Integrand size = 21, antiderivative size = 98 \[ \int \sec ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {(6 A+5 C) \text {arctanh}(\sin (c+d x))}{16 d}+\frac {(6 A+5 C) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {(6 A+5 C) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {C \sec ^5(c+d x) \tan (c+d x)}{6 d} \] Output:
1/16*(6*A+5*C)*arctanh(sin(d*x+c))/d+1/16*(6*A+5*C)*sec(d*x+c)*tan(d*x+c)/ d+1/24*(6*A+5*C)*sec(d*x+c)^3*tan(d*x+c)/d+1/6*C*sec(d*x+c)^5*tan(d*x+c)/d
Time = 0.02 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.40 \[ \int \sec ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 A \text {arctanh}(\sin (c+d x))}{8 d}+\frac {5 C \text {arctanh}(\sin (c+d x))}{16 d}+\frac {3 A \sec (c+d x) \tan (c+d x)}{8 d}+\frac {5 C \sec (c+d x) \tan (c+d x)}{16 d}+\frac {A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {5 C \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {C \sec ^5(c+d x) \tan (c+d x)}{6 d} \] Input:
Integrate[Sec[c + d*x]^5*(A + C*Sec[c + d*x]^2),x]
Output:
(3*A*ArcTanh[Sin[c + d*x]])/(8*d) + (5*C*ArcTanh[Sin[c + d*x]])/(16*d) + ( 3*A*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (5*C*Sec[c + d*x]*Tan[c + d*x])/(16 *d) + (A*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (5*C*Sec[c + d*x]^3*Tan[c + d*x])/(24*d) + (C*Sec[c + d*x]^5*Tan[c + d*x])/(6*d)
Time = 0.45 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4534, 3042, 4255, 3042, 4255, 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.
\(\displaystyle \int \sec ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^5 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 4534 |
\(\displaystyle \frac {1}{6} (6 A+5 C) \int \sec ^5(c+d x)dx+\frac {C \tan (c+d x) \sec ^5(c+d x)}{6 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{6} (6 A+5 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx+\frac {C \tan (c+d x) \sec ^5(c+d x)}{6 d}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {1}{6} (6 A+5 C) \left (\frac {3}{4} \int \sec ^3(c+d x)dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {C \tan (c+d x) \sec ^5(c+d x)}{6 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{6} (6 A+5 C) \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {C \tan (c+d x) \sec ^5(c+d x)}{6 d}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {1}{6} (6 A+5 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {C \tan (c+d x) \sec ^5(c+d x)}{6 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{6} (6 A+5 C) \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {C \tan (c+d x) \sec ^5(c+d x)}{6 d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {1}{6} (6 A+5 C) \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {C \tan (c+d x) \sec ^5(c+d x)}{6 d}\) |
Input:
Int[Sec[c + d*x]^5*(A + C*Sec[c + d*x]^2),x]
Output:
(C*Sec[c + d*x]^5*Tan[c + d*x])/(6*d) + ((6*A + 5*C)*((Sec[c + d*x]^3*Tan[ c + d*x])/(4*d) + (3*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4))/6
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
Time = 0.80 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.10
method | result | size |
derivativedivides | \(\frac {A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+C \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\) | \(108\) |
default | \(\frac {A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+C \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\) | \(108\) |
parts | \(\frac {A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {C \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\) | \(110\) |
parallelrisch | \(\frac {-270 \left (\frac {\cos \left (6 d x +6 c \right )}{15}+\frac {2 \cos \left (4 d x +4 c \right )}{5}+\cos \left (2 d x +2 c \right )+\frac {2}{3}\right ) \left (A +\frac {5 C}{6}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+270 \left (\frac {\cos \left (6 d x +6 c \right )}{15}+\frac {2 \cos \left (4 d x +4 c \right )}{5}+\cos \left (2 d x +2 c \right )+\frac {2}{3}\right ) \left (A +\frac {5 C}{6}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (204 A +170 C \right ) \sin \left (3 d x +3 c \right )+\left (36 A +30 C \right ) \sin \left (5 d x +5 c \right )+168 \left (A +\frac {33 C}{14}\right ) \sin \left (d x +c \right )}{48 d \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right )}\) | \(193\) |
