Integrand size = 25, antiderivative size = 145 \[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {39 a \log (1-\sin (c+d x))}{16 d}-\frac {9 a \log (1+\sin (c+d x))}{16 d}-\frac {a \sin (c+d x)}{d}-\frac {a \sin ^2(c+d x)}{2 d}+\frac {a^5}{8 d \left (a^2-a^2 \sin (c+d x)\right )^2}-\frac {5 a^5}{4 d \left (a^4-a^4 \sin (c+d x)\right )}-\frac {a^5}{8 d \left (a^4+a^4 \sin (c+d x)\right )} \] Output:
-39/16*a*ln(1-sin(d*x+c))/d-9/16*a*ln(1+sin(d*x+c))/d-a*sin(d*x+c)/d-1/2*a *sin(d*x+c)^2/d+1/8*a^5/d/(a^2-a^2*sin(d*x+c))^2-5/4*a^5/d/(a^4-a^4*sin(d* x+c))-1/8*a^5/d/(a^4+a^4*sin(d*x+c))
Time = 0.47 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.01 \[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {15 a \text {arctanh}(\sin (c+d x))}{8 d}-\frac {a \left (12 \log (\cos (c+d x))+6 \sec ^2(c+d x)-\sec ^4(c+d x)+2 \sin ^2(c+d x)\right )}{4 d}+\frac {15 a \sec (c+d x) \tan (c+d x)}{8 d}-\frac {15 a \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {5 a \sec (c+d x) \tan ^3(c+d x)}{d}-\frac {a \sin (c+d x) \tan ^4(c+d x)}{d} \] Input:
Integrate[Sin[c + d*x]*(a + a*Sin[c + d*x])*Tan[c + d*x]^5,x]
Output:
(15*a*ArcTanh[Sin[c + d*x]])/(8*d) - (a*(12*Log[Cos[c + d*x]] + 6*Sec[c + d*x]^2 - Sec[c + d*x]^4 + 2*Sin[c + d*x]^2))/(4*d) + (15*a*Sec[c + d*x]*Ta n[c + d*x])/(8*d) - (15*a*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (5*a*Sec[c + d*x]*Tan[c + d*x]^3)/d - (a*Sin[c + d*x]*Tan[c + d*x]^4)/d
Time = 0.33 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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.
| \(\displaystyle \int \sin (c+d x) \tan ^5(c+d x) (a \sin (c+d x)+a) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^6 (a \sin (c+d x)+a)}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {a^5 \int \frac {\sin ^6(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a^6 \sin ^6(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{a d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (\frac {a^4}{4 (a-a \sin (c+d x))^3}-\frac {5 a^3}{4 (a-a \sin (c+d x))^2}+\frac {a^3}{8 (\sin (c+d x) a+a)^2}+\frac {39 a^2}{16 (a-a \sin (c+d x))}-\frac {9 a^2}{16 (\sin (c+d x) a+a)}-\sin (c+d x) a-a\right )d(a \sin (c+d x))}{a d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {a^4}{8 (a-a \sin (c+d x))^2}-\frac {5 a^3}{4 (a-a \sin (c+d x))}-\frac {a^3}{8 (a \sin (c+d x)+a)}-\frac {1}{2} a^2 \sin ^2(c+d x)-a^2 \sin (c+d x)-\frac {39}{16} a^2 \log (a-a \sin (c+d x))-\frac {9}{16} a^2 \log (a \sin (c+d x)+a)}{a d}\) |
Input:
Int[Sin[c + d*x]*(a + a*Sin[c + d*x])*Tan[c + d*x]^5,x]
Output:
((-39*a^2*Log[a - a*Sin[c + d*x]])/16 - (9*a^2*Log[a + a*Sin[c + d*x]])/16 - a^2*Sin[c + d*x] - (a^2*Sin[c + d*x]^2)/2 + a^4/(8*(a - a*Sin[c + d*x]) ^2) - (5*a^3)/(4*(a - a*Sin[c + d*x])) - a^3/(8*(a + a*Sin[c + d*x])))/(a* 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 = 1.41 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.15
| method | result | size |
| derivativedivides | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{8}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{6}}{2}-\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {3 \sin \left (d x +c \right )^{2}}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+a \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(167\) |
| default | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{8}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{6}}{2}-\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {3 \sin \left (d x +c \right )^{2}}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+a \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(167\) |
