Integrand size = 29, antiderivative size = 171 \[ \int \frac {\csc ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\csc (c+d x)}{a d}-\frac {(3 a+4 b) \log (1-\sin (c+d x))}{4 (a+b)^2 d}-\frac {b \log (\sin (c+d x))}{a^2 d}+\frac {(3 a-4 b) \log (1+\sin (c+d x))}{4 (a-b)^2 d}+\frac {b^5 \log (a+b \sin (c+d x))}{a^2 \left (a^2-b^2\right )^2 d}+\frac {1}{4 (a+b) d (1-\sin (c+d x))}-\frac {1}{4 (a-b) d (1+\sin (c+d x))} \]
-csc(d*x+c)/a/d-1/4*(3*a+4*b)*ln(1-sin(d*x+c))/(a+b)^2/d-b*ln(sin(d*x+c))/ a^2/d+1/4*(3*a-4*b)*ln(1+sin(d*x+c))/(a-b)^2/d+b^5*ln(a+b*sin(d*x+c))/a^2/ (a^2-b^2)^2/d+1/4/(a+b)/d/(1-sin(d*x+c))-1/4/(a-b)/d/(1+sin(d*x+c))
Time = 0.54 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.02 \[ \int \frac {\csc ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\csc (c+d x) (a+b \sin (c+d x)) \left (\frac {4 \csc (c+d x)}{a}+\frac {(3 a+4 b) \log (1-\sin (c+d x))}{(a+b)^2}+\frac {4 b \log (\sin (c+d x))}{a^2}-\frac {(3 a-4 b) \log (1+\sin (c+d x))}{(a-b)^2}-\frac {4 b^5 \log (a+b \sin (c+d x))}{a^2 (a-b)^2 (a+b)^2}+\frac {1}{(a+b) (-1+\sin (c+d x))}+\frac {1}{(a-b) (1+\sin (c+d x))}\right )}{4 d (b+a \csc (c+d x))} \]
-1/4*(Csc[c + d*x]*(a + b*Sin[c + d*x])*((4*Csc[c + d*x])/a + ((3*a + 4*b) *Log[1 - Sin[c + d*x]])/(a + b)^2 + (4*b*Log[Sin[c + d*x]])/a^2 - ((3*a - 4*b)*Log[1 + Sin[c + d*x]])/(a - b)^2 - (4*b^5*Log[a + b*Sin[c + d*x]])/(a ^2*(a - b)^2*(a + b)^2) + 1/((a + b)*(-1 + Sin[c + d*x])) + 1/((a - b)*(1 + Sin[c + d*x]))))/(d*(b + a*Csc[c + d*x]))
Time = 0.44 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3316, 27, 615, 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.
| \(\displaystyle \int \frac {\csc ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^2 \cos (c+d x)^3 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^3 \int \frac {\csc ^2(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^5 \int \frac {\csc ^2(c+d x)}{b^2 (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 615 |
| \(\displaystyle \frac {b^5 \int \left (\frac {\csc ^2(c+d x)}{a b^6}-\frac {\csc (c+d x)}{a^2 b^5}+\frac {3 a+4 b}{4 b^5 (a+b)^2 (b-b \sin (c+d x))}+\frac {1}{a^2 (a-b)^2 (a+b)^2 (a+b \sin (c+d x))}+\frac {3 a-4 b}{4 (a-b)^2 b^5 (\sin (c+d x) b+b)}+\frac {1}{4 b^4 (a+b) (b-b \sin (c+d x))^2}+\frac {1}{4 (a-b) b^4 (\sin (c+d x) b+b)^2}\right )d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^5 \left (-\frac {\log (b \sin (c+d x))}{a^2 b^4}+\frac {\log (a+b \sin (c+d x))}{a^2 \left (a^2-b^2\right )^2}-\frac {\csc (c+d x)}{a b^5}-\frac {(3 a+4 b) \log (b-b \sin (c+d x))}{4 b^5 (a+b)^2}+\frac {(3 a-4 b) \log (b \sin (c+d x)+b)}{4 b^5 (a-b)^2}+\frac {1}{4 b^4 (a+b) (b-b \sin (c+d x))}-\frac {1}{4 b^4 (a-b) (b \sin (c+d x)+b)}\right )}{d}\) |
(b^5*(-(Csc[c + d*x]/(a*b^5)) - Log[b*Sin[c + d*x]]/(a^2*b^4) - ((3*a + 4* b)*Log[b - b*Sin[c + d*x]])/(4*b^5*(a + b)^2) + Log[a + b*Sin[c + d*x]]/(a ^2*(a^2 - b^2)^2) + ((3*a - 4*b)*Log[b + b*Sin[c + d*x]])/(4*(a - b)^2*b^5 ) + 1/(4*b^4*(a + b)*(b - b*Sin[c + d*x])) - 1/(4*(a - b)*b^4*(b + b*Sin[c + d*x]))))/d
3.14.50.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 0.92 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.89
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (3 a -4 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}+\frac {b^{5} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} a^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-3 a -4 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}-\frac {1}{a \sin \left (d x +c \right )}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2}}}{d}\) | \(152\) |
| default | \(\frac {-\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (3 a -4 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}+\frac {b^{5} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} a^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-3 a -4 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}-\frac {1}{a \sin \left (d x +c \right )}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2}}}{d}\) | \(152\) |
