Integrand size = 29, antiderivative size = 72 \[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {5 \text {arctanh}(\cos (c+d x))}{2 a^3 d}-\frac {4 \cot (c+d x)}{a^3 d}-\frac {\cot ^3(c+d x)}{3 a^3 d}+\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^3 d} \] Output:
5/2*arctanh(cos(d*x+c))/a^3/d-4*cot(d*x+c)/a^3/d-1/3*cot(d*x+c)^3/a^3/d+3/ 2*cot(d*x+c)*csc(d*x+c)/a^3/d
Time = 2.95 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.60 \[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\csc ^3(c+d x) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6 \left (30 \cos (c+d x)-22 \cos (3 (c+d x))-60 \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin ^3(c+d x)-18 \sin (2 (c+d x))\right )}{24 a^3 d (1+\sin (c+d x))^3} \] Input:
Integrate[(Cos[c + d*x]^2*Cot[c + d*x]^4)/(a + a*Sin[c + d*x])^3,x]
Output:
-1/24*(Csc[c + d*x]^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6*(30*Cos[c + d*x] - 22*Cos[3*(c + d*x)] - 60*(Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x) /2]])*Sin[c + d*x]^3 - 18*Sin[2*(c + d*x)]))/(a^3*d*(1 + Sin[c + d*x])^3)
Time = 0.46 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3348, 3042, 3236, 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.
\(\displaystyle \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^6}{\sin (c+d x)^4 (a \sin (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3348 |
\(\displaystyle \frac {\int \csc ^4(c+d x) (a-a \sin (c+d x))^3dx}{a^6}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(a-a \sin (c+d x))^3}{\sin (c+d x)^4}dx}{a^6}\) |
\(\Big \downarrow \) 3236 |
\(\displaystyle \frac {\int \left (a^3 \csc ^4(c+d x)-3 a^3 \csc ^3(c+d x)+3 a^3 \csc ^2(c+d x)-a^3 \csc (c+d x)\right )dx}{a^6}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {5 a^3 \text {arctanh}(\cos (c+d x))}{2 d}-\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {4 a^3 \cot (c+d x)}{d}+\frac {3 a^3 \cot (c+d x) \csc (c+d x)}{2 d}}{a^6}\) |
Input:
Int[(Cos[c + d*x]^2*Cot[c + d*x]^4)/(a + a*Sin[c + d*x])^3,x]
Output:
((5*a^3*ArcTanh[Cos[c + d*x]])/(2*d) - (4*a^3*Cot[c + d*x])/d - (a^3*Cot[c + d*x]^3)/(3*d) + (3*a^3*Cot[c + d*x]*Csc[c + d*x])/(2*d))/a^6
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_.), x_Symbol] :> Int[ExpandTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IGt Q[m, 0] && RationalQ[n]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[a^(2*m) Int[(d* Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 0]
Time = 2.75 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.36
method | result | size |
derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-20 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {3}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {15}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{8 d \,a^{3}}\) | \(98\) |
default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-20 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {3}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {15}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{8 d \,a^{3}}\) | \(98\) |
risch | \(-\frac {18 i {\mathrm e}^{4 i \left (d x +c \right )}+9 \,{\mathrm e}^{5 i \left (d x +c \right )}-48 i {\mathrm e}^{2 i \left (d x +c \right )}+22 i-9 \,{\mathrm e}^{i \left (d x +c \right )}}{3 a^{3} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d \,a^{3}}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d \,a^{3}}\) | \(112\) |
Input:
int(cos(d*x+c)^2*cot(d*x+c)^4/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/8/d/a^3*(1/3*tan(1/2*d*x+1/2*c)^3-3*tan(1/2*d*x+1/2*c)^2+15*tan(1/2*d*x+ 1/2*c)-20*ln(tan(1/2*d*x+1/2*c))-1/3/tan(1/2*d*x+1/2*c)^3+3/tan(1/2*d*x+1/ 2*c)^2-15/tan(1/2*d*x+1/2*c))
Time = 0.09 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.71 \[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {44 \, \cos \left (d x + c\right )^{3} - 15 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 15 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 18 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 48 \, \cos \left (d x + c\right )}{12 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="frica s")
Output:
-1/12*(44*cos(d*x + c)^3 - 15*(cos(d*x + c)^2 - 1)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 15*(cos(d*x + c)^2 - 1)*log(-1/2*cos(d*x + c) + 1/2)*s in(d*x + c) + 18*cos(d*x + c)*sin(d*x + c) - 48*cos(d*x + c))/((a^3*d*cos( d*x + c)^2 - a^3*d)*sin(d*x + c))
\[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\int \frac {\cos ^{2}{\left (c + d x \right )} \cot ^{4}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(cos(d*x+c)**2*cot(d*x+c)**4/(a+a*sin(d*x+c))**3,x)
Output:
Integral(cos(c + d*x)**2*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
Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (66) = 132\).
Time = 0.04 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.12 \[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {\frac {45 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {9 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} + \frac {{\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {45 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{3}}{a^{3} \sin \left (d x + c\right )^{3}}}{24 \, d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="maxim a")
Output:
1/24*((45*sin(d*x + c)/(cos(d*x + c) + 1) - 9*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^3 - 60*log(sin(d*x + c)/( cos(d*x + c) + 1))/a^3 + (9*sin(d*x + c)/(cos(d*x + c) + 1) - 45*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)*(cos(d*x + c) + 1)^3/(a^3*sin(d*x + c)^3)) /d
Time = 0.20 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.78 \[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {60 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {110 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 45 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} - \frac {a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 45 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{9}}}{24 \, d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="giac" )
Output:
-1/24*(60*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - (110*tan(1/2*d*x + 1/2*c)^3 - 45*tan(1/2*d*x + 1/2*c)^2 + 9*tan(1/2*d*x + 1/2*c) - 1)/(a^3*tan(1/2*d* x + 1/2*c)^3) - (a^6*tan(1/2*d*x + 1/2*c)^3 - 9*a^6*tan(1/2*d*x + 1/2*c)^2 + 45*a^6*tan(1/2*d*x + 1/2*c))/a^9)/d
Time = 33.40 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.65 \[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^3\,d}-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}-\frac {5\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a^3\,d}+\frac {15\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^3\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {1}{3}\right )}{8\,a^3\,d} \] Input:
int((cos(c + d*x)^2*cot(c + d*x)^4)/(a + a*sin(c + d*x))^3,x)
Output:
tan(c/2 + (d*x)/2)^3/(24*a^3*d) - (3*tan(c/2 + (d*x)/2)^2)/(8*a^3*d) - (5* log(tan(c/2 + (d*x)/2)))/(2*a^3*d) + (15*tan(c/2 + (d*x)/2))/(8*a^3*d) - ( cot(c/2 + (d*x)/2)^3*(15*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2) + 1/3 ))/(8*a^3*d)
Time = 0.16 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.04 \[ \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {-22 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+9 \cos \left (d x +c \right ) \sin \left (d x +c \right )-2 \cos \left (d x +c \right )-15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3}}{6 \sin \left (d x +c \right )^{3} a^{3} d} \] Input:
int(cos(d*x+c)^2*cot(d*x+c)^4/(a+a*sin(d*x+c))^3,x)
Output:
( - 22*cos(c + d*x)*sin(c + d*x)**2 + 9*cos(c + d*x)*sin(c + d*x) - 2*cos( c + d*x) - 15*log(tan((c + d*x)/2))*sin(c + d*x)**3)/(6*sin(c + d*x)**3*a* *3*d)