Integrand size = 19, antiderivative size = 53 \[ \int \frac {\cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\log (\sin (c+d x))}{a^2 d}-\frac {\log (a+b \sin (c+d x))}{a^2 d}+\frac {1}{a d (a+b \sin (c+d x))} \] Output:
ln(sin(d*x+c))/a^2/d-ln(a+b*sin(d*x+c))/a^2/d+1/a/d/(a+b*sin(d*x+c))
Time = 0.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.79 \[ \int \frac {\cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\log (\sin (c+d x))-\log (a+b \sin (c+d x))+\frac {a}{a+b \sin (c+d x)}}{a^2 d} \] Input:
Integrate[Cot[c + d*x]/(a + b*Sin[c + d*x])^2,x]
Output:
(Log[Sin[c + d*x]] - Log[a + b*Sin[c + d*x]] + a/(a + b*Sin[c + d*x]))/(a^ 2*d)
Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3200, 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.
\(\displaystyle \int \frac {\cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (c+d x) (a+b \sin (c+d x))^2}dx\) |
\(\Big \downarrow \) 3200 |
\(\displaystyle \frac {\int \frac {\csc (c+d x)}{b (a+b \sin (c+d x))^2}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\int \left (\frac {\csc (c+d x)}{a^2 b}-\frac {1}{a^2 (a+b \sin (c+d x))}-\frac {1}{a (a+b \sin (c+d x))^2}\right )d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\log (b \sin (c+d x))}{a^2}-\frac {\log (a+b \sin (c+d x))}{a^2}+\frac {1}{a (a+b \sin (c+d x))}}{d}\) |
Input:
Int[Cot[c + d*x]/(a + b*Sin[c + d*x])^2,x]
Output:
(Log[b*Sin[c + d*x]]/a^2 - Log[a + b*Sin[c + d*x]]/a^2 + 1/(a*(a + b*Sin[c + d*x])))/d
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b ^2, 0] && IntegerQ[(p + 1)/2]
Time = 0.58 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.92
method | result | size |
derivativedivides | \(\frac {\frac {\ln \left (\sin \left (d x +c \right )\right )}{a^{2}}-\frac {\ln \left (a +b \sin \left (d x +c \right )\right )}{a^{2}}+\frac {1}{a \left (a +b \sin \left (d x +c \right )\right )}}{d}\) | \(49\) |
default | \(\frac {\frac {\ln \left (\sin \left (d x +c \right )\right )}{a^{2}}-\frac {\ln \left (a +b \sin \left (d x +c \right )\right )}{a^{2}}+\frac {1}{a \left (a +b \sin \left (d x +c \right )\right )}}{d}\) | \(49\) |
risch | \(\frac {2 i {\mathrm e}^{i \left (d x +c \right )}}{a d \left ({\mathrm e}^{2 i \left (d x +c \right )} b -b +2 i {\mathrm e}^{i \left (d x +c \right )} a \right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{a^{2} d}\) | \(105\) |
Input:
int(cot(d*x+c)/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(1/a^2*ln(sin(d*x+c))-1/a^2*ln(a+b*sin(d*x+c))+1/a/(a+b*sin(d*x+c)))
Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.30 \[ \int \frac {\cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {{\left (b \sin \left (d x + c\right ) + a\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (b \sin \left (d x + c\right ) + a\right )} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) - a}{a^{2} b d \sin \left (d x + c\right ) + a^{3} d} \] Input:
integrate(cot(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="fricas")
Output:
-((b*sin(d*x + c) + a)*log(b*sin(d*x + c) + a) - (b*sin(d*x + c) + a)*log( -1/2*sin(d*x + c)) - a)/(a^2*b*d*sin(d*x + c) + a^3*d)
\[ \int \frac {\cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int \frac {\cot {\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(cot(d*x+c)/(a+b*sin(d*x+c))**2,x)
Output:
Integral(cot(c + d*x)/(a + b*sin(c + d*x))**2, x)
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.89 \[ \int \frac {\cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {1}{a b \sin \left (d x + c\right ) + a^{2}} - \frac {\log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2}} + \frac {\log \left (\sin \left (d x + c\right )\right )}{a^{2}}}{d} \] Input:
integrate(cot(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="maxima")
Output:
(1/(a*b*sin(d*x + c) + a^2) - log(b*sin(d*x + c) + a)/a^2 + log(sin(d*x + c))/a^2)/d
Time = 0.13 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.04 \[ \int \frac {\cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} d} + \frac {\log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2} d} + \frac {1}{{\left (b \sin \left (d x + c\right ) + a\right )} a d} \] Input:
integrate(cot(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="giac")
Output:
-log(abs(b*sin(d*x + c) + a))/(a^2*d) + log(abs(sin(d*x + c)))/(a^2*d) + 1 /((b*sin(d*x + c) + a)*a*d)
Time = 17.84 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.98 \[ \int \frac {\cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{a^2\,d}-\frac {2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3+2\,b\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \] Input:
int(cot(c + d*x)/(a + b*sin(c + d*x))^2,x)
Output:
log(tan(c/2 + (d*x)/2))/(a^2*d) - log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c /2 + (d*x)/2)^2)/(a^2*d) - (2*b*tan(c/2 + (d*x)/2))/(d*(a^3*tan(c/2 + (d*x )/2)^2 + a^3 + 2*a^2*b*tan(c/2 + (d*x)/2)))
Time = 0.16 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.25 \[ \int \frac {\cot (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) \sin \left (d x +c \right ) b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +a}{a^{2} d \left (\sin \left (d x +c \right ) b +a \right )} \] Input:
int(cot(d*x+c)/(a+b*sin(d*x+c))^2,x)
Output:
( - log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)*b - log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a + log(tan((c + d* x)/2))*sin(c + d*x)*b + log(tan((c + d*x)/2))*a + a)/(a**2*d*(sin(c + d*x) *b + a))