Integrand size = 27, antiderivative size = 85 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \csc (c+d x)}{a^2 d}-\frac {\csc ^2(c+d x)}{2 a^2 d}+\frac {3 \log (\sin (c+d x))}{a^2 d}-\frac {3 \log (1+\sin (c+d x))}{a^2 d}+\frac {1}{d \left (a^2+a^2 \sin (c+d x)\right )} \] Output:
2*csc(d*x+c)/a^2/d-1/2*csc(d*x+c)^2/a^2/d+3*ln(sin(d*x+c))/a^2/d-3*ln(1+si n(d*x+c))/a^2/d+1/d/(a^2+a^2*sin(d*x+c))
Time = 0.14 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.72 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {4 \csc (c+d x)-\csc ^2(c+d x)+6 \log (\sin (c+d x))-6 \log (1+\sin (c+d x))+\frac {2}{1+\sin (c+d x)}}{2 a^2 d} \] Input:
Integrate[(Cot[c + d*x]*Csc[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]
Output:
(4*Csc[c + d*x] - Csc[c + d*x]^2 + 6*Log[Sin[c + d*x]] - 6*Log[1 + Sin[c + d*x]] + 2/(1 + Sin[c + d*x]))/(2*a^2*d)
Time = 0.31 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3312, 27, 54, 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.
\(\displaystyle \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)}{\sin (c+d x)^3 (a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \frac {\csc ^3(c+d x)}{(\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^2 \int \frac {\csc ^3(c+d x)}{a^3 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {a^2 \int \left (\frac {\csc ^3(c+d x)}{a^5}-\frac {2 \csc ^2(c+d x)}{a^5}+\frac {3 \csc (c+d x)}{a^5}-\frac {3}{a^4 (\sin (c+d x) a+a)}-\frac {1}{a^3 (\sin (c+d x) a+a)^2}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \left (-\frac {\csc ^2(c+d x)}{2 a^4}+\frac {2 \csc (c+d x)}{a^4}+\frac {3 \log (a \sin (c+d x))}{a^4}-\frac {3 \log (a \sin (c+d x)+a)}{a^4}+\frac {1}{a^3 (a \sin (c+d x)+a)}\right )}{d}\) |
Input:
Int[(Cot[c + d*x]*Csc[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]
Output:
(a^2*((2*Csc[c + d*x])/a^4 - Csc[c + d*x]^2/(2*a^4) + (3*Log[a*Sin[c + d*x ]])/a^4 - (3*Log[a + a*Sin[c + d*x]])/a^4 + 1/(a^3*(a + a*Sin[c + d*x])))) /d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f) Su bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 0.97 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.58
method | result | size |
derivativedivides | \(-\frac {\frac {\csc \left (d x +c \right )^{2}}{2}-2 \csc \left (d x +c \right )+3 \ln \left (1+\csc \left (d x +c \right )\right )+\frac {1}{1+\csc \left (d x +c \right )}}{d \,a^{2}}\) | \(49\) |
default | \(-\frac {\frac {\csc \left (d x +c \right )^{2}}{2}-2 \csc \left (d x +c \right )+3 \ln \left (1+\csc \left (d x +c \right )\right )+\frac {1}{1+\csc \left (d x +c \right )}}{d \,a^{2}}\) | \(49\) |
risch | \(\frac {2 i \left (3 i {\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{5 i \left (d x +c \right )}-3 i {\mathrm e}^{2 i \left (d x +c \right )}-4 \,{\mathrm e}^{3 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} d \,a^{2}}-\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{2}}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{2}}\) | \(137\) |
Input:
int(cot(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
-1/d/a^2*(1/2*csc(d*x+c)^2-2*csc(d*x+c)+3*ln(1+csc(d*x+c))+1/(1+csc(d*x+c) ))
Time = 0.09 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.73 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {6 \, \cos \left (d x + c\right )^{2} + 6 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 6 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \sin \left (d x + c\right ) - 5}{2 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d + {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="fricas" )
Output:
1/2*(6*cos(d*x + c)^2 + 6*(cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin(d*x + c) - 1)*log(1/2*sin(d*x + c)) - 6*(cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)* sin(d*x + c) - 1)*log(sin(d*x + c) + 1) - 3*sin(d*x + c) - 5)/(a^2*d*cos(d *x + c)^2 - a^2*d + (a^2*d*cos(d*x + c)^2 - a^2*d)*sin(d*x + c))
\[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\cot {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)**2/(a+a*sin(d*x+c))**2,x)
Output:
Integral(cot(c + d*x)*csc(c + d*x)**2/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2
Time = 0.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {6 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) - 1}{a^{2} \sin \left (d x + c\right )^{3} + a^{2} \sin \left (d x + c\right )^{2}} - \frac {6 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {6 \, \log \left (\sin \left (d x + c\right )\right )}{a^{2}}}{2 \, d} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="maxima" )
Output:
1/2*((6*sin(d*x + c)^2 + 3*sin(d*x + c) - 1)/(a^2*sin(d*x + c)^3 + a^2*sin (d*x + c)^2) - 6*log(sin(d*x + c) + 1)/a^2 + 6*log(sin(d*x + c))/a^2)/d
Time = 0.16 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.95 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {3 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} d} + \frac {3 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2} d} + \frac {6 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) - 1}{2 \, a^{2} d {\left (\sin \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )^{2}} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
-3*log(abs(sin(d*x + c) + 1))/(a^2*d) + 3*log(abs(sin(d*x + c)))/(a^2*d) + 1/2*(6*sin(d*x + c)^2 + 3*sin(d*x + c) - 1)/(a^2*d*(sin(d*x + c) + 1)*sin (d*x + c)^2)
Time = 18.18 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.98 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^2\,d}-\frac {6\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^2\,d}+\frac {-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{2}}{d\,\left (4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2\,d} \] Input:
int(cot(c + d*x)/(sin(c + d*x)^2*(a + a*sin(c + d*x))^2),x)
Output:
(3*log(tan(c/2 + (d*x)/2)))/(a^2*d) - tan(c/2 + (d*x)/2)^2/(8*a^2*d) - (6* log(tan(c/2 + (d*x)/2) + 1))/(a^2*d) + (3*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 + 8*a^2*tan(c/2 + (d*x)/2)^3 + 4*a^2*tan(c/2 + (d*x)/2)^4)) + tan(c/ 2 + (d*x)/2)/(a^2*d)
Time = 0.16 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.65 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3}-24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}-3 \sin \left (d x +c \right )^{3}+9 \sin \left (d x +c \right )^{2}+6 \sin \left (d x +c \right )-2}{4 \sin \left (d x +c \right )^{2} a^{2} d \left (\sin \left (d x +c \right )+1\right )} \] Input:
int(cot(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c))^2,x)
Output:
( - 24*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3 - 24*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 12*log(tan((c + d*x)/2))*sin(c + d*x)**3 + 12*log( tan((c + d*x)/2))*sin(c + d*x)**2 - 3*sin(c + d*x)**3 + 9*sin(c + d*x)**2 + 6*sin(c + d*x) - 2)/(4*sin(c + d*x)**2*a**2*d*(sin(c + d*x) + 1))