Integrand size = 29, antiderivative size = 116 \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {2 x}{a^2}+\frac {9 \text {arctanh}(\cos (c+d x))}{8 a^2 d}-\frac {\cos (c+d x)}{a^2 d}-\frac {2 \cot (c+d x)}{a^2 d}+\frac {2 \cot ^3(c+d x)}{3 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d} \] Output:
-2*x/a^2+9/8*arctanh(cos(d*x+c))/a^2/d-cos(d*x+c)/a^2/d-2*cot(d*x+c)/a^2/d +2/3*cot(d*x+c)^3/a^2/d+1/8*cot(d*x+c)*csc(d*x+c)/a^2/d-1/4*cot(d*x+c)*csc (d*x+c)^3/a^2/d
Time = 3.58 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.89 \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\left (\csc \left (\frac {1}{2} (c+d x)\right )+\sec \left (\frac {1}{2} (c+d x)\right )\right )^4 \left (192 \cot (c+d x)+\csc ^2\left (\frac {1}{2} (c+d x)\right ) (128-6 \csc (c+d x))+\csc ^4\left (\frac {1}{2} (c+d x)\right ) (-8+3 \csc (c+d x))+8 \left (3 \csc (c+d x) \left (16 (c+d x)-9 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+9 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-(7+8 \cos (c+d x)) \sec ^4\left (\frac {1}{2} (c+d x)\right )+3 \csc ^3(c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )-6 \csc ^5(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin ^5(c+d x)}{3072 a^2 d (1+\sin (c+d x))^2} \] Input:
Integrate[(Cos[c + d*x]^3*Cot[c + d*x]^5)/(a + a*Sin[c + d*x])^2,x]
Output:
-1/3072*((Csc[(c + d*x)/2] + Sec[(c + d*x)/2])^4*(192*Cot[c + d*x] + Csc[( c + d*x)/2]^2*(128 - 6*Csc[c + d*x]) + Csc[(c + d*x)/2]^4*(-8 + 3*Csc[c + d*x]) + 8*(3*Csc[c + d*x]*(16*(c + d*x) - 9*Log[Cos[(c + d*x)/2]] + 9*Log[ Sin[(c + d*x)/2]]) - (7 + 8*Cos[c + d*x])*Sec[(c + d*x)/2]^4 + 3*Csc[c + d *x]^3*Sin[(c + d*x)/2]^2 - 6*Csc[c + d*x]^5*Sin[(c + d*x)/2]^4))*Sin[c + d *x]^5)/(a^2*d*(1 + Sin[c + d*x])^2)
Time = 0.55 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3354, 3042, 3351, 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.
| \(\displaystyle \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^8}{\sin (c+d x)^5 (a \sin (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {\int \cot ^4(c+d x) \csc (c+d x) (a-a \sin (c+d x))^2dx}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\cos (c+d x)^4 (a-a \sin (c+d x))^2}{\sin (c+d x)^5}dx}{a^4}\) |
| \(\Big \downarrow \) 3351 |
| \(\displaystyle \frac {\int \left (\csc ^5(c+d x) a^6-2 \csc ^4(c+d x) a^6-\csc ^3(c+d x) a^6+4 \csc ^2(c+d x) a^6-\csc (c+d x) a^6+\sin (c+d x) a^6-2 a^6\right )dx}{a^8}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {9 a^6 \text {arctanh}(\cos (c+d x))}{8 d}-\frac {a^6 \cos (c+d x)}{d}+\frac {2 a^6 \cot ^3(c+d x)}{3 d}-\frac {2 a^6 \cot (c+d x)}{d}-\frac {a^6 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac {a^6 \cot (c+d x) \csc (c+d x)}{8 d}-2 a^6 x}{a^8}\) |
Input:
Int[(Cos[c + d*x]^3*Cot[c + d*x]^5)/(a + a*Sin[c + d*x])^2,x]
Output:
(-2*a^6*x + (9*a^6*ArcTanh[Cos[c + d*x]])/(8*d) - (a^6*Cos[c + d*x])/d - ( 2*a^6*Cot[c + d*x])/d + (2*a^6*Cot[c + d*x]^3)/(3*d) + (a^6*Cot[c + d*x]*C sc[c + d*x])/(8*d) - (a^6*Cot[c + d*x]*Csc[c + d*x]^3)/(4*d))/a^8
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[1/a^p Int[Expan dTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x])^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && In tegersQ[m, n, p/2] && ((GtQ[m, 0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (G tQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 0]))
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(a/g)^(2* m) Int[(g*Cos[e + f*x])^(2*m + p)*((d*Sin[e + f*x])^n/(a - b*Sin[e + f*x] )^m), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]
Time = 4.09 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.09
| method | result | size |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {32}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-64 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {4}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {20}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-18 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d \,a^{2}}\) | \(127\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {32}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-64 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {4}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {20}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-18 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d \,a^{2}}\) | \(127\) |
| risch | \(-\frac {2 x}{a^{2}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{2}}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{2}}-\frac {96 i {\mathrm e}^{6 i \left (d x +c \right )}+3 \,{\mathrm e}^{7 i \left (d x +c \right )}-192 i {\mathrm e}^{4 i \left (d x +c \right )}+21 \,{\mathrm e}^{5 i \left (d x +c \right )}+160 i {\mathrm e}^{2 i \left (d x +c \right )}+21 \,{\mathrm e}^{3 i \left (d x +c \right )}-64 i+3 \,{\mathrm e}^{i \left (d x +c \right )}}{12 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d \,a^{2}}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d \,a^{2}}\) | \(186\) |
Input:
int(cos(d*x+c)^3*cot(d*x+c)^5/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/16/d/a^2*(1/4*tan(1/2*d*x+1/2*c)^4-4/3*tan(1/2*d*x+1/2*c)^3+20*tan(1/2*d *x+1/2*c)-32/(1+tan(1/2*d*x+1/2*c)^2)-64*arctan(tan(1/2*d*x+1/2*c))-1/4/ta n(1/2*d*x+1/2*c)^4+4/3/tan(1/2*d*x+1/2*c)^3-20/tan(1/2*d*x+1/2*c)-18*ln(ta n(1/2*d*x+1/2*c)))
Time = 0.10 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.61 \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {96 \, d x \cos \left (d x + c\right )^{4} + 48 \, \cos \left (d x + c\right )^{5} - 192 \, d x \cos \left (d x + c\right )^{2} - 90 \, \cos \left (d x + c\right )^{3} + 96 \, d x - 27 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 27 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 32 \, {\left (4 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 54 \, \cos \left (d x + c\right )}{48 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d\right )}} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
-1/48*(96*d*x*cos(d*x + c)^4 + 48*cos(d*x + c)^5 - 192*d*x*cos(d*x + c)^2 - 90*cos(d*x + c)^3 + 96*d*x - 27*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)* log(1/2*cos(d*x + c) + 1/2) + 27*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*l og(-1/2*cos(d*x + c) + 1/2) - 32*(4*cos(d*x + c)^3 - 3*cos(d*x + c))*sin(d *x + c) + 54*cos(d*x + c))/(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2 + a^2*d)
Timed out. \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**3*cot(d*x+c)**5/(a+a*sin(d*x+c))**2,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 263 vs. \(2 (108) = 216\).
