))**n,x)
Output:
Timed out
\[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4} \,d x } \] Input:
integrate(csc(d*x+c)^4*(a+b*sec(d*x+c))^n,x, algorithm="maxima")
Output:
integrate((b*sec(d*x + c) + a)^n*csc(d*x + c)^4, x)
\[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4} \,d x } \] Input:
integrate(csc(d*x+c)^4*(a+b*sec(d*x+c))^n,x, algorithm="giac")
Output:
integrate((b*sec(d*x + c) + a)^n*csc(d*x + c)^4, x)
Timed out. \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^n}{{\sin \left (c+d\,x\right )}^4} \,d x \] Input:
int((a + b/cos(c + d*x))^n/sin(c + d*x)^4,x)
Output:
int((a + b/cos(c + d*x))^n/sin(c + d*x)^4, x)
\[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx=\int \left (\sec \left (d x +c \right ) b +a \right )^{n} \csc \left (d x +c \right )^{4}d x \] Input:
int(csc(d*x+c)^4*(a+b*sec(d*x+c))^n,x)
Output:
int((sec(c + d*x)*b + a)**n*csc(c + d*x)**4,x)