Integrand size = 21, antiderivative size = 98 \[ \int \sec ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {(6 a+5 b) \text {arctanh}(\sin (e+f x))}{16 f}+\frac {(6 a+5 b) \sec (e+f x) \tan (e+f x)}{16 f}+\frac {(6 a+5 b) \sec ^3(e+f x) \tan (e+f x)}{24 f}+\frac {b \sec ^5(e+f x) \tan (e+f x)}{6 f} \] Output:
1/16*(6*a+5*b)*arctanh(sin(f*x+e))/f+1/16*(6*a+5*b)*sec(f*x+e)*tan(f*x+e)/ f+1/24*(6*a+5*b)*sec(f*x+e)^3*tan(f*x+e)/f+1/6*b*sec(f*x+e)^5*tan(f*x+e)/f
Time = 0.02 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.40 \[ \int \sec ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {3 a \text {arctanh}(\sin (e+f x))}{8 f}+\frac {5 b \text {arctanh}(\sin (e+f x))}{16 f}+\frac {3 a \sec (e+f x) \tan (e+f x)}{8 f}+\frac {5 b \sec (e+f x) \tan (e+f x)}{16 f}+\frac {a \sec ^3(e+f x) \tan (e+f x)}{4 f}+\frac {5 b \sec ^3(e+f x) \tan (e+f x)}{24 f}+\frac {b \sec ^5(e+f x) \tan (e+f x)}{6 f} \] Input:
Integrate[Sec[e + f*x]^5*(a + b*Sec[e + f*x]^2),x]
Output:
(3*a*ArcTanh[Sin[e + f*x]])/(8*f) + (5*b*ArcTanh[Sin[e + f*x]])/(16*f) + ( 3*a*Sec[e + f*x]*Tan[e + f*x])/(8*f) + (5*b*Sec[e + f*x]*Tan[e + f*x])/(16 *f) + (a*Sec[e + f*x]^3*Tan[e + f*x])/(4*f) + (5*b*Sec[e + f*x]^3*Tan[e + f*x])/(24*f) + (b*Sec[e + f*x]^5*Tan[e + f*x])/(6*f)
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.
| \(\displaystyle \int \sec ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (e+f x+\frac {\pi }{2}\right )^5 \left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {1}{6} (6 a+5 b) \int \sec ^5(e+f x)dx+\frac {b \tan (e+f x) \sec ^5(e+f x)}{6 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} (6 a+5 b) \int \csc \left (e+f x+\frac {\pi }{2}\right )^5dx+\frac {b \tan (e+f x) \sec ^5(e+f x)}{6 f}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{6} (6 a+5 b) \left (\frac {3}{4} \int \sec ^3(e+f x)dx+\frac {\tan (e+f x) \sec ^3(e+f x)}{4 f}\right )+\frac {b \tan (e+f x) \sec ^5(e+f x)}{6 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} (6 a+5 b) \left (\frac {3}{4} \int \csc \left (e+f x+\frac {\pi }{2}\right )^3dx+\frac {\tan (e+f x) \sec ^3(e+f x)}{4 f}\right )+\frac {b \tan (e+f x) \sec ^5(e+f x)}{6 f}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{6} (6 a+5 b) \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (e+f x)dx+\frac {\tan (e+f x) \sec (e+f x)}{2 f}\right )+\frac {\tan (e+f x) \sec ^3(e+f x)}{4 f}\right )+\frac {b \tan (e+f x) \sec ^5(e+f x)}{6 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} (6 a+5 b) \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (e+f x+\frac {\pi }{2}\right )dx+\frac {\tan (e+f x) \sec (e+f x)}{2 f}\right )+\frac {\tan (e+f x) \sec ^3(e+f x)}{4 f}\right )+\frac {b \tan (e+f x) \sec ^5(e+f x)}{6 f}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{6} (6 a+5 b) \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (e+f x))}{2 f}+\frac {\tan (e+f x) \sec (e+f x)}{2 f}\right )+\frac {\tan (e+f x) \sec ^3(e+f x)}{4 f}\right )+\frac {b \tan (e+f x) \sec ^5(e+f x)}{6 f}\) |
Input:
Int[Sec[e + f*x]^5*(a + b*Sec[e + f*x]^2),x]
Output:
(b*Sec[e + f*x]^5*Tan[e + f*x])/(6*f) + ((6*a + 5*b)*((Sec[e + f*x]^3*Tan[ e + f*x])/(4*f) + (3*(ArcTanh[Sin[e + f*x]]/(2*f) + (Sec[e + f*x]*Tan[e + f*x])/(2*f)))/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 = 1.78 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.10
| method | result | size |
