\(\int \csc ^3(c+d x) (a+b \sec (c+d x))^n \, dx\) [272]

Optimal result

Integrand size = 21, antiderivative size = 180 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^n \, dx=\frac {(a-b (1+n)) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \sec (c+d x)}{a-b}\right ) (a+b \sec (c+d x))^{1+n}}{4 (a-b)^2 d (1+n)}-\frac {(a+b+b n) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \sec (c+d x)}{a+b}\right ) (a+b \sec (c+d x))^{1+n}}{4 (a+b)^2 d (1+n)}+\frac {\cot ^2(c+d x) (b-a \sec (c+d x)) (a+b \sec (c+d x))^{1+n}}{2 \left (a^2-b^2\right ) d} \] Output:

1/4*(a-b*(1+n))*hypergeom([1, 1+n],[2+n],(a+b*sec(d*x+c))/(a-b))*(a+b*sec( 
d*x+c))^(1+n)/(a-b)^2/d/(1+n)-1/4*(b*n+a+b)*hypergeom([1, 1+n],[2+n],(a+b* 
sec(d*x+c))/(a+b))*(a+b*sec(d*x+c))^(1+n)/(a+b)^2/d/(1+n)+1/2*cot(d*x+c)^2 
*(b-a*sec(d*x+c))*(a+b*sec(d*x+c))^(1+n)/(a^2-b^2)/d
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1520\) vs. \(2(180)=360\).

Time = 16.55 (sec) , antiderivative size = 1520, normalized size of antiderivative = 8.44 \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^n \, dx =\text {Too large to display} \] Input:

Integrate[Csc[c + d*x]^3*(a + b*Sec[c + d*x])^n,x]
 

Output:

((a + b)*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^n*(((-a + b)*Cos[c + d*x]*Sec[( 
c + d*x)/2]^2)/b)^n*(a + b*Sec[c + d*x])^n*((1 - Tan[(c + d*x)/2]^2)^(-1)) 
^n*(1 - Tan[(c + d*x)/2]^4)^n*(b + (a - a*Tan[(c + d*x)/2]^2)/(1 + Tan[(c 
+ d*x)/2]^2))^n*(-((Hypergeometric2F1[n, 1 + n, 2 + n, (a + b - a*Tan[(c + 
 d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(2*b)]*(((a - b)*(-1 + Tan[(c + d*x)/2] 
^2))/b)^n*(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)^(1 + n))/( 
(a - b)*(1 + n)*(2 - 2*Tan[(c + d*x)/2]^2)^n)) + (Hypergeometric2F1[1, -n, 
 1 - n, ((a + b)*(-1 + Tan[(c + d*x)/2]^2))/(a*(-1 + Tan[(c + d*x)/2]^2) - 
 b*(1 + Tan[(c + d*x)/2]^2))]*(a + b + (-a + b)*Tan[(c + d*x)/2]^2)^n)/(n* 
(1 - Tan[(c + d*x)/2]^2)^n) - (Cot[(c + d*x)/2]^2*(1 - Tan[(c + d*x)/2]^2) 
^(1 - n)*(a + b + (-a + b)*Tan[(c + d*x)/2]^2)^(1 + n))/(a + b) - (2^n*Hyp 
ergeometric2F1[-n, -n, 1 - n, ((a - b)*(-1 + Tan[(c + d*x)/2]^2))/(2*b)]*( 
a + b + (-a + b)*Tan[(c + d*x)/2]^2)^n)/(n*(1 - Tan[(c + d*x)/2]^2)^n*((a 
+ b + (-a + b)*Tan[(c + d*x)/2]^2)/b)^n) - ((Hypergeometric2F1[n, 1 + n, 2 
 + n, (a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(2*b)]*(((a - 
b)*(-1 + Tan[(c + d*x)/2]^2))/b)^n*(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[( 
c + d*x)/2]^2)^(1 + n))/((1 + n)*(2 - 2*Tan[(c + d*x)/2]^2)^n) + ((a - b)* 
Hypergeometric2F1[1, -n, 1 - n, ((a + b)*(-1 + Tan[(c + d*x)/2]^2))/(a*(-1 
 + Tan[(c + d*x)/2]^2) - b*(1 + Tan[(c + d*x)/2]^2))]*(a + b + (-a + b)*Ta 
n[(c + d*x)/2]^2)^n)/(n*(1 - Tan[(c + d*x)/2]^2)^n) - (2^n*(a - b)*Hype...
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.24, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 4362, 198, 2009}

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.

