Integrand size = 19, antiderivative size = 97 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\log (\sin (c+d x))}{a^4 d}-\frac {\log (1+\sin (c+d x))}{a^4 d}+\frac {1}{3 a d (a+a \sin (c+d x))^3}+\frac {1}{2 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {1}{d \left (a^4+a^4 \sin (c+d x)\right )} \] Output:
ln(sin(d*x+c))/a^4/d-ln(1+sin(d*x+c))/a^4/d+1/3/a/d/(a+a*sin(d*x+c))^3+1/2 /d/(a^2+a^2*sin(d*x+c))^2+1/d/(a^4+a^4*sin(d*x+c))
Time = 0.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.64 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {6 \log (\sin (c+d x))-6 \log (1+\sin (c+d x))+\frac {11+15 \sin (c+d x)+6 \sin ^2(c+d x)}{(1+\sin (c+d x))^3}}{6 a^4 d} \] Input:
Integrate[Cot[c + d*x]/(a + a*Sin[c + d*x])^4,x]
Output:
(6*Log[Sin[c + d*x]] - 6*Log[1 + Sin[c + d*x]] + (11 + 15*Sin[c + d*x] + 6 *Sin[c + d*x]^2)/(1 + Sin[c + d*x])^3)/(6*a^4*d)
Time = 0.27 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3186, 54, 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.
\(\displaystyle \int \frac {\cot (c+d x)}{(a \sin (c+d x)+a)^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (c+d x) (a \sin (c+d x)+a)^4}dx\) |
\(\Big \downarrow \) 3186 |
\(\displaystyle \frac {\int \frac {\csc (c+d x)}{a (\sin (c+d x) a+a)^4}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\int \left (\frac {\csc (c+d x)}{a^5}-\frac {1}{a^4 (\sin (c+d x) a+a)}-\frac {1}{a^3 (\sin (c+d x) a+a)^2}-\frac {1}{a^2 (\sin (c+d x) a+a)^3}-\frac {1}{a (\sin (c+d x) a+a)^4}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\log (a \sin (c+d x))}{a^4}-\frac {\log (a \sin (c+d x)+a)}{a^4}+\frac {1}{a^3 (a \sin (c+d x)+a)}+\frac {1}{2 a^2 (a \sin (c+d x)+a)^2}+\frac {1}{3 a (a \sin (c+d x)+a)^3}}{d}\) |
Input:
Int[Cot[c + d*x]/(a + a*Sin[c + d*x])^4,x]
Output:
(Log[a*Sin[c + d*x]]/a^4 - Log[a + a*Sin[c + d*x]]/a^4 + 1/(3*a*(a + a*Sin [c + d*x])^3) + 1/(2*a^2*(a + a*Sin[c + d*x])^2) + 1/(a^3*(a + a*Sin[c + d *x])))/d
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
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 = 3.94 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.63
method | result | size |
derivativedivides | \(\frac {\ln \left (\sin \left (d x +c \right )\right )+\frac {1}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {1}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {1}{1+\sin \left (d x +c \right )}-\ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{4}}\) | \(61\) |
default | \(\frac {\ln \left (\sin \left (d x +c \right )\right )+\frac {1}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {1}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {1}{1+\sin \left (d x +c \right )}-\ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{4}}\) | \(61\) |
risch | \(\frac {2 i \left (-28 \,{\mathrm e}^{3 i \left (d x +c \right )}-15 i {\mathrm e}^{2 i \left (d x +c \right )}+15 i {\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{5 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6}}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{4}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{4}}\) | \(123\) |
Input:
int(cot(d*x+c)/(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
1/d/a^4*(ln(sin(d*x+c))+1/3/(1+sin(d*x+c))^3+1/2/(1+sin(d*x+c))^2+1/(1+sin (d*x+c))-ln(1+sin(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.57 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {6 \, \cos \left (d x + c\right )^{2} + 6 \, {\left (3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 6 \, {\left (3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, \sin \left (d x + c\right ) - 17}{6 \, {\left (3 \, a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d + {\left (a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(cot(d*x+c)/(a+a*sin(d*x+c))^4,x, algorithm="fricas")
Output:
