Integrand size = 27, antiderivative size = 119 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \log (\sin (c+d x))}{d}+\frac {2 a^2 \sin (c+d x)}{d}-\frac {a^2 \sin ^2(c+d x)}{2 d}-\frac {4 a^2 \sin ^3(c+d x)}{3 d}-\frac {a^2 \sin ^4(c+d x)}{4 d}+\frac {2 a^2 \sin ^5(c+d x)}{5 d}+\frac {a^2 \sin ^6(c+d x)}{6 d} \] Output:
a^2*ln(sin(d*x+c))/d+2*a^2*sin(d*x+c)/d-1/2*a^2*sin(d*x+c)^2/d-4/3*a^2*sin (d*x+c)^3/d-1/4*a^2*sin(d*x+c)^4/d+2/5*a^2*sin(d*x+c)^5/d+1/6*a^2*sin(d*x+ c)^6/d
Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.66 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \left (60 \log (\sin (c+d x))+120 \sin (c+d x)-30 \sin ^2(c+d x)-80 \sin ^3(c+d x)-15 \sin ^4(c+d x)+24 \sin ^5(c+d x)+10 \sin ^6(c+d x)\right )}{60 d} \] Input:
Integrate[Cos[c + d*x]^4*Cot[c + d*x]*(a + a*Sin[c + d*x])^2,x]
Output:
(a^2*(60*Log[Sin[c + d*x]] + 120*Sin[c + d*x] - 30*Sin[c + d*x]^2 - 80*Sin [c + d*x]^3 - 15*Sin[c + d*x]^4 + 24*Sin[c + d*x]^5 + 10*Sin[c + d*x]^6))/ (60*d)
Time = 0.30 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.90, 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.
\(\displaystyle \int \cos ^4(c+d x) \cot (c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5 (a \sin (c+d x)+a)^2}{\sin (c+d x)}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \csc (c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\csc (c+d x) (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4}{a}d(a \sin (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\sin ^5(c+d x) a^5+2 \sin ^4(c+d x) a^5-\sin ^3(c+d x) a^5-4 \sin ^2(c+d x) a^5+\csc (c+d x) a^5-\sin (c+d x) a^5+2 a^5\right )d(a \sin (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{6} a^6 \sin ^6(c+d x)+\frac {2}{5} a^6 \sin ^5(c+d x)-\frac {1}{4} a^6 \sin ^4(c+d x)-\frac {4}{3} a^6 \sin ^3(c+d x)-\frac {1}{2} a^6 \sin ^2(c+d x)+2 a^6 \sin (c+d x)+a^6 \log (a \sin (c+d x))}{a^4 d}\) |
Input:
Int[Cos[c + d*x]^4*Cot[c + d*x]*(a + a*Sin[c + d*x])^2,x]
Output:
(a^6*Log[a*Sin[c + d*x]] + 2*a^6*Sin[c + d*x] - (a^6*Sin[c + d*x]^2)/2 - ( 4*a^6*Sin[c + d*x]^3)/3 - (a^6*Sin[c + d*x]^4)/4 + (2*a^6*Sin[c + d*x]^5)/ 5 + (a^6*Sin[c + d*x]^6)/6)/(a^4*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 = 4.14 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.69
method | result | size |
derivativedivides | \(\frac {-\frac {a^{2} \cos \left (d x +c \right )^{6}}{6}+\frac {2 a^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+a^{2} \left (\frac {\cos \left (d x +c \right )^{4}}{4}+\frac {\cos \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(82\) |
default | \(\frac {-\frac {a^{2} \cos \left (d x +c \right )^{6}}{6}+\frac {2 a^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+a^{2} \left (\frac {\cos \left (d x +c \right )^{4}}{4}+\frac {\cos \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\) | \(82\) |
risch | \(-i a^{2} x +\frac {19 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{128 d}+\frac {19 a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{128 d}-\frac {2 i a^{2} c}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {5 a^{2} \sin \left (d x +c \right )}{4 d}-\frac {a^{2} \cos \left (6 d x +6 c \right )}{192 d}+\frac {a^{2} \sin \left (5 d x +5 c \right )}{40 d}+\frac {5 a^{2} \sin \left (3 d x +3 c \right )}{24 d}\) | \(137\) |
Input:
int(cos(d*x+c)^4*cot(d*x+c)*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/6*a^2*cos(d*x+c)^6+2/5*a^2*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin (d*x+c)+a^2*(1/4*cos(d*x+c)^4+1/2*cos(d*x+c)^2+ln(sin(d*x+c))))
Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.83 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {10 \, a^{2} \cos \left (d x + c\right )^{6} - 15 \, a^{2} \cos \left (d x + c\right )^{4} - 30 \, a^{2} \cos \left (d x + c\right )^{2} - 60 \, a^{2} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 8 \, {\left (3 \, a^{2} \cos \left (d x + c\right )^{4} + 4 \, a^{2} \cos \left (d x + c\right )^{2} + 8 \, a^{2}\right )} \sin \left (d x + c\right )}{60 \, d} \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="fricas" )
