Integrand size = 21, antiderivative size = 82 \[ \int \frac {\cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3 \text {arctanh}(\cos (c+d x))}{8 a d}-\frac {\cot ^5(c+d x)}{5 a d}-\frac {3 \cot (c+d x) \csc (c+d x)}{8 a d}+\frac {\cot ^3(c+d x) \csc (c+d x)}{4 a d} \] Output:
3/8*arctanh(cos(d*x+c))/a/d-1/5*cot(d*x+c)^5/a/d-3/8*cot(d*x+c)*csc(d*x+c) /a/d+1/4*cot(d*x+c)^3*csc(d*x+c)/a/d
Leaf count is larger than twice the leaf count of optimal. \(189\) vs. \(2(82)=164\).
Time = 0.82 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.30 \[ \int \frac {\cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\csc ^5(c+d x) \left (80 \cos (c+d x)+40 \cos (3 (c+d x))+8 \cos (5 (c+d x))-150 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+150 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+20 \sin (2 (c+d x))+75 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-75 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-50 \sin (4 (c+d x))-15 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))+15 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (5 (c+d x))\right )}{640 a d} \] Input:
Integrate[Cot[c + d*x]^6/(a + a*Sin[c + d*x]),x]
Output:
-1/640*(Csc[c + d*x]^5*(80*Cos[c + d*x] + 40*Cos[3*(c + d*x)] + 8*Cos[5*(c + d*x)] - 150*Log[Cos[(c + d*x)/2]]*Sin[c + d*x] + 150*Log[Sin[(c + d*x)/ 2]]*Sin[c + d*x] + 20*Sin[2*(c + d*x)] + 75*Log[Cos[(c + d*x)/2]]*Sin[3*(c + d*x)] - 75*Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 50*Sin[4*(c + d*x)] - 15*Log[Cos[(c + d*x)/2]]*Sin[5*(c + d*x)] + 15*Log[Sin[(c + d*x)/2]]*Si n[5*(c + d*x)]))/(a*d)
Time = 0.55 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3185, 3042, 3087, 15, 3091, 3042, 3091, 3042, 4257}
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.
| \(\displaystyle \int \frac {\cot ^6(c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (c+d x)^6 (a \sin (c+d x)+a)}dx\) |
| \(\Big \downarrow \) 3185 |
| \(\displaystyle \frac {\int \cot ^4(c+d x) \csc ^2(c+d x)dx}{a}-\frac {\int \cot ^4(c+d x) \csc (c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sec \left (c+d x-\frac {\pi }{2}\right )^2 \tan \left (c+d x-\frac {\pi }{2}\right )^4dx}{a}-\frac {\int \sec \left (c+d x-\frac {\pi }{2}\right ) \tan \left (c+d x-\frac {\pi }{2}\right )^4dx}{a}\) |
| \(\Big \downarrow \) 3087 |
| \(\displaystyle \frac {\int \cot ^4(c+d x)d(-\cot (c+d x))}{a d}-\frac {\int \sec \left (c+d x-\frac {\pi }{2}\right ) \tan \left (c+d x-\frac {\pi }{2}\right )^4dx}{a}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle -\frac {\int \sec \left (c+d x-\frac {\pi }{2}\right ) \tan \left (c+d x-\frac {\pi }{2}\right )^4dx}{a}-\frac {\cot ^5(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle -\frac {-\frac {3}{4} \int \cot ^2(c+d x) \csc (c+d x)dx-\frac {\cot ^3(c+d x) \csc (c+d x)}{4 d}}{a}-\frac {\cot ^5(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {3}{4} \int \sec \left (c+d x-\frac {\pi }{2}\right ) \tan \left (c+d x-\frac {\pi }{2}\right )^2dx-\frac {\cot ^3(c+d x) \csc (c+d x)}{4 d}}{a}-\frac {\cot ^5(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle -\frac {-\frac {3}{4} \left (-\frac {1}{2} \int \csc (c+d x)dx-\frac {\cot (c+d x) \csc (c+d x)}{2 d}\right )-\frac {\cot ^3(c+d x) \csc (c+d x)}{4 d}}{a}-\frac {\cot ^5(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {3}{4} \left (-\frac {1}{2} \int \csc (c+d x)dx-\frac {\cot (c+d x) \csc (c+d x)}{2 d}\right )-\frac {\cot ^3(c+d x) \csc (c+d x)}{4 d}}{a}-\frac {\cot ^5(c+d x)}{5 a d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle -\frac {-\frac {3}{4} \left (\frac {\text {arctanh}(\cos (c+d x))}{2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 d}\right )-\frac {\cot ^3(c+d x) \csc (c+d x)}{4 d}}{a}-\frac {\cot ^5(c+d x)}{5 a d}\) |
Input:
Int[Cot[c + d*x]^6/(a + a*Sin[c + d*x]),x]
Output:
-1/5*Cot[c + d*x]^5/(a*d) - (-1/4*(Cot[c + d*x]^3*Csc[c + d*x])/d - (3*(Ar cTanh[Cos[c + d*x]]/(2*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*d)))/4)/a
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> Simp[1/f Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n - 1) /2] && LtQ[0, n, m - 1])
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1)) Int[(a*Sec[e + f*x])^m*( b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & & NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[1/a Int[Sec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x ] - Simp[1/(b*g) Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /; Fre eQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Result contains complex when optimal does not.
