Integrand size = 23, antiderivative size = 51 \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a x}{2}-\frac {a \text {arctanh}(\cos (c+d x))}{d}+\frac {a \cos (c+d x)}{d}+\frac {a \cos (c+d x) \sin (c+d x)}{2 d} \] Output:
1/2*a*x-a*arctanh(cos(d*x+c))/d+a*cos(d*x+c)/d+1/2*a*cos(d*x+c)*sin(d*x+c) /d
Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.45 \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a (c+d x)}{2 d}+\frac {a \cos (c+d x)}{d}-\frac {a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {a \sin (2 (c+d x))}{4 d} \] Input:
Integrate[Cos[c + d*x]*Cot[c + d*x]*(a + a*Sin[c + d*x]),x]
Output:
(a*(c + d*x))/(2*d) + (a*Cos[c + d*x])/d - (a*Log[Cos[(c + d*x)/2]])/d + ( a*Log[Sin[(c + d*x)/2]])/d + (a*Sin[2*(c + d*x)])/(4*d)
Time = 0.33 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.98, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 3317, 3042, 25, 3072, 262, 219, 3115, 24}
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.
\(\displaystyle \int \cos (c+d x) \cot (c+d x) (a \sin (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^2 (a \sin (c+d x)+a)}{\sin (c+d x)}dx\) |
\(\Big \downarrow \) 3317 |
\(\displaystyle a \int \cos ^2(c+d x)dx+a \int \cos (c+d x) \cot (c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+a \int -\sin \left (c+d x+\frac {\pi }{2}\right ) \tan \left (c+d x+\frac {\pi }{2}\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle a \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-a \int \sin \left (\frac {1}{2} (2 c+\pi )+d x\right ) \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx\) |
\(\Big \downarrow \) 3072 |
\(\displaystyle a \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {a \int \frac {\cos ^2(c+d x)}{1-\cos ^2(c+d x)}d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle a \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {a \left (\int \frac {1}{1-\cos ^2(c+d x)}d\cos (c+d x)-\cos (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle a \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {a (\text {arctanh}(\cos (c+d x))-\cos (c+d x))}{d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle a \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {a (\text {arctanh}(\cos (c+d x))-\cos (c+d x))}{d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle a \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {a (\text {arctanh}(\cos (c+d x))-\cos (c+d x))}{d}\) |
Input:
Int[Cos[c + d*x]*Cot[c + d*x]*(a + a*Sin[c + d*x]),x]
Output:
-((a*(ArcTanh[Cos[c + d*x]] - Cos[c + d*x]))/d) + a*(x/2 + (Cos[c + d*x]*S in[c + d*x])/(2*d))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ (ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)], x ]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[(g*Co s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[(g*Cos[e + f*x])^ p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.08
method | result | size |
derivativedivides | \(\frac {a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\) | \(55\) |
default | \(\frac {a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\) | \(55\) |
risch | \(\frac {a x}{2}+\frac {a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}+\frac {a \sin \left (2 d x +2 c \right )}{4 d}\) | \(86\) |
Input:
int(cos(d*x+c)*cot(d*x+c)*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(a*(1/2*sin(d*x+c)*cos(d*x+c)+1/2*d*x+1/2*c)+a*(cos(d*x+c)+ln(csc(d*x+ c)-cot(d*x+c))))
Time = 0.09 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.18 \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a d x + a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 2 \, a \cos \left (d x + c\right ) - a \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + a \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, d} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
1/2*(a*d*x + a*cos(d*x + c)*sin(d*x + c) + 2*a*cos(d*x + c) - a*log(1/2*co s(d*x + c) + 1/2) + a*log(-1/2*cos(d*x + c) + 1/2))/d
\[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=a \left (\int \cos {\left (c + d x \right )} \cot {\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )} \cos {\left (c + d x \right )} \cot {\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cos(d*x+c)*cot(d*x+c)*(a+a*sin(d*x+c)),x)
Output:
a*(Integral(cos(c + d*x)*cot(c + d*x), x) + Integral(sin(c + d*x)*cos(c + d*x)*cot(c + d*x), x))
Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.12 \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a + 2 \, a {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{4 \, d} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*a + 2*a*(2*cos(d*x + c) - log(cos(d* x + c) + 1) + log(cos(d*x + c) - 1)))/d
Time = 0.15 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.71 \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {{\left (d x + c\right )} a + 2 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/2*((d*x + c)*a + 2*a*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(a*tan(1/2*d*x + 1/2*c)^3 - 2*a*tan(1/2*d*x + 1/2*c)^2 - a*tan(1/2*d*x + 1/2*c) - 2*a)/(ta n(1/2*d*x + 1/2*c)^2 + 1)^2)/d
Time = 18.18 (sec) , antiderivative size = 160, normalized size of antiderivative = 3.14 \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+2\,a}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {a\,\mathrm {atan}\left (\frac {a^2}{2\,a^2-a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^2-a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \] Input:
int(cos(c + d*x)*cot(c + d*x)*(a + a*sin(c + d*x)),x)
Output:
(2*a + a*tan(c/2 + (d*x)/2) + 2*a*tan(c/2 + (d*x)/2)^2 - a*tan(c/2 + (d*x) /2)^3)/(d*(2*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 + 1)) + (a*log(ta n(c/2 + (d*x)/2)))/d + (a*atan(a^2/(2*a^2 - a^2*tan(c/2 + (d*x)/2)) + (2*a ^2*tan(c/2 + (d*x)/2))/(2*a^2 - a^2*tan(c/2 + (d*x)/2))))/d
Time = 0.16 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.88 \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+c +d x -2\right )}{2 d} \] Input:
int(cos(d*x+c)*cot(d*x+c)*(a+a*sin(d*x+c)),x)
Output:
(a*(cos(c + d*x)*sin(c + d*x) + 2*cos(c + d*x) + 2*log(tan((c + d*x)/2)) + c + d*x - 2))/(2*d)