Integrand size = 27, antiderivative size = 59 \[ \int \frac {\cos ^2(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\log (\sin (c+d x))}{a d}+\frac {\left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{a b^2 d}-\frac {\sin (c+d x)}{b d} \] Output:
ln(sin(d*x+c))/a/d+(a^2-b^2)*ln(a+b*sin(d*x+c))/a/b^2/d-sin(d*x+c)/b/d
Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90 \[ \int \frac {\cos ^2(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b^2 \log (\sin (c+d x))+\left (a^2-b^2\right ) \log (a+b \sin (c+d x))-a b \sin (c+d x)}{a b^2 d} \] Input:
Integrate[(Cos[c + d*x]^2*Cot[c + d*x])/(a + b*Sin[c + d*x]),x]
Output:
(b^2*Log[Sin[c + d*x]] + (a^2 - b^2)*Log[a + b*Sin[c + d*x]] - a*b*Sin[c + d*x])/(a*b^2*d)
Time = 0.31 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3316, 27, 522, 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.
\(\displaystyle \int \frac {\cos ^2(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^3}{\sin (c+d x) (a+b \sin (c+d x))}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {\int \frac {\csc (c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )}{a+b \sin (c+d x)}d(b \sin (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc (c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )}{b (a+b \sin (c+d x))}d(b \sin (c+d x))}{b^2 d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \frac {\int \left (\frac {a^2-b^2}{a (a+b \sin (c+d x))}+\frac {b \csc (c+d x)}{a}-1\right )d(b \sin (c+d x))}{b^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{a}+\frac {b^2 \log (b \sin (c+d x))}{a}-b \sin (c+d x)}{b^2 d}\) |
Input:
Int[(Cos[c + d*x]^2*Cot[c + d*x])/(a + b*Sin[c + d*x]),x]
Output:
((b^2*Log[b*Sin[c + d*x]])/a + ((a^2 - b^2)*Log[a + b*Sin[c + d*x]])/a - b *Sin[c + d*x])/(b^2*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 0.92 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.93
method | result | size |
derivativedivides | \(\frac {-\frac {\sin \left (d x +c \right )}{b}+\frac {\ln \left (\sin \left (d x +c \right )\right )}{a}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a \,b^{2}}}{d}\) | \(55\) |
default | \(\frac {-\frac {\sin \left (d x +c \right )}{b}+\frac {\ln \left (\sin \left (d x +c \right )\right )}{a}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a \,b^{2}}}{d}\) | \(55\) |
risch | \(-\frac {i x a}{b^{2}}+\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 b d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 b d}-\frac {2 i a c}{b^{2} d}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{2} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}\) | \(148\) |
Input:
int(cos(d*x+c)^2*cot(d*x+c)/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(-sin(d*x+c)/b+1/a*ln(sin(d*x+c))+(a^2-b^2)/a/b^2*ln(a+b*sin(d*x+c)))
Time = 0.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.93 \[ \int \frac {\cos ^2(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b^{2} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) - a b \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a b^{2} d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
(b^2*log(-1/2*sin(d*x + c)) - a*b*sin(d*x + c) + (a^2 - b^2)*log(b*sin(d*x + c) + a))/(a*b^2*d)
\[ \int \frac {\cos ^2(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\cos ^{2}{\left (c + d x \right )} \cot {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:
integrate(cos(d*x+c)**2*cot(d*x+c)/(a+b*sin(d*x+c)),x)
Output:
Integral(cos(c + d*x)**2*cot(c + d*x)/(a + b*sin(c + d*x)), x)
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^2(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {\log \left (\sin \left (d x + c\right )\right )}{a} - \frac {\sin \left (d x + c\right )}{b} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a b^{2}}}{d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
(log(sin(d*x + c))/a - sin(d*x + c)/b + (a^2 - b^2)*log(b*sin(d*x + c) + a )/(a*b^2))/d
Time = 0.18 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^2(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a d} - \frac {\sin \left (d x + c\right )}{b d} + \frac {{\left (a^{2} - b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a b^{2} d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
log(abs(sin(d*x + c)))/(a*d) - sin(d*x + c)/(b*d) + (a^2 - b^2)*log(abs(b* sin(d*x + c) + a))/(a*b^2*d)
Time = 19.82 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.66 \[ \int \frac {\cos ^2(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}-\frac {\sin \left (c+d\,x\right )}{b\,d}+\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (\frac {a}{b^2}-\frac {1}{a}\right )}{d}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{b^2\,d} \] Input:
int((cos(c + d*x)^2*cot(c + d*x))/(a + b*sin(c + d*x)),x)
Output:
log(tan(c/2 + (d*x)/2))/(a*d) - sin(c + d*x)/(b*d) + (log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2)*(a/b^2 - 1/a))/d - (a*log(tan(c/2 + ( d*x)/2)^2 + 1))/(b^2*d)
Time = 4.84 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.02 \[ \int \frac {\cos ^2(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) b^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}-\sin \left (d x +c \right ) a b}{a \,b^{2} d} \] Input:
int(cos(d*x+c)^2*cot(d*x+c)/(a+b*sin(d*x+c)),x)
Output:
( - log(tan((c + d*x)/2)**2 + 1)*a**2 + log(tan((c + d*x)/2)**2*a + 2*tan( (c + d*x)/2)*b + a)*a**2 - log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)* b + a)*b**2 + log(tan((c + d*x)/2))*b**2 - sin(c + d*x)*a*b)/(a*b**2*d)