Integrand size = 23, antiderivative size = 108 \[ \int \frac {\cot ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(2 a+3 b) \csc ^2(e+f x)}{2 (a+b)^2 f}-\frac {\csc ^4(e+f x)}{4 (a+b) f}+\frac {b^3 \log \left (b+a \cos ^2(e+f x)\right )}{2 a (a+b)^3 f}+\frac {\left (a^2+3 a b+3 b^2\right ) \log (\sin (e+f x))}{(a+b)^3 f} \] Output:
1/2*(2*a+3*b)*csc(f*x+e)^2/(a+b)^2/f-1/4*csc(f*x+e)^4/(a+b)/f+1/2*b^3*ln(b +a*cos(f*x+e)^2)/a/(a+b)^3/f+(a^2+3*a*b+3*b^2)*ln(sin(f*x+e))/(a+b)^3/f
Time = 0.45 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.28 \[ \int \frac {\cot ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b+a \cos (2 e+2 f x)) \left (\frac {2 (2 a+3 b) \csc ^2(e+f x)}{(a+b)^2}-\frac {\csc ^4(e+f x)}{a+b}+\frac {4 \left (a^2+3 a b+3 b^2\right ) \log (\sin (e+f x))}{(a+b)^3}+\frac {2 b^3 \log \left (a+b-a \sin ^2(e+f x)\right )}{a (a+b)^3}\right ) \sec ^2(e+f x)}{8 f \left (a+b \sec ^2(e+f x)\right )} \] Input:
Integrate[Cot[e + f*x]^5/(a + b*Sec[e + f*x]^2),x]
Output:
((a + 2*b + a*Cos[2*e + 2*f*x])*((2*(2*a + 3*b)*Csc[e + f*x]^2)/(a + b)^2 - Csc[e + f*x]^4/(a + b) + (4*(a^2 + 3*a*b + 3*b^2)*Log[Sin[e + f*x]])/(a + b)^3 + (2*b^3*Log[a + b - a*Sin[e + f*x]^2])/(a*(a + b)^3))*Sec[e + f*x] ^2)/(8*f*(a + b*Sec[e + f*x]^2))
Time = 0.36 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4626, 354, 99, 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.
| \(\displaystyle \int \frac {\cot ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (e+f x)^5 \left (a+b \sec (e+f x)^2\right )}dx\) |
| \(\Big \downarrow \) 4626 |
| \(\displaystyle -\frac {\int \frac {\cos ^7(e+f x)}{\left (1-\cos ^2(e+f x)\right )^3 \left (a \cos ^2(e+f x)+b\right )}d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle -\frac {\int \frac {\cos ^6(e+f x)}{\left (1-\cos ^2(e+f x)\right )^3 \left (a \cos ^2(e+f x)+b\right )}d\cos ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle -\frac {\int \left (-\frac {b^3}{(a+b)^3 \left (a \cos ^2(e+f x)+b\right )}+\frac {-a^2-3 b a-3 b^2}{(a+b)^3 \left (\cos ^2(e+f x)-1\right )}+\frac {-2 a-3 b}{(a+b)^2 \left (\cos ^2(e+f x)-1\right )^2}-\frac {1}{(a+b) \left (\cos ^2(e+f x)-1\right )^3}\right )d\cos ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {-\frac {\left (a^2+3 a b+3 b^2\right ) \log \left (1-\cos ^2(e+f x)\right )}{(a+b)^3}-\frac {b^3 \log \left (a \cos ^2(e+f x)+b\right )}{a (a+b)^3}-\frac {2 a+3 b}{(a+b)^2 \left (1-\cos ^2(e+f x)\right )}+\frac {1}{2 (a+b) \left (1-\cos ^2(e+f x)\right )^2}}{2 f}\) |
Input:
Int[Cot[e + f*x]^5/(a + b*Sec[e + f*x]^2),x]
Output:
-1/2*(1/(2*(a + b)*(1 - Cos[e + f*x]^2)^2) - (2*a + 3*b)/((a + b)^2*(1 - C os[e + f*x]^2)) - ((a^2 + 3*a*b + 3*b^2)*Log[1 - Cos[e + f*x]^2])/(a + b)^ 3 - (b^3*Log[b + a*Cos[e + f*x]^2])/(a*(a + b)^3))/f
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> Module[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-(f *ff^(m + n*p - 1))^(-1) Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff* x)^n)^p/x^(m + n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n] && IntegerQ[p]
Time = 1.72 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.67