Time = 0.80 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.63
| method | result | size |
| risch | \(\frac {-40 i {\mathrm e}^{8 i \left (d x +c \right )}+25 \,{\mathrm e}^{9 i \left (d x +c \right )}-10 \,{\mathrm e}^{7 i \left (d x +c \right )}-80 i {\mathrm e}^{4 i \left (d x +c \right )}+10 \,{\mathrm e}^{3 i \left (d x +c \right )}-8 i-25 \,{\mathrm e}^{i \left (d x +c \right )}}{20 a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d a}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d a}\) | \(134\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}-\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}\) | \(148\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {1}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {1}{5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}-\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}\) | \(148\) |
Input:
int(cot(d*x+c)^6/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/20*(-40*I*exp(8*I*(d*x+c))+25*exp(9*I*(d*x+c))-10*exp(7*I*(d*x+c))-80*I* exp(4*I*(d*x+c))+10*exp(3*I*(d*x+c))-8*I-25*exp(I*(d*x+c)))/a/d/(exp(2*I*( d*x+c))-1)^5+3/8/d/a*ln(exp(I*(d*x+c))+1)-3/8/d/a*ln(exp(I*(d*x+c))-1)
Leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (74) = 148\).
Time = 0.09 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.89 \[ \int \frac {\cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {16 \, \cos \left (d x + c\right )^{5} - 15 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 15 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 10 \, {\left (5 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{80 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cot(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/80*(16*cos(d*x + c)^5 - 15*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log( 1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 15*(cos(d*x + c)^4 - 2*cos(d*x + c) ^2 + 1)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 10*(5*cos(d*x + c)^3 - 3*cos(d*x + c))*sin(d*x + c))/((a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^2 + a*d)*sin(d*x + c))
\[ \int \frac {\cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cot ^{6}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(cot(d*x+c)**6/(a+a*sin(d*x+c)),x)
Output:
Integral(cot(c + d*x)**6/(sin(c + d*x) + 1), x)/a
Leaf count of result is larger than twice the leaf count of optimal. 234 vs. \(2 (74) = 148\).
Time = 0.04 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.85 \[ \int \frac {\cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\frac {20 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {5 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {2 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a} - \frac {120 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {{\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {20 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 2\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{5}}{a \sin \left (d x + c\right )^{5}}}{320 \, d} \] Input:
integrate(cot(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
1/320*((20*sin(d*x + c)/(cos(d*x + c) + 1) + 40*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 5*sin(d*x + c)^4/(cos (d*x + c) + 1)^4 + 2*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a - 120*log(sin( d*x + c)/(cos(d*x + c) + 1))/a + (5*sin(d*x + c)/(cos(d*x + c) + 1) + 10*s in(d*x + c)^2/(cos(d*x + c) + 1)^2 - 40*sin(d*x + c)^3/(cos(d*x + c) + 1)^ 3 - 20*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 2)*(cos(d*x + c) + 1)^5/(a*si n(d*x + c)^5))/d
Leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (74) = 148\).
Time = 0.20 (sec) , antiderivative size = 187, normalized size of antiderivative = 2.28 \[ \int \frac {\cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {120 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} - \frac {2 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 10 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 20 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{5}} - \frac {274 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 20 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{320 \, d} \] Input:
integrate(cot(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/320*(120*log(abs(tan(1/2*d*x + 1/2*c)))/a - (2*a^4*tan(1/2*d*x + 1/2*c) ^5 - 5*a^4*tan(1/2*d*x + 1/2*c)^4 - 10*a^4*tan(1/2*d*x + 1/2*c)^3 + 40*a^4 *tan(1/2*d*x + 1/2*c)^2 + 20*a^4*tan(1/2*d*x + 1/2*c))/a^5 - (274*tan(1/2* d*x + 1/2*c)^5 - 20*tan(1/2*d*x + 1/2*c)^4 - 40*tan(1/2*d*x + 1/2*c)^3 + 1 0*tan(1/2*d*x + 1/2*c)^2 + 5*tan(1/2*d*x + 1/2*c) - 2)/(a*tan(1/2*d*x + 1/ 2*c)^5))/d
Time = 26.66 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.23 \[ \int \frac {\cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{32\,a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,a\,d}-\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,a\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,a\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {1}{5}\right )}{32\,a\,d} \] Input:
int(cot(c + d*x)^6/(a + a*sin(c + d*x)),x)
Output:
tan(c/2 + (d*x)/2)^2/(8*a*d) - tan(c/2 + (d*x)/2)^3/(32*a*d) - tan(c/2 + ( d*x)/2)^4/(64*a*d) + tan(c/2 + (d*x)/2)^5/(160*a*d) - (3*log(tan(c/2 + (d* x)/2)))/(8*a*d) + tan(c/2 + (d*x)/2)/(16*a*d) - (cot(c/2 + (d*x)/2)^5*(4*t an(c/2 + (d*x)/2)^3 - tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d*x)/2)/2 + 2*tan( c/2 + (d*x)/2)^4 + 1/5))/(32*a*d)
Time = 0.15 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.30 \[ \int \frac {\cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-25 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+10 \cos \left (d x +c \right ) \sin \left (d x +c \right )-8 \cos \left (d x +c \right )-15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5}}{40 \sin \left (d x +c \right )^{5} a d} \] Input:
int(cot(d*x+c)^6/(a+a*sin(d*x+c)),x)
Output:
( - 8*cos(c + d*x)*sin(c + d*x)**4 - 25*cos(c + d*x)*sin(c + d*x)**3 + 16* cos(c + d*x)*sin(c + d*x)**2 + 10*cos(c + d*x)*sin(c + d*x) - 8*cos(c + d* x) - 15*log(tan((c + d*x)/2))*sin(c + d*x)**5)/(40*sin(c + d*x)**5*a*d)