Integrand size = 21, antiderivative size = 51 \[ \int \frac {\cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\cot ^4(c+d x)}{4 a d}-\frac {\csc (c+d x)}{a d}+\frac {\csc ^3(c+d x)}{3 a d} \] Output:
-1/4*cot(d*x+c)^4/a/d-csc(d*x+c)/a/d+1/3*csc(d*x+c)^3/a/d
Time = 0.06 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.59 \[ \int \frac {\cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {(-1+\csc (c+d x))^3 (5+3 \csc (c+d x))}{12 a d} \] Input:
Integrate[Cot[c + d*x]^5/(a + a*Sin[c + d*x]),x]
Output:
-1/12*((-1 + Csc[c + d*x])^3*(5 + 3*Csc[c + d*x]))/(a*d)
Time = 0.40 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.92, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3185, 3042, 25, 3086, 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.
\(\displaystyle \int \frac {\cot ^5(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (c+d x)^5 (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3185 |
\(\displaystyle \frac {\int \cot ^3(c+d x) \csc ^2(c+d x)dx}{a}-\frac {\int \cot ^3(c+d x) \csc (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int -\sec \left (c+d x-\frac {\pi }{2}\right )^2 \tan \left (c+d x-\frac {\pi }{2}\right )^3dx}{a}-\frac {\int -\sec \left (c+d x-\frac {\pi }{2}\right ) \tan \left (c+d x-\frac {\pi }{2}\right )^3dx}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right ) \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^3dx}{a}-\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^3dx}{a}\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle \frac {\int \left (\csc ^2(c+d x)-1\right )d\csc (c+d x)}{a d}-\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^3dx}{a}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{3} \csc ^3(c+d x)-\csc (c+d x)}{a d}-\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^3dx}{a}\) |
\(\Big \downarrow \) 3087 |
\(\displaystyle \frac {\frac {1}{3} \csc ^3(c+d x)-\csc (c+d x)}{a d}-\frac {\int -\cot ^3(c+d x)d(-\cot (c+d x))}{a d}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {\frac {1}{3} \csc ^3(c+d x)-\csc (c+d x)}{a d}-\frac {\cot ^4(c+d x)}{4 a d}\) |
Input:
Int[Cot[c + d*x]^5/(a + a*Sin[c + d*x]),x]
Output:
-1/4*Cot[c + d*x]^4/(a*d) + (-Csc[c + d*x] + Csc[c + d*x]^3/3)/(a*d)
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
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[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[1/a Int[Sec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x ] - Simp[1/(b*g) Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /; Fre eQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]
Time = 1.09 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96
method | result | size |
derivativedivides | \(\frac {\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{\sin \left (d x +c \right )}+\frac {1}{2 \sin \left (d x +c \right )^{2}}}{d a}\) | \(49\) |
default | \(\frac {\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{\sin \left (d x +c \right )}+\frac {1}{2 \sin \left (d x +c \right )^{2}}}{d a}\) | \(49\) |
risch | \(-\frac {2 i \left (-3 i {\mathrm e}^{6 i \left (d x +c \right )}+3 \,{\mathrm e}^{7 i \left (d x +c \right )}-5 \,{\mathrm e}^{5 i \left (d x +c \right )}-3 i {\mathrm e}^{2 i \left (d x +c \right )}+5 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}\) | \(92\) |
Input:
int(cot(d*x+c)^5/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(1/3/sin(d*x+c)^3-1/4/sin(d*x+c)^4-1/sin(d*x+c)+1/2/sin(d*x+c)^2)
Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.24 \[ \int \frac {\cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {6 \, \cos \left (d x + c\right )^{2} - 4 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) - 3}{12 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )}} \] Input:
integrate(cot(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/12*(6*cos(d*x + c)^2 - 4*(3*cos(d*x + c)^2 - 2)*sin(d*x + c) - 3)/(a*d* cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^2 + a*d)
\[ \int \frac {\cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cot ^{5}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(cot(d*x+c)**5/(a+a*sin(d*x+c)),x)
Output:
Integral(cot(c + d*x)**5/(sin(c + d*x) + 1), x)/a
Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.90 \[ \int \frac {\cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {12 \, \sin \left (d x + c\right )^{3} - 6 \, \sin \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right ) + 3}{12 \, a d \sin \left (d x + c\right )^{4}} \] Input:
integrate(cot(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/12*(12*sin(d*x + c)^3 - 6*sin(d*x + c)^2 - 4*sin(d*x + c) + 3)/(a*d*sin (d*x + c)^4)
Time = 0.14 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.90 \[ \int \frac {\cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {12 \, \sin \left (d x + c\right )^{3} - 6 \, \sin \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right ) + 3}{12 \, a d \sin \left (d x + c\right )^{4}} \] Input:
integrate(cot(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/12*(12*sin(d*x + c)^3 - 6*sin(d*x + c)^2 - 4*sin(d*x + c) + 3)/(a*d*sin (d*x + c)^4)
Time = 17.74 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.88 \[ \int \frac {\cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-{\sin \left (c+d\,x\right )}^3+\frac {{\sin \left (c+d\,x\right )}^2}{2}+\frac {\sin \left (c+d\,x\right )}{3}-\frac {1}{4}}{a\,d\,{\sin \left (c+d\,x\right )}^4} \] Input:
int(cot(c + d*x)^5/(a + a*sin(c + d*x)),x)
Output:
(sin(c + d*x)/3 + sin(c + d*x)^2/2 - sin(c + d*x)^3 - 1/4)/(a*d*sin(c + d* x)^4)
Time = 0.16 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.10 \[ \int \frac {\cot ^5(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-15 \sin \left (d x +c \right )^{4}-96 \sin \left (d x +c \right )^{3}+48 \sin \left (d x +c \right )^{2}+32 \sin \left (d x +c \right )-24}{96 \sin \left (d x +c \right )^{4} a d} \] Input:
int(cot(d*x+c)^5/(a+a*sin(d*x+c)),x)
Output:
( - 15*sin(c + d*x)**4 - 96*sin(c + d*x)**3 + 48*sin(c + d*x)**2 + 32*sin( c + d*x) - 24)/(96*sin(c + d*x)**4*a*d)