Integrand size = 27, antiderivative size = 114 \[ \int \cos ^5(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d}-\frac {3 a \sin (c+d x)}{d}-\frac {3 a \sin ^2(c+d x)}{2 d}+\frac {a \sin ^3(c+d x)}{d}+\frac {3 a \sin ^4(c+d x)}{4 d}-\frac {a \sin ^5(c+d x)}{5 d}-\frac {a \sin ^6(c+d x)}{6 d} \] Output:
-a*csc(d*x+c)/d+a*ln(sin(d*x+c))/d-3*a*sin(d*x+c)/d-3/2*a*sin(d*x+c)^2/d+a *sin(d*x+c)^3/d+3/4*a*sin(d*x+c)^4/d-1/5*a*sin(d*x+c)^5/d-1/6*a*sin(d*x+c) ^6/d
Time = 0.06 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00 \[ \int \cos ^5(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d}-\frac {3 a \sin (c+d x)}{d}-\frac {3 a \sin ^2(c+d x)}{2 d}+\frac {a \sin ^3(c+d x)}{d}+\frac {3 a \sin ^4(c+d x)}{4 d}-\frac {a \sin ^5(c+d x)}{5 d}-\frac {a \sin ^6(c+d x)}{6 d} \] Input:
Integrate[Cos[c + d*x]^5*Cot[c + d*x]^2*(a + a*Sin[c + d*x]),x]
Output:
-((a*Csc[c + d*x])/d) + (a*Log[Sin[c + d*x]])/d - (3*a*Sin[c + d*x])/d - ( 3*a*Sin[c + d*x]^2)/(2*d) + (a*Sin[c + d*x]^3)/d + (3*a*Sin[c + d*x]^4)/(4 *d) - (a*Sin[c + d*x]^5)/(5*d) - (a*Sin[c + d*x]^6)/(6*d)
Time = 0.32 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.01, 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.
| \(\displaystyle \int \cos ^5(c+d x) \cot ^2(c+d x) (a \sin (c+d x)+a) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^7 (a \sin (c+d x)+a)}{\sin (c+d x)^2}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {\int \csc ^2(c+d x) (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a^7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\csc ^2(c+d x) (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^4}{a^2}d(a \sin (c+d x))}{a^5 d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (-\sin ^5(c+d x) a^5-\sin ^4(c+d x) a^5+3 \sin ^3(c+d x) a^5+\csc ^2(c+d x) a^5+3 \sin ^2(c+d x) a^5+\csc (c+d x) a^5-3 \sin (c+d x) a^5-3 a^5\right )d(a \sin (c+d x))}{a^5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {1}{6} a^6 \sin ^6(c+d x)-\frac {1}{5} a^6 \sin ^5(c+d x)+\frac {3}{4} a^6 \sin ^4(c+d x)+a^6 \sin ^3(c+d x)-\frac {3}{2} a^6 \sin ^2(c+d x)-3 a^6 \sin (c+d x)-a^6 \csc (c+d x)+a^6 \log (a \sin (c+d x))}{a^5 d}\) |
Input:
Int[Cos[c + d*x]^5*Cot[c + d*x]^2*(a + a*Sin[c + d*x]),x]
Output:
(-(a^6*Csc[c + d*x]) + a^6*Log[a*Sin[c + d*x]] - 3*a^6*Sin[c + d*x] - (3*a ^6*Sin[c + d*x]^2)/2 + a^6*Sin[c + d*x]^3 + (3*a^6*Sin[c + d*x]^4)/4 - (a^ 6*Sin[c + d*x]^5)/5 - (a^6*Sin[c + d*x]^6)/6)/(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 = 6.97 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.92
| method | result | size |
| derivativedivides | \(\frac {a \left (\frac {\cos \left (d x +c \right )^{6}}{6}+\frac {\cos \left (d x +c \right )^{4}}{4}+\frac {\cos \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+a \left (-\frac {\cos \left (d x +c \right )^{8}}{\sin \left (d x +c \right )}-\left (\frac {16}{5}+\cos \left (d x +c \right )^{6}+\frac {6 \cos \left (d x +c \right )^{4}}{5}+\frac {8 \cos \left (d x +c \right )^{2}}{5}\right ) \sin \left (d x +c \right )\right )}{d}\) | \(105\) |
