Integrand size = 29, antiderivative size = 134 \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {5 x}{2 a}-\frac {5 \text {arctanh}(\cos (c+d x))}{2 a d}+\frac {2 \cos (c+d x)}{a d}+\frac {\cos ^3(c+d x)}{3 a d}+\frac {2 \cot (c+d x)}{a d}-\frac {\cot ^3(c+d x)}{3 a d}+\frac {\cot (c+d x) \csc (c+d x)}{2 a d}+\frac {\cos (c+d x) \sin (c+d x)}{2 a d} \] Output:
5/2*x/a-5/2*arctanh(cos(d*x+c))/a/d+2*cos(d*x+c)/a/d+1/3*cos(d*x+c)^3/a/d+ 2*cot(d*x+c)/a/d-1/3*cot(d*x+c)^3/a/d+1/2*cot(d*x+c)*csc(d*x+c)/a/d+1/2*co s(d*x+c)*sin(d*x+c)/a/d
Time = 2.11 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.47 \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\csc ^3(c+d x) \left (-30 \cos (c+d x)+65 \cos (3 (c+d x))-3 \cos (5 (c+d x))-180 c \sin (c+d x)-180 d x \sin (c+d x)+180 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-180 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-75 \sin (2 (c+d x))+60 c \sin (3 (c+d x))+60 d x \sin (3 (c+d x))-60 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))+60 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))+24 \sin (4 (c+d x))+\sin (6 (c+d x))\right )}{96 a d} \] Input:
Integrate[(Cos[c + d*x]^4*Cot[c + d*x]^4)/(a + a*Sin[c + d*x]),x]
Output:
-1/96*(Csc[c + d*x]^3*(-30*Cos[c + d*x] + 65*Cos[3*(c + d*x)] - 3*Cos[5*(c + d*x)] - 180*c*Sin[c + d*x] - 180*d*x*Sin[c + d*x] + 180*Log[Cos[(c + d* x)/2]]*Sin[c + d*x] - 180*Log[Sin[(c + d*x)/2]]*Sin[c + d*x] - 75*Sin[2*(c + d*x)] + 60*c*Sin[3*(c + d*x)] + 60*d*x*Sin[3*(c + d*x)] - 60*Log[Cos[(c + d*x)/2]]*Sin[3*(c + d*x)] + 60*Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] + 24*Sin[4*(c + d*x)] + Sin[6*(c + d*x)]))/(a*d)
Time = 0.54 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.99, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {3042, 3318, 3042, 25, 3071, 252, 254, 2009, 3072, 252, 254, 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.
| \(\displaystyle \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^8}{\sin (c+d x)^4 (a \sin (c+d x)+a)}dx\) |
| \(\Big \downarrow \) 3318 |
| \(\displaystyle \frac {\int \cos ^2(c+d x) \cot ^4(c+d x)dx}{a}-\frac {\int \cos ^3(c+d x) \cot ^3(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \tan \left (c+d x+\frac {\pi }{2}\right )^4dx}{a}-\frac {\int -\sin \left (c+d x+\frac {\pi }{2}\right )^3 \tan \left (c+d x+\frac {\pi }{2}\right )^3dx}{a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \tan \left (c+d x+\frac {\pi }{2}\right )^4dx}{a}+\frac {\int \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )^3dx}{a}\) |
| \(\Big \downarrow \) 3071 |
| \(\displaystyle \frac {\int \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )^3dx}{a}-\frac {\int \frac {\cot ^6(c+d x)}{\left (\cot ^2(c+d x)+1\right )^2}d\cot (c+d x)}{a d}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {\int \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )^3dx}{a}-\frac {\frac {5}{2} \int \frac {\cot ^4(c+d x)}{\cot ^2(c+d x)+1}d\cot (c+d x)-\frac {\cot ^5(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}}{a d}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle \frac {\int \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )^3dx}{a}-\frac {\frac {5}{2} \int \left (\cot ^2(c+d x)+\frac {1}{\cot ^2(c+d x)+1}-1\right )d\cot (c+d x)-\frac {\cot ^5(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}}{a d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\int \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )^3dx}{a}-\frac {\frac {5}{2} \left (\arctan (\cot (c+d x))+\frac {1}{3} \cot ^3(c+d x)-\cot (c+d x)\right )-\frac {\cot ^5(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}}{a d}\) |
| \(\Big \downarrow \) 3072 |
| \(\displaystyle \frac {\int \frac {\cos ^6(c+d x)}{\left (1-\cos ^2(c+d x)\right )^2}d\cos (c+d x)}{a d}-\frac {\frac {5}{2} \left (\arctan (\cot (c+d x))+\frac {1}{3} \cot ^3(c+d x)-\cot (c+d x)\right )-\frac {\cot ^5(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}}{a d}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {\frac {\cos ^5(c+d x)}{2 \left (1-\cos ^2(c+d x)\right )}-\frac {5}{2} \int \frac {\cos ^4(c+d x)}{1-\cos ^2(c+d x)}d\cos (c+d x)}{a d}-\frac {\frac {5}{2} \left (\arctan (\cot (c+d x))+\frac {1}{3} \cot ^3(c+d x)-\cot (c+d x)\right )-\frac {\cot ^5(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}}{a d}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle \frac {\frac {\cos ^5(c+d x)}{2 \left (1-\cos ^2(c+d x)\right )}-\frac {5}{2} \int \left (-\cos ^2(c+d x)+\frac {1}{1-\cos ^2(c+d x)}-1\right )d\cos (c+d x)}{a d}-\frac {\frac {5}{2} \left (\arctan (\cot (c+d x))+\frac {1}{3} \cot ^3(c+d x)-\cot (c+d x)\right )-\frac {\cot ^5(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}}{a d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\cos ^5(c+d x)}{2 \left (1-\cos ^2(c+d x)\right )}-\frac {5}{2} \left (\text {arctanh}(\cos (c+d x))-\frac {1}{3} \cos ^3(c+d x)-\cos (c+d x)\right )}{a d}-\frac {\frac {5}{2} \left (\arctan (\cot (c+d x))+\frac {1}{3} \cot ^3(c+d x)-\cot (c+d x)\right )-\frac {\cot ^5(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}}{a d}\) |
