Integrand size = 27, antiderivative size = 83 \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d}-\frac {2 a \sin (c+d x)}{d}-\frac {a \sin ^2(c+d x)}{d}+\frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^4(c+d x)}{4 d} \] Output:
-a*csc(d*x+c)/d+a*ln(sin(d*x+c))/d-2*a*sin(d*x+c)/d-a*sin(d*x+c)^2/d+1/3*a *sin(d*x+c)^3/d+1/4*a*sin(d*x+c)^4/d
Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00 \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}+\frac {a \log (\sin (c+d x))}{d}-\frac {2 a \sin (c+d x)}{d}-\frac {a \sin ^2(c+d x)}{d}+\frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^4(c+d x)}{4 d} \] Input:
Integrate[Cos[c + d*x]^3*Cot[c + d*x]^2*(a + a*Sin[c + d*x]),x]
Output:
-((a*Csc[c + d*x])/d) + (a*Log[Sin[c + d*x]])/d - (2*a*Sin[c + d*x])/d - ( a*Sin[c + d*x]^2)/d + (a*Sin[c + d*x]^3)/(3*d) + (a*Sin[c + d*x]^4)/(4*d)
Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3315, 27, 99, 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.
\(\displaystyle \int \cos ^3(c+d x) \cot ^2(c+d x) (a \sin (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5 (a \sin (c+d x)+a)}{\sin (c+d x)^2}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \csc ^2(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^3d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc ^2(c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^3}{a^2}d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\sin ^3(c+d x) a^3+\csc ^2(c+d x) a^3+\sin ^2(c+d x) a^3+\csc (c+d x) a^3-2 \sin (c+d x) a^3-2 a^3\right )d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{4} a^4 \sin ^4(c+d x)+\frac {1}{3} a^4 \sin ^3(c+d x)-a^4 \sin ^2(c+d x)-2 a^4 \sin (c+d x)-a^4 \csc (c+d x)+a^4 \log (a \sin (c+d x))}{a^3 d}\) |
Input:
Int[Cos[c + d*x]^3*Cot[c + d*x]^2*(a + a*Sin[c + d*x]),x]
Output:
(-(a^4*Csc[c + d*x]) + a^4*Log[a*Sin[c + d*x]] - 2*a^4*Sin[c + d*x] - a^4* Sin[c + d*x]^2 + (a^4*Sin[c + d*x]^3)/3 + (a^4*Sin[c + d*x]^4)/4)/(a^3*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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 = 2.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.02
method | result | size |
derivativedivides | \(\frac {a \left (\frac {\cos \left (d x +c \right )^{4}}{4}+\frac {\cos \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+a \left (-\frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )\right )}{d}\) | \(85\) |
default | \(\frac {a \left (\frac {\cos \left (d x +c \right )^{4}}{4}+\frac {\cos \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+a \left (-\frac {\cos \left (d x +c \right )^{6}}{\sin \left (d x +c \right )}-\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )\right )}{d}\) | \(85\) |
risch | \(-i a x +\frac {3 a \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 d}+\frac {7 i a \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {7 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {3 a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 d}-\frac {2 i a c}{d}-\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {a \cos \left (4 d x +4 c \right )}{32 d}-\frac {a \sin \left (3 d x +3 c \right )}{12 d}\) | \(153\) |
Input:
int(cos(d*x+c)^3*cot(d*x+c)^2*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(a*(1/4*cos(d*x+c)^4+1/2*cos(d*x+c)^2+ln(sin(d*x+c)))+a*(-1/sin(d*x+c) *cos(d*x+c)^6-(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)))
Time = 0.10 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.10 \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {32 \, a \cos \left (d x + c\right )^{4} + 128 \, a \cos \left (d x + c\right )^{2} + 96 \, a \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 3 \, {\left (8 \, a \cos \left (d x + c\right )^{4} + 16 \, a \cos \left (d x + c\right )^{2} - 11 \, a\right )} \sin \left (d x + c\right ) - 256 \, a}{96 \, d \sin \left (d x + c\right )} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