| parallelrisch | \(\frac {4 b^{5} \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-6 \left (a +\frac {4 b}{3}\right ) \left (a -b \right )^{2} \left (1+\cos \left (2 d x +2 c \right )\right ) a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 \left (a +b \right ) \left (\left (a +b \right ) \left (a -\frac {4 b}{3}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {\left (\frac {4 b \left (a -b \right ) \left (a +b \right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (a^{2}-\frac {2 b^{2}}{3}\right ) \cos \left (2 d x +2 c \right )+\frac {a^{2}}{3}-\frac {2 b^{2}}{3}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {2 b a \left (-1+\cos \left (2 d x +2 c \right )\right )}{3}\right ) a \right ) \left (a -b \right )}{2}\right )}{4 \left (a -b \right )^{2} \left (a +b \right )^{2} a^{2} d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(266\) |
| norman | \(\frac {-\frac {1}{2 a d}-\frac {\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}-\frac {2 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{2}-b^{2}\right )}+\frac {\left (3 a^{2}-b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a \left (a^{2}-b^{2}\right )}+\frac {\left (3 a^{2}-b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a \left (a^{2}-b^{2}\right )}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {b^{5} \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{a^{2} d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d}+\frac {\left (3 a -4 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (3 a +4 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 \left (a^{2}+2 a b +b^{2}\right ) d}\) | \(315\) |
| risch | \(\frac {2 i b x}{a^{2}-2 a b +b^{2}}-\frac {3 i a c}{2 \left (a^{2}-2 a b +b^{2}\right ) d}-\frac {2 i b^{5} x}{a^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {2 i b c}{\left (a^{2}+2 a b +b^{2}\right ) d}-\frac {i \left (3 a^{2} {\mathrm e}^{5 i \left (d x +c \right )}-2 b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+2 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-4 b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-2 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 a^{2} {\mathrm e}^{i \left (d x +c \right )}-2 b^{2} {\mathrm e}^{i \left (d x +c \right )}+2 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left (a^{2}-b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {3 i a c}{2 d \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 i b^{5} c}{a^{2} d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {2 i b x}{a^{2}}+\frac {2 i b c}{a^{2} d}+\frac {2 i b c}{d \left (a^{2}-2 a b +b^{2}\right )}-\frac {3 i a x}{2 \left (a^{2}-2 a b +b^{2}\right )}+\frac {2 i b x}{a^{2}+2 a b +b^{2}}+\frac {3 i a x}{2 \left (a^{2}+2 a b +b^{2}\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a}{2 \left (a^{2}+2 a b +b^{2}\right ) d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{\left (a^{2}+2 a b +b^{2}\right ) d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a}{2 \left (a^{2}-2 a b +b^{2}\right ) d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{\left (a^{2}-2 a b +b^{2}\right ) d}+\frac {b^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a^{2} d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}\) | \(597\) |
1/d*(-1/(4*a-4*b)/(1+sin(d*x+c))+1/4*(3*a-4*b)/(a-b)^2*ln(1+sin(d*x+c))+b^ 5/(a+b)^2/(a-b)^2/a^2*ln(a+b*sin(d*x+c))-1/(4*a+4*b)/(sin(d*x+c)-1)+1/4/(a +b)^2*(-3*a-4*b)*ln(sin(d*x+c)-1)-1/a/sin(d*x+c)-1/a^2*b*ln(sin(d*x+c)))
Time = 0.94 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.68 \[ \int \frac {\csc ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {4 \, b^{5} \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) \sin \left (d x + c\right ) + 2 \, a^{5} - 2 \, a^{3} b^{2} - 4 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + {\left (3 \, a^{5} + 2 \, a^{4} b - 5 \, a^{3} b^{2} - 4 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - {\left (3 \, a^{5} - 2 \, a^{4} b - 5 \, a^{3} b^{2} + 4 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 2 \, {\left (3 \, a^{5} - 5 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{4} b - a^{2} b^{3}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )} \]