| derivativedivides | \(\frac {a \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )+b \left (-\left (-\frac {\sec \left (f x +e \right )^{5}}{6}-\frac {5 \sec \left (f x +e \right )^{3}}{24}-\frac {5 \sec \left (f x +e \right )}{16}\right ) \tan \left (f x +e \right )+\frac {5 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{16}\right )}{f}\) | \(108\) |
| default | \(\frac {a \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )+b \left (-\left (-\frac {\sec \left (f x +e \right )^{5}}{6}-\frac {5 \sec \left (f x +e \right )^{3}}{24}-\frac {5 \sec \left (f x +e \right )}{16}\right ) \tan \left (f x +e \right )+\frac {5 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{16}\right )}{f}\) | \(108\) |
| parts | \(\frac {a \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )}{f}+\frac {b \left (-\left (-\frac {\sec \left (f x +e \right )^{5}}{6}-\frac {5 \sec \left (f x +e \right )^{3}}{24}-\frac {5 \sec \left (f x +e \right )}{16}\right ) \tan \left (f x +e \right )+\frac {5 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{16}\right )}{f}\) | \(110\) |
| parallelrisch | \(\frac {-270 \left (\frac {\cos \left (6 f x +6 e \right )}{15}+\frac {2 \cos \left (4 f x +4 e \right )}{5}+\cos \left (2 f x +2 e \right )+\frac {2}{3}\right ) \left (a +\frac {5 b}{6}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )+270 \left (\frac {\cos \left (6 f x +6 e \right )}{15}+\frac {2 \cos \left (4 f x +4 e \right )}{5}+\cos \left (2 f x +2 e \right )+\frac {2}{3}\right ) \left (a +\frac {5 b}{6}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )+\left (204 a +170 b \right ) \sin \left (3 f x +3 e \right )+\left (36 a +30 b \right ) \sin \left (5 f x +5 e \right )+168 \left (a +\frac {33 b}{14}\right ) \sin \left (f x +e \right )}{48 f \left (\cos \left (6 f x +6 e \right )+6 \cos \left (4 f x +4 e \right )+15 \cos \left (2 f x +2 e \right )+10\right )}\) | \(193\) |
| norman | \(\frac {\frac {\left (2 a +15 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{4 f}+\frac {\left (2 a +15 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{4 f}+\frac {\left (10 a +11 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{8 f}+\frac {\left (10 a +11 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{8 f}-\frac {\left (42 a -5 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{24 f}-\frac {\left (42 a -5 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{24 f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{6}}-\frac {\left (6 a +5 b \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{16 f}+\frac {\left (6 a +5 b \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{16 f}\) | \(203\) |
| risch | \(-\frac {i {\mathrm e}^{i \left (f x +e \right )} \left (18 a \,{\mathrm e}^{10 i \left (f x +e \right )}+15 b \,{\mathrm e}^{10 i \left (f x +e \right )}+102 a \,{\mathrm e}^{8 i \left (f x +e \right )}+85 b \,{\mathrm e}^{8 i \left (f x +e \right )}+84 a \,{\mathrm e}^{6 i \left (f x +e \right )}+198 b \,{\mathrm e}^{6 i \left (f x +e \right )}-84 a \,{\mathrm e}^{4 i \left (f x +e \right )}-198 b \,{\mathrm e}^{4 i \left (f x +e \right )}-102 a \,{\mathrm e}^{2 i \left (f x +e \right )}-85 b \,{\mathrm e}^{2 i \left (f x +e \right )}-18 a -15 b \right )}{24 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{6}}-\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) a}{8 f}-\frac {5 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) b}{16 f}+\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) a}{8 f}+\frac {5 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) b}{16 f}\) | \(255\) |
Input:
int(sec(f*x+e)^5*(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