| default | \(\frac {a \left (\frac {\cos \left (d x +c \right )^{6}}{6}+\frac {\cos \left (d x +c \right )^{4}}{4}+\frac {\cos \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+a \left (-\frac {\cos \left (d x +c \right )^{8}}{\sin \left (d x +c \right )}-\left (\frac {16}{5}+\cos \left (d x +c \right )^{6}+\frac {6 \cos \left (d x +c \right )^{4}}{5}+\frac {8 \cos \left (d x +c \right )^{2}}{5}\right ) \sin \left (d x +c \right )\right )}{d}\) | \(105\) |
| risch | \(-\frac {19 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{16 d}-i a x +\frac {29 a \,{\mathrm e}^{2 i \left (d x +c \right )}}{128 d}+\frac {19 i a \,{\mathrm e}^{i \left (d x +c \right )}}{16 d}+\frac {29 a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{128 d}-\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {2 i a c}{d}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {a \cos \left (6 d x +6 c \right )}{192 d}-\frac {a \sin \left (5 d x +5 c \right )}{80 d}+\frac {a \cos \left (4 d x +4 c \right )}{16 d}-\frac {3 a \sin \left (3 d x +3 c \right )}{16 d}\) | \(183\) |
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*(1/6*cos(d*x+c)^6+1/4*cos(d*x+c)^4+1/2*cos(d*x+c)^2+ln(sin(d*x+c))) +a*(-1/sin(d*x+c)*cos(d*x+c)^8-(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos (d*x+c)^2)*sin(d*x+c)))
Time = 0.10 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.99 \[ \int \cos ^5(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {48 \, a \cos \left (d x + c\right )^{6} + 96 \, a \cos \left (d x + c\right )^{4} + 384 \, a \cos \left (d x + c\right )^{2} + 240 \, a \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 5 \, {\left (8 \, a \cos \left (d x + c\right )^{6} + 12 \, a \cos \left (d x + c\right )^{4} + 24 \, a \cos \left (d x + c\right )^{2} - 19 \, a\right )} \sin \left (d x + c\right ) - 768 \, a}{240 \, d \sin \left (d x + c\right )} \] Input:
integrate(cos(d*x+c)^5*cot(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
1/240*(48*a*cos(d*x + c)^6 + 96*a*cos(d*x + c)^4 + 384*a*cos(d*x + c)^2 + 240*a*log(1/2*sin(d*x + c))*sin(d*x + c) + 5*(8*a*cos(d*x + c)^6 + 12*a*co s(d*x + c)^4 + 24*a*cos(d*x + c)^2 - 19*a)*sin(d*x + c) - 768*a)/(d*sin(d* x + c))
\[ \int \cos ^5(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=a \left (\int \cos ^{5}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cos(d*x+c)**5*cot(d*x+c)**2*(a+a*sin(d*x+c)),x)
Output:
a*(Integral(cos(c + d*x)**5*cot(c + d*x)**2, x) + Integral(sin(c + d*x)*co s(c + d*x)**5*cot(c + d*x)**2, x))
Time = 0.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.80 \[ \int \cos ^5(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {10 \, a \sin \left (d x + c\right )^{6} + 12 \, a \sin \left (d x + c\right )^{5} - 45 \, a \sin \left (d x + c\right )^{4} - 60 \, a \sin \left (d x + c\right )^{3} + 90 \, a \sin \left (d x + c\right )^{2} - 60 \, a \log \left (\sin \left (d x + c\right )\right ) + 180 \, a \sin \left (d x + c\right ) + \frac {60 \, a}{\sin \left (d x + c\right )}}{60 \, d} \] Input:
integrate(cos(d*x+c)^5*cot(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
-1/60*(10*a*sin(d*x + c)^6 + 12*a*sin(d*x + c)^5 - 45*a*sin(d*x + c)^4 - 6 0*a*sin(d*x + c)^3 + 90*a*sin(d*x + c)^2 - 60*a*log(sin(d*x + c)) + 180*a* sin(d*x + c) + 60*a/sin(d*x + c))/d
