Integrand size = 27, antiderivative size = 63 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc (c+d x)}{a d}-\frac {\csc ^2(c+d x)}{2 a d}+\frac {\log (\sin (c+d x))}{a d}-\frac {\log (1+\sin (c+d x))}{a d} \] Output:
csc(d*x+c)/a/d-1/2*csc(d*x+c)^2/a/d+ln(sin(d*x+c))/a/d-ln(1+sin(d*x+c))/a/ d
Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc (c+d x)}{a d}-\frac {\csc ^2(c+d x)}{2 a d}+\frac {\log (\sin (c+d x))}{a d}-\frac {\log (1+\sin (c+d x))}{a d} \] Input:
Integrate[(Cot[c + d*x]*Csc[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
Output:
Csc[c + d*x]/(a*d) - Csc[c + d*x]^2/(2*a*d) + Log[Sin[c + d*x]]/(a*d) - Lo g[1 + Sin[c + d*x]]/(a*d)
Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.98, 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.
\(\displaystyle \int \frac {\cot (c+d x) \csc ^2(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)}{\sin (c+d x)^3 (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \frac {\csc ^3(c+d x)}{\sin (c+d x) a+a}d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^2 \int \frac {\csc ^3(c+d x)}{a^3 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {a^2 \int \left (\frac {\csc ^3(c+d x)}{a^4}-\frac {\csc ^2(c+d x)}{a^4}+\frac {\csc (c+d x)}{a^4}-\frac {1}{a^3 (\sin (c+d x) a+a)}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \left (-\frac {\csc ^2(c+d x)}{2 a^3}+\frac {\csc (c+d x)}{a^3}+\frac {\log (a \sin (c+d x))}{a^3}-\frac {\log (a \sin (c+d x)+a)}{a^3}\right )}{d}\) |
Input:
Int[(Cot[c + d*x]*Csc[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
Output:
(a^2*(Csc[c + d*x]/a^3 - Csc[c + d*x]^2/(2*a^3) + Log[a*Sin[c + d*x]]/a^3 - Log[a + a*Sin[c + d*x]]/a^3))/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 = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.59
method | result | size |
derivativedivides | \(-\frac {\frac {\csc \left (d x +c \right )^{2}}{2}-\csc \left (d x +c \right )+\ln \left (1+\csc \left (d x +c \right )\right )}{d a}\) | \(37\) |
default | \(-\frac {\frac {\csc \left (d x +c \right )^{2}}{2}-\csc \left (d x +c \right )+\ln \left (1+\csc \left (d x +c \right )\right )}{d a}\) | \(37\) |
risch | \(\frac {2 i \left (-i {\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}\) | \(97\) |
Input:
int(cot(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
-1/d/a*(1/2*csc(d*x+c)^2-csc(d*x+c)+ln(1+csc(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.14 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {2 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 2 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, \sin \left (d x + c\right ) + 1}{2 \, {\left (a d \cos \left (d x + c\right )^{2} - a d\right )}} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
1/2*(2*(cos(d*x + c)^2 - 1)*log(1/2*sin(d*x + c)) - 2*(cos(d*x + c)^2 - 1) *log(sin(d*x + c) + 1) - 2*sin(d*x + c) + 1)/(a*d*cos(d*x + c)^2 - a*d)
\[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cot {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)**2/(a+a*sin(d*x+c)),x)
Output:
Integral(cot(c + d*x)*csc(c + d*x)**2/(sin(c + d*x) + 1), x)/a
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.87 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {2 \, \log \left (\sin \left (d x + c\right )\right )}{a} - \frac {2 \, \sin \left (d x + c\right ) - 1}{a \sin \left (d x + c\right )^{2}}}{2 \, d} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/2*(2*log(sin(d*x + c) + 1)/a - 2*log(sin(d*x + c))/a - (2*sin(d*x + c) - 1)/(a*sin(d*x + c)^2))/d
Time = 0.19 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.95 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a d} + \frac {\log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a d} + \frac {2 \, \sin \left (d x + c\right ) - 1}{2 \, a d \sin \left (d x + c\right )^{2}} \] Input:
integrate(cot(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-log(abs(sin(d*x + c) + 1))/(a*d) + log(abs(sin(d*x + c)))/(a*d) + 1/2*(2* sin(d*x + c) - 1)/(a*d*sin(d*x + c)^2)
Time = 18.02 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.68 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}-\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{2}\right )}{4\,a\,d} \] Input:
int(cot(c + d*x)/(sin(c + d*x)^2*(a + a*sin(c + d*x))),x)
Output:
log(tan(c/2 + (d*x)/2))/(a*d) - tan(c/2 + (d*x)/2)^2/(8*a*d) - (2*log(tan( c/2 + (d*x)/2) + 1))/(a*d) + tan(c/2 + (d*x)/2)/(2*a*d) + (cot(c/2 + (d*x) /2)^2*(2*tan(c/2 + (d*x)/2) - 1/2))/(4*a*d)
Time = 0.16 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.21 \[ \int \frac {\cot (c+d x) \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}+\sin \left (d x +c \right )^{2}+4 \sin \left (d x +c \right )-2}{4 \sin \left (d x +c \right )^{2} a d} \] Input:
int(cot(d*x+c)*csc(d*x+c)^2/(a+a*sin(d*x+c)),x)
Output:
( - 8*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 4*log(tan((c + d*x)/2))* sin(c + d*x)**2 + sin(c + d*x)**2 + 4*sin(c + d*x) - 2)/(4*sin(c + d*x)**2 *a*d)