Integrand size = 21, antiderivative size = 120 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {12 \csc (c+d x)}{a^4 d}-\frac {4 \csc ^2(c+d x)}{a^4 d}+\frac {4 \csc ^3(c+d x)}{3 a^4 d}-\frac {\csc ^4(c+d x)}{4 a^4 d}+\frac {16 \log (\sin (c+d x))}{a^4 d}-\frac {16 \log (1+\sin (c+d x))}{a^4 d}+\frac {4}{d \left (a^4+a^4 \sin (c+d x)\right )} \] Output:
12*csc(d*x+c)/a^4/d-4*csc(d*x+c)^2/a^4/d+4/3*csc(d*x+c)^3/a^4/d-1/4*csc(d* x+c)^4/a^4/d+16*ln(sin(d*x+c))/a^4/d-16*ln(1+sin(d*x+c))/a^4/d+4/d/(a^4+a^ 4*sin(d*x+c))
Time = 0.80 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.68 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {144 \csc (c+d x)-48 \csc ^2(c+d x)+16 \csc ^3(c+d x)-3 \csc ^4(c+d x)+192 \log (\sin (c+d x))-192 \log (1+\sin (c+d x))+\frac {48}{1+\sin (c+d x)}}{12 a^4 d} \] Input:
Integrate[Cot[c + d*x]^5/(a + a*Sin[c + d*x])^4,x]
Output:
(144*Csc[c + d*x] - 48*Csc[c + d*x]^2 + 16*Csc[c + d*x]^3 - 3*Csc[c + d*x] ^4 + 192*Log[Sin[c + d*x]] - 192*Log[1 + Sin[c + d*x]] + 48/(1 + Sin[c + d *x]))/(12*a^4*d)
Time = 0.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3186, 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.
| \(\displaystyle \int \frac {\cot ^5(c+d x)}{(a \sin (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (c+d x)^5 (a \sin (c+d x)+a)^4}dx\) |
| \(\Big \downarrow \) 3186 |
| \(\displaystyle \frac {\int \frac {\csc ^5(c+d x) (a-a \sin (c+d x))^2}{a^5 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (\frac {\csc ^5(c+d x)}{a^5}-\frac {4 \csc ^4(c+d x)}{a^5}+\frac {8 \csc ^3(c+d x)}{a^5}-\frac {12 \csc ^2(c+d x)}{a^5}+\frac {16 \csc (c+d x)}{a^5}-\frac {16}{a^4 (\sin (c+d x) a+a)}-\frac {4}{a^3 (\sin (c+d x) a+a)^2}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {\csc ^4(c+d x)}{4 a^4}+\frac {4 \csc ^3(c+d x)}{3 a^4}-\frac {4 \csc ^2(c+d x)}{a^4}+\frac {12 \csc (c+d x)}{a^4}+\frac {16 \log (a \sin (c+d x))}{a^4}-\frac {16 \log (a \sin (c+d x)+a)}{a^4}+\frac {4}{a^3 (a \sin (c+d x)+a)}}{d}\) |
Input:
Int[Cot[c + d*x]^5/(a + a*Sin[c + d*x])^4,x]
Output:
((12*Csc[c + d*x])/a^4 - (4*Csc[c + d*x]^2)/a^4 + (4*Csc[c + d*x]^3)/(3*a^ 4) - Csc[c + d*x]^4/(4*a^4) + (16*Log[a*Sin[c + d*x]])/a^4 - (16*Log[a + a *Sin[c + d*x]])/a^4 + 4/(a^3*(a + a*Sin[c + d*x])))/d
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[((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 = 16.78 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.68
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{4 \sin \left (d x +c \right )^{4}}+\frac {4}{3 \sin \left (d x +c \right )^{3}}-\frac {4}{\sin \left (d x +c \right )^{2}}+\frac {12}{\sin \left (d x +c \right )}+16 \ln \left (\sin \left (d x +c \right )\right )+\frac {4}{1+\sin \left (d x +c \right )}-16 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{4}}\) | \(81\) |
| default | \(\frac {-\frac {1}{4 \sin \left (d x +c \right )^{4}}+\frac {4}{3 \sin \left (d x +c \right )^{3}}-\frac {4}{\sin \left (d x +c \right )^{2}}+\frac {12}{\sin \left (d x +c \right )}+16 \ln \left (\sin \left (d x +c \right )\right )+\frac {4}{1+\sin \left (d x +c \right )}-16 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{4}}\) | \(81\) |
