Integrand size = 29, antiderivative size = 62 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\csc (c+d x)}{a d}-\frac {\log (\sin (c+d x))}{a d}-\frac {\sin (c+d x)}{a d}+\frac {\sin ^2(c+d x)}{2 a d} \] Output:
-csc(d*x+c)/a/d-ln(sin(d*x+c))/a/d-sin(d*x+c)/a/d+1/2*sin(d*x+c)^2/a/d
Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.74 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {2 \csc (c+d x)+2 \log (\sin (c+d x))+2 \sin (c+d x)-\sin ^2(c+d x)}{2 a d} \] Input:
Integrate[(Cos[c + d*x]^3*Cot[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
Output:
-1/2*(2*Csc[c + d*x] + 2*Log[Sin[c + d*x]] + 2*Sin[c + d*x] - Sin[c + d*x] ^2)/(a*d)
Time = 0.32 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.95, 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.
\(\displaystyle \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5}{\sin (c+d x)^2 (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \csc ^2(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc ^2(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)}{a^2}d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 84 |
\(\displaystyle \frac {\int \left (a \csc ^2(c+d x)-a \csc (c+d x)-a+a \sin (c+d x)\right )d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{2} a^2 \sin ^2(c+d x)-a^2 \sin (c+d x)-a^2 \csc (c+d x)-a^2 \log (a \sin (c+d x))}{a^3 d}\) |
Input:
Int[(Cos[c + d*x]^3*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^3*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 = 0.77 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.74
method | result | size |
derivativedivides | \(\frac {\frac {\sin \left (d x +c \right )^{2}}{2}-\sin \left (d x +c \right )-\ln \left (\sin \left (d x +c \right )\right )-\frac {1}{\sin \left (d x +c \right )}}{d a}\) | \(46\) |
default | \(\frac {\frac {\sin \left (d x +c \right )^{2}}{2}-\sin \left (d x +c \right )-\ln \left (\sin \left (d x +c \right )\right )-\frac {1}{\sin \left (d x +c \right )}}{d a}\) | \(46\) |
risch | \(\frac {i x}{a}-\frac {{\mathrm e}^{2 i \left (d x +c \right )}}{8 a d}+\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 d a}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 d a}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{8 a d}+\frac {2 i c}{a d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )}}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}\) | \(140\) |
Input:
int(cos(d*x+c)^3*cot(d*x+c)^2/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(1/2*sin(d*x+c)^2-sin(d*x+c)-ln(sin(d*x+c))-1/sin(d*x+c))
Time = 0.09 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.05 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {4 \, \cos \left (d x + c\right )^{2} - {\left (2 \, \cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 4 \, \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 8}{4 \, a d \sin \left (d x + c\right )} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
1/4*(4*cos(d*x + c)^2 - (2*cos(d*x + c)^2 - 1)*sin(d*x + c) - 4*log(1/2*si n(d*x + c))*sin(d*x + c) - 8)/(a*d*sin(d*x + c))
Timed out. \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**3*cot(d*x+c)**2/(a+a*sin(d*x+c)),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.84 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )}{a} - \frac {2 \, \log \left (\sin \left (d x + c\right )\right )}{a} - \frac {2}{a \sin \left (d x + c\right )}}{2 \, d} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
1/2*((sin(d*x + c)^2 - 2*sin(d*x + c))/a - 2*log(sin(d*x + c))/a - 2/(a*si n(d*x + c)))/d
Time = 0.12 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin \left (d x + c\right )^{2} - \frac {2}{\sin \left (d x + c\right )} - 2 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 2 \, \sin \left (d x + c\right )}{2 \, a d} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/2*(sin(d*x + c)^2 - 2/sin(d*x + c) - 2*log(abs(sin(d*x + c))) - 2*sin(d* x + c))/(a*d)
Time = 32.72 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.35 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1}{d\,\left (2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d} \] Input:
int((cos(c + d*x)^3*cot(c + d*x)^2)/(a + a*sin(c + d*x)),x)
Output:
log(tan(c/2 + (d*x)/2)^2 + 1)/(a*d) - (6*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^3 + 5*tan(c/2 + (d*x)/2)^4 + 1)/(d*(2*a*tan(c/2 + (d*x)/2) + 4* a*tan(c/2 + (d*x)/2)^3 + 2*a*tan(c/2 + (d*x)/2)^5)) - tan(c/2 + (d*x)/2)/( 2*a*d) - log(tan(c/2 + (d*x)/2))/(a*d)
Time = 156.15 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.23 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )+\sin \left (d x +c \right )^{3}-2 \sin \left (d x +c \right )^{2}-2}{2 \sin \left (d x +c \right ) a d} \] Input:
int(cos(d*x+c)^3*cot(d*x+c)^2/(a+a*sin(d*x+c)),x)
Output:
(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)*a*d)