Integrand size = 23, antiderivative size = 51 \[ \int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\frac {b x}{2}-\frac {a \text {arctanh}(\cos (c+d x))}{d}+\frac {a \cos (c+d x)}{d}+\frac {b \cos (c+d x) \sin (c+d x)}{2 d} \]
Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.45 \[ \int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\frac {b (c+d x)}{2 d}+\frac {a \cos (c+d x)}{d}-\frac {a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {b \sin (2 (c+d x))}{4 d} \]
(b*(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 + (b*Sin[2*(c + d*x)])/(4*d)
Time = 0.31 (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.
\(\displaystyle \int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^2 (a+b \sin (c+d x))}{\sin (c+d x)}dx\) |
\(\Big \downarrow \) 3317 |
\(\displaystyle a \int \cos (c+d x) \cot (c+d x)dx+b \int \cos ^2(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int -\sin \left (c+d x+\frac {\pi }{2}\right ) \tan \left (c+d x+\frac {\pi }{2}\right )dx+b \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle b \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-a \int \sin \left (\frac {1}{2} (2 c+\pi )+d x\right ) \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx\) |
\(\Big \downarrow \) 3072 |
\(\displaystyle b \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {a \int \frac {\cos ^2(c+d x)}{1-\cos ^2(c+d x)}d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle b \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {a \left (\int \frac {1}{1-\cos ^2(c+d x)}d\cos (c+d x)-\cos (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle b \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {a (\text {arctanh}(\cos (c+d x))-\cos (c+d x))}{d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle b \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {a (\text {arctanh}(\cos (c+d x))-\cos (c+d x))}{d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle b \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {a (\text {arctanh}(\cos (c+d x))-\cos (c+d x))}{d}\) |
-((a*(ArcTanh[Cos[c + d*x]] - Cos[c + d*x]))/d) + b*(x/2 + (Cos[c + d*x]*S in[c + d*x])/(2*d))
3.11.53.3.1 Defintions of rubi rules used
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 = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.94
method | result | size |
parallelrisch | \(\frac {2 b x d +4 a \cos \left (d x +c \right )+b \sin \left (2 d x +2 c \right )+4 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 a}{4 d}\) | \(48\) |
derivativedivides | \(\frac {a \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(55\) |
default | \(\frac {a \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(55\) |
risch | \(\frac {b x}{2}+\frac {a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {b \sin \left (2 d x +2 c \right )}{4 d}\) | \(86\) |
norman | \(\frac {\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+b x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {b x}{2}-\frac {b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {b x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(132\) |
Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.18 \[ \int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\frac {b d x + b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 2 \, a \cos \left (d x + c\right ) - a \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + a \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, d} \]
1/2*(b*d*x + b*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
\[ \int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx \]
Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.12 \[ \int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} b + 2 \, a {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{4 \, d} \]
1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*b + 2*a*(2*cos(d*x + c) - log(cos(d* x + c) + 1) + log(cos(d*x + c) - 1)))/d
Time = 0.31 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.71 \[ \int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\frac {{\left (d x + c\right )} b + 2 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]
1/2*((d*x + c)*b + 2*a*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(b*tan(1/2*d*x + 1/2*c)^3 - 2*a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c) - 2*a)/(ta n(1/2*d*x + 1/2*c)^2 + 1)^2)/d
Time = 10.81 (sec) , antiderivative size = 157, normalized size of antiderivative = 3.08 \[ \int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x)) \, dx=\frac {b\,\mathrm {atan}\left (\frac {b^2}{2\,a\,b-b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {2\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,b-b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {-b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+2\,a}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]