Integrand size = 27, antiderivative size = 99 \[ \int \frac {\cos ^6(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\log (\sin (c+d x))}{a d}-\frac {\sin (c+d x)}{a d}-\frac {\sin ^2(c+d x)}{a d}+\frac {2 \sin ^3(c+d x)}{3 a d}+\frac {\sin ^4(c+d x)}{4 a d}-\frac {\sin ^5(c+d x)}{5 a d} \] Output:
ln(sin(d*x+c))/a/d-sin(d*x+c)/a/d-sin(d*x+c)^2/a/d+2/3*sin(d*x+c)^3/a/d+1/ 4*sin(d*x+c)^4/a/d-1/5*sin(d*x+c)^5/a/d
Time = 0.09 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^6(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {60 \log (\sin (c+d x))-60 \sin (c+d x)-60 \sin ^2(c+d x)+40 \sin ^3(c+d x)+15 \sin ^4(c+d x)-12 \sin ^5(c+d x)}{60 a d} \] Input:
Integrate[(Cos[c + d*x]^6*Cot[c + d*x])/(a + a*Sin[c + d*x]),x]
Output:
(60*Log[Sin[c + d*x]] - 60*Sin[c + d*x] - 60*Sin[c + d*x]^2 + 40*Sin[c + d *x]^3 + 15*Sin[c + d*x]^4 - 12*Sin[c + d*x]^5)/(60*a*d)
Time = 0.31 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, 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.
\(\displaystyle \int \frac {\cos ^6(c+d x) \cot (c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^7}{\sin (c+d x) (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \csc (c+d x) (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc (c+d x) (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}{a}d(a \sin (c+d x))}{a^6 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (-\sin ^4(c+d x) a^4+\sin ^3(c+d x) a^4+2 \sin ^2(c+d x) a^4+\csc (c+d x) a^4-2 \sin (c+d x) a^4-a^4\right )d(a \sin (c+d x))}{a^6 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{5} a^5 \sin ^5(c+d x)+\frac {1}{4} a^5 \sin ^4(c+d x)+\frac {2}{3} a^5 \sin ^3(c+d x)-a^5 \sin ^2(c+d x)-a^5 \sin (c+d x)+a^5 \log (a \sin (c+d x))}{a^6 d}\) |
Input:
Int[(Cos[c + d*x]^6*Cot[c + d*x])/(a + a*Sin[c + d*x]),x]
Output:
(a^5*Log[a*Sin[c + d*x]] - a^5*Sin[c + d*x] - a^5*Sin[c + d*x]^2 + (2*a^5* Sin[c + d*x]^3)/3 + (a^5*Sin[c + d*x]^4)/4 - (a^5*Sin[c + d*x]^5)/5)/(a^6* 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 = 2.63 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.65
method | result | size |
derivativedivides | \(\frac {-\frac {\sin \left (d x +c \right )^{5}}{5}+\frac {\sin \left (d x +c \right )^{4}}{4}+\frac {2 \sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )^{2}-\sin \left (d x +c \right )+\ln \left (\sin \left (d x +c \right )\right )}{d a}\) | \(64\) |
default | \(\frac {-\frac {\sin \left (d x +c \right )^{5}}{5}+\frac {\sin \left (d x +c \right )^{4}}{4}+\frac {2 \sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )^{2}-\sin \left (d x +c \right )+\ln \left (\sin \left (d x +c \right )\right )}{d a}\) | \(64\) |
risch | \(-\frac {i x}{a}+\frac {3 \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 a d}+\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 a d}-\frac {2 i c}{a d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}-\frac {5 \sin \left (d x +c \right )}{8 a d}-\frac {\sin \left (5 d x +5 c \right )}{80 d a}+\frac {\cos \left (4 d x +4 c \right )}{32 a d}-\frac {5 \sin \left (3 d x +3 c \right )}{48 d a}\) | \(137\) |
Input:
int(cos(d*x+c)^6*cot(d*x+c)/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-1/5*sin(d*x+c)^5+1/4*sin(d*x+c)^4+2/3*sin(d*x+c)^3-sin(d*x+c)^2-si n(d*x+c)+ln(sin(d*x+c)))
Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^6(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {15 \, \cos \left (d x + c\right )^{4} + 30 \, \cos \left (d x + c\right )^{2} - 4 \, {\left (3 \, \cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) + 60 \, \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right )}{60 \, a d} \] Input:
integrate(cos(d*x+c)^6*cot(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
1/60*(15*cos(d*x + c)^4 + 30*cos(d*x + c)^2 - 4*(3*cos(d*x + c)^4 + 4*cos( d*x + c)^2 + 8)*sin(d*x + c) + 60*log(1/2*sin(d*x + c)))/(a*d)
\[ \int \frac {\cos ^6(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cos ^{6}{\left (c + d x \right )} \cot {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(cos(d*x+c)**6*cot(d*x+c)/(a+a*sin(d*x+c)),x)
Output:
Integral(cos(c + d*x)**6*cot(c + d*x)/(sin(c + d*x) + 1), x)/a
Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.72 \[ \int \frac {\cos ^6(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {12 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} - 40 \, \sin \left (d x + c\right )^{3} + 60 \, \sin \left (d x + c\right )^{2} + 60 \, \sin \left (d x + c\right )}{a} - \frac {60 \, \log \left (\sin \left (d x + c\right )\right )}{a}}{60 \, d} \] Input:
integrate(cos(d*x+c)^6*cot(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/60*((12*sin(d*x + c)^5 - 15*sin(d*x + c)^4 - 40*sin(d*x + c)^3 + 60*sin (d*x + c)^2 + 60*sin(d*x + c))/a - 60*log(sin(d*x + c))/a)/d
Time = 0.14 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.68 \[ \int \frac {\cos ^6(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {12 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} - 40 \, \sin \left (d x + c\right )^{3} + 60 \, \sin \left (d x + c\right )^{2} - 60 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 60 \, \sin \left (d x + c\right )}{60 \, a d} \] Input:
integrate(cos(d*x+c)^6*cot(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/60*(12*sin(d*x + c)^5 - 15*sin(d*x + c)^4 - 40*sin(d*x + c)^3 + 60*sin( d*x + c)^2 - 60*log(abs(sin(d*x + c))) + 60*sin(d*x + c))/(a*d)
Time = 31.52 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.41 \[ \int \frac {\cos ^6(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a\,d}-\frac {\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{a\,d}-\frac {8\,\sin \left (c+d\,x\right )}{15\,a\,d}+\frac {{\cos \left (c+d\,x\right )}^2}{2\,a\,d}+\frac {{\cos \left (c+d\,x\right )}^4}{4\,a\,d}-\frac {4\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{15\,a\,d}-\frac {{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{5\,a\,d} \] Input:
int((cos(c + d*x)^6*cot(c + d*x))/(a + a*sin(c + d*x)),x)
Output:
log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))/(a*d) - log(1/cos(c/2 + (d*x)/2 )^2)/(a*d) - (8*sin(c + d*x))/(15*a*d) + cos(c + d*x)^2/(2*a*d) + cos(c + d*x)^4/(4*a*d) - (4*cos(c + d*x)^2*sin(c + d*x))/(15*a*d) - (cos(c + d*x)^ 4*sin(c + d*x))/(5*a*d)
\[ \int \frac {\cos ^6(c+d x) \cot (c+d x)}{a+a \sin (c+d x)} \, dx=\int \frac {\cos \left (d x +c \right )^{6} \cot \left (d x +c \right )}{\sin \left (d x +c \right ) a +a}d x \] Input:
int(cos(d*x+c)^6*cot(d*x+c)/(a+a*sin(d*x+c)),x)
Output:
int(cos(d*x+c)^6*cot(d*x+c)/(a+a*sin(d*x+c)),x)