Integrand size = 27, antiderivative size = 64 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc (c+d x)}{a d}+\frac {\csc ^2(c+d x)}{2 a d}-\frac {\csc ^3(c+d x)}{3 a d}+\frac {\log (\sin (c+d x))}{a d} \] Output:
csc(d*x+c)/a/d+1/2*csc(d*x+c)^2/a/d-1/3*csc(d*x+c)^3/a/d+ln(sin(d*x+c))/a/ d
Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.75 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {6 \csc (c+d x)+3 \csc ^2(c+d x)-2 \csc ^3(c+d x)+6 \log (\sin (c+d x))}{6 a d} \] Input:
Integrate[(Cos[c + d*x]*Cot[c + d*x]^4)/(a + a*Sin[c + d*x]),x]
Output:
(6*Csc[c + d*x] + 3*Csc[c + d*x]^2 - 2*Csc[c + d*x]^3 + 6*Log[Sin[c + d*x] ])/(6*a*d)
Time = 0.31 (sec) , antiderivative size = 47, 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, 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.
\(\displaystyle \int \frac {\cos (c+d x) \cot ^4(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5}{\sin (c+d x)^4 (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \csc ^4(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc ^4(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)}{a^4}d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 84 |
\(\displaystyle \frac {\int \left (\frac {\csc ^4(c+d x)}{a}-\frac {\csc ^3(c+d x)}{a}-\frac {\csc ^2(c+d x)}{a}+\frac {\csc (c+d x)}{a}\right )d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\log (a \sin (c+d x))-\frac {1}{3} \csc ^3(c+d x)+\frac {1}{2} \csc ^2(c+d x)+\csc (c+d x)}{a d}\) |
Input:
Int[(Cos[c + d*x]*Cot[c + d*x]^4)/(a + a*Sin[c + d*x]),x]
Output:
(Csc[c + d*x] + Csc[c + d*x]^2/2 - Csc[c + d*x]^3/3 + 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[((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 = 0.47 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.69
method | result | size |
derivativedivides | \(\frac {\ln \left (\sin \left (d x +c \right )\right )-\frac {1}{3 \sin \left (d x +c \right )^{3}}+\frac {1}{2 \sin \left (d x +c \right )^{2}}+\frac {1}{\sin \left (d x +c \right )}}{d a}\) | \(44\) |
default | \(\frac {\ln \left (\sin \left (d x +c \right )\right )-\frac {1}{3 \sin \left (d x +c \right )^{3}}+\frac {1}{2 \sin \left (d x +c \right )^{2}}+\frac {1}{\sin \left (d x +c \right )}}{d a}\) | \(44\) |
risch | \(-\frac {i x}{a}-\frac {2 i c}{a d}+\frac {2 i \left (3 \,{\mathrm e}^{5 i \left (d x +c \right )}-2 \,{\mathrm e}^{3 i \left (d x +c \right )}+3 i {\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}-3 i {\mathrm e}^{2 i \left (d x +c \right )}\right )}{3 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}\) | \(118\) |
Input:
int(cos(d*x+c)*cot(d*x+c)^4/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(ln(sin(d*x+c))-1/3/sin(d*x+c)^3+1/2/sin(d*x+c)^2+1/sin(d*x+c))
Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.17 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {6 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 6 \, \cos \left (d x + c\right )^{2} - 3 \, \sin \left (d x + c\right ) - 4}{6 \, {\left (a d \cos \left (d x + c\right )^{2} - a d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
1/6*(6*(cos(d*x + c)^2 - 1)*log(1/2*sin(d*x + c))*sin(d*x + c) + 6*cos(d*x + c)^2 - 3*sin(d*x + c) - 4)/((a*d*cos(d*x + c)^2 - a*d)*sin(d*x + c))
\[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cos {\left (c + d x \right )} \cot ^{4}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)**4/(a+a*sin(d*x+c)),x)
Output:
Integral(cos(c + d*x)*cot(c + d*x)**4/(sin(c + d*x) + 1), x)/a
Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.78 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {6 \, \log \left (\sin \left (d x + c\right )\right )}{a} + \frac {6 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) - 2}{a \sin \left (d x + c\right )^{3}}}{6 \, d} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
1/6*(6*log(sin(d*x + c))/a + (6*sin(d*x + c)^2 + 3*sin(d*x + c) - 2)/(a*si n(d*x + c)^3))/d
Time = 0.15 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.75 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {6 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) - 2}{\sin \left (d x + c\right )^{3}} + 6 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{6 \, a d} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/6*((6*sin(d*x + c)^2 + 3*sin(d*x + c) - 2)/sin(d*x + c)^3 + 6*log(abs(si n(d*x + c))))/(a*d)
Time = 32.22 (sec) , antiderivative size = 138, normalized size of antiderivative = 2.16 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}+\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a\,d}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{3}\right )}{8\,a\,d} \] Input:
int((cos(c + d*x)*cot(c + d*x)^4)/(a + a*sin(c + d*x)),x)
Output:
tan(c/2 + (d*x)/2)^2/(8*a*d) - tan(c/2 + (d*x)/2)^3/(24*a*d) + log(tan(c/2 + (d*x)/2))/(a*d) + (3*tan(c/2 + (d*x)/2))/(8*a*d) - log(tan(c/2 + (d*x)/ 2)^2 + 1)/(a*d) + (cot(c/2 + (d*x)/2)^3*(tan(c/2 + (d*x)/2) + 3*tan(c/2 + (d*x)/2)^2 - 1/3))/(8*a*d)
Time = 5.05 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.41 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{3}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3}-3 \sin \left (d x +c \right )^{3}+12 \sin \left (d x +c \right )^{2}+6 \sin \left (d x +c \right )-4}{12 \sin \left (d x +c \right )^{3} a d} \] Input:
int(cos(d*x+c)*cot(d*x+c)^4/(a+a*sin(d*x+c)),x)
Output:
( - 12*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**3 + 12*log(tan((c + d*x) /2))*sin(c + d*x)**3 - 3*sin(c + d*x)**3 + 12*sin(c + d*x)**2 + 6*sin(c + d*x) - 4)/(12*sin(c + d*x)**3*a*d)