Time = 0.11 (sec) , antiderivative size = 263, normalized size of antiderivative = 2.27 \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {\frac {16 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {224 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {384 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {240 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 3}{\frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {240 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{a^{2}} - \frac {768 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {216 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{192 \, d} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
1/192*((16*sin(d*x + c)/(cos(d*x + c) + 1) - 3*sin(d*x + c)^2/(cos(d*x + c ) + 1)^2 - 224*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 384*sin(d*x + c)^4/(c os(d*x + c) + 1)^4 - 240*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 3)/(a^2*sin (d*x + c)^4/(cos(d*x + c) + 1)^4 + a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 ) + (240*sin(d*x + c)/(cos(d*x + c) + 1) - 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)/a^2 - 768*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 - 216*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^ 2)/d
Time = 0.16 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.37 \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {384 \, {\left (d x + c\right )}}{a^{2}} + \frac {216 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {384}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{2}} - \frac {450 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 240 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 16 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}} - \frac {3 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 16 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 240 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{8}}}{192 \, d} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
-1/192*(384*(d*x + c)/a^2 + 216*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + 384/( (tan(1/2*d*x + 1/2*c)^2 + 1)*a^2) - (450*tan(1/2*d*x + 1/2*c)^4 - 240*tan( 1/2*d*x + 1/2*c)^3 + 16*tan(1/2*d*x + 1/2*c) - 3)/(a^2*tan(1/2*d*x + 1/2*c )^4) - (3*a^6*tan(1/2*d*x + 1/2*c)^4 - 16*a^6*tan(1/2*d*x + 1/2*c)^3 + 240 *a^6*tan(1/2*d*x + 1/2*c))/a^8)/d
Time = 31.18 (sec) , antiderivative size = 232, normalized size of antiderivative = 2.00 \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12\,a^2\,d}+\frac {4\,\mathrm {atan}\left (\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-9}+\frac {16}{16\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-9}\right )}{a^2\,d}-\frac {9\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,a^2\,d}-\frac {20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {56\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}-\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {1}{4}}{d\,\left (16\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}+\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^2\,d} \] Input:
int((cos(c + d*x)^3*cot(c + d*x)^5)/(a + a*sin(c + d*x))^2,x)
Output:
tan(c/2 + (d*x)/2)^4/(64*a^2*d) - tan(c/2 + (d*x)/2)^3/(12*a^2*d) + (4*ata n((9*tan(c/2 + (d*x)/2))/(16*tan(c/2 + (d*x)/2) - 9) + 16/(16*tan(c/2 + (d *x)/2) - 9)))/(a^2*d) - (9*log(tan(c/2 + (d*x)/2)))/(8*a^2*d) - (tan(c/2 + (d*x)/2)^2/4 - (4*tan(c/2 + (d*x)/2))/3 + (56*tan(c/2 + (d*x)/2)^3)/3 + 3 2*tan(c/2 + (d*x)/2)^4 + 20*tan(c/2 + (d*x)/2)^5 + 1/4)/(d*(16*a^2*tan(c/2 + (d*x)/2)^4 + 16*a^2*tan(c/2 + (d*x)/2)^6)) + (5*tan(c/2 + (d*x)/2))/(4* a^2*d)
Time = 20.28 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.11 \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-64 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+16 \cos \left (d x +c \right ) \sin \left (d x +c \right )-6 \cos \left (d x +c \right )-27 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4}-48 \sin \left (d x +c \right )^{4} d x +24 \sin \left (d x +c \right )^{4}}{24 \sin \left (d x +c \right )^{4} a^{2} d} \] Input:
int(cos(d*x+c)^3*cot(d*x+c)^5/(a+a*sin(d*x+c))^2,x)
Output:
( - 24*cos(c + d*x)*sin(c + d*x)**4 - 64*cos(c + d*x)*sin(c + d*x)**3 + 3* cos(c + d*x)*sin(c + d*x)**2 + 16*cos(c + d*x)*sin(c + d*x) - 6*cos(c + d* x) - 27*log(tan((c + d*x)/2))*sin(c + d*x)**4 - 48*sin(c + d*x)**4*d*x + 2 4*sin(c + d*x)**4)/(24*sin(c + d*x)**4*a**2*d)