Integrand size = 27, antiderivative size = 156 \[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {(2 a+3 b) \log (1-\sin (c+d x))}{4 (a+b)^2 d}+\frac {\log (\sin (c+d x))}{a d}-\frac {(2 a-3 b) \log (1+\sin (c+d x))}{4 (a-b)^2 d}-\frac {b^4 \log (a+b \sin (c+d x))}{a \left (a^2-b^2\right )^2 d}+\frac {1}{4 (a+b) d (1-\sin (c+d x))}+\frac {1}{4 (a-b) d (1+\sin (c+d x))} \] Output:
-1/4*(2*a+3*b)*ln(1-sin(d*x+c))/(a+b)^2/d+ln(sin(d*x+c))/a/d-1/4*(2*a-3*b) *ln(1+sin(d*x+c))/(a-b)^2/d-b^4*ln(a+b*sin(d*x+c))/a/(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 = 0.76 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.97 \[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b^4 \left (-\frac {(2 a+3 b) \log (1-\sin (c+d x))}{b^4 (a+b)^2}+\frac {4 \log (\sin (c+d x))}{a b^4}-\frac {(2 a-3 b) \log (1+\sin (c+d x))}{(a-b)^2 b^4}-\frac {4 \log (a+b \sin (c+d x))}{a (a-b)^2 (a+b)^2}-\frac {1}{b^4 (a+b) (-1+\sin (c+d x))}+\frac {1}{(a-b) b^4 (1+\sin (c+d x))}\right )}{4 d} \] Input:
Integrate[(Csc[c + d*x]*Sec[c + d*x]^3)/(a + b*Sin[c + d*x]),x]
Output:
(b^4*(-(((2*a + 3*b)*Log[1 - Sin[c + d*x]])/(b^4*(a + b)^2)) + (4*Log[Sin[ c + d*x]])/(a*b^4) - ((2*a - 3*b)*Log[1 + Sin[c + d*x]])/((a - b)^2*b^4) - (4*Log[a + b*Sin[c + d*x]])/(a*(a - b)^2*(a + b)^2) - 1/(b^4*(a + b)*(-1 + Sin[c + d*x])) + 1/((a - b)*b^4*(1 + Sin[c + d*x]))))/(4*d)
Time = 0.46 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, 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.
| \(\displaystyle \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x) \cos (c+d x)^3 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^3 \int \frac {\csc (c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^4 \int \frac {\csc (c+d x)}{b (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 615 |
| \(\displaystyle \frac {b^4 \int \left (\frac {3 b-2 a}{4 (a-b)^2 b^4 (\sin (c+d x) b+b)}+\frac {\csc (c+d x)}{a b^5}+\frac {2 a+3 b}{4 b^4 (a+b)^2 (b-b \sin (c+d x))}-\frac {1}{a (a-b)^2 (a+b)^2 (a+b \sin (c+d x))}+\frac {1}{4 b^3 (a+b) (b-b \sin (c+d x))^2}-\frac {1}{4 (a-b) b^3 (\sin (c+d x) b+b)^2}\right )d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^4 \left (-\frac {\log (a+b \sin (c+d x))}{a \left (a^2-b^2\right )^2}+\frac {\log (b \sin (c+d x))}{a b^4}-\frac {(2 a+3 b) \log (b-b \sin (c+d x))}{4 b^4 (a+b)^2}-\frac {(2 a-3 b) \log (b \sin (c+d x)+b)}{4 b^4 (a-b)^2}+\frac {1}{4 b^3 (a+b) (b-b \sin (c+d x))}+\frac {1}{4 b^3 (a-b) (b \sin (c+d x)+b)}\right )}{d}\) |
Input:
Int[(Csc[c + d*x]*Sec[c + d*x]^3)/(a + b*Sin[c + d*x]),x]
Output:
(b^4*(Log[b*Sin[c + d*x]]/(a*b^4) - ((2*a + 3*b)*Log[b - b*Sin[c + d*x]])/ (4*b^4*(a + b)^2) - Log[a + b*Sin[c + d*x]]/(a*(a^2 - b^2)^2) - ((2*a - 3* b)*Log[b + b*Sin[c + d*x]])/(4*(a - b)^2*b^4) + 1/(4*b^3*(a + b)*(b - b*Si n[c + d*x])) + 1/(4*(a - b)*b^3*(b + b*Sin[c + d*x]))))/d
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 = 1.00 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.88
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-2 a -3 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}-\frac {b^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} a}+\frac {\ln \left (\sin \left (d x +c \right )\right )}{a}+\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (-2 a +3 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}}{d}\) | \(137\) |
| default | \(\frac {-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-2 a -3 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}-\frac {b^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} a}+\frac {\ln \left (\sin \left (d x +c \right )\right )}{a}+\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (-2 a +3 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}}{d}\) | \(137\) |
| parallelrisch | \(\frac {-b^{4} \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (a +\frac {3 b}{2}\right ) \left (a -b \right )^{2} a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right ) \left (a -\frac {3 b}{2}\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (\left (a -b \right ) \left (a +b \right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {a \left (a \cos \left (2 d x +2 c \right )+2 b \sin \left (d x +c \right )-a \right )}{2}\right ) \left (a -b \right )\right ) \left (a +b \right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} a d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(218\) |
