Integrand size = 29, antiderivative size = 73 \[ \int \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 x}{a^2}-\frac {\text {arctanh}(\cos (c+d x))}{2 a^2 d}+\frac {\cos (c+d x)}{a^2 d}+\frac {2 \cot (c+d x)}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d} \] Output:
2*x/a^2-1/2*arctanh(cos(d*x+c))/a^2/d+cos(d*x+c)/a^2/d+2*cot(d*x+c)/a^2/d- 1/2*cot(d*x+c)*csc(d*x+c)/a^2/d
Time = 2.15 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.84 \[ \int \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4 \left (16 (c+d x)+8 \cos (c+d x)+8 \cot \left (\frac {1}{2} (c+d x)\right )-\csc ^2\left (\frac {1}{2} (c+d x)\right )-4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\sec ^2\left (\frac {1}{2} (c+d x)\right )-8 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{8 d (a+a \sin (c+d x))^2} \] Input:
Integrate[(Cos[c + d*x]^3*Cot[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]
Output:
((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*(16*(c + d*x) + 8*Cos[c + d*x] + 8*Cot[(c + d*x)/2] - Csc[(c + d*x)/2]^2 - 4*Log[Cos[(c + d*x)/2]] + 4*Log[ Sin[(c + d*x)/2]] + Sec[(c + d*x)/2]^2 - 8*Tan[(c + d*x)/2]))/(8*d*(a + a* Sin[c + d*x])^2)
Time = 0.50 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.05, 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.
\(\displaystyle \int \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^6}{\sin (c+d x)^3 (a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3354 |
\(\displaystyle \frac {\int \cot ^2(c+d x) \csc (c+d x) (a-a \sin (c+d x))^2dx}{a^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\cos (c+d x)^2 (a-a \sin (c+d x))^2}{\sin (c+d x)^3}dx}{a^4}\) |
\(\Big \downarrow \) 3351 |
\(\displaystyle \frac {\int \left (\csc ^3(c+d x) a^4-2 \csc ^2(c+d x) a^4-\sin (c+d x) a^4+2 a^4\right )dx}{a^6}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {a^4 \text {arctanh}(\cos (c+d x))}{2 d}+\frac {a^4 \cos (c+d x)}{d}+\frac {2 a^4 \cot (c+d x)}{d}-\frac {a^4 \cot (c+d x) \csc (c+d x)}{2 d}+2 a^4 x}{a^6}\) |
Input:
Int[(Cos[c + d*x]^3*Cot[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]
Output:
(2*a^4*x - (a^4*ArcTanh[Cos[c + d*x]])/(2*d) + (a^4*Cos[c + d*x])/d + (2*a ^4*Cot[c + d*x])/d - (a^4*Cot[c + d*x]*Csc[c + d*x])/(2*d))/a^6
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 = 1.84 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.38
method | result | size |
derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {8}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+16 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{2}}\) | \(101\) |
default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {8}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+16 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{2}}\) | \(101\) |
risch | \(\frac {2 x}{a^{2}}+\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{2}}+\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{2}}+\frac {{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}+4 i {\mathrm e}^{2 i \left (d x +c \right )}-4 i}{a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d \,a^{2}}\) | \(135\) |
Input:
int(cos(d*x+c)^3*cot(d*x+c)^3/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/4/d/a^2*(1/2*tan(1/2*d*x+1/2*c)^2-4*tan(1/2*d*x+1/2*c)-1/2/tan(1/2*d*x+1 /2*c)^2+4/tan(1/2*d*x+1/2*c)+2*ln(tan(1/2*d*x+1/2*c))+8/(1+tan(1/2*d*x+1/2 *c)^2)+16*arctan(tan(1/2*d*x+1/2*c)))
Time = 0.09 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.62 \[ \int \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {8 \, d x \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right )^{3} - 8 \, d x - {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 8 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
1/4*(8*d*x*cos(d*x + c)^2 + 4*cos(d*x + c)^3 - 8*d*x - (cos(d*x + c)^2 - 1 )*log(1/2*cos(d*x + c) + 1/2) + (cos(d*x + c)^2 - 1)*log(-1/2*cos(d*x + c) + 1/2) - 8*cos(d*x + c)*sin(d*x + c) - 2*cos(d*x + c))/(a^2*d*cos(d*x + c )^2 - a^2*d)
Timed out. \[ \int \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**3*cot(d*x+c)**3/(a+a*sin(d*x+c))**2,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (69) = 138\).
Time = 0.14 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.79 \[ \int \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {\frac {8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {15 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {8 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 1}{\frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {\frac {8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2}} + \frac {32 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {4 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{8 \, d} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
1/8*((8*sin(d*x + c)/(cos(d*x + c) + 1) + 15*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 8*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 1)/(a^2*sin(d*x + c)^2/(c os(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) - (8*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/a^2 + 32*a rctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + 4*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d
Time = 0.16 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.75 \[ \int \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {16 \, {\left (d x + c\right )}}{a^{2}} + \frac {4 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{4}} + \frac {16}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{2}} - \frac {6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
1/8*(16*(d*x + c)/a^2 + 4*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + (a^2*tan(1/ 2*d*x + 1/2*c)^2 - 8*a^2*tan(1/2*d*x + 1/2*c))/a^4 + 16/((tan(1/2*d*x + 1/ 2*c)^2 + 1)*a^2) - (6*tan(1/2*d*x + 1/2*c)^2 - 8*tan(1/2*d*x + 1/2*c) + 1) /(a^2*tan(1/2*d*x + 1/2*c)^2))/d
Time = 32.62 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.55 \[ \int \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^2\,d}-\frac {4\,\mathrm {atan}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-4}+\frac {16}{16\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-4}\right )}{a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a^2\,d}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{2}}{d\,\left (4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2\,d} \] Input:
int((cos(c + d*x)^3*cot(c + d*x)^3)/(a + a*sin(c + d*x))^2,x)
Output:
tan(c/2 + (d*x)/2)^2/(8*a^2*d) - (4*atan((4*tan(c/2 + (d*x)/2))/(16*tan(c/ 2 + (d*x)/2) - 4) + 16/(16*tan(c/2 + (d*x)/2) - 4)))/(a^2*d) + log(tan(c/2 + (d*x)/2))/(2*a^2*d) + (4*tan(c/2 + (d*x)/2) + (15*tan(c/2 + (d*x)/2)^2) /2 + 4*tan(c/2 + (d*x)/2)^3 - 1/2)/(d*(4*a^2*tan(c/2 + (d*x)/2)^2 + 4*a^2* tan(c/2 + (d*x)/2)^4)) - tan(c/2 + (d*x)/2)/(a^2*d)
Time = 0.43 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.33 \[ \int \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+16 \cos \left (d x +c \right ) \sin \left (d x +c \right )-4 \cos \left (d x +c \right )+4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}+16 \sin \left (d x +c \right )^{2} d x -7 \sin \left (d x +c \right )^{2}}{8 \sin \left (d x +c \right )^{2} a^{2} d} \] Input:
int(cos(d*x+c)^3*cot(d*x+c)^3/(a+a*sin(d*x+c))^2,x)
Output:
(8*cos(c + d*x)*sin(c + d*x)**2 + 16*cos(c + d*x)*sin(c + d*x) - 4*cos(c + d*x) + 4*log(tan((c + d*x)/2))*sin(c + d*x)**2 + 16*sin(c + d*x)**2*d*x - 7*sin(c + d*x)**2)/(8*sin(c + d*x)**2*a**2*d)