norman | \(\frac {\frac {\left (2 A +15 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}+\frac {\left (2 A +15 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {\left (10 A +11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {\left (10 A +11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{8 d}-\frac {\left (42 A -5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 d}-\frac {\left (42 A -5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{24 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{6}}-\frac {\left (6 A +5 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16 d}+\frac {\left (6 A +5 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 d}\) | \(203\) |
risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (18 A \,{\mathrm e}^{10 i \left (d x +c \right )}+15 C \,{\mathrm e}^{10 i \left (d x +c \right )}+102 A \,{\mathrm e}^{8 i \left (d x +c \right )}+85 C \,{\mathrm e}^{8 i \left (d x +c \right )}+84 A \,{\mathrm e}^{6 i \left (d x +c \right )}+198 C \,{\mathrm e}^{6 i \left (d x +c \right )}-84 A \,{\mathrm e}^{4 i \left (d x +c \right )}-198 C \,{\mathrm e}^{4 i \left (d x +c \right )}-102 A \,{\mathrm e}^{2 i \left (d x +c \right )}-85 C \,{\mathrm e}^{2 i \left (d x +c \right )}-18 A -15 C \right )}{24 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{8 d}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{16 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{8 d}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{16 d}\) | \(255\) |
Input:
int(sec(d*x+c)^5*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)
Output:
1/d*(A*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+t an(d*x+c)))+C*(-(-1/6*sec(d*x+c)^5-5/24*sec(d*x+c)^3-5/16*sec(d*x+c))*tan( d*x+c)+5/16*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.09 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.16 \[ \int \sec ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (6 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (6 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, {\left (6 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (6 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, C\right )} \sin \left (d x + c\right )}{96 \, d \cos \left (d x + c\right )^{6}} \] Input:
integrate(sec(d*x+c)^5*(A+C*sec(d*x+c)^2),x, algorithm="fricas")
Output:
1/96*(3*(6*A + 5*C)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 3*(6*A + 5*C)*c os(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*(3*(6*A + 5*C)*cos(d*x + c)^4 + 2 *(6*A + 5*C)*cos(d*x + c)^2 + 8*C)*sin(d*x + c))/(d*cos(d*x + c)^6)
\[ \int \sec ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{5}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**5*(A+C*sec(d*x+c)**2),x)
Output:
Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**5, x)
Time = 0.03 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.29 \[ \int \sec ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (6 \, A + 5 \, C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (6 \, A + 5 \, C\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (6 \, A + 5 \, C\right )} \sin \left (d x + c\right )^{5} - 8 \, {\left (6 \, A + 5 \, C\right )} \sin \left (d x + c\right )^{3} + 3 \, {\left (10 \, A + 11 \, C\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1}}{96 \, d} \] Input:
integrate(sec(d*x+c)^5*(A+C*sec(d*x+c)^2),x, algorithm="maxima")
Output:
1/96*(3*(6*A + 5*C)*log(sin(d*x + c) + 1) - 3*(6*A + 5*C)*log(sin(d*x + c) - 1) - 2*(3*(6*A + 5*C)*sin(d*x + c)^5 - 8*(6*A + 5*C)*sin(d*x + c)^3 + 3 *(10*A + 11*C)*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(d* x + c)^2 - 1))/d