1/4*(4*b^5*cos(d*x + c)^2*log(b*sin(d*x + c) + a)*sin(d*x + c) + 2*a^5 - 2 *a^3*b^2 - 4*(a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c)^2*log(1/2*sin(d*x + c) )*sin(d*x + c) + (3*a^5 + 2*a^4*b - 5*a^3*b^2 - 4*a^2*b^3)*cos(d*x + c)^2* log(sin(d*x + c) + 1)*sin(d*x + c) - (3*a^5 - 2*a^4*b - 5*a^3*b^2 + 4*a^2* b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1)*sin(d*x + c) - 2*(3*a^5 - 5*a^3 *b^2 + 2*a*b^4)*cos(d*x + c)^2 - 2*(a^4*b - a^2*b^3)*sin(d*x + c))/((a^6 - 2*a^4*b^2 + a^2*b^4)*d*cos(d*x + c)^2*sin(d*x + c))
\[ \int \frac {\csc ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\csc ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.17 \[ \int \frac {\csc ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {4 \, b^{5} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}} + \frac {{\left (3 \, a - 4 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (3 \, a + 4 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, {\left (a b \sin \left (d x + c\right ) - {\left (3 \, a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{2} + 2 \, a^{2} - 2 \, b^{2}\right )}}{{\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )^{3} - {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )} - \frac {4 \, b \log \left (\sin \left (d x + c\right )\right )}{a^{2}}}{4 \, d} \]
1/4*(4*b^5*log(b*sin(d*x + c) + a)/(a^6 - 2*a^4*b^2 + a^2*b^4) + (3*a - 4* b)*log(sin(d*x + c) + 1)/(a^2 - 2*a*b + b^2) - (3*a + 4*b)*log(sin(d*x + c ) - 1)/(a^2 + 2*a*b + b^2) + 2*(a*b*sin(d*x + c) - (3*a^2 - 2*b^2)*sin(d*x + c)^2 + 2*a^2 - 2*b^2)/((a^3 - a*b^2)*sin(d*x + c)^3 - (a^3 - a*b^2)*sin (d*x + c)) - 4*b*log(sin(d*x + c))/a^2)/d
Time = 0.67 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.63 \[ \int \frac {\csc ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {12 \, b^{6} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}} + \frac {3 \, {\left (3 \, a - 4 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {3 \, {\left (3 \, a + 4 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {12 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2}} + \frac {2 \, {\left (2 \, b^{5} \sin \left (d x + c\right )^{3} - 9 \, a^{5} \sin \left (d x + c\right )^{2} + 15 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} - 6 \, a b^{4} \sin \left (d x + c\right )^{2} + 3 \, a^{4} b \sin \left (d x + c\right ) - 3 \, a^{2} b^{3} \sin \left (d x + c\right ) - 2 \, b^{5} \sin \left (d x + c\right ) + 6 \, a^{5} - 12 \, a^{3} b^{2} + 6 \, a b^{4}\right )}}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} {\left (\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )\right )}}}{12 \, d} \]
1/12*(12*b^6*log(abs(b*sin(d*x + c) + a))/(a^6*b - 2*a^4*b^3 + a^2*b^5) + 3*(3*a - 4*b)*log(abs(sin(d*x + c) + 1))/(a^2 - 2*a*b + b^2) - 3*(3*a + 4* b)*log(abs(sin(d*x + c) - 1))/(a^2 + 2*a*b + b^2) - 12*b*log(abs(sin(d*x + c)))/a^2 + 2*(2*b^5*sin(d*x + c)^3 - 9*a^5*sin(d*x + c)^2 + 15*a^3*b^2*si n(d*x + c)^2 - 6*a*b^4*sin(d*x + c)^2 + 3*a^4*b*sin(d*x + c) - 3*a^2*b^3*s in(d*x + c) - 2*b^5*sin(d*x + c) + 6*a^5 - 12*a^3*b^2 + 6*a*b^4)/((a^6 - 2 *a^4*b^2 + a^2*b^4)*(sin(d*x + c)^3 - sin(d*x + c))))/d
Time = 12.41 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.14 \[ \int \frac {\csc ^2(c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (3\,a-4\,b\right )}{4\,d\,{\left (a-b\right )}^2}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {b}{4\,{\left (a+b\right )}^2}+\frac {3}{4\,\left (a+b\right )}\right )}{d}-\frac {\frac {1}{a}+\frac {b\,\sin \left (c+d\,x\right )}{2\,\left (a^2-b^2\right )}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (3\,a^2-2\,b^2\right )}{2\,a\,\left (a^2-b^2\right )}}{d\,\left (\sin \left (c+d\,x\right )-{\sin \left (c+d\,x\right )}^3\right )}-\frac {b\,\ln \left (\sin \left (c+d\,x\right )\right )}{a^2\,d}+\frac {b^5\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{a^2\,d\,{\left (a^2-b^2\right )}^2} \]
(log(sin(c + d*x) + 1)*(3*a - 4*b))/(4*d*(a - b)^2) - (log(sin(c + d*x) - 1)*(b/(4*(a + b)^2) + 3/(4*(a + b))))/d - (1/a + (b*sin(c + d*x))/(2*(a^2 - b^2)) - (sin(c + d*x)^2*(3*a^2 - 2*b^2))/(2*a*(a^2 - b^2)))/(d*(sin(c + d*x) - sin(c + d*x)^3)) - (b*log(sin(c + d*x)))/(a^2*d) + (b^5*log(a + b*s in(c + d*x)))/(a^2*d*(a^2 - b^2)^2)