\(\displaystyle \int \csc ^3(c+d x) (a+b \sec (c+d x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^n}{\cos \left (c+d x-\frac {\pi }{2}\right )^3}dx\)

\(\Big \downarrow \) 4362

\(\displaystyle -\frac {\int \frac {\sec ^2(c+d x) (a+b \sec (c+d x))^n}{(1-\sec (c+d x))^2 (\sec (c+d x)+1)^2}d(-\sec (c+d x))}{d}\)

\(\Big \downarrow \) 198

\(\displaystyle -\frac {\int \left (\frac {(a+b \sec (c+d x))^n}{2 \left (\sec ^2(c+d x)-1\right )}+\frac {(a+b \sec (c+d x))^n}{4 (-\sec (c+d x)-1)^2}+\frac {(a+b \sec (c+d x))^n}{4 (1-\sec (c+d x))^2}\right )d(-\sec (c+d x))}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {-\frac {(a+b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \sec (c+d x)}{a-b}\right )}{4 (n+1) (a-b)}+\frac {(a+b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \sec (c+d x)}{a+b}\right )}{4 (n+1) (a+b)}-\frac {b (a+b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {a+b \sec (c+d x)}{a-b}\right )}{4 (n+1) (a-b)^2}-\frac {b (a+b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {a+b \sec (c+d x)}{a+b}\right )}{4 (n+1) (a+b)^2}}{d}\)

Input:

Int[Csc[c + d*x]^3*(a + b*Sec[c + d*x])^n,x]
 

Output:

-((-1/4*(Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Sec[c + d*x])/(a - b)]* 
(a + b*Sec[c + d*x])^(1 + n))/((a - b)*(1 + n)) + (Hypergeometric2F1[1, 1 
+ n, 2 + n, (a + b*Sec[c + d*x])/(a + b)]*(a + b*Sec[c + d*x])^(1 + n))/(4 
*(a + b)*(1 + n)) - (b*Hypergeometric2F1[2, 1 + n, 2 + n, (a + b*Sec[c + d 
*x])/(a - b)]*(a + b*Sec[c + d*x])^(1 + n))/(4*(a - b)^2*(1 + n)) - (b*Hyp 
ergeometric2F1[2, 1 + n, 2 + n, (a + b*Sec[c + d*x])/(a + b)]*(a + b*Sec[c 
 + d*x])^(1 + n))/(4*(a + b)^2*(1 + n)))/d)
 

Defintions of rubi rules used

rule 198

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_))^(q_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c 
 + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, 
 m, n}, x] && IntegersQ[p, q]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4362

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m 
_), x_Symbol] :> Simp[-f^(-1)   Subst[Int[(-1 + x)^((p - 1)/2)*(1 + x)^((p 
- 1)/2)*((a + b*x)^m/x^(p + 1)), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, e 
, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
 

Maple [F]

\[\int \csc \left (d x +c \right )^{3} \left (a +b \sec \left (d x +c \right )\right )^{n}d x\]

Input:

int(csc(d*x+c)^3*(a+b*sec(d*x+c))^n,x)
 

Output:

int(csc(d*x+c)^3*(a+b*sec(d*x+c))^n,x)
 

Fricas [F]

\[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^n \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{3} \,d x } \] Input:

integrate(csc(d*x+c)^3*(a+b*sec(d*x+c))^n,x, algorithm="fricas")
 

Output:

integral((b*sec(d*x + c) + a)^n*csc(d*x + c)^3, x)
 

Sympy [F(-1)]

Timed out. \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^n \, dx=\text {Timed out} \] Input:

integrate(csc(d*x+c)**3*(a+b*sec(d*x+c))**n,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^n \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{3} \,d x } \] Input:

integrate(csc(d*x+c)^3*(a+b*sec(d*x+c))^n,x, algorithm="maxima")
 

Output:

integrate((b*sec(d*x + c) + a)^n*csc(d*x + c)^3, x)
 

Giac [F]

\[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^n \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{3} \,d x } \] Input:

integrate(csc(d*x+c)^3*(a+b*sec(d*x+c))^n,x, algorithm="giac")
 

Output:

integrate((b*sec(d*x + c) + a)^n*csc(d*x + c)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^n \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^n}{{\sin \left (c+d\,x\right )}^3} \,d x \] Input:

int((a + b/cos(c + d*x))^n/sin(c + d*x)^3,x)
 

Output:

int((a + b/cos(c + d*x))^n/sin(c + d*x)^3, x)
 

Reduce [F]

\[ \int \csc ^3(c+d x) (a+b \sec (c+d x))^n \, dx=\int \left (\sec \left (d x +c \right ) b +a \right )^{n} \csc \left (d x +c \right )^{3}d x \] Input:

int(csc(d*x+c)^3*(a+b*sec(d*x+c))^n,x)
 

Output:

int((sec(c + d*x)*b + a)**n*csc(c + d*x)**3,x)