Integrand size = 25, antiderivative size = 56 \[ \int \cos ^2(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \log (\sin (c+d x))}{d}+\frac {a \sin (c+d x)}{d}-\frac {a \sin ^2(c+d x)}{2 d}-\frac {a \sin ^3(c+d x)}{3 d} \] Output:
a*ln(sin(d*x+c))/d+a*sin(d*x+c)/d-1/2*a*sin(d*x+c)^2/d-1/3*a*sin(d*x+c)^3/ d
Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00 \[ \int \cos ^2(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \log (\sin (c+d x))}{d}+\frac {a \sin (c+d x)}{d}-\frac {a \sin ^2(c+d x)}{2 d}-\frac {a \sin ^3(c+d x)}{3 d} \] Input:
Integrate[Cos[c + d*x]^2*Cot[c + d*x]*(a + a*Sin[c + d*x]),x]
Output:
(a*Log[Sin[c + d*x]])/d + (a*Sin[c + d*x])/d - (a*Sin[c + d*x]^2)/(2*d) - (a*Sin[c + d*x]^3)/(3*d)
Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3315, 27, 84, 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.
\(\displaystyle \int \cos ^2(c+d x) \cot (c+d x) (a \sin (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^3 (a \sin (c+d x)+a)}{\sin (c+d x)}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \csc (c+d x) (a-a \sin (c+d x)) (\sin (c+d x) a+a)^2d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc (c+d x) (a-a \sin (c+d x)) (\sin (c+d x) a+a)^2}{a}d(a \sin (c+d x))}{a^2 d}\) |
\(\Big \downarrow \) 84 |
\(\displaystyle \frac {\int \left (-\sin ^2(c+d x) a^2+\csc (c+d x) a^2-\sin (c+d x) a^2+a^2\right )d(a \sin (c+d x))}{a^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{3} a^3 \sin ^3(c+d x)-\frac {1}{2} a^3 \sin ^2(c+d x)+a^3 \sin (c+d x)+a^3 \log (a \sin (c+d x))}{a^2 d}\) |
Input:
Int[Cos[c + d*x]^2*Cot[c + d*x]*(a + a*Sin[c + d*x]),x]
Output:
(a^3*Log[a*Sin[c + d*x]] + a^3*Sin[c + d*x] - (a^3*Sin[c + d*x]^2)/2 - (a^ 3*Sin[c + d*x]^3)/3)/(a^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[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] && !(ILtQ[n + p + 2, 0 ] && GtQ[n + 2*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 + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 0.66 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.80
method | result | size |
derivativedivides | \(\frac {\frac {a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a \left (\frac {\cos \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(45\) |
default | \(\frac {\frac {a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a \left (\frac {\cos \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(45\) |
risch | \(-i a x +\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {2 i a c}{d}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {3 a \sin \left (d x +c \right )}{4 d}+\frac {a \sin \left (3 d x +3 c \right )}{12 d}\) | \(89\) |
Input:
int(cos(d*x+c)^2*cot(d*x+c)*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/3*a*(2+cos(d*x+c)^2)*sin(d*x+c)+a*(1/2*cos(d*x+c)^2+ln(sin(d*x+c))) )
Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.91 \[ \int \cos ^2(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {3 \, a \cos \left (d x + c\right )^{2} + 6 \, a \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 2 \, {\left (a \cos \left (d x + c\right )^{2} + 2 \, a\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
1/6*(3*a*cos(d*x + c)^2 + 6*a*log(1/2*sin(d*x + c)) + 2*(a*cos(d*x + c)^2 + 2*a)*sin(d*x + c))/d
\[ \int \cos ^2(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=a \left (\int \cos ^{2}{\left (c + d x \right )} \cot {\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} \cot {\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cos(d*x+c)**2*cot(d*x+c)*(a+a*sin(d*x+c)),x)
Output:
a*(Integral(cos(c + d*x)**2*cot(c + d*x), x) + Integral(sin(c + d*x)*cos(c + d*x)**2*cot(c + d*x), x))
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.84 \[ \int \cos ^2(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {2 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - 6 \, a \log \left (\sin \left (d x + c\right )\right ) - 6 \, a \sin \left (d x + c\right )}{6 \, d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/6*(2*a*sin(d*x + c)^3 + 3*a*sin(d*x + c)^2 - 6*a*log(sin(d*x + c)) - 6* a*sin(d*x + c))/d
Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.86 \[ \int \cos ^2(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {2 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - 6 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 6 \, a \sin \left (d x + c\right )}{6 \, d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/6*(2*a*sin(d*x + c)^3 + 3*a*sin(d*x + c)^2 - 6*a*log(abs(sin(d*x + c))) - 6*a*sin(d*x + c))/d
Time = 17.65 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.64 \[ \int \cos ^2(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}+\frac {a\,{\cos \left (c+d\,x\right )}^2}{2\,d}+\frac {2\,a\,\sin \left (c+d\,x\right )}{3\,d}+\frac {a\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d} \] Input:
int(cos(c + d*x)^2*cot(c + d*x)*(a + a*sin(c + d*x)),x)
Output:
(a*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (a*log(1/cos(c/2 + (d*x )/2)^2))/d + (a*cos(c + d*x)^2)/(2*d) + (2*a*sin(c + d*x))/(3*d) + (a*cos( c + d*x)^2*sin(c + d*x))/(3*d)
Time = 0.17 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.12 \[ \int \cos ^2(c+d x) \cot (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \sin \left (d x +c \right )^{3}-3 \sin \left (d x +c \right )^{2}+6 \sin \left (d x +c \right )\right )}{6 d} \] Input:
int(cos(d*x+c)^2*cot(d*x+c)*(a+a*sin(d*x+c)),x)
Output:
(a*( - 6*log(tan((c + d*x)/2)**2 + 1) + 6*log(tan((c + d*x)/2)) - 2*sin(c + d*x)**3 - 3*sin(c + d*x)**2 + 6*sin(c + d*x)))/(6*d)