Input:
Int[(Cos[c + d*x]^4*Cot[c + d*x]^4)/(a + a*Sin[c + d*x]),x]
Output:
(Cos[c + d*x]^5/(2*(1 - Cos[c + d*x]^2)) - (5*(ArcTanh[Cos[c + d*x]] - Cos [c + d*x] - Cos[c + d*x]^3/3))/2)/(a*d) - (-1/2*Cot[c + d*x]^5/(1 + Cot[c + d*x]^2) + (5*(ArcTan[Cot[c + d*x]] - Cot[c + d*x] + Cot[c + d*x]^3/3))/2 )/(a*d)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[b*(ff/f) Subst[I nt[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, b*(Tan[e + f*x]/ff)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ (ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)], x ]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g^2/a Int [(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[g^2/(b*d) Int [(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0]
Time = 3.59 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.32
| method | result | size |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+48 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+64 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {112}{3}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+40 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {9}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+20 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}\) | \(177\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+48 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+64 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {112}{3}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+40 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {9}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+20 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}\) | \(177\) |
| risch | \(\frac {5 x}{2 a}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 d a}+\frac {9 \,{\mathrm e}^{i \left (d x +c \right )}}{8 a d}+\frac {9 \,{\mathrm e}^{-i \left (d x +c \right )}}{8 a d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d a}-\frac {-18 i {\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{5 i \left (d x +c \right )}+24 i {\mathrm e}^{2 i \left (d x +c \right )}-14 i-3 \,{\mathrm e}^{i \left (d x +c \right )}}{3 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d a}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d a}+\frac {\cos \left (3 d x +3 c \right )}{12 a d}\) | \(205\) |
Input:
int(cos(d*x+c)^4*cot(d*x+c)^4/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/8/d/a*(1/3*tan(1/2*d*x+1/2*c)^3-tan(1/2*d*x+1/2*c)^2-9*tan(1/2*d*x+1/2*c )+16*(-1/2*tan(1/2*d*x+1/2*c)^5+3*tan(1/2*d*x+1/2*c)^4+4*tan(1/2*d*x+1/2*c )^2+1/2*tan(1/2*d*x+1/2*c)+7/3)/(1+tan(1/2*d*x+1/2*c)^2)^3+40*arctan(tan(1 /2*d*x+1/2*c))-1/3/tan(1/2*d*x+1/2*c)^3+1/tan(1/2*d*x+1/2*c)^2+9/tan(1/2*d *x+1/2*c)+20*ln(tan(1/2*d*x+1/2*c)))
Time = 0.09 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.25 \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {6 \, \cos \left (d x + c\right )^{5} - 40 \, \cos \left (d x + c\right )^{3} + 15 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 15 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 2 \, {\left (2 \, \cos \left (d x + c\right )^{5} + 15 \, d x \cos \left (d x + c\right )^{2} + 10 \, \cos \left (d x + c\right )^{3} - 15 \, d x - 15 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 30 \, \cos \left (d x + c\right )}{12 \, {\left (a d \cos \left (d x + c\right )^{2} - a d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
-1/12*(6*cos(d*x + c)^5 - 40*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*co s(d*x + c) + 1/2)*sin(d*x + c) - 2*(2*cos(d*x + c)^5 + 15*d*x*cos(d*x + c) ^2 + 10*cos(d*x + c)^3 - 15*d*x - 15*cos(d*x + c))*sin(d*x + c) + 30*cos(d *x + c))/((a*d*cos(d*x + c)^2 - a*d)*sin(d*x + c))
\[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cos ^{4}{\left (c + d x \right )} \cot ^{4}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(cos(d*x+c)**4*cot(d*x+c)**4/(a+a*sin(d*x+c)),x)
Output:
Integral(cos(c + d*x)**4*cot(c + d*x)**4/(sin(c + d*x) + 1), x)/a
Leaf count of result is larger than twice the leaf count of optimal. 362 vs. \(2 (122) = 244\).