1/6*(6*cos(d*x + c)^2 + 6*(3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 4)*sin(d*x + c) - 4)*log(1/2*sin(d*x + c)) - 6*(3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 4)*sin(d*x + c) - 4)*log(sin(d*x + c) + 1) - 15*sin(d*x + c) - 17)/(3*a^4 *d*cos(d*x + c)^2 - 4*a^4*d + (a^4*d*cos(d*x + c)^2 - 4*a^4*d)*sin(d*x + c ))
\[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\int \frac {\cot {\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} + 4 \sin ^{3}{\left (c + d x \right )} + 6 \sin ^{2}{\left (c + d x \right )} + 4 \sin {\left (c + d x \right )} + 1}\, dx}{a^{4}} \] Input:
integrate(cot(d*x+c)/(a+a*sin(d*x+c))**4,x)
Output:
Integral(cot(c + d*x)/(sin(c + d*x)**4 + 4*sin(c + d*x)**3 + 6*sin(c + d*x )**2 + 4*sin(c + d*x) + 1), x)/a**4
Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\frac {6 \, \sin \left (d x + c\right )^{2} + 15 \, \sin \left (d x + c\right ) + 11}{a^{4} \sin \left (d x + c\right )^{3} + 3 \, a^{4} \sin \left (d x + c\right )^{2} + 3 \, a^{4} \sin \left (d x + c\right ) + a^{4}} - \frac {6 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} + \frac {6 \, \log \left (\sin \left (d x + c\right )\right )}{a^{4}}}{6 \, d} \] Input:
integrate(cot(d*x+c)/(a+a*sin(d*x+c))^4,x, algorithm="maxima")
Output:
1/6*((6*sin(d*x + c)^2 + 15*sin(d*x + c) + 11)/(a^4*sin(d*x + c)^3 + 3*a^4 *sin(d*x + c)^2 + 3*a^4*sin(d*x + c) + a^4) - 6*log(sin(d*x + c) + 1)/a^4 + 6*log(sin(d*x + c))/a^4)/d
Time = 0.16 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4} d} + \frac {\log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{4} d} + \frac {6 \, \sin \left (d x + c\right )^{2} + 15 \, \sin \left (d x + c\right ) + 11}{6 \, a^{4} d {\left (\sin \left (d x + c\right ) + 1\right )}^{3}} \] Input:
integrate(cot(d*x+c)/(a+a*sin(d*x+c))^4,x, algorithm="giac")
Output:
-log(abs(sin(d*x + c) + 1))/(a^4*d) + log(abs(sin(d*x + c)))/(a^4*d) + 1/6 *(6*sin(d*x + c)^2 + 15*sin(d*x + c) + 11)/(a^4*d*(sin(d*x + c) + 1)^3)
Time = 18.76 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.12 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^4\,d}-\frac {6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {80\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+15\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+20\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+15\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a^4\right )} \] Input:
int(cot(c + d*x)/(a + a*sin(c + d*x))^4,x)
Output:
log(tan(c/2 + (d*x)/2))/(a^4*d) - (2*log(tan(c/2 + (d*x)/2) + 1))/(a^4*d) - (6*tan(c/2 + (d*x)/2) + 18*tan(c/2 + (d*x)/2)^2 + (80*tan(c/2 + (d*x)/2) ^3)/3 + 18*tan(c/2 + (d*x)/2)^4 + 6*tan(c/2 + (d*x)/2)^5)/(d*(15*a^4*tan(c /2 + (d*x)/2)^2 + 20*a^4*tan(c/2 + (d*x)/2)^3 + 15*a^4*tan(c/2 + (d*x)/2)^ 4 + 6*a^4*tan(c/2 + (d*x)/2)^5 + a^4*tan(c/2 + (d*x)/2)^6 + a^4 + 6*a^4*ta n(c/2 + (d*x)/2)))
Time = 0.16 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.14 \[ \int \frac {\cot (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3}-36 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}-36 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3}+18 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}+18 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 \sin \left (d x +c \right )^{3}-9 \sin \left (d x +c \right )^{2}+6}{6 a^{4} d \left (\sin \left (d x +c \right )^{3}+3 \sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )+1\right )} \] Input:
int(cot(d*x+c)/(a+a*sin(d*x+c))^4,x)
Output:
( - 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3 - 36*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 - 36*log(tan((c + d*x)/2) + 1)*sin(c + d*x) - 12*log (tan((c + d*x)/2) + 1) + 6*log(tan((c + d*x)/2))*sin(c + d*x)**3 + 18*log( tan((c + d*x)/2))*sin(c + d*x)**2 + 18*log(tan((c + d*x)/2))*sin(c + d*x) + 6*log(tan((c + d*x)/2)) - 5*sin(c + d*x)**3 - 9*sin(c + d*x)**2 + 6)/(6* a**4*d*(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1))