Integrand size = 29, antiderivative size = 95 \[ \int \frac {\cos ^5(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\csc (c+d x)}{a d}-\frac {\log (\sin (c+d x))}{a d}-\frac {2 \sin (c+d x)}{a d}+\frac {\sin ^2(c+d x)}{a d}+\frac {\sin ^3(c+d x)}{3 a d}-\frac {\sin ^4(c+d x)}{4 a d} \] Output:
-csc(d*x+c)/a/d-ln(sin(d*x+c))/a/d-2*sin(d*x+c)/a/d+sin(d*x+c)^2/a/d+1/3*s in(d*x+c)^3/a/d-1/4*sin(d*x+c)^4/a/d
Time = 0.15 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^5(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {12 \csc (c+d x)+12 \log (\sin (c+d x))+24 \sin (c+d x)-12 \sin ^2(c+d x)-4 \sin ^3(c+d x)+3 \sin ^4(c+d x)}{12 a d} \] Input:
Integrate[(Cos[c + d*x]^5*Cot[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
Output:
-1/12*(12*Csc[c + d*x] + 12*Log[Sin[c + d*x]] + 24*Sin[c + d*x] - 12*Sin[c + d*x]^2 - 4*Sin[c + d*x]^3 + 3*Sin[c + d*x]^4)/(a*d)
Time = 0.33 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3315, 27, 99, 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.
\(\displaystyle \int \frac {\cos ^5(c+d x) \cot ^2(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^7}{\sin (c+d x)^2 (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \csc ^2(c+d x) (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc ^2(c+d x) (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}{a^2}d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (-\sin ^3(c+d x) a^3+\csc ^2(c+d x) a^3+\sin ^2(c+d x) a^3-\csc (c+d x) a^3+2 \sin (c+d x) a^3-2 a^3\right )d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{4} a^4 \sin ^4(c+d x)+\frac {1}{3} a^4 \sin ^3(c+d x)+a^4 \sin ^2(c+d x)-2 a^4 \sin (c+d x)-a^4 \csc (c+d x)-a^4 \log (a \sin (c+d x))}{a^5 d}\) |
Input:
Int[(Cos[c + d*x]^5*Cot[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
Output:
(-(a^4*Csc[c + d*x]) - a^4*Log[a*Sin[c + d*x]] - 2*a^4*Sin[c + d*x] + a^4* Sin[c + d*x]^2 + (a^4*Sin[c + d*x]^3)/3 - (a^4*Sin[c + d*x]^4)/4)/(a^5*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 2.87 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.67
method | result | size |
derivativedivides | \(\frac {-\frac {\sin \left (d x +c \right )^{4}}{4}+\frac {\sin \left (d x +c \right )^{3}}{3}+\sin \left (d x +c \right )^{2}-2 \sin \left (d x +c \right )-\frac {1}{\sin \left (d x +c \right )}-\ln \left (\sin \left (d x +c \right )\right )}{d a}\) | \(64\) |
default | \(\frac {-\frac {\sin \left (d x +c \right )^{4}}{4}+\frac {\sin \left (d x +c \right )^{3}}{3}+\sin \left (d x +c \right )^{2}-2 \sin \left (d x +c \right )-\frac {1}{\sin \left (d x +c \right )}-\ln \left (\sin \left (d x +c \right )\right )}{d a}\) | \(64\) |
risch | \(\frac {i x}{a}-\frac {3 \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 a d}+\frac {7 i {\mathrm e}^{i \left (d x +c \right )}}{8 a d}-\frac {7 i {\mathrm e}^{-i \left (d x +c \right )}}{8 d a}-\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 a d}+\frac {2 i c}{a d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )}}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}-\frac {\cos \left (4 d x +4 c \right )}{32 a d}-\frac {\sin \left (3 d x +3 c \right )}{12 d a}\) | \(174\) |
Input:
int(cos(d*x+c)^5*cot(d*x+c)^2/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-1/4*sin(d*x+c)^4+1/3*sin(d*x+c)^3+sin(d*x+c)^2-2*sin(d*x+c)-1/sin( d*x+c)-ln(sin(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^5(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {32 \, \cos \left (d x + c\right )^{4} + 128 \, \cos \left (d x + c\right )^{2} - 3 \, {\left (8 \, \cos \left (d x + c\right )^{4} + 16 \, \cos \left (d x + c\right )^{2} - 11\right )} \sin \left (d x + c\right ) - 96 \, \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 256}{96 \, a d \sin \left (d x + c\right )} \] Input:
integrate(cos(d*x+c)^5*cot(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
1/96*(32*cos(d*x + c)^4 + 128*cos(d*x + c)^2 - 3*(8*cos(d*x + c)^4 + 16*co s(d*x + c)^2 - 11)*sin(d*x + c) - 96*log(1/2*sin(d*x + c))*sin(d*x + c) - 256)/(a*d*sin(d*x + c))
Timed out. \[ \int \frac {\cos ^5(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**5*cot(d*x+c)**2/(a+a*sin(d*x+c)),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^5(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {3 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3} - 12 \, \sin \left (d x + c\right )^{2} + 24 \, \sin \left (d x + c\right )}{a} + \frac {12 \, \log \left (\sin \left (d x + c\right )\right )}{a} + \frac {12}{a \sin \left (d x + c\right )}}{12 \, d} \] Input:
integrate(cos(d*x+c)^5*cot(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
-1/12*((3*sin(d*x + c)^4 - 4*sin(d*x + c)^3 - 12*sin(d*x + c)^2 + 24*sin(d *x + c))/a + 12*log(sin(d*x + c))/a + 12/(a*sin(d*x + c)))/d
Time = 0.12 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^5(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3} - 12 \, \sin \left (d x + c\right )^{2} + \frac {12}{\sin \left (d x + c\right )} + 12 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 24 \, \sin \left (d x + c\right )}{12 \, a d} \] Input:
integrate(cos(d*x+c)^5*cot(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/12*(3*sin(d*x + c)^4 - 4*sin(d*x + c)^3 - 12*sin(d*x + c)^2 + 12/sin(d* x + c) + 12*log(abs(sin(d*x + c))) + 24*sin(d*x + c))/(a*d)
Time = 31.87 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.86 \[ \int \frac {\cos ^5(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a\,d}-\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{a\,d}+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{a\,d}-\frac {4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{a\,d}+\frac {\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{a\,d}-\frac {\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a\,d}+\frac {20\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3\,a\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3\,a\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3\,a\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {9\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )} \] Input:
int((cos(c + d*x)^5*cot(c + d*x)^2)/(a + a*sin(c + d*x)),x)
Output:
(4*cos(c/2 + (d*x)/2)^2)/(a*d) - (8*cos(c/2 + (d*x)/2)^4)/(a*d) + (8*cos(c /2 + (d*x)/2)^6)/(a*d) - (4*cos(c/2 + (d*x)/2)^8)/(a*d) + log(1/cos(c/2 + (d*x)/2)^2)/(a*d) - log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))/(a*d) + (20 *cos(c/2 + (d*x)/2)^3)/(3*a*d*sin(c/2 + (d*x)/2)) - (16*cos(c/2 + (d*x)/2) ^5)/(3*a*d*sin(c/2 + (d*x)/2)) + (8*cos(c/2 + (d*x)/2)^7)/(3*a*d*sin(c/2 + (d*x)/2)) - (9*cos(c/2 + (d*x)/2))/(2*a*d*sin(c/2 + (d*x)/2)) - sin(c/2 + (d*x)/2)/(2*a*d*cos(c/2 + (d*x)/2))
\[ \int \frac {\cos ^5(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\int \frac {\cos \left (d x +c \right )^{5} \cot \left (d x +c \right )^{2}}{\sin \left (d x +c \right ) a +a}d x \] Input:
int(cos(d*x+c)^5*cot(d*x+c)^2/(a+a*sin(d*x+c)),x)
Output:
int(cos(d*x+c)^5*cot(d*x+c)^2/(a+a*sin(d*x+c)),x)