1/96*(32*a*cos(d*x + c)^4 + 128*a*cos(d*x + c)^2 + 96*a*log(1/2*sin(d*x + c))*sin(d*x + c) + 3*(8*a*cos(d*x + c)^4 + 16*a*cos(d*x + c)^2 - 11*a)*sin (d*x + c) - 256*a)/(d*sin(d*x + c))
\[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=a \left (\int \cos ^{3}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cos(d*x+c)**3*cot(d*x+c)**2*(a+a*sin(d*x+c)),x)
Output:
a*(Integral(cos(c + d*x)**3*cot(c + d*x)**2, x) + Integral(sin(c + d*x)*co s(c + d*x)**3*cot(c + d*x)**2, x))
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.83 \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {3 \, a \sin \left (d x + c\right )^{4} + 4 \, a \sin \left (d x + c\right )^{3} - 12 \, a \sin \left (d x + c\right )^{2} + 12 \, a \log \left (\sin \left (d x + c\right )\right ) - 24 \, a \sin \left (d x + c\right ) - \frac {12 \, a}{\sin \left (d x + c\right )}}{12 \, d} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
1/12*(3*a*sin(d*x + c)^4 + 4*a*sin(d*x + c)^3 - 12*a*sin(d*x + c)^2 + 12*a *log(sin(d*x + c)) - 24*a*sin(d*x + c) - 12*a/sin(d*x + c))/d
Time = 0.14 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.84 \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {3 \, a \sin \left (d x + c\right )^{4} + 4 \, a \sin \left (d x + c\right )^{3} - 12 \, a \sin \left (d x + c\right )^{2} + 12 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 24 \, a \sin \left (d x + c\right ) - \frac {12 \, a}{\sin \left (d x + c\right )}}{12 \, d} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/12*(3*a*sin(d*x + c)^4 + 4*a*sin(d*x + c)^3 - 12*a*sin(d*x + c)^2 + 12*a *log(abs(sin(d*x + c))) - 24*a*sin(d*x + c) - 12*a/sin(d*x + c))/d
Time = 18.32 (sec) , antiderivative size = 250, normalized size of antiderivative = 3.01 \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}-\frac {4\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{d}+\frac {8\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d}-\frac {8\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{d}+\frac {4\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d}-\frac {9\,a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {20\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {16\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {8\,a\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3\,d\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )} \] Input:
int(cos(c + d*x)^3*cot(c + d*x)^2*(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 - (4*a*cos(c/2 + (d*x)/2)^2)/d + (8*a*cos(c/2 + (d*x)/2)^4)/d - (8*a*cos(c/2 + (d*x)/2)^6)/d + (4*a*cos(c/2 + (d*x)/2)^8)/d - (9*a*cos(c/ 2 + (d*x)/2))/(2*d*sin(c/2 + (d*x)/2)) - (a*sin(c/2 + (d*x)/2))/(2*d*cos(c /2 + (d*x)/2)) + (20*a*cos(c/2 + (d*x)/2)^3)/(3*d*sin(c/2 + (d*x)/2)) - (1 6*a*cos(c/2 + (d*x)/2)^5)/(3*d*sin(c/2 + (d*x)/2)) + (8*a*cos(c/2 + (d*x)/ 2)^7)/(3*d*sin(c/2 + (d*x)/2))
Time = 0.15 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.16 \[ \int \cos ^3(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )+3 \sin \left (d x +c \right )^{5}+4 \sin \left (d x +c \right )^{4}-12 \sin \left (d x +c \right )^{3}-24 \sin \left (d x +c \right )^{2}-12\right )}{12 \sin \left (d x +c \right ) d} \] Input:
int(cos(d*x+c)^3*cot(d*x+c)^2*(a+a*sin(d*x+c)),x)
Output:
(a*( - 12*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x) + 12*log(tan((c + d*x) /2))*sin(c + d*x) + 3*sin(c + d*x)**5 + 4*sin(c + d*x)**4 - 12*sin(c + d*x )**3 - 24*sin(c + d*x)**2 - 12))/(12*sin(c + d*x)*d)