Integrand size = 27, antiderivative size = 100 \[ \int \frac {\cos (c+d x) \cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\csc (c+d x)}{a d}-\frac {\csc ^2(c+d x)}{a d}+\frac {2 \csc ^3(c+d x)}{3 a d}+\frac {\csc ^4(c+d x)}{4 a d}-\frac {\csc ^5(c+d x)}{5 a d}-\frac {\log (\sin (c+d x))}{a d} \] Output:
-csc(d*x+c)/a/d-csc(d*x+c)^2/a/d+2/3*csc(d*x+c)^3/a/d+1/4*csc(d*x+c)^4/a/d -1/5*csc(d*x+c)^5/a/d-ln(sin(d*x+c))/a/d
Time = 0.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.68 \[ \int \frac {\cos (c+d x) \cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {60 \csc (c+d x)+60 \csc ^2(c+d x)-40 \csc ^3(c+d x)-15 \csc ^4(c+d x)+12 \csc ^5(c+d x)+60 \log (\sin (c+d x))}{60 a d} \] Input:
Integrate[(Cos[c + d*x]*Cot[c + d*x]^6)/(a + a*Sin[c + d*x]),x]
Output:
-1/60*(60*Csc[c + d*x] + 60*Csc[c + d*x]^2 - 40*Csc[c + d*x]^3 - 15*Csc[c + d*x]^4 + 12*Csc[c + d*x]^5 + 60*Log[Sin[c + d*x]])/(a*d)
Time = 0.32 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.73, 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.
\(\displaystyle \int \frac {\cos (c+d x) \cot ^6(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^7}{\sin (c+d x)^6 (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \csc ^6(c+d x) (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc ^6(c+d x) (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}{a^6}d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\frac {\csc ^6(c+d x)}{a}-\frac {\csc ^5(c+d x)}{a}-\frac {2 \csc ^4(c+d x)}{a}+\frac {2 \csc ^3(c+d x)}{a}+\frac {\csc ^2(c+d x)}{a}-\frac {\csc (c+d x)}{a}\right )d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\log (a \sin (c+d x))-\frac {1}{5} \csc ^5(c+d x)+\frac {1}{4} \csc ^4(c+d x)+\frac {2}{3} \csc ^3(c+d x)-\csc ^2(c+d x)-\csc (c+d x)}{a d}\) |
Input:
Int[(Cos[c + d*x]*Cot[c + d*x]^6)/(a + a*Sin[c + d*x]),x]
Output:
(-Csc[c + d*x] - Csc[c + d*x]^2 + (2*Csc[c + d*x]^3)/3 + Csc[c + d*x]^4/4 - Csc[c + d*x]^5/5 - Log[a*Sin[c + d*x]])/(a*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 = 0.67 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.68
method | result | size |
derivativedivides | \(\frac {\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{\sin \left (d x +c \right )^{2}}-\ln \left (\sin \left (d x +c \right )\right )+\frac {2}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{5 \sin \left (d x +c \right )^{5}}-\frac {1}{\sin \left (d x +c \right )}}{d a}\) | \(68\) |
default | \(\frac {\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{\sin \left (d x +c \right )^{2}}-\ln \left (\sin \left (d x +c \right )\right )+\frac {2}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{5 \sin \left (d x +c \right )^{5}}-\frac {1}{\sin \left (d x +c \right )}}{d a}\) | \(68\) |
risch | \(\frac {i x}{a}+\frac {2 i c}{a d}-\frac {2 i \left (15 \,{\mathrm e}^{9 i \left (d x +c \right )}-20 \,{\mathrm e}^{7 i \left (d x +c \right )}+30 i {\mathrm e}^{8 i \left (d x +c \right )}+58 \,{\mathrm e}^{5 i \left (d x +c \right )}-60 i {\mathrm e}^{6 i \left (d x +c \right )}-20 \,{\mathrm e}^{3 i \left (d x +c \right )}+60 i {\mathrm e}^{4 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}-30 i {\mathrm e}^{2 i \left (d x +c \right )}\right )}{15 a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}\) | \(165\) |
Input:
int(cos(d*x+c)*cot(d*x+c)^6/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(1/4/sin(d*x+c)^4-1/sin(d*x+c)^2-ln(sin(d*x+c))+2/3/sin(d*x+c)^3-1/5 /sin(d*x+c)^5-1/sin(d*x+c))