Time = 0.12 (sec) , antiderivative size = 362, normalized size of antiderivative = 2.70 \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {\frac {27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a} - \frac {\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {24 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {121 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {102 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {201 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {80 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {147 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {3 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - 1}{\frac {a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {a \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{24 \, d} \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
-1/24*((27*sin(d*x + c)/(cos(d*x + c) + 1) + 3*sin(d*x + c)^2/(cos(d*x + c ) + 1)^2 - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a - (3*sin(d*x + c)/(cos(d *x + c) + 1) + 24*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 121*sin(d*x + c)^3 /(cos(d*x + c) + 1)^3 + 102*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 201*sin( d*x + c)^5/(cos(d*x + c) + 1)^5 + 80*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 147*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 1)/(a*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*a*sin(d*x + c)^5/(c os(d*x + c) + 1)^5 + 3*a*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + a*sin(d*x + c)^9/(cos(d*x + c) + 1)^9) - 120*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/ a - 60*log(sin(d*x + c)/(cos(d*x + c) + 1))/a)/d
Time = 0.17 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.70 \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {180 \, {\left (d x + c\right )}}{a} + \frac {180 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} + \frac {3 \, {\left (a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 27 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{3}} - \frac {110 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 111 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 240 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 273 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 306 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 253 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}^{3} a}}{72 \, d} \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/72*(180*(d*x + c)/a + 180*log(abs(tan(1/2*d*x + 1/2*c)))/a + 3*(a^2*tan( 1/2*d*x + 1/2*c)^3 - 3*a^2*tan(1/2*d*x + 1/2*c)^2 - 27*a^2*tan(1/2*d*x + 1 /2*c))/a^3 - (110*tan(1/2*d*x + 1/2*c)^9 - 9*tan(1/2*d*x + 1/2*c)^8 - 111* tan(1/2*d*x + 1/2*c)^7 - 240*tan(1/2*d*x + 1/2*c)^6 - 273*tan(1/2*d*x + 1/ 2*c)^5 - 306*tan(1/2*d*x + 1/2*c)^4 - 253*tan(1/2*d*x + 1/2*c)^3 - 72*tan( 1/2*d*x + 1/2*c)^2 - 9*tan(1/2*d*x + 1/2*c) + 3)/((tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c))^3*a))/d
Time = 31.82 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.16 \[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}-\frac {5\,\mathrm {atan}\left (\frac {25\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{25\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-25}+\frac {25}{25\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-25}\right )}{a\,d}+\frac {5\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+49\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {80\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+67\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+34\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {121\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{3}}{d\,\left (8\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+24\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+24\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}-\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a\,d} \] Input:
int((cos(c + d*x)^4*cot(c + d*x)^4)/(a + a*sin(c + d*x)),x)
Output:
tan(c/2 + (d*x)/2)^3/(24*a*d) - tan(c/2 + (d*x)/2)^2/(8*a*d) - (5*atan((25 *tan(c/2 + (d*x)/2))/(25*tan(c/2 + (d*x)/2) - 25) + 25/(25*tan(c/2 + (d*x) /2) - 25)))/(a*d) + (5*log(tan(c/2 + (d*x)/2)))/(2*a*d) + (tan(c/2 + (d*x) /2) + 8*tan(c/2 + (d*x)/2)^2 + (121*tan(c/2 + (d*x)/2)^3)/3 + 34*tan(c/2 + (d*x)/2)^4 + 67*tan(c/2 + (d*x)/2)^5 + (80*tan(c/2 + (d*x)/2)^6)/3 + 49*t an(c/2 + (d*x)/2)^7 + tan(c/2 + (d*x)/2)^8 - 1/3)/(d*(8*a*tan(c/2 + (d*x)/ 2)^3 + 24*a*tan(c/2 + (d*x)/2)^5 + 24*a*tan(c/2 + (d*x)/2)^7 + 8*a*tan(c/2 + (d*x)/2)^9)) - (9*tan(c/2 + (d*x)/2))/(8*a*d)
\[ \int \frac {\cos ^4(c+d x) \cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\int \frac {\cos \left (d x +c \right )^{4} \cot \left (d x +c \right )^{4}}{\sin \left (d x +c \right ) a +a}d x \] Input:
int(cos(d*x+c)^4*cot(d*x+c)^4/(a+a*sin(d*x+c)),x)
Output:
int(cos(d*x+c)^4*cot(d*x+c)^4/(a+a*sin(d*x+c)),x)