Integrand size = 21, antiderivative size = 65 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \csc (c+d x)}{a^2 d}-\frac {\csc ^2(c+d x)}{2 a^2 d}+\frac {2 \log (\sin (c+d x))}{a^2 d}-\frac {2 \log (1+\sin (c+d x))}{a^2 d} \] Output:
2*csc(d*x+c)/a^2/d-1/2*csc(d*x+c)^2/a^2/d+2*ln(sin(d*x+c))/a^2/d-2*ln(1+si n(d*x+c))/a^2/d
Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.75 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {4 \csc (c+d x)-\csc ^2(c+d x)+4 \log (\sin (c+d x))-4 \log (1+\sin (c+d x))}{2 a^2 d} \] Input:
Integrate[Cot[c + d*x]^3/(a + a*Sin[c + d*x])^2,x]
Output:
(4*Csc[c + d*x] - Csc[c + d*x]^2 + 4*Log[Sin[c + d*x]] - 4*Log[1 + Sin[c + d*x]])/(2*a^2*d)
Time = 0.26 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3186, 86, 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.
\(\displaystyle \int \frac {\cot ^3(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (c+d x)^3 (a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3186 |
\(\displaystyle \frac {\int \frac {\csc ^3(c+d x) (a-a \sin (c+d x))}{a^3 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (\frac {\csc ^3(c+d x)}{a^3}-\frac {2 \csc ^2(c+d x)}{a^3}+\frac {2 \csc (c+d x)}{a^3}-\frac {2}{a^2 (\sin (c+d x) a+a)}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {\csc ^2(c+d x)}{2 a^2}+\frac {2 \csc (c+d x)}{a^2}+\frac {2 \log (a \sin (c+d x))}{a^2}-\frac {2 \log (a \sin (c+d x)+a)}{a^2}}{d}\) |
Input:
Int[Cot[c + d*x]^3/(a + a*Sin[c + d*x])^2,x]
Output:
((2*Csc[c + d*x])/a^2 - Csc[c + d*x]^2/(2*a^2) + (2*Log[a*Sin[c + d*x]])/a ^2 - (2*Log[a + a*Sin[c + d*x]])/a^2)/d
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[x^p*((a + x)^(m - (p + 1)/2)/(a - x) ^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && E qQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]
Time = 2.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.75
method | result | size |
derivativedivides | \(\frac {-2 \ln \left (1+\sin \left (d x +c \right )\right )-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\frac {2}{\sin \left (d x +c \right )}+2 \ln \left (\sin \left (d x +c \right )\right )}{d \,a^{2}}\) | \(49\) |
default | \(\frac {-2 \ln \left (1+\sin \left (d x +c \right )\right )-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\frac {2}{\sin \left (d x +c \right )}+2 \ln \left (\sin \left (d x +c \right )\right )}{d \,a^{2}}\) | \(49\) |
risch | \(\frac {2 i \left (-i {\mathrm e}^{2 i \left (d x +c \right )}+2 \,{\mathrm e}^{3 i \left (d x +c \right )}-2 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{2}}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}\) | \(100\) |
Input:
int(cot(d*x+c)^3/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d/a^2*(-2*ln(1+sin(d*x+c))-1/2/sin(d*x+c)^2+2/sin(d*x+c)+2*ln(sin(d*x+c) ))
Time = 0.16 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.17 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 4 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, \sin \left (d x + c\right ) + 1}{2 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}} \] Input:
integrate(cot(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="fricas")
Output:
1/2*(4*(cos(d*x + c)^2 - 1)*log(1/2*sin(d*x + c)) - 4*(cos(d*x + c)^2 - 1) *log(sin(d*x + c) + 1) - 4*sin(d*x + c) + 1)/(a^2*d*cos(d*x + c)^2 - a^2*d )
\[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\cot ^{3}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(cot(d*x+c)**3/(a+a*sin(d*x+c))**2,x)
Output:
Integral(cot(c + d*x)**3/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.85 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {4 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} - \frac {4 \, \log \left (\sin \left (d x + c\right )\right )}{a^{2}} - \frac {4 \, \sin \left (d x + c\right ) - 1}{a^{2} \sin \left (d x + c\right )^{2}}}{2 \, d} \] Input:
integrate(cot(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxima")
Output:
-1/2*(4*log(sin(d*x + c) + 1)/a^2 - 4*log(sin(d*x + c))/a^2 - (4*sin(d*x + c) - 1)/(a^2*sin(d*x + c)^2))/d
Time = 0.13 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {2 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} d} + \frac {2 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2} d} + \frac {4 \, \sin \left (d x + c\right ) - 1}{2 \, a^{2} d \sin \left (d x + c\right )^{2}} \] Input:
integrate(cot(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
-2*log(abs(sin(d*x + c) + 1))/(a^2*d) + 2*log(abs(sin(d*x + c)))/(a^2*d) + 1/2*(4*sin(d*x + c) - 1)/(a^2*d*sin(d*x + c)^2)
Time = 17.69 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.58 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^2\,d}-\frac {4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^2\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{8}\right )}{a^2\,d} \] Input:
int(cot(c + d*x)^3/(a + a*sin(c + d*x))^2,x)
Output:
(2*log(tan(c/2 + (d*x)/2)))/(a^2*d) - tan(c/2 + (d*x)/2)^2/(8*a^2*d) - (4* log(tan(c/2 + (d*x)/2) + 1))/(a^2*d) + tan(c/2 + (d*x)/2)/(a^2*d) + (cot(c /2 + (d*x)/2)^2*(tan(c/2 + (d*x)/2) - 1/8))/(a^2*d)
Time = 0.18 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.17 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-16 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}+\sin \left (d x +c \right )^{2}+8 \sin \left (d x +c \right )-2}{4 \sin \left (d x +c \right )^{2} a^{2} d} \] Input:
int(cot(d*x+c)^3/(a+a*sin(d*x+c))^2,x)
Output:
( - 16*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 8*log(tan((c + d*x)/2)) *sin(c + d*x)**2 + sin(c + d*x)**2 + 8*sin(c + d*x) - 2)/(4*sin(c + d*x)** 2*a**2*d)