3.2.50 \(\int \csc ^3(c+d x) (a+a \sec (c+d x))^n \, dx\) [150]

3.2.50.1 Optimal result

Integrand size = 21, antiderivative size = 112 \[ \int \csc ^3(c+d x) (a+a \sec (c+d x))^n \, dx=-\frac {a (2-n) (a+a \sec (c+d x))^{-1+n}}{4 d (1-n)}+\frac {a (a+a \sec (c+d x))^{-1+n}}{2 d (1-\sec (c+d x))}-\frac {(2+n) \operatorname {Hypergeometric2F1}\left (1,n,1+n,\frac {1}{2} (1+\sec (c+d x))\right ) (a+a \sec (c+d x))^n}{8 d n} \]

output

-1/4*a*(2-n)*(a+a*sec(d*x+c))^(-1+n)/d/(1-n)+1/2*a*(a+a*sec(d*x+c))^(-1+n) 
/d/(1-sec(d*x+c))-1/8*(2+n)*hypergeom([1, n],[1+n],1/2+1/2*sec(d*x+c))*(a+ 
a*sec(d*x+c))^n/d/n
 

3.2.50.2 Mathematica [A] (verified)

Time = 1.07 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.60 \[ \int \csc ^3(c+d x) (a+a \sec (c+d x))^n \, dx=\frac {\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x))^{-n} (a (1+\sec (c+d x)))^n \left (2^{1+n} \operatorname {Hypergeometric2F1}\left (1,1-n,2-n,\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^n+2^n \operatorname {Hypergeometric2F1}\left (2,1-n,2-n,\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^n+(1+\sec (c+d x))^n\right )}{8 d (-1+n)} \]

input

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

output

(Cos[c + d*x]*Sec[(c + d*x)/2]^2*(a*(1 + Sec[c + d*x]))^n*(2^(1 + n)*Hyper 
geometric2F1[1, 1 - n, 2 - n, Cos[c + d*x]*Sec[(c + d*x)/2]^2]*(Cos[(c + d 
*x)/2]^2*Sec[c + d*x])^n + 2^n*Hypergeometric2F1[2, 1 - n, 2 - n, Cos[c + 
d*x]*Sec[(c + d*x)/2]^2]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^n + (1 + Sec[c 
+ d*x])^n))/(8*d*(-1 + n)*(1 + Sec[c + d*x])^n)
 

3.2.50.3 Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4361, 27, 100, 25, 27, 88, 78}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4361

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 88

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

\(\Big \downarrow \) 78

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

input

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

output

-((a^2*(-1/2*(a + a*Sec[c + d*x])^(-1 + n)/(a*(1 - Sec[c + d*x])) + (((2 - 
 n)*(a + a*Sec[c + d*x])^(-1 + n))/(2*a*(1 - n)) + ((2 + n)*Hypergeometric 
2F1[1, n, 1 + n, (1 + Sec[c + d*x])/2]*(a + a*Sec[c + d*x])^n)/(4*a^2*n))/ 
2))/d)
 

3.2.50.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 78

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b 
*c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m 
 + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] 
 &&  !IntegerQ[m] && IntegerQ[n]
 

rule 88

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], 
 x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimpl 
erQ[p, 1]
 

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

rule 3042

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

rule 4361

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

3.2.50.4 Maple [F]

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

input

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

output

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

3.2.50.5 Fricas [F]

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

input

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

output

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

3.2.50.6 Sympy [F]

\[ \int \csc ^3(c+d x) (a+a \sec (c+d x))^n \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n} \csc ^{3}{\left (c + d x \right )}\, dx \]

input

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

output

Integral((a*(sec(c + d*x) + 1))**n*csc(c + d*x)**3, x)
 

3.2.50.7 Maxima [F]

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

input

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

output

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

3.2.50.8 Giac [F]

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

input

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

output

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

3.2.50.9 Mupad [F(-1)]

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

input

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

output

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