| norman | \(\frac {-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (a^{2}-b^{2}\right )}-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d \left (a^{2}-b^{2}\right )}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d \left (a^{2}-b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (2 a -3 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (2 a +3 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \left (a^{2}+2 a b +b^{2}\right )}-\frac {b^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{a d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}\) | \(248\) |
| risch | \(\frac {2 i b^{4} c}{a d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {3 i b x}{2 \left (a^{2}-2 a b +b^{2}\right )}+\frac {i a c}{d \left (a^{2}-2 a b +b^{2}\right )}+\frac {3 i b x}{2 \left (a^{2}+2 a b +b^{2}\right )}+\frac {i \left (-2 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{3 i \left (d x +c \right )}-b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left (a^{2}-b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {2 i x}{a}+\frac {i a x}{a^{2}+2 a b +b^{2}}-\frac {3 i b c}{2 d \left (a^{2}-2 a b +b^{2}\right )}+\frac {i a x}{a^{2}-2 a b +b^{2}}+\frac {i a c}{d \left (a^{2}+2 a b +b^{2}\right )}+\frac {3 i b c}{2 d \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 i b^{4} x}{a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 i c}{a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a}{d \left (a^{2}-2 a b +b^{2}\right )}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a}{d \left (a^{2}+2 a b +b^{2}\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{2 d \left (a^{2}+2 a b +b^{2}\right )}-\frac {b^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}\) | \(502\) |
Input:
int(csc(d*x+c)*sec(d*x+c)^3/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(-1/(4*a+4*b)/(sin(d*x+c)-1)+1/4/(a+b)^2*(-2*a-3*b)*ln(sin(d*x+c)-1)-b ^4/(a+b)^2/(a-b)^2/a*ln(a+b*sin(d*x+c))+1/a*ln(sin(d*x+c))+1/(4*a-4*b)/(1+ sin(d*x+c))+1/4/(a-b)^2*(-2*a+3*b)*ln(1+sin(d*x+c)))
Time = 0.38 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.37 \[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {4 \, b^{4} \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \, a^{4} + 2 \, a^{2} b^{2} - 4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) + {\left (2 \, a^{4} + a^{3} b - 4 \, a^{2} b^{2} - 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, a^{4} - a^{3} b - 4 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \cos \left (d x + c\right )^{2}} \] Input:
integrate(csc(d*x+c)*sec(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/4*(4*b^4*cos(d*x + c)^2*log(b*sin(d*x + c) + a) - 2*a^4 + 2*a^2*b^2 - 4 *(a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^2*log(-1/2*sin(d*x + c)) + (2*a^4 + a^3*b - 4*a^2*b^2 - 3*a*b^3)*cos(d*x + c)^2*log(sin(d*x + c) + 1) + (2*a^4 - a^3*b - 4*a^2*b^2 + 3*a*b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2* (a^3*b - a*b^3)*sin(d*x + c))/((a^5 - 2*a^3*b^2 + a*b^4)*d*cos(d*x + c)^2)
\[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\csc {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:
integrate(csc(d*x+c)*sec(d*x+c)**3/(a+b*sin(d*x+c)),x)
Output:
Integral(csc(c + d*x)*sec(c + d*x)**3/(a + b*sin(c + d*x)), x)
Time = 0.04 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00 \[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {4 \, b^{4} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4}} + \frac {{\left (2 \, a - 3 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {{\left (2 \, a + 3 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {2 \, {\left (b \sin \left (d x + c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} + b^{2}} - \frac {4 \, \log \left (\sin \left (d x + c\right )\right )}{a}}{4 \, d} \] Input:
integrate(csc(d*x+c)*sec(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/4*(4*b^4*log(b*sin(d*x + c) + a)/(a^5 - 2*a^3*b^2 + a*b^4) + (2*a - 3*b )*log(sin(d*x + c) + 1)/(a^2 - 2*a*b + b^2) + (2*a + 3*b)*log(sin(d*x + c) - 1)/(a^2 + 2*a*b + b^2) - 2*(b*sin(d*x + c) - a)/((a^2 - b^2)*sin(d*x + c)^2 - a^2 + b^2) - 4*log(sin(d*x + c))/a)/d
Time = 0.18 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.29 \[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {b^{5} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{5} b d - 2 \, a^{3} b^{3} d + a b^{5} d} - \frac {{\left (2 \, a + 3 \, b\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{4 \, {\left (a^{2} d + 2 \, a b d + b^{2} d\right )}} - \frac {{\left (2 \, a - 3 \, b\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{4 \, {\left (a^{2} d - 2 \, a b d + b^{2} d\right )}} + \frac {\log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a d} - \frac {a^{3} - a b^{2} - {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a + b\right )}^{2} {\left (a - b\right )}^{2} d {\left (\sin \left (d x + c\right ) + 1\right )} {\left (\sin \left (d x + c\right ) - 1\right )}} \] Input:
integrate(csc(d*x+c)*sec(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
-b^5*log(abs(b*sin(d*x + c) + a))/(a^5*b*d - 2*a^3*b^3*d + a*b^5*d) - 1/4* (2*a + 3*b)*log(abs(-sin(d*x + c) + 1))/(a^2*d + 2*a*b*d + b^2*d) - 1/4*(2 *a - 3*b)*log(abs(-sin(d*x + c) - 1))/(a^2*d - 2*a*b*d + b^2*d) + log(abs( sin(d*x + c)))/(a*d) - 1/2*(a^3 - a*b^2 - (a^2*b - b^3)*sin(d*x + c))/((a + b)^2*(a - b)^2*d*(sin(d*x + c) + 1)*(sin(d*x + c) - 1))
Time = 19.96 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.09 \[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {b}{4\,{\left (a-b\right )}^2}-\frac {1}{2\,\left (a-b\right )}\right )}{d}-\frac {\frac {a}{2\,\left (a^2-b^2\right )}-\frac {b\,\sin \left (c+d\,x\right )}{2\,\left (a^2-b^2\right )}}{d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )}+\frac {\ln \left (\sin \left (c+d\,x\right )\right )}{a\,d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {b}{4\,{\left (a+b\right )}^2}+\frac {1}{2\,\left (a+b\right )}\right )}{d}-\frac {b^4\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{a\,d\,{\left (a^2-b^2\right )}^2} \] Input:
int(1/(cos(c + d*x)^3*sin(c + d*x)*(a + b*sin(c + d*x))),x)
Output:
(log(sin(c + d*x) + 1)*(b/(4*(a - b)^2) - 1/(2*(a - b))))/d - (a/(2*(a^2 - b^2)) - (b*sin(c + d*x))/(2*(a^2 - b^2)))/(d*(sin(c + d*x)^2 - 1)) + log( sin(c + d*x))/(a*d) - (log(sin(c + d*x) - 1)*(b/(4*(a + b)^2) + 1/(2*(a + b))))/d - (b^4*log(a + b*sin(c + d*x)))/(a*d*(a^2 - b^2)^2)
Time = 0.18 (sec) , antiderivative size = 682, normalized size of antiderivative = 4.37 \[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(csc(d*x+c)*sec(d*x+c)^3/(a+b*sin(d*x+c)),x)
Output:
( - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**4 + log(tan((c + d*x)/2 ) - 1)*sin(c + d*x)**2*a**3*b + 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)** 2*a**2*b**2 - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**3 + 2*log(t an((c + d*x)/2) - 1)*a**4 - log(tan((c + d*x)/2) - 1)*a**3*b - 4*log(tan(( c + d*x)/2) - 1)*a**2*b**2 + 3*log(tan((c + d*x)/2) - 1)*a*b**3 - 2*log(ta n((c + d*x)/2) + 1)*sin(c + d*x)**2*a**4 - log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3*b + 4*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b**2 + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b**3 + 2*log(tan((c + d*x) /2) + 1)*a**4 + log(tan((c + d*x)/2) + 1)*a**3*b - 4*log(tan((c + d*x)/2) + 1)*a**2*b**2 - 3*log(tan((c + d*x)/2) + 1)*a*b**3 - 2*log(tan((c + d*x)/ 2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**2*b**4 + 2*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*b**4 + 2*log(tan((c + d*x)/2))*si n(c + d*x)**2*a**4 - 4*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**2*b**2 + 2 *log(tan((c + d*x)/2))*sin(c + d*x)**2*b**4 - 2*log(tan((c + d*x)/2))*a**4 + 4*log(tan((c + d*x)/2))*a**2*b**2 - 2*log(tan((c + d*x)/2))*b**4 + sin( c + d*x)**2*a**4 - sin(c + d*x)**2*a**2*b**2 + sin(c + d*x)*a**3*b - sin(c + d*x)*a*b**3 - 2*a**4 + 2*a**2*b**2)/(2*a*d*(sin(c + d*x)**2*a**4 - 2*si n(c + d*x)**2*a**2*b**2 + sin(c + d*x)**2*b**4 - a**4 + 2*a**2*b**2 - b**4 ))