Integrand size = 21, antiderivative size = 54 \[ \int \frac {\cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {\cot (c+d x)}{a^2 d}-\frac {2 \cot (c+d x)}{a^2 d (1+\csc (c+d x))} \] Output:
2*arctanh(cos(d*x+c))/a^2/d-cot(d*x+c)/a^2/d-2*cot(d*x+c)/a^2/d/(1+csc(d*x +c))
Leaf count is larger than twice the leaf count of optimal. \(216\) vs. \(2(54)=108\).
Time = 1.03 (sec) , antiderivative size = 216, normalized size of antiderivative = 4.00 \[ \int \frac {\cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\csc \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (\cos \left (\frac {3}{2} (c+d x)\right ) \left (5+2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \left (-3-2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+2 \left (-2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\cos (c+d x) \left (1-2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )\right ) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 a^2 d (1+\sin (c+d x))^2} \] Input:
Integrate[Cot[c + d*x]^2/(a + a*Sin[c + d*x])^2,x]
Output:
-1/4*(Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/ 2])^3*(Cos[(3*(c + d*x))/2]*(5 + 2*Log[Cos[(c + d*x)/2]] - 2*Log[Sin[(c + d*x)/2]]) + Cos[(c + d*x)/2]*(-3 - 2*Log[Cos[(c + d*x)/2]] + 2*Log[Sin[(c + d*x)/2]]) + 2*(-2*Log[Cos[(c + d*x)/2]] + 2*Log[Sin[(c + d*x)/2]] + Cos[ c + d*x]*(1 - 2*Log[Cos[(c + d*x)/2]] + 2*Log[Sin[(c + d*x)/2]]))*Sin[(c + d*x)/2]))/(a^2*d*(1 + Sin[c + d*x])^2)
Time = 0.30 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.91, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 3188, 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.
\(\displaystyle \int \frac {\cot ^2(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (c+d x)^2 (a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3188 |
\(\displaystyle \frac {\int \left (\csc ^2(c+d x)-2 \csc (c+d x)-\frac {2}{\csc (c+d x)+1}+2\right )dx}{a^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {2 \text {arctanh}(\cos (c+d x))}{d}-\frac {\cot (c+d x)}{d}-\frac {2 \cot (c+d x)}{d (\csc (c+d x)+1)}}{a^2}\) |
Input:
Int[Cot[c + d*x]^2/(a + a*Sin[c + d*x])^2,x]
Output:
((2*ArcTanh[Cos[c + d*x]])/d - Cot[c + d*x]/d - (2*Cot[c + d*x])/(d*(1 + C sc[c + d*x])))/a^2
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_ ), x_Symbol] :> Simp[a^p Int[ExpandIntegrand[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^(m - p/2)/(a - b*Sin[e + f*x])^(p/2)), x], x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])
Time = 0.90 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.09
method | result | size |
derivativedivides | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\) | \(59\) |
default | \(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\) | \(59\) |
risch | \(-\frac {2 \left (-3+i {\mathrm e}^{i \left (d x +c \right )}+2 \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) d \,a^{2}}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{2}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{2}}\) | \(102\) |
Input:
int(cot(d*x+c)^2/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/2/d/a^2*(tan(1/2*d*x+1/2*c)-8/(tan(1/2*d*x+1/2*c)+1)-1/tan(1/2*d*x+1/2*c )-4*ln(tan(1/2*d*x+1/2*c)))
Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (54) = 108\).
Time = 0.08 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.96 \[ \int \frac {\cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (3 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) + \cos \left (d x + c\right ) - 2}{a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d - {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cot(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="fricas")
Output:
(3*cos(d*x + c)^2 + (cos(d*x + c)^2 - (cos(d*x + c) + 1)*sin(d*x + c) - 1) *log(1/2*cos(d*x + c) + 1/2) - (cos(d*x + c)^2 - (cos(d*x + c) + 1)*sin(d* x + c) - 1)*log(-1/2*cos(d*x + c) + 1/2) + (3*cos(d*x + c) + 2)*sin(d*x + c) + cos(d*x + c) - 2)/(a^2*d*cos(d*x + c)^2 - a^2*d - (a^2*d*cos(d*x + c) + a^2*d)*sin(d*x + c))
\[ \int \frac {\cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\cot ^{2}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(cot(d*x+c)**2/(a+a*sin(d*x+c))**2,x)
Output:
Integral(cot(c + d*x)**2/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2
Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (54) = 108\).
Time = 0.04 (sec) , antiderivative size = 116, normalized size of antiderivative = 2.15 \[ \int \frac {\cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1}{\frac {a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}} + \frac {4 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {\sin \left (d x + c\right )}{a^{2} {\left (\cos \left (d x + c\right ) + 1\right )}}}{2 \, d} \] Input:
integrate(cot(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="maxima")
Output:
-1/2*((9*sin(d*x + c)/(cos(d*x + c) + 1) + 1)/(a^2*sin(d*x + c)/(cos(d*x + c) + 1) + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2) + 4*log(sin(d*x + c)/( cos(d*x + c) + 1))/a^2 - sin(d*x + c)/(a^2*(cos(d*x + c) + 1)))/d
Time = 0.17 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.67 \[ \int \frac {\cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {4 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} - \frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} a^{2}}}{2 \, d} \] Input:
integrate(cot(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
-1/2*(4*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - tan(1/2*d*x + 1/2*c)/a^2 - (2 *tan(1/2*d*x + 1/2*c)^2 - 7*tan(1/2*d*x + 1/2*c) - 1)/((tan(1/2*d*x + 1/2* c)^2 + tan(1/2*d*x + 1/2*c))*a^2))/d
Time = 17.68 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.61 \[ \int \frac {\cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^2\,d}-\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1}{d\,\left (2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d} \] Input:
int(cot(c + d*x)^2/(a + a*sin(c + d*x))^2,x)
Output:
tan(c/2 + (d*x)/2)/(2*a^2*d) - (9*tan(c/2 + (d*x)/2) + 1)/(d*(2*a^2*tan(c/ 2 + (d*x)/2)^2 + 2*a^2*tan(c/2 + (d*x)/2))) - (2*log(tan(c/2 + (d*x)/2)))/ (a^2*d)
Time = 0.16 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.89 \[ \int \frac {\cot ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \] Input:
int(cot(d*x+c)^2/(a+a*sin(d*x+c))^2,x)
Output:
( - 4*log(tan((c + d*x)/2))*tan((c + d*x)/2)**2 - 4*log(tan((c + d*x)/2))* tan((c + d*x)/2) + tan((c + d*x)/2)**3 + 10*tan((c + d*x)/2)**2 - 1)/(2*ta n((c + d*x)/2)*a**2*d*(tan((c + d*x)/2) + 1))