| risch | \(\frac {4 i \left (24 i {\mathrm e}^{8 i \left (d x +c \right )}+24 \,{\mathrm e}^{9 i \left (d x +c \right )}-85 i {\mathrm e}^{6 i \left (d x +c \right )}-80 \,{\mathrm e}^{7 i \left (d x +c \right )}+85 i {\mathrm e}^{4 i \left (d x +c \right )}+106 \,{\mathrm e}^{5 i \left (d x +c \right )}-24 i {\mathrm e}^{2 i \left (d x +c \right )}-80 \,{\mathrm e}^{3 i \left (d x +c \right )}+24 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} d \,a^{4}}-\frac {32 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{4}}+\frac {16 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{4}}\) | \(183\) |
Input:
int(cot(d*x+c)^5/(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
1/d/a^4*(-1/4/sin(d*x+c)^4+4/3/sin(d*x+c)^3-4/sin(d*x+c)^2+12/sin(d*x+c)+1 6*ln(sin(d*x+c))+4/(1+sin(d*x+c))-16*ln(1+sin(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (116) = 232\).
Time = 0.09 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.96 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {192 \, \cos \left (d x + c\right )^{4} - 352 \, \cos \left (d x + c\right )^{2} + 192 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 192 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (96 \, \cos \left (d x + c\right )^{2} - 109\right )} \sin \left (d x + c\right ) + 157}{12 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} - 2 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d + {\left (a^{4} d \cos \left (d x + c\right )^{4} - 2 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(cot(d*x+c)^5/(a+a*sin(d*x+c))^4,x, algorithm="fricas")
Output:
1/12*(192*cos(d*x + c)^4 - 352*cos(d*x + c)^2 + 192*(cos(d*x + c)^4 - 2*co s(d*x + c)^2 + (cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*sin(d*x + c) + 1)*l og(1/2*sin(d*x + c)) - 192*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + (cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*sin(d*x + c) + 1)*log(sin(d*x + c) + 1) - (9 6*cos(d*x + c)^2 - 109)*sin(d*x + c) + 157)/(a^4*d*cos(d*x + c)^4 - 2*a^4* d*cos(d*x + c)^2 + a^4*d + (a^4*d*cos(d*x + c)^4 - 2*a^4*d*cos(d*x + c)^2 + a^4*d)*sin(d*x + c))
\[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\int \frac {\cot ^{5}{\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} + 4 \sin ^{3}{\left (c + d x \right )} + 6 \sin ^{2}{\left (c + d x \right )} + 4 \sin {\left (c + d x \right )} + 1}\, dx}{a^{4}} \] Input:
integrate(cot(d*x+c)**5/(a+a*sin(d*x+c))**4,x)
Output:
Integral(cot(c + d*x)**5/(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 = 100, normalized size of antiderivative = 0.83 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\frac {192 \, \sin \left (d x + c\right )^{4} + 96 \, \sin \left (d x + c\right )^{3} - 32 \, \sin \left (d x + c\right )^{2} + 13 \, \sin \left (d x + c\right ) - 3}{a^{4} \sin \left (d x + c\right )^{5} + a^{4} \sin \left (d x + c\right )^{4}} - \frac {192 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} + \frac {192 \, \log \left (\sin \left (d x + c\right )\right )}{a^{4}}}{12 \, d} \] Input:
integrate(cot(d*x+c)^5/(a+a*sin(d*x+c))^4,x, algorithm="maxima")
Output:
1/12*((192*sin(d*x + c)^4 + 96*sin(d*x + c)^3 - 32*sin(d*x + c)^2 + 13*sin (d*x + c) - 3)/(a^4*sin(d*x + c)^5 + a^4*sin(d*x + c)^4) - 192*log(sin(d*x + c) + 1)/a^4 + 192*log(sin(d*x + c))/a^4)/d
