\(\int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx\) [276]

Optimal result

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)
 

Mathematica [B] (warning: unable to verify)

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
 

Rubi [F]

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
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4365

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]
 

Maple [F]

\[\int \csc \left (d x +c \right )^{4} \left (a +b \sec \left (d x +c \right )\right )^{n}d x\]

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)
 

Fricas [F]

\[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4} \,d x } \] 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)
 

Sympy [F(-1)]

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
 

Maxima [F]

\[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4} \,d x } \] 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)
 

Giac [F]

\[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4} \,d x } \] 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)
 

Mupad [F(-1)]

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)
 

Reduce [F]

\[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^n \, dx=\int \left (\sec \left (d x +c \right ) b +a \right )^{n} \csc \left (d x +c \right )^{4}d x \] 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)