| parts | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {a \left (\frac {\sin \left (d x +c \right )^{8}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{6}}{2}-\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {3 \sin \left (d x +c \right )^{2}}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(169\) |
| risch | \(3 i a x +\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {6 i a c}{d}+\frac {i a \left (6 i {\mathrm e}^{4 i \left (d x +c \right )}+9 \,{\mathrm e}^{5 i \left (d x +c \right )}-6 i {\mathrm e}^{2 i \left (d x +c \right )}+22 \,{\mathrm e}^{3 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} d}-\frac {9 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {39 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}\) | \(208\) |
Input:
int(sin(d*x+c)*(a+a*sin(d*x+c))*tan(d*x+c)^5,x,method=_RETURNVERBOSE)
Output:
1/d*(a*(1/4*sin(d*x+c)^8/cos(d*x+c)^4-1/2*sin(d*x+c)^8/cos(d*x+c)^2-1/2*si n(d*x+c)^6-3/4*sin(d*x+c)^4-3/2*sin(d*x+c)^2-3*ln(cos(d*x+c)))+a*(1/4*sin( d*x+c)^7/cos(d*x+c)^4-3/8*sin(d*x+c)^7/cos(d*x+c)^2-3/8*sin(d*x+c)^5-5/8*s in(d*x+c)^3-15/8*sin(d*x+c)+15/8*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.16 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.19 \[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {8 \, a \cos \left (d x + c\right )^{4} + 6 \, a \cos \left (d x + c\right )^{2} - 9 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 39 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (4 \, a \cos \left (d x + c\right )^{4} + 6 \, a \cos \left (d x + c\right )^{2} - 3 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate(sin(d*x+c)*(a+a*sin(d*x+c))*tan(d*x+c)^5,x, algorithm="fricas")
Output:
1/16*(8*a*cos(d*x + c)^4 + 6*a*cos(d*x + c)^2 - 9*(a*cos(d*x + c)^2*sin(d* x + c) - a*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 39*(a*cos(d*x + c)^2*si n(d*x + c) - a*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) + 2*(4*a*cos(d*x + c )^4 + 6*a*cos(d*x + c)^2 - 3*a)*sin(d*x + c) + 2*a)/(d*cos(d*x + c)^2*sin( d*x + c) - d*cos(d*x + c)^2)
\[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=a \left (\int \sin {\left (c + d x \right )} \tan ^{5}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \tan ^{5}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sin(d*x+c)*(a+a*sin(d*x+c))*tan(d*x+c)**5,x)
Output:
a*(Integral(sin(c + d*x)*tan(c + d*x)**5, x) + Integral(sin(c + d*x)**2*ta n(c + d*x)**5, x))
Time = 0.03 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.73 \[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {8 \, a \sin \left (d x + c\right )^{2} + 9 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) + 39 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) + 16 \, a \sin \left (d x + c\right ) - \frac {2 \, {\left (9 \, a \sin \left (d x + c\right )^{2} + 3 \, a \sin \left (d x + c\right ) - 10 \, a\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )^{2} - \sin \left (d x + c\right ) + 1}}{16 \, d} \] Input:
integrate(sin(d*x+c)*(a+a*sin(d*x+c))*tan(d*x+c)^5,x, algorithm="maxima")
Output:
-1/16*(8*a*sin(d*x + c)^2 + 9*a*log(sin(d*x + c) + 1) + 39*a*log(sin(d*x + c) - 1) + 16*a*sin(d*x + c) - 2*(9*a*sin(d*x + c)^2 + 3*a*sin(d*x + c) - 10*a)/(sin(d*x + c)^3 - sin(d*x + c)^2 - sin(d*x + c) + 1))/d
Time = 0.21 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.72 \[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {1}{16} \, a {\left (\frac {9 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{d} + \frac {39 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} + \frac {8 \, {\left (d \sin \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right )\right )}}{d^{2}} - \frac {2 \, {\left (9 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) - 10\right )}}{d {\left (\sin \left (d x + c\right ) + 1\right )} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}}\right )} \] Input:
