Integrand size = 29, antiderivative size = 97 \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {x}{a^3}+\frac {13 \text {arctanh}(\cos (c+d x))}{8 a^3 d}+\frac {\cot (c+d x)}{a^3 d}+\frac {\cot ^3(c+d x)}{a^3 d}-\frac {11 \cot (c+d x) \csc (c+d x)}{8 a^3 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^3 d} \] Output:
x/a^3+13/8*arctanh(cos(d*x+c))/a^3/d+cot(d*x+c)/a^3/d+cot(d*x+c)^3/a^3/d-1 1/8*cot(d*x+c)*csc(d*x+c)/a^3/d-1/4*cot(d*x+c)*csc(d*x+c)^3/a^3/d
Time = 4.20 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.70 \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6 \left (-22 \csc ^2\left (\frac {1}{2} (c+d x)\right )+22 \sec ^2\left (\frac {1}{2} (c+d x)\right )+\sec ^4\left (\frac {1}{2} (c+d x)\right )+8 \left (8 c+8 d x+13 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-13 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-8 \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )\right )+\csc ^4\left (\frac {1}{2} (c+d x)\right ) (-1+4 \sin (c+d x))\right )}{64 a^3 d (1+\sin (c+d x))^3} \] Input:
Integrate[(Cos[c + d*x]^3*Cot[c + d*x]^5)/(a + a*Sin[c + d*x])^3,x]
Output:
((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6*(-22*Csc[(c + d*x)/2]^2 + 22*Sec[ (c + d*x)/2]^2 + Sec[(c + d*x)/2]^4 + 8*(8*c + 8*d*x + 13*Log[Cos[(c + d*x )/2]] - 13*Log[Sin[(c + d*x)/2]] - 8*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4) + Csc[(c + d*x)/2]^4*(-1 + 4*Sin[c + d*x])))/(64*a^3*d*(1 + Sin[c + d*x])^3)
Time = 0.59 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.04, 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, 3352, 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.
\(\displaystyle \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^8}{\sin (c+d x)^5 (a \sin (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3354 |
\(\displaystyle \frac {\int \cot ^2(c+d x) \csc ^3(c+d x) (a-a \sin (c+d x))^3dx}{a^6}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\cos (c+d x)^2 (a-a \sin (c+d x))^3}{\sin (c+d x)^5}dx}{a^6}\) |
\(\Big \downarrow \) 3352 |
\(\displaystyle \frac {\int \left (\cot ^2(c+d x) \csc ^3(c+d x) a^3-\cot ^2(c+d x) a^3-3 \cot ^2(c+d x) \csc ^2(c+d x) a^3+3 \cot ^2(c+d x) \csc (c+d x) a^3\right )dx}{a^6}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {13 a^3 \text {arctanh}(\cos (c+d x))}{8 d}+\frac {a^3 \cot ^3(c+d x)}{d}+\frac {a^3 \cot (c+d x)}{d}-\frac {a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {11 a^3 \cot (c+d x) \csc (c+d x)}{8 d}+a^3 x}{a^6}\) |
Input:
Int[(Cos[c + d*x]^3*Cot[c + d*x]^5)/(a + a*Sin[c + d*x])^3,x]
Output:
(a^3*x + (13*a^3*ArcTanh[Cos[c + d*x]])/(8*d) + (a^3*Cot[c + d*x])/d + (a^ 3*Cot[c + d*x]^3)/d - (11*a^3*Cot[c + d*x]*Csc[c + d*x])/(8*d) - (a^3*Cot[ c + d*x]*Csc[c + d*x]^3)/(4*d))/a^6
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig [(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 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 = 1.46 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.40
method | result | size |
derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {6}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-26 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+32 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d \,a^{3}}\) | \(136\) |
default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {6}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-26 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+32 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d \,a^{3}}\) | \(136\) |
risch | \(\frac {x}{a^{3}}+\frac {11 \,{\mathrm e}^{7 i \left (d x +c \right )}-19 \,{\mathrm e}^{5 i \left (d x +c \right )}-16 i {\mathrm e}^{6 i \left (d x +c \right )}-19 \,{\mathrm e}^{3 i \left (d x +c \right )}+11 \,{\mathrm e}^{i \left (d x +c \right )}+16 i {\mathrm e}^{2 i \left (d x +c \right )}}{4 a^{3} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d \,a^{3}}-\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d \,a^{3}}\) | \(137\) |
Input:
int(cos(d*x+c)^3*cot(d*x+c)^5/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/16/d/a^3*(1/4*tan(1/2*d*x+1/2*c)^4-2*tan(1/2*d*x+1/2*c)^3+6*tan(1/2*d*x+ 1/2*c)^2-2*tan(1/2*d*x+1/2*c)-1/4/tan(1/2*d*x+1/2*c)^4+2/tan(1/2*d*x+1/2*c )^3-6/tan(1/2*d*x+1/2*c)^2+2/tan(1/2*d*x+1/2*c)-26*ln(tan(1/2*d*x+1/2*c))+ 32*arctan(tan(1/2*d*x+1/2*c)))
Time = 0.09 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.69 \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {16 \, d x \cos \left (d x + c\right )^{4} - 32 \, d x \cos \left (d x + c\right )^{2} + 22 \, \cos \left (d x + c\right )^{3} + 16 \, d x + 13 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 13 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 16 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 26 \, \cos \left (d x + c\right )}{16 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )}} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="frica s")
Output:
1/16*(16*d*x*cos(d*x + c)^4 - 32*d*x*cos(d*x + c)^2 + 22*cos(d*x + c)^3 + 16*d*x + 13*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(1/2*cos(d*x + c) + 1/2) - 13*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(-1/2*cos(d*x + c) + 1/2) + 16*cos(d*x + c)*sin(d*x + c) - 26*cos(d*x + c))/(a^3*d*cos(d*x + c )^4 - 2*a^3*d*cos(d*x + c)^2 + a^3*d)
Timed out. \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**3*cot(d*x+c)**5/(a+a*sin(d*x+c))**3,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (91) = 182\).