Output:
-1/60*(10*a^2*cos(d*x + c)^6 - 15*a^2*cos(d*x + c)^4 - 30*a^2*cos(d*x + c) ^2 - 60*a^2*log(1/2*sin(d*x + c)) - 8*(3*a^2*cos(d*x + c)^4 + 4*a^2*cos(d* x + c)^2 + 8*a^2)*sin(d*x + c))/d
\[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=a^{2} \left (\int \cos ^{4}{\left (c + d x \right )} \cot {\left (c + d x \right )}\, dx + \int 2 \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )} \cot {\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )} \cot {\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cos(d*x+c)**4*cot(d*x+c)*(a+a*sin(d*x+c))**2,x)
Output:
a**2*(Integral(cos(c + d*x)**4*cot(c + d*x), x) + Integral(2*sin(c + d*x)* cos(c + d*x)**4*cot(c + d*x), x) + Integral(sin(c + d*x)**2*cos(c + d*x)** 4*cot(c + d*x), x))
Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.79 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {10 \, a^{2} \sin \left (d x + c\right )^{6} + 24 \, a^{2} \sin \left (d x + c\right )^{5} - 15 \, a^{2} \sin \left (d x + c\right )^{4} - 80 \, a^{2} \sin \left (d x + c\right )^{3} - 30 \, a^{2} \sin \left (d x + c\right )^{2} + 60 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) + 120 \, a^{2} \sin \left (d x + c\right )}{60 \, d} \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="maxima" )
Output:
1/60*(10*a^2*sin(d*x + c)^6 + 24*a^2*sin(d*x + c)^5 - 15*a^2*sin(d*x + c)^ 4 - 80*a^2*sin(d*x + c)^3 - 30*a^2*sin(d*x + c)^2 + 60*a^2*log(sin(d*x + c )) + 120*a^2*sin(d*x + c))/d
Time = 0.15 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.80 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {10 \, a^{2} \sin \left (d x + c\right )^{6} + 24 \, a^{2} \sin \left (d x + c\right )^{5} - 15 \, a^{2} \sin \left (d x + c\right )^{4} - 80 \, a^{2} \sin \left (d x + c\right )^{3} - 30 \, a^{2} \sin \left (d x + c\right )^{2} + 60 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 120 \, a^{2} \sin \left (d x + c\right )}{60 \, d} \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
1/60*(10*a^2*sin(d*x + c)^6 + 24*a^2*sin(d*x + c)^5 - 15*a^2*sin(d*x + c)^ 4 - 80*a^2*sin(d*x + c)^3 - 30*a^2*sin(d*x + c)^2 + 60*a^2*log(abs(sin(d*x + c))) + 120*a^2*sin(d*x + c))/d
Time = 18.79 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.11 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {5\,a^2\,\sin \left (c+d\,x\right )}{4\,d}-\frac {a^2\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}+\frac {a^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {19\,a^2\,\cos \left (2\,c+2\,d\,x\right )}{64\,d}-\frac {a^2\,\cos \left (6\,c+6\,d\,x\right )}{192\,d}+\frac {5\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{24\,d}+\frac {a^2\,\sin \left (5\,c+5\,d\,x\right )}{40\,d} \] Input:
int(cos(c + d*x)^4*cot(c + d*x)*(a + a*sin(c + d*x))^2,x)
Output:
(5*a^2*sin(c + d*x))/(4*d) - (a^2*log(1/cos(c/2 + (d*x)/2)^2))/d + (a^2*lo g(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (19*a^2*cos(2*c + 2*d*x))/(6 4*d) - (a^2*cos(6*c + 6*d*x))/(192*d) + (5*a^2*sin(3*c + 3*d*x))/(24*d) + (a^2*sin(5*c + 5*d*x))/(40*d)
Time = 0.16 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.80 \[ \int \cos ^4(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (-60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )+60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 \sin \left (d x +c \right )^{6}+24 \sin \left (d x +c \right )^{5}-15 \sin \left (d x +c \right )^{4}-80 \sin \left (d x +c \right )^{3}-30 \sin \left (d x +c \right )^{2}+120 \sin \left (d x +c \right )\right )}{60 d} \] Input:
int(cos(d*x+c)^4*cot(d*x+c)*(a+a*sin(d*x+c))^2,x)
Output:
(a**2*( - 60*log(tan((c + d*x)/2)**2 + 1) + 60*log(tan((c + d*x)/2)) + 10* sin(c + d*x)**6 + 24*sin(c + d*x)**5 - 15*sin(c + d*x)**4 - 80*sin(c + d*x )**3 - 30*sin(c + d*x)**2 + 120*sin(c + d*x)))/(60*d)