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{2 \left (8 a +8 b \right ) \left (1+\cos \left (f x +e \right )\right )^{2}}-\frac {-7 a -11 b}{16 \left (a +b \right )^{2} \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (a^{2}+3 a b +3 b^{2}\right ) \ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{3}}+\frac {b^{3} \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{2 \left (a +b \right )^{3} a}-\frac {1}{2 \left (8 a +8 b \right ) \left (-1+\cos \left (f x +e \right )\right )^{2}}-\frac {7 a +11 b}{16 \left (a +b \right )^{2} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (a^{2}+3 a b +3 b^{2}\right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{3}}}{f}\) | \(180\) |
| default | \(\frac {-\frac {1}{2 \left (8 a +8 b \right ) \left (1+\cos \left (f x +e \right )\right )^{2}}-\frac {-7 a -11 b}{16 \left (a +b \right )^{2} \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (a^{2}+3 a b +3 b^{2}\right ) \ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{3}}+\frac {b^{3} \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{2 \left (a +b \right )^{3} a}-\frac {1}{2 \left (8 a +8 b \right ) \left (-1+\cos \left (f x +e \right )\right )^{2}}-\frac {7 a +11 b}{16 \left (a +b \right )^{2} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (a^{2}+3 a b +3 b^{2}\right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{3}}}{f}\) | \(180\) |
| risch | \(\frac {i x}{a}-\frac {2 i a^{2} x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}-\frac {2 i a^{2} e}{f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {6 i a b x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}-\frac {6 i a b e}{f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {6 i b^{2} x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}-\frac {6 i b^{2} e}{f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {2 i b^{3} x}{a \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {2 i b^{3} e}{a f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {2 \left (2 a \,{\mathrm e}^{6 i \left (f x +e \right )}+3 b \,{\mathrm e}^{6 i \left (f x +e \right )}-2 a \,{\mathrm e}^{4 i \left (f x +e \right )}-4 b \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+3 b \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{f \left (a +b \right )^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) a^{2}}{f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) a b}{f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) b^{2}}{f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {b^{3} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 a f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}\) | \(529\) |
Input:
int(cot(f*x+e)^5/(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
1/f*(-1/2/(8*a+8*b)/(1+cos(f*x+e))^2-1/16*(-7*a-11*b)/(a+b)^2/(1+cos(f*x+e ))+1/2*(a^2+3*a*b+3*b^2)/(a+b)^3*ln(1+cos(f*x+e))+1/2*b^3/(a+b)^3/a*ln(b+a *cos(f*x+e)^2)-1/2/(8*a+8*b)/(-1+cos(f*x+e))^2-1/16*(7*a+11*b)/(a+b)^2/(-1 +cos(f*x+e))+1/2*(a^2+3*a*b+3*b^2)/(a+b)^3*ln(-1+cos(f*x+e)))
Leaf count of result is larger than twice the leaf count of optimal. 265 vs. \(2 (102) = 204\).
