Integrand size = 19, antiderivative size = 49 \[ \int (d \cos (a+b x))^n \csc ^5(a+b x) \, dx=-\frac {(d \cos (a+b x))^{1+n} \operatorname {Hypergeometric2F1}\left (3,\frac {1+n}{2},\frac {3+n}{2},\cos ^2(a+b x)\right )}{b d (1+n)} \] Output:
-(d*cos(b*x+a))^(1+n)*hypergeom([3, 1/2+1/2*n],[3/2+1/2*n],cos(b*x+a)^2)/b /d/(1+n)
Leaf count is larger than twice the leaf count of optimal. \(244\) vs. \(2(49)=98\).
Time = 2.09 (sec) , antiderivative size = 244, normalized size of antiderivative = 4.98 \[ \int (d \cos (a+b x))^n \csc ^5(a+b x) \, dx=-\frac {2^{-5-n} \cos (a+b x) (d \cos (a+b x))^n \left (3\ 2^{1+n} \operatorname {Hypergeometric2F1}(1,1+n,2+n,\cos (a+b x))+3\ 2^{1+n} \operatorname {Hypergeometric2F1}(2,1+n,2+n,\cos (a+b x))+2^{2+n} \operatorname {Hypergeometric2F1}(3,1+n,2+n,\cos (a+b x))+2 \operatorname {Hypergeometric2F1}\left (-1+n,1+n,2+n,\frac {1}{2} \cos (a+b x) \sec ^2\left (\frac {1}{2} (a+b x)\right )\right ) \sec ^2\left (\frac {1}{2} (a+b x)\right )^{1+n}+3 \operatorname {Hypergeometric2F1}\left (n,1+n,2+n,\frac {1}{2} \cos (a+b x) \sec ^2\left (\frac {1}{2} (a+b x)\right )\right ) \sec ^2\left (\frac {1}{2} (a+b x)\right )^{1+n}+3 \operatorname {Hypergeometric2F1}\left (1+n,1+n,2+n,\frac {1}{2} \cos (a+b x) \sec ^2\left (\frac {1}{2} (a+b x)\right )\right ) \sec ^2\left (\frac {1}{2} (a+b x)\right )^{1+n}\right )}{b (1+n)} \] Input:
Integrate[(d*Cos[a + b*x])^n*Csc[a + b*x]^5,x]
Output:
-((2^(-5 - n)*Cos[a + b*x]*(d*Cos[a + b*x])^n*(3*2^(1 + n)*Hypergeometric2 F1[1, 1 + n, 2 + n, Cos[a + b*x]] + 3*2^(1 + n)*Hypergeometric2F1[2, 1 + n , 2 + n, Cos[a + b*x]] + 2^(2 + n)*Hypergeometric2F1[3, 1 + n, 2 + n, Cos[ a + b*x]] + 2*Hypergeometric2F1[-1 + n, 1 + n, 2 + n, (Cos[a + b*x]*Sec[(a + b*x)/2]^2)/2]*(Sec[(a + b*x)/2]^2)^(1 + n) + 3*Hypergeometric2F1[n, 1 + n, 2 + n, (Cos[a + b*x]*Sec[(a + b*x)/2]^2)/2]*(Sec[(a + b*x)/2]^2)^(1 + n) + 3*Hypergeometric2F1[1 + n, 1 + n, 2 + n, (Cos[a + b*x]*Sec[(a + b*x)/ 2]^2)/2]*(Sec[(a + b*x)/2]^2)^(1 + n)))/(b*(1 + n)))
Time = 0.23 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3045, 27, 278}
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.
\(\displaystyle \int \csc ^5(a+b x) (d \cos (a+b x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(d \cos (a+b x))^n}{\sin (a+b x)^5}dx\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {\int \frac {d^6 (d \cos (a+b x))^n}{\left (d^2-d^2 \cos ^2(a+b x)\right )^3}d(d \cos (a+b x))}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {d^5 \int \frac {(d \cos (a+b x))^n}{\left (d^2-d^2 \cos ^2(a+b x)\right )^3}d(d \cos (a+b x))}{b}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle -\frac {(d \cos (a+b x))^{n+1} \operatorname {Hypergeometric2F1}\left (3,\frac {n+1}{2},\frac {n+3}{2},\cos ^2(a+b x)\right )}{b d (n+1)}\) |
Input:
Int[(d*Cos[a + b*x])^n*Csc[a + b*x]^5,x]
Output:
-(((d*Cos[a + b*x])^(1 + n)*Hypergeometric2F1[3, (1 + n)/2, (3 + n)/2, Cos [a + b*x]^2])/(b*d*(1 + n)))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> Simp[-(a*f)^(-1) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
\[\int \left (d \cos \left (b x +a \right )\right )^{n} \csc \left (b x +a \right )^{5}d x\]
Input:
int((d*cos(b*x+a))^n*csc(b*x+a)^5,x)
Output:
int((d*cos(b*x+a))^n*csc(b*x+a)^5,x)
\[ \int (d \cos (a+b x))^n \csc ^5(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{n} \csc \left (b x + a\right )^{5} \,d x } \] Input:
integrate((d*cos(b*x+a))^n*csc(b*x+a)^5,x, algorithm="fricas")
Output:
integral((d*cos(b*x + a))^n*csc(b*x + a)^5, x)
Timed out. \[ \int (d \cos (a+b x))^n \csc ^5(a+b x) \, dx=\text {Timed out} \] Input:
integrate((d*cos(b*x+a))**n*csc(b*x+a)**5,x)
Output:
Timed out
\[ \int (d \cos (a+b x))^n \csc ^5(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{n} \csc \left (b x + a\right )^{5} \,d x } \] Input:
integrate((d*cos(b*x+a))^n*csc(b*x+a)^5,x, algorithm="maxima")
Output:
integrate((d*cos(b*x + a))^n*csc(b*x + a)^5, x)
\[ \int (d \cos (a+b x))^n \csc ^5(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{n} \csc \left (b x + a\right )^{5} \,d x } \] Input:
integrate((d*cos(b*x+a))^n*csc(b*x+a)^5,x, algorithm="giac")
Output:
integrate((d*cos(b*x + a))^n*csc(b*x + a)^5, x)
Timed out. \[ \int (d \cos (a+b x))^n \csc ^5(a+b x) \, dx=\int \frac {{\left (d\,\cos \left (a+b\,x\right )\right )}^n}{{\sin \left (a+b\,x\right )}^5} \,d x \] Input:
int((d*cos(a + b*x))^n/sin(a + b*x)^5,x)
Output:
int((d*cos(a + b*x))^n/sin(a + b*x)^5, x)
\[ \int (d \cos (a+b x))^n \csc ^5(a+b x) \, dx=d^{n} \left (\int \cos \left (b x +a \right )^{n} \csc \left (b x +a \right )^{5}d x \right ) \] Input:
int((d*cos(b*x+a))^n*csc(b*x+a)^5,x)
Output:
d**n*int(cos(a + b*x)**n*csc(a + b*x)**5,x)