Time = 0.11 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.25 \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {\frac {8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {24 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {8 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{a^{3}} - \frac {128 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} + \frac {104 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac {{\left (\frac {8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {24 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {8 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{4}}{a^{3} \sin \left (d x + c\right )^{4}}}{64 \, d} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="maxim a")
Output:
-1/64*((8*sin(d*x + c)/(cos(d*x + c) + 1) - 24*sin(d*x + c)^2/(cos(d*x + c ) + 1)^2 + 8*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - sin(d*x + c)^4/(cos(d*x + c) + 1)^4)/a^3 - 128*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 + 104* log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 - (8*sin(d*x + c)/(cos(d*x + c) + 1) - 24*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 8*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 1)*(cos(d*x + c) + 1)^4/(a^3*sin(d*x + c)^4))/d
Time = 0.24 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.71 \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {192 \, {\left (d x + c\right )}}{a^{3}} - \frac {312 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {650 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}} + \frac {3 \, {\left (a^{9} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 \, a^{9} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, a^{9} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a^{9} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{12}}}{192 \, d} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="giac" )
Output:
1/192*(192*(d*x + c)/a^3 - 312*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + (650*t an(1/2*d*x + 1/2*c)^4 + 24*tan(1/2*d*x + 1/2*c)^3 - 72*tan(1/2*d*x + 1/2*c )^2 + 24*tan(1/2*d*x + 1/2*c) - 3)/(a^3*tan(1/2*d*x + 1/2*c)^4) + 3*(a^9*t an(1/2*d*x + 1/2*c)^4 - 8*a^9*tan(1/2*d*x + 1/2*c)^3 + 24*a^9*tan(1/2*d*x + 1/2*c)^2 - 8*a^9*tan(1/2*d*x + 1/2*c))/a^12)/d
Time = 31.86 (sec) , antiderivative size = 315, normalized size of antiderivative = 3.25 \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+8\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-24\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+24\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+128\,\mathrm {atan}\left (\frac {8\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-13\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{13\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+104\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4} \] Input:
int((cos(c + d*x)^3*cot(c + d*x)^5)/(a + a*sin(c + d*x))^3,x)
Output:
-(cos(c/2 + (d*x)/2)^8 - sin(c/2 + (d*x)/2)^8 + 8*cos(c/2 + (d*x)/2)*sin(c /2 + (d*x)/2)^7 - 8*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2) - 24*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^6 + 8*cos(c/2 + (d*x)/2)^3*sin(c/2 + (d*x)/ 2)^5 - 8*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^3 + 24*cos(c/2 + (d*x)/2) ^6*sin(c/2 + (d*x)/2)^2 + 128*atan((8*cos(c/2 + (d*x)/2) - 13*sin(c/2 + (d *x)/2))/(13*cos(c/2 + (d*x)/2) + 8*sin(c/2 + (d*x)/2)))*cos(c/2 + (d*x)/2) ^4*sin(c/2 + (d*x)/2)^4 + 104*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*c os(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^4)/(64*a^3*d*cos(c/2 + (d*x)/2)^4*s in(c/2 + (d*x)/2)^4)
Time = 0.17 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.90 \[ \int \frac {\cos ^3(c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {-11 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+8 \cos \left (d x +c \right ) \sin \left (d x +c \right )-2 \cos \left (d x +c \right )-13 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4}+8 \sin \left (d x +c \right )^{4} d x}{8 \sin \left (d x +c \right )^{4} a^{3} d} \] Input:
int(cos(d*x+c)^3*cot(d*x+c)^5/(a+a*sin(d*x+c))^3,x)
Output:
( - 11*cos(c + d*x)*sin(c + d*x)**2 + 8*cos(c + d*x)*sin(c + d*x) - 2*cos( c + d*x) - 13*log(tan((c + d*x)/2))*sin(c + d*x)**4 + 8*sin(c + d*x)**4*d* x)/(8*sin(c + d*x)**4*a**3*d)