Integrand size = 25, antiderivative size = 53 \[ \int \cos (c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d}-\frac {a \sin (c+d x)}{d}-\frac {a \sin ^2(c+d x)}{2 d} \] Output:
-a*csc(d*x+c)/d+a*ln(sin(d*x+c))/d-a*sin(d*x+c)/d-1/2*a*sin(d*x+c)^2/d
Time = 0.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00 \[ \int \cos (c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d}-\frac {a \sin (c+d x)}{d}-\frac {a \sin ^2(c+d x)}{2 d} \] Input:
Integrate[Cos[c + d*x]*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 - (a*Sin[c + d*x])/d - (a* Sin[c + d*x]^2)/(2*d)
Time = 0.26 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3315, 27, 84, 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.
\(\displaystyle \int \cos (c+d x) \cot ^2(c+d x) (a \sin (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^3 (a \sin (c+d x)+a)}{\sin (c+d x)^2}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \csc ^2(c+d x) (a-a \sin (c+d x)) (\sin (c+d x) a+a)^2d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc ^2(c+d x) (a-a \sin (c+d x)) (\sin (c+d x) a+a)^2}{a^2}d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 84 |
\(\displaystyle \frac {\int \left (a \csc ^2(c+d x)+a \csc (c+d x)-a-a \sin (c+d x)\right )d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{2} a^2 \sin ^2(c+d x)-a^2 \sin (c+d x)-a^2 \csc (c+d x)+a^2 \log (a \sin (c+d x))}{a d}\) |
Input:
Int[Cos[c + d*x]*Cot[c + d*x]^2*(a + a*Sin[c + d*x]),x]
Output:
(-(a^2*Csc[c + d*x]) + a^2*Log[a*Sin[c + d*x]] - a^2*Sin[c + d*x] - (a^2*S in[c + d*x]^2)/2)/(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[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] && !(ILtQ[n + p + 2, 0 ] && GtQ[n + 2*p, 0])
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 = 0.66 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.23
method | result | size |
derivativedivides | \(\frac {a \left (\frac {\cos \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+a \left (-\frac {\cos \left (d x +c \right )^{4}}{\sin \left (d x +c \right )}-\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )\right )}{d}\) | \(65\) |
default | \(\frac {a \left (\frac {\cos \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+a \left (-\frac {\cos \left (d x +c \right )^{4}}{\sin \left (d x +c \right )}-\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )\right )}{d}\) | \(65\) |
risch | \(-i a x +\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {2 i a c}{d}-\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(123\) |
Input:
int(cos(d*x+c)*cot(d*x+c)^2*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(a*(1/2*cos(d*x+c)^2+ln(sin(d*x+c)))+a*(-1/sin(d*x+c)*cos(d*x+c)^4-(2+ cos(d*x+c)^2)*sin(d*x+c)))
Time = 0.10 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.28 \[ \int \cos (c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {4 \, a \cos \left (d x + c\right )^{2} + 4 \, a \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + {\left (2 \, a \cos \left (d x + c\right )^{2} - a\right )} \sin \left (d x + c\right ) - 8 \, a}{4 \, d \sin \left (d x + c\right )} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
1/4*(4*a*cos(d*x + c)^2 + 4*a*log(1/2*sin(d*x + c))*sin(d*x + c) + (2*a*co s(d*x + c)^2 - a)*sin(d*x + c) - 8*a)/(d*sin(d*x + c))
\[ \int \cos (c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=a \left (\int \cos {\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )} \cos {\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cos(d*x+c)*cot(d*x+c)**2*(a+a*sin(d*x+c)),x)
Output:
a*(Integral(cos(c + d*x)*cot(c + d*x)**2, x) + Integral(sin(c + d*x)*cos(c + d*x)*cot(c + d*x)**2, x))
Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.87 \[ \int \cos (c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \sin \left (d x + c\right )^{2} - 2 \, a \log \left (\sin \left (d x + c\right )\right ) + 2 \, a \sin \left (d x + c\right ) + \frac {2 \, a}{\sin \left (d x + c\right )}}{2 \, d} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/2*(a*sin(d*x + c)^2 - 2*a*log(sin(d*x + c)) + 2*a*sin(d*x + c) + 2*a/si n(d*x + c))/d
Time = 0.14 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.89 \[ \int \cos (c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \sin \left (d x + c\right )^{2} - 2 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 2 \, a \sin \left (d x + c\right ) + \frac {2 \, a}{\sin \left (d x + c\right )}}{2 \, d} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/2*(a*sin(d*x + c)^2 - 2*a*log(abs(sin(d*x + c))) + 2*a*sin(d*x + c) + 2 *a/sin(d*x + c))/d
Time = 17.69 (sec) , antiderivative size = 140, normalized size of antiderivative = 2.64 \[ \int \cos (c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \] Input:
int(cos(c + d*x)*cot(c + d*x)^2*(a + a*sin(c + d*x)),x)
Output:
(a*log(tan(c/2 + (d*x)/2)))/d - (a*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (a*t an(c/2 + (d*x)/2))/(2*d) - (a + 6*a*tan(c/2 + (d*x)/2)^2 + 4*a*tan(c/2 + ( d*x)/2)^3 + 5*a*tan(c/2 + (d*x)/2)^4)/(d*(2*tan(c/2 + (d*x)/2) + 4*tan(c/2 + (d*x)/2)^3 + 2*tan(c/2 + (d*x)/2)^5))
Time = 0.18 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.43 \[ \int \cos (c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )-\sin \left (d x +c \right )^{3}-2 \sin \left (d x +c \right )^{2}-2\right )}{2 \sin \left (d x +c \right ) d} \] Input:
int(cos(d*x+c)*cot(d*x+c)^2*(a+a*sin(d*x+c)),x)
Output:
(a*( - 2*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x) + 2*log(tan((c + d*x)/2 ))*sin(c + d*x) - sin(c + d*x)**3 - 2*sin(c + d*x)**2 - 2))/(2*sin(c + d*x )*d)