Time = 0.09 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.18 \[ \int \frac {\cos (c+d x) \cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {60 \, \cos \left (d x + c\right )^{4} + 60 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 80 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (4 \, \cos \left (d x + c\right )^{2} - 3\right )} \sin \left (d x + c\right ) + 32}{60 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/60*(60*cos(d*x + c)^4 + 60*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log( 1/2*sin(d*x + c))*sin(d*x + c) - 80*cos(d*x + c)^2 - 15*(4*cos(d*x + c)^2 - 3)*sin(d*x + c) + 32)/((a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^2 + a*d) *sin(d*x + c))
\[ \int \frac {\cos (c+d x) \cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cos {\left (c + d x \right )} \cot ^{6}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)**6/(a+a*sin(d*x+c)),x)
Output:
Integral(cos(c + d*x)*cot(c + d*x)**6/(sin(c + d*x) + 1), x)/a
Time = 0.04 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.70 \[ \int \frac {\cos (c+d x) \cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {60 \, \log \left (\sin \left (d x + c\right )\right )}{a} + \frac {60 \, \sin \left (d x + c\right )^{4} + 60 \, \sin \left (d x + c\right )^{3} - 40 \, \sin \left (d x + c\right )^{2} - 15 \, \sin \left (d x + c\right ) + 12}{a \sin \left (d x + c\right )^{5}}}{60 \, d} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/60*(60*log(sin(d*x + c))/a + (60*sin(d*x + c)^4 + 60*sin(d*x + c)^3 - 4 0*sin(d*x + c)^2 - 15*sin(d*x + c) + 12)/(a*sin(d*x + c)^5))/d
Time = 0.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.68 \[ \int \frac {\cos (c+d x) \cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {60 \, \sin \left (d x + c\right )^{4} + 60 \, \sin \left (d x + c\right )^{3} - 40 \, \sin \left (d x + c\right )^{2} - 15 \, \sin \left (d x + c\right ) + 12}{\sin \left (d x + c\right )^{5}} + 60 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{60 \, a d} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/60*((60*sin(d*x + c)^4 + 60*sin(d*x + c)^3 - 40*sin(d*x + c)^2 - 15*sin (d*x + c) + 12)/sin(d*x + c)^5 + 60*log(abs(sin(d*x + c))))/(a*d)
Time = 31.42 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.04 \[ \int \frac {\cos (c+d x) \cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,a\,d}-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,a\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,a\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}-\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,a\,d}+\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {1}{5}\right )}{32\,a\,d} \] Input:
int((cos(c + d*x)*cot(c + d*x)^6)/(a + a*sin(c + d*x)),x)
Output:
(5*tan(c/2 + (d*x)/2)^3)/(96*a*d) - (3*tan(c/2 + (d*x)/2)^2)/(16*a*d) + ta n(c/2 + (d*x)/2)^4/(64*a*d) - tan(c/2 + (d*x)/2)^5/(160*a*d) - log(tan(c/2 + (d*x)/2))/(a*d) - (5*tan(c/2 + (d*x)/2))/(16*a*d) + log(tan(c/2 + (d*x) /2)^2 + 1)/(a*d) - (cot(c/2 + (d*x)/2)^5*(6*tan(c/2 + (d*x)/2)^3 - (5*tan( c/2 + (d*x)/2)^2)/3 - tan(c/2 + (d*x)/2)/2 + 10*tan(c/2 + (d*x)/2)^4 + 1/5 ))/(32*a*d)
Time = 59.58 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.10 \[ \int \frac {\cos (c+d x) \cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {480 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{5}-480 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5}+195 \sin \left (d x +c \right )^{5}-480 \sin \left (d x +c \right )^{4}-480 \sin \left (d x +c \right )^{3}+320 \sin \left (d x +c \right )^{2}+120 \sin \left (d x +c \right )-96}{480 \sin \left (d x +c \right )^{5} a d} \] Input:
int(cos(d*x+c)*cot(d*x+c)^6/(a+a*sin(d*x+c)),x)
Output:
(480*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**5 - 480*log(tan((c + d*x)/ 2))*sin(c + d*x)**5 + 195*sin(c + d*x)**5 - 480*sin(c + d*x)**4 - 480*sin( c + d*x)**3 + 320*sin(c + d*x)**2 + 120*sin(c + d*x) - 96)/(480*sin(c + d* x)**5*a*d)