Time = 0.14 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.81 \[ \int \cos ^5(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {10 \, a \sin \left (d x + c\right )^{6} + 12 \, a \sin \left (d x + c\right )^{5} - 45 \, a \sin \left (d x + c\right )^{4} - 60 \, a \sin \left (d x + c\right )^{3} + 90 \, a \sin \left (d x + c\right )^{2} - 60 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 180 \, a \sin \left (d x + c\right ) + \frac {60 \, a}{\sin \left (d x + c\right )}}{60 \, d} \] Input:
integrate(cos(d*x+c)^5*cot(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/60*(10*a*sin(d*x + c)^6 + 12*a*sin(d*x + c)^5 - 45*a*sin(d*x + c)^4 - 6 0*a*sin(d*x + c)^3 + 90*a*sin(d*x + c)^2 - 60*a*log(abs(sin(d*x + c))) + 1 80*a*sin(d*x + c) + 60*a/sin(d*x + c))/d
Time = 34.11 (sec) , antiderivative size = 340, normalized size of antiderivative = 2.98 \[ \int \cos ^5(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}-\frac {6\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{d}+\frac {18\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d}-\frac {104\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3\,d}+\frac {44\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d}-\frac {32\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{d}+\frac {32\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{3\,d}-\frac {13\,a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {14\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {112\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {136\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{5\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {96\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{5\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {32\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{5\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )} \] Input:
int(cos(c + d*x)^5*cot(c + d*x)^2*(a + a*sin(c + d*x)),x)
Output:
(a*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (a*log(1/cos(c/2 + (d*x )/2)^2))/d - (6*a*cos(c/2 + (d*x)/2)^2)/d + (18*a*cos(c/2 + (d*x)/2)^4)/d - (104*a*cos(c/2 + (d*x)/2)^6)/(3*d) + (44*a*cos(c/2 + (d*x)/2)^8)/d - (32 *a*cos(c/2 + (d*x)/2)^10)/d + (32*a*cos(c/2 + (d*x)/2)^12)/(3*d) - (13*a*c os(c/2 + (d*x)/2))/(2*d*sin(c/2 + (d*x)/2)) - (a*sin(c/2 + (d*x)/2))/(2*d* cos(c/2 + (d*x)/2)) + (14*a*cos(c/2 + (d*x)/2)^3)/(d*sin(c/2 + (d*x)/2)) - (112*a*cos(c/2 + (d*x)/2)^5)/(5*d*sin(c/2 + (d*x)/2)) + (136*a*cos(c/2 + (d*x)/2)^7)/(5*d*sin(c/2 + (d*x)/2)) - (96*a*cos(c/2 + (d*x)/2)^9)/(5*d*si n(c/2 + (d*x)/2)) + (32*a*cos(c/2 + (d*x)/2)^11)/(5*d*sin(c/2 + (d*x)/2))
Time = 0.16 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.02 \[ \int \cos ^5(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (-60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )+60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )-10 \sin \left (d x +c \right )^{7}-12 \sin \left (d x +c \right )^{6}+45 \sin \left (d x +c \right )^{5}+60 \sin \left (d x +c \right )^{4}-90 \sin \left (d x +c \right )^{3}-180 \sin \left (d x +c \right )^{2}-60\right )}{60 \sin \left (d x +c \right ) d} \] Input:
int(cos(d*x+c)^5*cot(d*x+c)^2*(a+a*sin(d*x+c)),x)
Output:
(a*( - 60*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x) + 60*log(tan((c + d*x) /2))*sin(c + d*x) - 10*sin(c + d*x)**7 - 12*sin(c + d*x)**6 + 45*sin(c + d *x)**5 + 60*sin(c + d*x)**4 - 90*sin(c + d*x)**3 - 180*sin(c + d*x)**2 - 6 0))/(60*sin(c + d*x)*d)