integrate(sin(d*x+c)*(a+a*sin(d*x+c))*tan(d*x+c)^5,x, algorithm="giac")
Output:
-1/16*a*(9*log(abs(sin(d*x + c) + 1))/d + 39*log(abs(sin(d*x + c) - 1))/d + 8*(d*sin(d*x + c)^2 + 2*d*sin(d*x + c))/d^2 - 2*(9*sin(d*x + c)^2 + 3*si n(d*x + c) - 10)/(d*(sin(d*x + c) + 1)*(sin(d*x + c) - 1)^2))
Time = 32.99 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.97 \[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {-\frac {15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}+7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}-\frac {7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}-\frac {15\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {9\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{8\,d}-\frac {39\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{8\,d}+\frac {3\,a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \] Input:
int(sin(c + d*x)*tan(c + d*x)^5*(a + a*sin(c + d*x)),x)
Output:
((3*a*tan(c/2 + (d*x)/2)^2)/2 - (15*a*tan(c/2 + (d*x)/2))/4 + 7*a*tan(c/2 + (d*x)/2)^3 - (7*a*tan(c/2 + (d*x)/2)^4)/2 + (11*a*tan(c/2 + (d*x)/2)^5)/ 2 - (7*a*tan(c/2 + (d*x)/2)^6)/2 + 7*a*tan(c/2 + (d*x)/2)^7 + (3*a*tan(c/2 + (d*x)/2)^8)/2 - (15*a*tan(c/2 + (d*x)/2)^9)/4)/(d*(tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)/2) - 2*tan(c/2 + (d*x)/2)^4 + 4*tan(c/2 + (d*x)/2)^5 - 2*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 - 2*tan(c/2 + (d*x)/2)^9 + tan(c/2 + (d*x)/2)^10 + 1)) - (9*a*log(tan(c/2 + (d*x)/2) + 1))/(8*d) - ( 39*a*log(tan(c/2 + (d*x)/2) - 1))/(8*d) + (3*a*log(tan(c/2 + (d*x)/2)^2 + 1))/d
Time = 0.21 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.21 \[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {a \left (24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{3}-24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}-24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )+24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )-39 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3}+39 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+39 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )-39 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3}+9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )-9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-4 \sin \left (d x +c \right )^{5}-4 \sin \left (d x +c \right )^{4}+7 \sin \left (d x +c \right )^{3}+18 \sin \left (d x +c \right )^{2}-15\right )}{8 d \left (\sin \left (d x +c \right )^{3}-\sin \left (d x +c \right )^{2}-\sin \left (d x +c \right )+1\right )} \] Input:
int(sin(d*x+c)*(a+a*sin(d*x+c))*tan(d*x+c)^5,x)
Output:
(a*(24*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**3 - 24*log(tan((c + d*x) /2)**2 + 1)*sin(c + d*x)**2 - 24*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x) + 24*log(tan((c + d*x)/2)**2 + 1) - 39*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3 + 39*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 + 39*log(tan((c + d *x)/2) - 1)*sin(c + d*x) - 39*log(tan((c + d*x)/2) - 1) - 9*log(tan((c + d *x)/2) + 1)*sin(c + d*x)**3 + 9*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 9*log(tan((c + d*x)/2) + 1)*sin(c + d*x) - 9*log(tan((c + d*x)/2) + 1) - 4*sin(c + d*x)**5 - 4*sin(c + d*x)**4 + 7*sin(c + d*x)**3 + 18*sin(c + d* x)**2 - 15))/(8*d*(sin(c + d*x)**3 - sin(c + d*x)**2 - sin(c + d*x) + 1))