Integrand size = 29, antiderivative size = 60 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc (c+d x)}{a d}-\frac {\csc ^2(c+d x)}{2 a d}-\frac {\log (\sin (c+d x))}{a d}+\frac {\sin (c+d x)}{a d} \] Output:
csc(d*x+c)/a/d-1/2*csc(d*x+c)^2/a/d-ln(sin(d*x+c))/a/d+sin(d*x+c)/a/d
Time = 0.10 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {2 \csc (c+d x)-\csc ^2(c+d x)-2 \log (\sin (c+d x))+2 \sin (c+d x)}{2 a d} \] Input:
Integrate[(Cos[c + d*x]^2*Cot[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
Output:
(2*Csc[c + d*x] - Csc[c + d*x]^2 - 2*Log[Sin[c + d*x]] + 2*Sin[c + d*x])/( 2*a*d)
Time = 0.32 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.82, 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, 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.
\(\displaystyle \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5}{\sin (c+d x)^3 (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \csc ^3(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc ^3(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)}{a^3}d(a \sin (c+d x))}{a^2 d}\) |
\(\Big \downarrow \) 84 |
\(\displaystyle \frac {\int \left (\csc ^3(c+d x)-\csc ^2(c+d x)-\csc (c+d x)+1\right )d(a \sin (c+d x))}{a^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a \sin (c+d x)-\frac {1}{2} a \csc ^2(c+d x)+a \csc (c+d x)-a \log (a \sin (c+d x))}{a^2 d}\) |
Input:
Int[(Cos[c + d*x]^2*Cot[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
Output:
(a*Csc[c + d*x] - (a*Csc[c + d*x]^2)/2 - a*Log[a*Sin[c + d*x]] + a*Sin[c + d*x])/(a^2*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 = 42, normalized size of antiderivative = 0.70
method | result | size |
derivativedivides | \(\frac {\sin \left (d x +c \right )-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\frac {1}{\sin \left (d x +c \right )}-\ln \left (\sin \left (d x +c \right )\right )}{d a}\) | \(42\) |
default | \(\frac {\sin \left (d x +c \right )-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\frac {1}{\sin \left (d x +c \right )}-\ln \left (\sin \left (d x +c \right )\right )}{d a}\) | \(42\) |
risch | \(\frac {i x}{a}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 d a}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 d a}+\frac {2 i c}{a d}+\frac {2 i \left (-i {\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}\) | \(130\) |
Input:
int(cos(d*x+c)^2*cot(d*x+c)^3/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(sin(d*x+c)-1/2/sin(d*x+c)^2+1/sin(d*x+c)-ln(sin(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {2 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 2 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) - 1}{2 \, {\left (a d \cos \left (d x + c\right )^{2} - a d\right )}} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
-1/2*(2*(cos(d*x + c)^2 - 1)*log(1/2*sin(d*x + c)) - 2*(cos(d*x + c)^2 - 2 )*sin(d*x + c) - 1)/(a*d*cos(d*x + c)^2 - a*d)
\[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cos ^{2}{\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(cos(d*x+c)**2*cot(d*x+c)**3/(a+a*sin(d*x+c)),x)
Output:
Integral(cos(c + d*x)**2*cot(c + d*x)**3/(sin(c + d*x) + 1), x)/a
Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, \log \left (\sin \left (d x + c\right )\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{a} - \frac {2 \, \sin \left (d x + c\right ) - 1}{a \sin \left (d x + c\right )^{2}}}{2 \, d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
-1/2*(2*log(sin(d*x + c))/a - 2*sin(d*x + c)/a - (2*sin(d*x + c) - 1)/(a*s in(d*x + c)^2))/d
Time = 0.14 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {2 \, \sin \left (d x + c\right ) - 1}{\sin \left (d x + c\right )^{2}} - 2 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 2 \, \sin \left (d x + c\right )}{2 \, a d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/2*((2*sin(d*x + c) - 1)/sin(d*x + c)^2 - 2*log(abs(sin(d*x + c))) + 2*si n(d*x + c))/(a*d)
Time = 31.95 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.50 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{2}}{d\,\left (4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d}+\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d} \] Input:
int((cos(c + d*x)^2*cot(c + d*x)^3)/(a + a*sin(c + d*x)),x)
Output:
(2*tan(c/2 + (d*x)/2) - tan(c/2 + (d*x)/2)^2/2 + 10*tan(c/2 + (d*x)/2)^3 - 1/2)/(d*(4*a*tan(c/2 + (d*x)/2)^2 + 4*a*tan(c/2 + (d*x)/2)^4)) - log(tan( c/2 + (d*x)/2))/(a*d) - tan(c/2 + (d*x)/2)^2/(8*a*d) + tan(c/2 + (d*x)/2)/ (2*a*d) + log(tan(c/2 + (d*x)/2)^2 + 1)/(a*d)
Time = 63.80 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.50 \[ \int \frac {\cos ^2(c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}-8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}+8 \sin \left (d x +c \right )^{3}+3 \sin \left (d x +c \right )^{2}+8 \sin \left (d x +c \right )-4}{8 \sin \left (d x +c \right )^{2} a d} \] Input:
int(cos(d*x+c)^2*cot(d*x+c)^3/(a+a*sin(d*x+c)),x)
Output:
(8*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2 - 8*log(tan((c + d*x)/2))* sin(c + d*x)**2 + 8*sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 8*sin(c + d*x) - 4)/(8*sin(c + d*x)**2*a*d)