Integrand size = 27, antiderivative size = 58 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {x}{a}+\frac {\text {arctanh}(\cos (c+d x))}{2 a d}+\frac {\cot (c+d x)}{a d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d} \] Output:
x/a+1/2*arctanh(cos(d*x+c))/a/d+cot(d*x+c)/a/d-1/2*cot(d*x+c)*csc(d*x+c)/a /d
Time = 0.90 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.76 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\left (\csc \left (\frac {1}{2} (c+d x)\right )+\sec \left (\frac {1}{2} (c+d x)\right )\right )^2 \left (\left (2 c+2 d x+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin ^2(c+d x)+\cos (c+d x) (-1+2 \sin (c+d x))\right )}{8 a d (1+\sin (c+d x))} \] Input:
Integrate[(Cos[c + d*x]*Cot[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
Output:
((Csc[(c + d*x)/2] + Sec[(c + d*x)/2])^2*((2*c + 2*d*x + Log[Cos[(c + d*x) /2]] - Log[Sin[(c + d*x)/2]])*Sin[c + d*x]^2 + Cos[c + d*x]*(-1 + 2*Sin[c + d*x])))/(8*a*d*(1 + Sin[c + d*x]))
Time = 0.46 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3042, 3318, 3042, 3091, 3042, 3954, 24, 4257}
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.
\(\displaystyle \int \frac {\cos (c+d x) \cot ^3(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^4}{\sin (c+d x)^3 (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3318 |
\(\displaystyle \frac {\int \cot ^2(c+d x) \csc (c+d x)dx}{a}-\frac {\int \cot ^2(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sec \left (c+d x-\frac {\pi }{2}\right ) \tan \left (c+d x-\frac {\pi }{2}\right )^2dx}{a}-\frac {\int \tan \left (c+d x+\frac {\pi }{2}\right )^2dx}{a}\) |
\(\Big \downarrow \) 3091 |
\(\displaystyle \frac {-\frac {1}{2} \int \csc (c+d x)dx-\frac {\cot (c+d x) \csc (c+d x)}{2 d}}{a}-\frac {\int \tan \left (c+d x+\frac {\pi }{2}\right )^2dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {1}{2} \int \csc (c+d x)dx-\frac {\cot (c+d x) \csc (c+d x)}{2 d}}{a}-\frac {\int \tan \left (c+d x+\frac {\pi }{2}\right )^2dx}{a}\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle \frac {-\frac {1}{2} \int \csc (c+d x)dx-\frac {\cot (c+d x) \csc (c+d x)}{2 d}}{a}-\frac {-\int 1dx-\frac {\cot (c+d x)}{d}}{a}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {-\frac {1}{2} \int \csc (c+d x)dx-\frac {\cot (c+d x) \csc (c+d x)}{2 d}}{a}-\frac {-\frac {\cot (c+d x)}{d}-x}{a}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\frac {\text {arctanh}(\cos (c+d x))}{2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 d}}{a}-\frac {-\frac {\cot (c+d x)}{d}-x}{a}\) |
Input:
Int[(Cos[c + d*x]*Cot[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
Output:
-((-x - Cot[c + d*x]/d)/a) + (ArcTanh[Cos[c + d*x]]/(2*d) - (Cot[c + d*x]* Csc[c + d*x])/(2*d))/a
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1)) Int[(a*Sec[e + f*x])^m*( b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & & NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
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]
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d *x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2 Int[(b*Tan[c + d*x])^(n - 2), x] , x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.39 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.45
method | result | size |
derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}\) | \(84\) |
default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}\) | \(84\) |
risch | \(\frac {x}{a}+\frac {{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}+2 i {\mathrm e}^{2 i \left (d x +c \right )}-2 i}{a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d a}\) | \(100\) |
Input:
int(cos(d*x+c)*cot(d*x+c)^3/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/4/d/a*(1/2*tan(1/2*d*x+1/2*c)^2-2*tan(1/2*d*x+1/2*c)+8*arctan(tan(1/2*d* x+1/2*c))-1/2/tan(1/2*d*x+1/2*c)^2+2/tan(1/2*d*x+1/2*c)-2*ln(tan(1/2*d*x+1 /2*c)))
Time = 0.09 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.79 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {4 \, d x \cos \left (d x + c\right )^{2} - 4 \, d x + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 4 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 2 \, \cos \left (d x + c\right )}{4 \, {\left (a d \cos \left (d x + c\right )^{2} - a d\right )}} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
1/4*(4*d*x*cos(d*x + c)^2 - 4*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) - 4*cos(d*x + c)*sin(d*x + c) + 2*cos(d*x + c))/(a*d*cos(d*x + c)^2 - a*d)
\[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cos {\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)**3/(a+a*sin(d*x+c)),x)
Output:
Integral(cos(c + d*x)*cot(c + d*x)**3/(sin(c + d*x) + 1), x)/a
Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (54) = 108\).
Time = 0.11 (sec) , antiderivative size = 138, normalized size of antiderivative = 2.38 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {\frac {4 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a} - \frac {16 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {4 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {{\left (\frac {4 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{a \sin \left (d x + c\right )^{2}}}{8 \, d} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/8*((4*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/a - 16*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + 4*log(sin(d*x + c )/(cos(d*x + c) + 1))/a - (4*sin(d*x + c)/(cos(d*x + c) + 1) - 1)*(cos(d*x + c) + 1)^2/(a*sin(d*x + c)^2))/d
Time = 0.15 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.78 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {8 \, {\left (d x + c\right )}}{a} - \frac {4 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} + \frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} + \frac {6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/8*(8*(d*x + c)/a - 4*log(abs(tan(1/2*d*x + 1/2*c)))/a + (a*tan(1/2*d*x + 1/2*c)^2 - 4*a*tan(1/2*d*x + 1/2*c))/a^2 + (6*tan(1/2*d*x + 1/2*c)^2 + 4* tan(1/2*d*x + 1/2*c) - 1)/(a*tan(1/2*d*x + 1/2*c)^2))/d
Time = 17.71 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.74 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}-\frac {2\,\mathrm {atan}\left (\frac {2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}-\frac {\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2\,a\,d}+\frac {\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d} \] Input:
int((cos(c + d*x)*cot(c + d*x)^3)/(a + a*sin(c + d*x)),x)
Output:
tan(c/2 + (d*x)/2)^2/(8*a*d) - (2*atan((2*cos(c/2 + (d*x)/2) - sin(c/2 + ( d*x)/2))/(cos(c/2 + (d*x)/2) + 2*sin(c/2 + (d*x)/2))))/(a*d) - cot(c/2 + ( d*x)/2)^2/(8*a*d) - log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))/(2*a*d) + c ot(c/2 + (d*x)/2)/(2*a*d) - tan(c/2 + (d*x)/2)/(2*a*d)
Time = 0.59 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.22 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {2 \cos \left (d x +c \right ) \sin \left (d x +c \right )-\cos \left (d x +c \right )-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}+2 \sin \left (d x +c \right )^{2} d x}{2 \sin \left (d x +c \right )^{2} a d} \] Input:
int(cos(d*x+c)*cot(d*x+c)^3/(a+a*sin(d*x+c)),x)
Output:
(2*cos(c + d*x)*sin(c + d*x) - cos(c + d*x) - log(tan((c + d*x)/2))*sin(c + d*x)**2 + 2*sin(c + d*x)**2*d*x)/(2*sin(c + d*x)**2*a*d)