1/f*(a*(-(-1/4*sec(f*x+e)^3-3/8*sec(f*x+e))*tan(f*x+e)+3/8*ln(sec(f*x+e)+t an(f*x+e)))+b*(-(-1/6*sec(f*x+e)^5-5/24*sec(f*x+e)^3-5/16*sec(f*x+e))*tan( f*x+e)+5/16*ln(sec(f*x+e)+tan(f*x+e))))
Time = 0.08 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.16 \[ \int \sec ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {3 \, {\left (6 \, a + 5 \, b\right )} \cos \left (f x + e\right )^{6} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, {\left (6 \, a + 5 \, b\right )} \cos \left (f x + e\right )^{6} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (3 \, {\left (6 \, a + 5 \, b\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (6 \, a + 5 \, b\right )} \cos \left (f x + e\right )^{2} + 8 \, b\right )} \sin \left (f x + e\right )}{96 \, f \cos \left (f x + e\right )^{6}} \] Input:
integrate(sec(f*x+e)^5*(a+b*sec(f*x+e)^2),x, algorithm="fricas")
Output:
1/96*(3*(6*a + 5*b)*cos(f*x + e)^6*log(sin(f*x + e) + 1) - 3*(6*a + 5*b)*c os(f*x + e)^6*log(-sin(f*x + e) + 1) + 2*(3*(6*a + 5*b)*cos(f*x + e)^4 + 2 *(6*a + 5*b)*cos(f*x + e)^2 + 8*b)*sin(f*x + e))/(f*cos(f*x + e)^6)
\[ \int \sec ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right ) \sec ^{5}{\left (e + f x \right )}\, dx \] Input:
integrate(sec(f*x+e)**5*(a+b*sec(f*x+e)**2),x)
Output:
Integral((a + b*sec(e + f*x)**2)*sec(e + f*x)**5, x)
Time = 0.03 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.29 \[ \int \sec ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {3 \, {\left (6 \, a + 5 \, b\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, {\left (6 \, a + 5 \, b\right )} \log \left (\sin \left (f x + e\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (6 \, a + 5 \, b\right )} \sin \left (f x + e\right )^{5} - 8 \, {\left (6 \, a + 5 \, b\right )} \sin \left (f x + e\right )^{3} + 3 \, {\left (10 \, a + 11 \, b\right )} \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{6} - 3 \, \sin \left (f x + e\right )^{4} + 3 \, \sin \left (f x + e\right )^{2} - 1}}{96 \, f} \] Input:
integrate(sec(f*x+e)^5*(a+b*sec(f*x+e)^2),x, algorithm="maxima")
Output:
1/96*(3*(6*a + 5*b)*log(sin(f*x + e) + 1) - 3*(6*a + 5*b)*log(sin(f*x + e) - 1) - 2*(3*(6*a + 5*b)*sin(f*x + e)^5 - 8*(6*a + 5*b)*sin(f*x + e)^3 + 3 *(10*a + 11*b)*sin(f*x + e))/(sin(f*x + e)^6 - 3*sin(f*x + e)^4 + 3*sin(f* x + e)^2 - 1))/f
Time = 0.14 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.23 \[ \int \sec ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {3 \, {\left (6 \, a + 5 \, b\right )} \log \left ({\left | \sin \left (f x + e\right ) + 1 \right |}\right ) - 3 \, {\left (6 \, a + 5 \, b\right )} \log \left ({\left | \sin \left (f x + e\right ) - 1 \right |}\right ) - \frac {2 \, {\left (18 \, a \sin \left (f x + e\right )^{5} + 15 \, b \sin \left (f x + e\right )^{5} - 48 \, a \sin \left (f x + e\right )^{3} - 40 \, b \sin \left (f x + e\right )^{3} + 30 \, a \sin \left (f x + e\right ) + 33 \, b \sin \left (f x + e\right )\right )}}{{\left (\sin \left (f x + e\right )^{2} - 1\right )}^{3}}}{96 \, f} \] Input:
integrate(sec(f*x+e)^5*(a+b*sec(f*x+e)^2),x, algorithm="giac")
Output:
1/96*(3*(6*a + 5*b)*log(abs(sin(f*x + e) + 1)) - 3*(6*a + 5*b)*log(abs(sin (f*x + e) - 1)) - 2*(18*a*sin(f*x + e)^5 + 15*b*sin(f*x + e)^5 - 48*a*sin( f*x + e)^3 - 40*b*sin(f*x + e)^3 + 30*a*sin(f*x + e) + 33*b*sin(f*x + e))/ (sin(f*x + e)^2 - 1)^3)/f