Time = 0.29 (sec) , antiderivative size = 265, normalized size of antiderivative = 2.45 \[ \int \frac {\cot ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {3 \, a^{3} + 8 \, a^{2} b + 5 \, a b^{2} - 2 \, {\left (2 \, a^{3} + 5 \, a^{2} b + 3 \, a b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (b^{3} \cos \left (f x + e\right )^{4} - 2 \, b^{3} \cos \left (f x + e\right )^{2} + b^{3}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) + 4 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{3} + 3 \, a^{2} b + 3 \, a b^{2} - 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (f x + e\right )\right )}{4 \, {\left ({\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} f\right )}} \] Input:
integrate(cot(f*x+e)^5/(a+b*sec(f*x+e)^2),x, algorithm="fricas")
Output:
1/4*(3*a^3 + 8*a^2*b + 5*a*b^2 - 2*(2*a^3 + 5*a^2*b + 3*a*b^2)*cos(f*x + e )^2 + 2*(b^3*cos(f*x + e)^4 - 2*b^3*cos(f*x + e)^2 + b^3)*log(a*cos(f*x + e)^2 + b) + 4*((a^3 + 3*a^2*b + 3*a*b^2)*cos(f*x + e)^4 + a^3 + 3*a^2*b + 3*a*b^2 - 2*(a^3 + 3*a^2*b + 3*a*b^2)*cos(f*x + e)^2)*log(1/2*sin(f*x + e) ))/((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f*cos(f*x + e)^4 - 2*(a^4 + 3*a^3* b + 3*a^2*b^2 + a*b^3)*f*cos(f*x + e)^2 + (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b ^3)*f)
\[ \int \frac {\cot ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\int \frac {\cot ^{5}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \] Input:
integrate(cot(f*x+e)**5/(a+b*sec(f*x+e)**2),x)
Output:
Integral(cot(e + f*x)**5/(a + b*sec(e + f*x)**2), x)
Time = 0.04 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.34 \[ \int \frac {\cot ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {2 \, b^{3} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}} + \frac {2 \, {\left (a^{2} + 3 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (2 \, a + 3 \, b\right )} \sin \left (f x + e\right )^{2} - a - b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sin \left (f x + e\right )^{4}}}{4 \, f} \] Input:
integrate(cot(f*x+e)^5/(a+b*sec(f*x+e)^2),x, algorithm="maxima")
Output:
1/4*(2*b^3*log(a*sin(f*x + e)^2 - a - b)/(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^ 3) + 2*(a^2 + 3*a*b + 3*b^2)*log(sin(f*x + e)^2)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + (2*(2*a + 3*b)*sin(f*x + e)^2 - a - b)/((a^2 + 2*a*b + b^2)*sin(f *x + e)^4))/f
Time = 0.14 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.50 \[ \int \frac {\cot ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {b^{3} \log \left ({\left | a \sin \left (f x + e\right )^{2} - a - b \right |}\right )}{2 \, {\left (a^{4} f + 3 \, a^{3} b f + 3 \, a^{2} b^{2} f + a b^{3} f\right )}} + \frac {{\left (a^{2} + 3 \, a b + 3 \, b^{2}\right )} \log \left ({\left | \sin \left (f x + e\right ) \right |}\right )}{a^{3} f + 3 \, a^{2} b f + 3 \, a b^{2} f + b^{3} f} + \frac {2 \, {\left (2 \, a^{2} + 5 \, a b + 3 \, b^{2}\right )} \sin \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}}{4 \, {\left (a + b\right )}^{3} f \sin \left (f x + e\right )^{4}} \] Input:
integrate(cot(f*x+e)^5/(a+b*sec(f*x+e)^2),x, algorithm="giac")
Output:
1/2*b^3*log(abs(a*sin(f*x + e)^2 - a - b))/(a^4*f + 3*a^3*b*f + 3*a^2*b^2* f + a*b^3*f) + (a^2 + 3*a*b + 3*b^2)*log(abs(sin(f*x + e)))/(a^3*f + 3*a^2 *b*f + 3*a*b^2*f + b^3*f) + 1/4*(2*(2*a^2 + 5*a*b + 3*b^2)*sin(f*x + e)^2 - a^2 - 2*a*b - b^2)/((a + b)^3*f*sin(f*x + e)^4)
