Integrand size = 27, antiderivative size = 55 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\cot ^4(c+d x)}{4 a d}+\frac {\csc ^3(c+d x)}{3 a d}-\frac {\csc ^5(c+d x)}{5 a d} \] Output:
1/4*cot(d*x+c)^4/a/d+1/3*csc(d*x+c)^3/a/d-1/5*csc(d*x+c)^5/a/d
Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.87 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc ^2(c+d x) \left (-30+20 \csc (c+d x)+15 \csc ^2(c+d x)-12 \csc ^3(c+d x)\right )}{60 a d} \] Input:
Integrate[(Cot[c + d*x]^5*Csc[c + d*x])/(a + a*Sin[c + d*x]),x]
Output:
(Csc[c + d*x]^2*(-30 + 20*Csc[c + d*x] + 15*Csc[c + d*x]^2 - 12*Csc[c + d* x]^3))/(60*a*d)
Time = 0.47 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {3042, 3314, 3042, 25, 3086, 25, 244, 2009, 3087, 15}
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.
\(\displaystyle \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5}{\sin (c+d x)^6 (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3314 |
\(\displaystyle \frac {\int \cot ^3(c+d x) \csc ^3(c+d x)dx}{a}-\frac {\int \cot ^3(c+d x) \csc ^2(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int -\sec \left (c+d x-\frac {\pi }{2}\right )^3 \tan \left (c+d x-\frac {\pi }{2}\right )^3dx}{a}-\frac {\int -\sec \left (c+d x-\frac {\pi }{2}\right )^2 \tan \left (c+d x-\frac {\pi }{2}\right )^3dx}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^3dx}{a}-\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^3 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^3dx}{a}\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle \frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^3dx}{a}-\frac {\int -\csc ^2(c+d x) \left (1-\csc ^2(c+d x)\right )d\csc (c+d x)}{a d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \csc ^2(c+d x) \left (1-\csc ^2(c+d x)\right )d\csc (c+d x)}{a d}+\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^3dx}{a}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \left (\csc ^2(c+d x)-\csc ^4(c+d x)\right )d\csc (c+d x)}{a d}+\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^3dx}{a}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^3dx}{a}-\frac {\frac {1}{5} \csc ^5(c+d x)-\frac {1}{3} \csc ^3(c+d x)}{a d}\) |
\(\Big \downarrow \) 3087 |
\(\displaystyle \frac {\int -\cot ^3(c+d x)d(-\cot (c+d x))}{a d}-\frac {\frac {1}{5} \csc ^5(c+d x)-\frac {1}{3} \csc ^3(c+d x)}{a d}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {\cot ^4(c+d x)}{4 a d}-\frac {\frac {1}{5} \csc ^5(c+d x)-\frac {1}{3} \csc ^3(c+d x)}{a d}\) |
Input:
Int[(Cot[c + d*x]^5*Csc[c + d*x])/(a + a*Sin[c + d*x]),x]
Output:
Cot[c + d*x]^4/(4*a*d) - (-1/3*Csc[c + d*x]^3 + Csc[c + d*x]^5/5)/(a*d)
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[a/f Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 ), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 ] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> Simp[1/f Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n - 1) /2] && LtQ[0, n, m - 1])
Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/(( a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/a Int[Cos[e + f *x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[1/(b*d) Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] & & IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n, -p]))
Time = 0.59 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {-\frac {\csc \left (d x +c \right )^{5}}{5}+\frac {\csc \left (d x +c \right )^{4}}{4}+\frac {\csc \left (d x +c \right )^{3}}{3}-\frac {\csc \left (d x +c \right )^{2}}{2}}{d a}\) | \(49\) |
default | \(\frac {-\frac {\csc \left (d x +c \right )^{5}}{5}+\frac {\csc \left (d x +c \right )^{4}}{4}+\frac {\csc \left (d x +c \right )^{3}}{3}-\frac {\csc \left (d x +c \right )^{2}}{2}}{d a}\) | \(49\) |
risch | \(\frac {-\frac {8 i {\mathrm e}^{7 i \left (d x +c \right )}}{3}+2 \,{\mathrm e}^{8 i \left (d x +c \right )}-\frac {16 i {\mathrm e}^{5 i \left (d x +c \right )}}{15}-2 \,{\mathrm e}^{6 i \left (d x +c \right )}-\frac {8 i {\mathrm e}^{3 i \left (d x +c \right )}}{3}+2 \,{\mathrm e}^{4 i \left (d x +c \right )}-2 \,{\mathrm e}^{2 i \left (d x +c \right )}}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}\) | \(103\) |
Input:
int(cot(d*x+c)^5*csc(d*x+c)/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-1/5*csc(d*x+c)^5+1/4*csc(d*x+c)^4+1/3*csc(d*x+c)^3-1/2*csc(d*x+c)^ 2)
Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.29 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {20 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (2 \, \cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 8}{60 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/60*(20*cos(d*x + c)^2 - 15*(2*cos(d*x + c)^2 - 1)*sin(d*x + c) - 8)/((a *d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^2 + a*d)*sin(d*x + c))
\[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cot ^{5}{\left (c + d x \right )} \csc {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(cot(d*x+c)**5*csc(d*x+c)/(a+a*sin(d*x+c)),x)
Output:
Integral(cot(c + d*x)**5*csc(c + d*x)/(sin(c + d*x) + 1), x)/a
Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {30 \, \sin \left (d x + c\right )^{3} - 20 \, \sin \left (d x + c\right )^{2} - 15 \, \sin \left (d x + c\right ) + 12}{60 \, a d \sin \left (d x + c\right )^{5}} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/60*(30*sin(d*x + c)^3 - 20*sin(d*x + c)^2 - 15*sin(d*x + c) + 12)/(a*d* sin(d*x + c)^5)
Time = 0.15 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {30 \, \sin \left (d x + c\right )^{3} - 20 \, \sin \left (d x + c\right )^{2} - 15 \, \sin \left (d x + c\right ) + 12}{60 \, a d \sin \left (d x + c\right )^{5}} \] Input:
integrate(cot(d*x+c)^5*csc(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/60*(30*sin(d*x + c)^3 - 20*sin(d*x + c)^2 - 15*sin(d*x + c) + 12)/(a*d* sin(d*x + c)^5)
Time = 34.73 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-30\,{\sin \left (c+d\,x\right )}^3+20\,{\sin \left (c+d\,x\right )}^2+15\,\sin \left (c+d\,x\right )-12}{60\,a\,d\,{\sin \left (c+d\,x\right )}^5} \] Input:
int(cot(c + d*x)^5/(sin(c + d*x)*(a + a*sin(c + d*x))),x)
Output:
(15*sin(c + d*x) + 20*sin(c + d*x)^2 - 30*sin(c + d*x)^3 - 12)/(60*a*d*sin (c + d*x)^5)
Time = 0.14 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.02 \[ \int \frac {\cot ^5(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {75 \sin \left (d x +c \right )^{5}-240 \sin \left (d x +c \right )^{3}+160 \sin \left (d x +c \right )^{2}+120 \sin \left (d x +c \right )-96}{480 \sin \left (d x +c \right )^{5} a d} \] Input:
int(cot(d*x+c)^5*csc(d*x+c)/(a+a*sin(d*x+c)),x)
Output:
(75*sin(c + d*x)**5 - 240*sin(c + d*x)**3 + 160*sin(c + d*x)**2 + 120*sin( c + d*x) - 96)/(480*sin(c + d*x)**5*a*d)