Time = 15.88 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.04 \[ \int \sec ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {\mathrm {atanh}\left (\sin \left (e+f\,x\right )\right )\,\left (\frac {3\,a}{8}+\frac {5\,b}{16}\right )}{f}-\frac {\left (\frac {3\,a}{8}+\frac {5\,b}{16}\right )\,{\sin \left (e+f\,x\right )}^5+\left (-a-\frac {5\,b}{6}\right )\,{\sin \left (e+f\,x\right )}^3+\left (\frac {5\,a}{8}+\frac {11\,b}{16}\right )\,\sin \left (e+f\,x\right )}{f\,\left ({\sin \left (e+f\,x\right )}^6-3\,{\sin \left (e+f\,x\right )}^4+3\,{\sin \left (e+f\,x\right )}^2-1\right )} \] Input:
int((a + b/cos(e + f*x)^2)/cos(e + f*x)^5,x)
Output:
(atanh(sin(e + f*x))*((3*a)/8 + (5*b)/16))/f - (sin(e + f*x)^5*((3*a)/8 + (5*b)/16) + sin(e + f*x)*((5*a)/8 + (11*b)/16) - sin(e + f*x)^3*(a + (5*b) /6))/(f*(3*sin(e + f*x)^2 - 3*sin(e + f*x)^4 + sin(e + f*x)^6 - 1))
Time = 0.17 (sec) , antiderivative size = 436, normalized size of antiderivative = 4.45 \[ \int \sec ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {-18 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \sin \left (f x +e \right )^{6} a -15 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \sin \left (f x +e \right )^{6} b +54 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \sin \left (f x +e \right )^{4} a +45 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \sin \left (f x +e \right )^{4} b -54 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \sin \left (f x +e \right )^{2} a -45 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \sin \left (f x +e \right )^{2} b +18 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) a +15 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) b +18 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \sin \left (f x +e \right )^{6} a +15 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \sin \left (f x +e \right )^{6} b -54 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \sin \left (f x +e \right )^{4} a -45 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \sin \left (f x +e \right )^{4} b +54 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \sin \left (f x +e \right )^{2} a +45 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \sin \left (f x +e \right )^{2} b -18 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a -15 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) b -18 \sin \left (f x +e \right )^{5} a -15 \sin \left (f x +e \right )^{5} b +48 \sin \left (f x +e \right )^{3} a +40 \sin \left (f x +e \right )^{3} b -30 \sin \left (f x +e \right ) a -33 \sin \left (f x +e \right ) b}{48 f \left (\sin \left (f x +e \right )^{6}-3 \sin \left (f x +e \right )^{4}+3 \sin \left (f x +e \right )^{2}-1\right )} \] Input:
int(sec(f*x+e)^5*(a+b*sec(f*x+e)^2),x)
Output:
( - 18*log(tan((e + f*x)/2) - 1)*sin(e + f*x)**6*a - 15*log(tan((e + f*x)/ 2) - 1)*sin(e + f*x)**6*b + 54*log(tan((e + f*x)/2) - 1)*sin(e + f*x)**4*a + 45*log(tan((e + f*x)/2) - 1)*sin(e + f*x)**4*b - 54*log(tan((e + f*x)/2 ) - 1)*sin(e + f*x)**2*a - 45*log(tan((e + f*x)/2) - 1)*sin(e + f*x)**2*b + 18*log(tan((e + f*x)/2) - 1)*a + 15*log(tan((e + f*x)/2) - 1)*b + 18*log (tan((e + f*x)/2) + 1)*sin(e + f*x)**6*a + 15*log(tan((e + f*x)/2) + 1)*si n(e + f*x)**6*b - 54*log(tan((e + f*x)/2) + 1)*sin(e + f*x)**4*a - 45*log( tan((e + f*x)/2) + 1)*sin(e + f*x)**4*b + 54*log(tan((e + f*x)/2) + 1)*sin (e + f*x)**2*a + 45*log(tan((e + f*x)/2) + 1)*sin(e + f*x)**2*b - 18*log(t an((e + f*x)/2) + 1)*a - 15*log(tan((e + f*x)/2) + 1)*b - 18*sin(e + f*x)* *5*a - 15*sin(e + f*x)**5*b + 48*sin(e + f*x)**3*a + 40*sin(e + f*x)**3*b - 30*sin(e + f*x)*a - 33*sin(e + f*x)*b)/(48*f*(sin(e + f*x)**6 - 3*sin(e + f*x)**4 + 3*sin(e + f*x)**2 - 1))