Time = 15.75 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.48 \[ \int \frac {\cot ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a^2+3\,a\,b+3\,b^2\right )}{f\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}-\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,a\,f}-\frac {\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )\,\left (\frac {b}{2\,{\left (a+b\right )}^2}+\frac {1}{2\,\left (a+b\right )}+\frac {b^2}{2\,{\left (a+b\right )}^3}-\frac {1}{2\,a}\right )}{f}-\frac {{\mathrm {cot}\left (e+f\,x\right )}^4\,\left (\frac {1}{4\,\left (a+b\right )}-\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (a+2\,b\right )}{2\,{\left (a+b\right )}^2}\right )}{f} \] Input:
int(cot(e + f*x)^5/(a + b/cos(e + f*x)^2),x)
Output:
(log(tan(e + f*x))*(3*a*b + a^2 + 3*b^2))/(f*(3*a*b^2 + 3*a^2*b + a^3 + b^ 3)) - log(tan(e + f*x)^2 + 1)/(2*a*f) - (log(a + b + b*tan(e + f*x)^2)*(b/ (2*(a + b)^2) + 1/(2*(a + b)) + b^2/(2*(a + b)^3) - 1/(2*a)))/f - (cot(e + f*x)^4*(1/(4*(a + b)) - (tan(e + f*x)^2*(a + 2*b))/(2*(a + b)^2)))/f
Time = 0.17 (sec) , antiderivative size = 414, normalized size of antiderivative = 3.83 \[ \int \frac {\cot ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {-32 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \sin \left (f x +e \right )^{4} a^{3}-96 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \sin \left (f x +e \right )^{4} a^{2} b -96 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \sin \left (f x +e \right )^{4} a \,b^{2}-32 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \sin \left (f x +e \right )^{4} b^{3}+16 \,\mathrm {log}\left (\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\sqrt {a +b}-2 \sqrt {a}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{4} b^{3}+16 \,\mathrm {log}\left (\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\sqrt {a +b}+2 \sqrt {a}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{4} b^{3}+32 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{4} a^{3}+96 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{4} a^{2} b +96 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{4} a \,b^{2}-13 \sin \left (f x +e \right )^{4} a^{3}-34 \sin \left (f x +e \right )^{4} a^{2} b -21 \sin \left (f x +e \right )^{4} a \,b^{2}+32 \sin \left (f x +e \right )^{2} a^{3}+80 \sin \left (f x +e \right )^{2} a^{2} b +48 \sin \left (f x +e \right )^{2} a \,b^{2}-8 a^{3}-16 a^{2} b -8 a \,b^{2}}{32 \sin \left (f x +e \right )^{4} a f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )} \] Input:
int(cot(f*x+e)^5/(a+b*sec(f*x+e)^2),x)
Output:
( - 32*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**4*a**3 - 96*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**4*a**2*b - 96*log(tan((e + f*x)/2)**2 + 1)* sin(e + f*x)**4*a*b**2 - 32*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**4*b **3 + 16*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan ((e + f*x)/2))*sin(e + f*x)**4*b**3 + 16*log(sqrt(a + b)*tan((e + f*x)/2)* *2 + sqrt(a + b) + 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)**4*b**3 + 32*l og(tan((e + f*x)/2))*sin(e + f*x)**4*a**3 + 96*log(tan((e + f*x)/2))*sin(e + f*x)**4*a**2*b + 96*log(tan((e + f*x)/2))*sin(e + f*x)**4*a*b**2 - 13*s in(e + f*x)**4*a**3 - 34*sin(e + f*x)**4*a**2*b - 21*sin(e + f*x)**4*a*b** 2 + 32*sin(e + f*x)**2*a**3 + 80*sin(e + f*x)**2*a**2*b + 48*sin(e + f*x)* *2*a*b**2 - 8*a**3 - 16*a**2*b - 8*a*b**2)/(32*sin(e + f*x)**4*a*f*(a**3 + 3*a**2*b + 3*a*b**2 + b**3))