Integrand size = 27, antiderivative size = 83 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {4 \csc (c+d x)}{a^3 d}+\frac {3 \csc ^2(c+d x)}{2 a^3 d}-\frac {\csc ^3(c+d x)}{3 a^3 d}-\frac {4 \log (\sin (c+d x))}{a^3 d}+\frac {4 \log (1+\sin (c+d x))}{a^3 d} \] Output:
-4*csc(d*x+c)/a^3/d+3/2*csc(d*x+c)^2/a^3/d-1/3*csc(d*x+c)^3/a^3/d-4*ln(sin (d*x+c))/a^3/d+4*ln(1+sin(d*x+c))/a^3/d
Time = 0.15 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.71 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {24 \csc (c+d x)-9 \csc ^2(c+d x)+2 \csc ^3(c+d x)+24 \log (\sin (c+d x))-24 \log (1+\sin (c+d x))}{6 a^3 d} \] Input:
Integrate[(Cos[c + d*x]*Cot[c + d*x]^4)/(a + a*Sin[c + d*x])^3,x]
Output:
-1/6*(24*Csc[c + d*x] - 9*Csc[c + d*x]^2 + 2*Csc[c + d*x]^3 + 24*Log[Sin[c + d*x]] - 24*Log[1 + Sin[c + d*x]])/(a^3*d)
Time = 0.32 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, 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.
\(\displaystyle \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5}{\sin (c+d x)^4 (a \sin (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \frac {\csc ^4(c+d x) (a-a \sin (c+d x))^2}{\sin (c+d x) a+a}d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc ^4(c+d x) (a-a \sin (c+d x))^2}{a^4 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\frac {\csc ^4(c+d x)}{a^3}-\frac {3 \csc ^3(c+d x)}{a^3}+\frac {4 \csc ^2(c+d x)}{a^3}-\frac {4 \csc (c+d x)}{a^3}+\frac {4}{a^2 (\sin (c+d x) a+a)}\right )d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {\csc ^3(c+d x)}{3 a^2}+\frac {3 \csc ^2(c+d x)}{2 a^2}-\frac {4 \csc (c+d x)}{a^2}-\frac {4 \log (a \sin (c+d x))}{a^2}+\frac {4 \log (a \sin (c+d x)+a)}{a^2}}{a d}\) |
Input:
Int[(Cos[c + d*x]*Cot[c + d*x]^4)/(a + a*Sin[c + d*x])^3,x]
Output:
((-4*Csc[c + d*x])/a^2 + (3*Csc[c + d*x]^2)/(2*a^2) - Csc[c + d*x]^3/(3*a^ 2) - (4*Log[a*Sin[c + d*x]])/a^2 + (4*Log[a + a*Sin[c + d*x]])/a^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[((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 = 3.41 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {-\frac {1}{3 \sin \left (d x +c \right )^{3}}+\frac {3}{2 \sin \left (d x +c \right )^{2}}-\frac {4}{\sin \left (d x +c \right )}-4 \ln \left (\sin \left (d x +c \right )\right )+4 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\) | \(59\) |
default | \(\frac {-\frac {1}{3 \sin \left (d x +c \right )^{3}}+\frac {3}{2 \sin \left (d x +c \right )^{2}}-\frac {4}{\sin \left (d x +c \right )}-4 \ln \left (\sin \left (d x +c \right )\right )+4 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\) | \(59\) |
risch | \(-\frac {2 i \left (12 \,{\mathrm e}^{5 i \left (d x +c \right )}-28 \,{\mathrm e}^{3 i \left (d x +c \right )}-9 i {\mathrm e}^{4 i \left (d x +c \right )}+12 \,{\mathrm e}^{i \left (d x +c \right )}+9 i {\mathrm e}^{2 i \left (d x +c \right )}\right )}{3 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {4 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{3}}+\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}\) | \(123\) |
Input:
int(cos(d*x+c)*cot(d*x+c)^4/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d/a^3*(-1/3/sin(d*x+c)^3+3/2/sin(d*x+c)^2-4/sin(d*x+c)-4*ln(sin(d*x+c))+ 4*ln(1+sin(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.28 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {24 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 24 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 24 \, \cos \left (d x + c\right )^{2} + 9 \, \sin \left (d x + c\right ) - 26}{6 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="fricas" )
Output:
-1/6*(24*(cos(d*x + c)^2 - 1)*log(1/2*sin(d*x + c))*sin(d*x + c) - 24*(cos (d*x + c)^2 - 1)*log(sin(d*x + c) + 1)*sin(d*x + c) + 24*cos(d*x + c)^2 + 9*sin(d*x + c) - 26)/((a^3*d*cos(d*x + c)^2 - a^3*d)*sin(d*x + c))
\[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\int \frac {\cos {\left (c + d x \right )} \cot ^{4}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)**4/(a+a*sin(d*x+c))**3,x)
Output:
Integral(cos(c + d*x)*cot(c + d*x)**4/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x)/a**3
Time = 0.06 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.78 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {24 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} - \frac {24 \, \log \left (\sin \left (d x + c\right )\right )}{a^{3}} - \frac {24 \, \sin \left (d x + c\right )^{2} - 9 \, \sin \left (d x + c\right ) + 2}{a^{3} \sin \left (d x + c\right )^{3}}}{6 \, d} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="maxima" )
Output:
1/6*(24*log(sin(d*x + c) + 1)/a^3 - 24*log(sin(d*x + c))/a^3 - (24*sin(d*x + c)^2 - 9*sin(d*x + c) + 2)/(a^3*sin(d*x + c)^3))/d
Time = 0.17 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.86 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {4 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} d} - \frac {4 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3} d} - \frac {24 \, \sin \left (d x + c\right )^{2} - 9 \, \sin \left (d x + c\right ) + 2}{6 \, a^{3} d \sin \left (d x + c\right )^{3}} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="giac")
Output:
4*log(abs(sin(d*x + c) + 1))/(a^3*d) - 4*log(abs(sin(d*x + c)))/(a^3*d) - 1/6*(24*sin(d*x + c)^2 - 9*sin(d*x + c) + 2)/(a^3*d*sin(d*x + c)^3)
Time = 33.04 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.67 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^3\,d}-\frac {4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}+\frac {8\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^3\,d}-\frac {17\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^3\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {1}{3}\right )}{8\,a^3\,d} \] Input:
int((cos(c + d*x)*cot(c + d*x)^4)/(a + a*sin(c + d*x))^3,x)
Output:
(3*tan(c/2 + (d*x)/2)^2)/(8*a^3*d) - tan(c/2 + (d*x)/2)^3/(24*a^3*d) - (4* log(tan(c/2 + (d*x)/2)))/(a^3*d) + (8*log(tan(c/2 + (d*x)/2) + 1))/(a^3*d) - (17*tan(c/2 + (d*x)/2))/(8*a^3*d) - (cot(c/2 + (d*x)/2)^3*(17*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2) + 1/3))/(8*a^3*d)
Time = 0.18 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.06 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {96 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3}-48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3}-9 \sin \left (d x +c \right )^{3}-48 \sin \left (d x +c \right )^{2}+18 \sin \left (d x +c \right )-4}{12 \sin \left (d x +c \right )^{3} a^{3} d} \] Input:
int(cos(d*x+c)*cot(d*x+c)^4/(a+a*sin(d*x+c))^3,x)
Output:
(96*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3 - 48*log(tan((c + d*x)/2))*s in(c + d*x)**3 - 9*sin(c + d*x)**3 - 48*sin(c + d*x)**2 + 18*sin(c + d*x) - 4)/(12*sin(c + d*x)**3*a**3*d)