Time = 0.14 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.84 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {16 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4} d} + \frac {16 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{4} d} + \frac {192 \, \sin \left (d x + c\right )^{4} + 96 \, \sin \left (d x + c\right )^{3} - 32 \, \sin \left (d x + c\right )^{2} + 13 \, \sin \left (d x + c\right ) - 3}{12 \, a^{4} d {\left (\sin \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )^{4}} \] Input:
integrate(cot(d*x+c)^5/(a+a*sin(d*x+c))^4,x, algorithm="giac")
Output:
-16*log(abs(sin(d*x + c) + 1))/(a^4*d) + 16*log(abs(sin(d*x + c)))/(a^4*d) + 1/12*(192*sin(d*x + c)^4 + 96*sin(d*x + c)^3 - 32*sin(d*x + c)^2 + 13*s in(d*x + c) - 3)/(a^4*d*(sin(d*x + c) + 1)*sin(d*x + c)^4)
Time = 35.00 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.94 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6\,a^4\,d}-\frac {17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a^4\,d}+\frac {16\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {32\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^4\,d}+\frac {-24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+191\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {218\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-\frac {143\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{12}+\frac {13\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{6}-\frac {1}{4}}{d\,\left (16\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+32\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+16\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}+\frac {13\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^4\,d} \] Input:
int(cot(c + d*x)^5/(a + a*sin(c + d*x))^4,x)
Output:
tan(c/2 + (d*x)/2)^3/(6*a^4*d) - (17*tan(c/2 + (d*x)/2)^2)/(16*a^4*d) - ta n(c/2 + (d*x)/2)^4/(64*a^4*d) + (16*log(tan(c/2 + (d*x)/2)))/(a^4*d) - (32 *log(tan(c/2 + (d*x)/2) + 1))/(a^4*d) + ((13*tan(c/2 + (d*x)/2))/6 - (143* tan(c/2 + (d*x)/2)^2)/12 + (218*tan(c/2 + (d*x)/2)^3)/3 + 191*tan(c/2 + (d *x)/2)^4 - 24*tan(c/2 + (d*x)/2)^5 - 1/4)/(d*(16*a^4*tan(c/2 + (d*x)/2)^4 + 32*a^4*tan(c/2 + (d*x)/2)^5 + 16*a^4*tan(c/2 + (d*x)/2)^6)) + (13*tan(c/ 2 + (d*x)/2))/(2*a^4*d)
Time = 0.16 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.33 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {-3072 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{5}-3072 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}+1536 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5}+1536 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4}-423 \sin \left (d x +c \right )^{5}+1113 \sin \left (d x +c \right )^{4}+768 \sin \left (d x +c \right )^{3}-256 \sin \left (d x +c \right )^{2}+104 \sin \left (d x +c \right )-24}{96 \sin \left (d x +c \right )^{4} a^{4} d \left (\sin \left (d x +c \right )+1\right )} \] Input:
int(cot(d*x+c)^5/(a+a*sin(d*x+c))^4,x)
Output:
( - 3072*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5 - 3072*log(tan((c + d*x )/2) + 1)*sin(c + d*x)**4 + 1536*log(tan((c + d*x)/2))*sin(c + d*x)**5 + 1 536*log(tan((c + d*x)/2))*sin(c + d*x)**4 - 423*sin(c + d*x)**5 + 1113*sin (c + d*x)**4 + 768*sin(c + d*x)**3 - 256*sin(c + d*x)**2 + 104*sin(c + d*x ) - 24)/(96*sin(c + d*x)**4*a**4*d*(sin(c + d*x) + 1))