Integrand size = 25, antiderivative size = 103 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}-\frac {b \log (\cos (c+d x))}{d}-\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}-\frac {b \tan ^2(c+d x)}{2 d}+\frac {a \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac {b \tan ^4(c+d x)}{4 d} \] Output:
3/8*a*arctanh(sin(d*x+c))/d-b*ln(cos(d*x+c))/d-3/8*a*sec(d*x+c)*tan(d*x+c) /d-1/2*b*tan(d*x+c)^2/d+1/4*a*sec(d*x+c)*tan(d*x+c)^3/d+1/4*b*tan(d*x+c)^4 /d
Time = 0.03 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.17 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}-\frac {b \log (\cos (c+d x))}{d}-\frac {b \sec ^2(c+d x)}{d}+\frac {b \sec ^4(c+d x)}{4 d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}-\frac {3 a \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a \sec (c+d x) \tan ^3(c+d x)}{d} \] Input:
Integrate[Sec[c + d*x]*(a + b*Sin[c + d*x])*Tan[c + d*x]^4,x]
Output:
(3*a*ArcTanh[Sin[c + d*x]])/(8*d) - (b*Log[Cos[c + d*x]])/d - (b*Sec[c + d *x]^2)/d + (b*Sec[c + d*x]^4)/(4*d) + (3*a*Sec[c + d*x]*Tan[c + d*x])/(8*d ) - (3*a*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a*Sec[c + d*x]*Tan[c + d*x] ^3)/d
Time = 0.71 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.05, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 3313, 3042, 3091, 3042, 3091, 3042, 3954, 3042, 3954, 3042, 3956, 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.
| \(\displaystyle \int \tan ^4(c+d x) \sec (c+d x) (a+b \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^4 (a+b \sin (c+d x))}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3313 |
| \(\displaystyle a \int \sec (c+d x) \tan ^4(c+d x)dx+b \int \tan ^5(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \sec (c+d x) \tan (c+d x)^4dx+b \int \tan (c+d x)^5dx\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle a \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \int \sec (c+d x) \tan ^2(c+d x)dx\right )+b \int \tan (c+d x)^5dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \int \sec (c+d x) \tan (c+d x)^2dx\right )+b \int \tan (c+d x)^5dx\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle a \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \sec (c+d x)dx\right )\right )+b \int \tan (c+d x)^5dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )\right )+b \int \tan (c+d x)^5dx\) |
| \(\Big \downarrow \) 3954 |
| \(\displaystyle a \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )\right )+b \left (\frac {\tan ^4(c+d x)}{4 d}-\int \tan ^3(c+d x)dx\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )\right )+b \left (\frac {\tan ^4(c+d x)}{4 d}-\int \tan (c+d x)^3dx\right )\) |
| \(\Big \downarrow \) 3954 |
| \(\displaystyle a \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )\right )+b \left (\int \tan (c+d x)dx+\frac {\tan ^4(c+d x)}{4 d}-\frac {\tan ^2(c+d x)}{2 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )\right )+b \left (\int \tan (c+d x)dx+\frac {\tan ^4(c+d x)}{4 d}-\frac {\tan ^2(c+d x)}{2 d}\right )\) |
| \(\Big \downarrow \) 3956 |
| \(\displaystyle a \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )\right )+b \left (\frac {\tan ^4(c+d x)}{4 d}-\frac {\tan ^2(c+d x)}{2 d}-\frac {\log (\cos (c+d x))}{d}\right )\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle a \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {\text {arctanh}(\sin (c+d x))}{2 d}\right )\right )+b \left (\frac {\tan ^4(c+d x)}{4 d}-\frac {\tan ^2(c+d x)}{2 d}-\frac {\log (\cos (c+d x))}{d}\right )\) |
Input:
Int[Sec[c + d*x]*(a + b*Sin[c + d*x])*Tan[c + d*x]^4,x]
Output:
b*(-(Log[Cos[c + d*x]]/d) - Tan[c + d*x]^2/(2*d) + Tan[c + d*x]^4/(4*d)) + a*((Sec[c + d*x]*Tan[c + d*x]^3)/(4*d) - (3*(-1/2*ArcTanh[Sin[c + d*x]]/d + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4)
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[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_ ) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[Cos[e + f*x]^ p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[Cos[e + f*x]^p*(d*Sin[e + f*x ])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2 ] && IntegerQ[n] && ((LtQ[p, 0] && NeQ[a^2 - b^2, 0]) || LtQ[0, n, p - 1] | | LtQ[p + 1, -n, 2*p + 1])
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d *x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2 Int[(b*Tan[c + d*x])^(n - 2), x] , x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.06 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.08
| method | result | size |
| derivativedivides | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(111\) |
| default | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(111\) |
| risch | \(i x b +\frac {2 i b c}{d}+\frac {i \left (5 a \,{\mathrm e}^{7 i \left (d x +c \right )}-3 a \,{\mathrm e}^{5 i \left (d x +c \right )}+16 i b \,{\mathrm e}^{6 i \left (d x +c \right )}+3 a \,{\mathrm e}^{3 i \left (d x +c \right )}+16 i b \,{\mathrm e}^{4 i \left (d x +c \right )}-5 a \,{\mathrm e}^{i \left (d x +c \right )}+16 i b \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{d}\) | \(198\) |
Input:
int(sec(d*x+c)*(a+b*sin(d*x+c))*tan(d*x+c)^4,x,method=_RETURNVERBOSE)
Output:
1/d*(a*(1/4*sin(d*x+c)^5/cos(d*x+c)^4-1/8*sin(d*x+c)^5/cos(d*x+c)^2-1/8*si n(d*x+c)^3-3/8*sin(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+b*(1/4*tan(d*x+c) ^4-1/2*tan(d*x+c)^2-ln(cos(d*x+c))))
Time = 0.09 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {{\left (3 \, a - 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, a + 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 16 \, b \cos \left (d x + c\right )^{2} - 2 \, {\left (5 \, a \cos \left (d x + c\right )^{2} - 2 \, a\right )} \sin \left (d x + c\right ) + 4 \, b}{16 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate(sec(d*x+c)*(a+b*sin(d*x+c))*tan(d*x+c)^4,x, algorithm="fricas")
Output:
1/16*((3*a - 8*b)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (3*a + 8*b)*cos(d *x + c)^4*log(-sin(d*x + c) + 1) - 16*b*cos(d*x + c)^2 - 2*(5*a*cos(d*x + c)^2 - 2*a)*sin(d*x + c) + 4*b)/(d*cos(d*x + c)^4)
\[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right ) \tan ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)*(a+b*sin(d*x+c))*tan(d*x+c)**4,x)
Output:
Integral((a + b*sin(c + d*x))*tan(c + d*x)**4*sec(c + d*x), x)
Time = 0.04 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.97 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {{\left (3 \, a - 8 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, a + 8 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + \frac {2 \, {\left (5 \, a \sin \left (d x + c\right )^{3} + 8 \, b \sin \left (d x + c\right )^{2} - 3 \, a \sin \left (d x + c\right ) - 6 \, b\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \] Input:
integrate(sec(d*x+c)*(a+b*sin(d*x+c))*tan(d*x+c)^4,x, algorithm="maxima")
Output:
1/16*((3*a - 8*b)*log(sin(d*x + c) + 1) - (3*a + 8*b)*log(sin(d*x + c) - 1 ) + 2*(5*a*sin(d*x + c)^3 + 8*b*sin(d*x + c)^2 - 3*a*sin(d*x + c) - 6*b)/( sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d
Time = 0.39 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.02 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {{\left (3 \, a - 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, d} - \frac {{\left (3 \, a + 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, d} + \frac {5 \, a \sin \left (d x + c\right )^{3} + 8 \, b \sin \left (d x + c\right )^{2} - 3 \, a \sin \left (d x + c\right ) - 6 \, b}{8 \, d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(sec(d*x+c)*(a+b*sin(d*x+c))*tan(d*x+c)^4,x, algorithm="giac")
Output:
1/16*(3*a - 8*b)*log(abs(sin(d*x + c) + 1))/d - 1/16*(3*a + 8*b)*log(abs(s in(d*x + c) - 1))/d + 1/8*(5*a*sin(d*x + c)^3 + 8*b*sin(d*x + c)^2 - 3*a*s in(d*x + c) - 6*b)/(d*(sin(d*x + c) + 1)^2*(sin(d*x + c) - 1)^2)
Time = 21.44 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.15 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {b\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-8\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {3\,a}{8}+b\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {3\,a}{8}-b\right )}{d} \] Input:
int((tan(c + d*x)^4*(a + b*sin(c + d*x)))/cos(c + d*x),x)
Output:
(b*log(tan(c/2 + (d*x)/2)^2 + 1))/d - ((3*a*tan(c/2 + (d*x)/2))/4 - (11*a* tan(c/2 + (d*x)/2)^3)/4 - (11*a*tan(c/2 + (d*x)/2)^5)/4 + (3*a*tan(c/2 + ( d*x)/2)^7)/4 + 2*b*tan(c/2 + (d*x)/2)^2 - 8*b*tan(c/2 + (d*x)/2)^4 + 2*b*t an(c/2 + (d*x)/2)^6)/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) - (log(tan(c/2 + (d*x )/2) - 1)*((3*a)/8 + b))/d + (log(tan(c/2 + (d*x)/2) + 1)*((3*a)/8 - b))/d
Time = 0.18 (sec) , antiderivative size = 381, normalized size of antiderivative = 3.70 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{4} b -16 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} b +8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a -8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} b +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a +16 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a -8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a -8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} b -6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +16 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +6 \sin \left (d x +c \right )^{4} b +5 \sin \left (d x +c \right )^{3} a -4 \sin \left (d x +c \right )^{2} b -3 \sin \left (d x +c \right ) a}{8 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)*(a+b*sin(d*x+c))*tan(d*x+c)^4,x)
Output:
(8*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**4*b - 16*log(tan((c + d*x)/2 )**2 + 1)*sin(c + d*x)**2*b + 8*log(tan((c + d*x)/2)**2 + 1)*b - 3*log(tan ((c + d*x)/2) - 1)*sin(c + d*x)**4*a - 8*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b + 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a + 16*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b - 3*log(tan((c + d*x)/2) - 1)*a - 8*log( tan((c + d*x)/2) - 1)*b + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a - 8*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b - 6*log(tan((c + d*x)/2) + 1 )*sin(c + d*x)**2*a + 16*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b + 3*l og(tan((c + d*x)/2) + 1)*a - 8*log(tan((c + d*x)/2) + 1)*b + 6*sin(c + d*x )**4*b + 5*sin(c + d*x)**3*a - 4*sin(c + d*x)**2*b - 3*sin(c + d*x)*a)/(8* d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))