Time = 0.32 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.23 \[ \int \sec ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (6 \, A + 5 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \, {\left (6 \, A + 5 \, C\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (18 \, A \sin \left (d x + c\right )^{5} + 15 \, C \sin \left (d x + c\right )^{5} - 48 \, A \sin \left (d x + c\right )^{3} - 40 \, C \sin \left (d x + c\right )^{3} + 30 \, A \sin \left (d x + c\right ) + 33 \, C \sin \left (d x + c\right )\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{3}}}{96 \, d} \] Input:
integrate(sec(d*x+c)^5*(A+C*sec(d*x+c)^2),x, algorithm="giac")
Output:
1/96*(3*(6*A + 5*C)*log(abs(sin(d*x + c) + 1)) - 3*(6*A + 5*C)*log(abs(sin (d*x + c) - 1)) - 2*(18*A*sin(d*x + c)^5 + 15*C*sin(d*x + c)^5 - 48*A*sin( d*x + c)^3 - 40*C*sin(d*x + c)^3 + 30*A*sin(d*x + c) + 33*C*sin(d*x + c))/ (sin(d*x + c)^2 - 1)^3)/d
Time = 0.22 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.04 \[ \int \sec ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (\frac {3\,A}{8}+\frac {5\,C}{16}\right )}{d}-\frac {\left (\frac {3\,A}{8}+\frac {5\,C}{16}\right )\,{\sin \left (c+d\,x\right )}^5+\left (-A-\frac {5\,C}{6}\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {5\,A}{8}+\frac {11\,C}{16}\right )\,\sin \left (c+d\,x\right )}{d\,\left ({\sin \left (c+d\,x\right )}^6-3\,{\sin \left (c+d\,x\right )}^4+3\,{\sin \left (c+d\,x\right )}^2-1\right )} \] Input:
int((A + C/cos(c + d*x)^2)/cos(c + d*x)^5,x)
Output:
(atanh(sin(c + d*x))*((3*A)/8 + (5*C)/16))/d - (sin(c + d*x)*((5*A)/8 + (1 1*C)/16) - sin(c + d*x)^3*(A + (5*C)/6) + sin(c + d*x)^5*((3*A)/8 + (5*C)/ 16))/(d*(3*sin(c + d*x)^2 - 3*sin(c + d*x)^4 + sin(c + d*x)^6 - 1))
Time = 0.16 (sec) , antiderivative size = 436, normalized size of antiderivative = 4.45 \[ \int \sec ^5(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {-18 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{6} a -15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{6} c +54 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a +45 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} c -54 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -45 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} c +18 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +18 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{6} a +15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{6} c -54 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a -45 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} c +54 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +45 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} c -18 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c -18 \sin \left (d x +c \right )^{5} a -15 \sin \left (d x +c \right )^{5} c +48 \sin \left (d x +c \right )^{3} a +40 \sin \left (d x +c \right )^{3} c -30 \sin \left (d x +c \right ) a -33 \sin \left (d x +c \right ) c}{48 d \left (\sin \left (d x +c \right )^{6}-3 \sin \left (d x +c \right )^{4}+3 \sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^5*(A+C*sec(d*x+c)^2),x)
Output:
( - 18*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6*a - 15*log(tan((c + d*x)/ 2) - 1)*sin(c + d*x)**6*c + 54*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a + 45*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*c - 54*log(tan((c + d*x)/2 ) - 1)*sin(c + d*x)**2*a - 45*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*c + 18*log(tan((c + d*x)/2) - 1)*a + 15*log(tan((c + d*x)/2) - 1)*c + 18*log (tan((c + d*x)/2) + 1)*sin(c + d*x)**6*a + 15*log(tan((c + d*x)/2) + 1)*si n(c + d*x)**6*c - 54*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a - 45*log( tan((c + d*x)/2) + 1)*sin(c + d*x)**4*c + 54*log(tan((c + d*x)/2) + 1)*sin (c + d*x)**2*a + 45*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*c - 18*log(t an((c + d*x)/2) + 1)*a - 15*log(tan((c + d*x)/2) + 1)*c - 18*sin(c + d*x)* *5*a - 15*sin(c + d*x)**5*c + 48*sin(c + d*x)**3*a + 40*sin(c + d*x)**3*c - 30*sin(c + d*x)*a - 33*sin(c + d*x)*c)/(48*d*(sin(c + d*x)**6 - 3*sin(c + d*x)**4 + 3*sin(c + d*x)**2 - 1))