Integrand size = 21, antiderivative size = 130 \[ \int \frac {\cot (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {b^3}{4 a^3 (a+b) f \left (b+a \cos ^2(e+f x)\right )^2}+\frac {b^2 (3 a+2 b)}{2 a^3 (a+b)^2 f \left (b+a \cos ^2(e+f x)\right )}+\frac {b \left (3 a^2+3 a b+b^2\right ) \log \left (b+a \cos ^2(e+f x)\right )}{2 a^3 (a+b)^3 f}+\frac {\log (\sin (e+f x))}{(a+b)^3 f} \] Output:
-1/4*b^3/a^3/(a+b)/f/(b+a*cos(f*x+e)^2)^2+1/2*b^2*(3*a+2*b)/a^3/(a+b)^2/f/ (b+a*cos(f*x+e)^2)+1/2*b*(3*a^2+3*a*b+b^2)*ln(b+a*cos(f*x+e)^2)/a^3/(a+b)^ 3/f+ln(sin(f*x+e))/(a+b)^3/f
Time = 0.83 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.22 \[ \int \frac {\cot (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {(a+2 b+a \cos (2 (e+f x)))^3 \sec ^6(e+f x) \left (4 \log (\sin (e+f x))+\frac {2 b \left (3 a^2+3 a b+b^2\right ) \log \left (a+b-a \sin ^2(e+f x)\right )}{a^3}-\frac {b^3 (a+b)^2}{a^3 \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {2 b^2 (a+b) (3 a+2 b)}{a^3 \left (a+b-a \sin ^2(e+f x)\right )}\right )}{32 (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^3} \] Input:
Integrate[Cot[e + f*x]/(a + b*Sec[e + f*x]^2)^3,x]
Output:
((a + 2*b + a*Cos[2*(e + f*x)])^3*Sec[e + f*x]^6*(4*Log[Sin[e + f*x]] + (2 *b*(3*a^2 + 3*a*b + b^2)*Log[a + b - a*Sin[e + f*x]^2])/a^3 - (b^3*(a + b) ^2)/(a^3*(a + b - a*Sin[e + f*x]^2)^2) + (2*b^2*(a + b)*(3*a + 2*b))/(a^3* (a + b - a*Sin[e + f*x]^2))))/(32*(a + b)^3*f*(a + b*Sec[e + f*x]^2)^3)
Time = 0.36 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, 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.
| \(\displaystyle \int \frac {\cot (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (e+f x) \left (a+b \sec (e+f x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4626 |
| \(\displaystyle -\frac {\int \frac {\cos ^7(e+f x)}{\left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )^3}d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle -\frac {\int \frac {\cos ^6(e+f x)}{\left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right )^3}d\cos ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle -\frac {\int \left (-\frac {b^3}{a^2 (a+b) \left (a \cos ^2(e+f x)+b\right )^3}+\frac {(3 a+2 b) b^2}{a^2 (a+b)^2 \left (a \cos ^2(e+f x)+b\right )^2}-\frac {\left (3 a^2+3 b a+b^2\right ) b}{a^2 (a+b)^3 \left (a \cos ^2(e+f x)+b\right )}-\frac {1}{(a+b)^3 \left (\cos ^2(e+f x)-1\right )}\right )d\cos ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {b^3}{2 a^3 (a+b) \left (a \cos ^2(e+f x)+b\right )^2}-\frac {b^2 (3 a+2 b)}{a^3 (a+b)^2 \left (a \cos ^2(e+f x)+b\right )}-\frac {b \left (3 a^2+3 a b+b^2\right ) \log \left (a \cos ^2(e+f x)+b\right )}{a^3 (a+b)^3}-\frac {\log \left (1-\cos ^2(e+f x)\right )}{(a+b)^3}}{2 f}\) |
Input:
Int[Cot[e + f*x]/(a + b*Sec[e + f*x]^2)^3,x]
Output:
-1/2*(b^3/(2*a^3*(a + b)*(b + a*Cos[e + f*x]^2)^2) - (b^2*(3*a + 2*b))/(a^ 3*(a + b)^2*(b + a*Cos[e + f*x]^2)) - Log[1 - Cos[e + f*x]^2]/(a + b)^3 - (b*(3*a^2 + 3*a*b + b^2)*Log[b + a*Cos[e + f*x]^2])/(a^3*(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 = 9.88 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.11
| method | result | size |
| derivativedivides | \(\frac {\frac {\ln \left (-1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{3}}+\frac {b \left (\frac {b \left (3 a^{2}+5 a b +2 b^{2}\right )}{a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )}+\frac {\left (3 a^{2}+3 a b +b^{2}\right ) \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{a^{3}}-\frac {b^{2} \left (a^{2}+2 a b +b^{2}\right )}{2 a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )^{2}}\right )}{2 \left (a +b \right )^{3}}+\frac {\ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{3}}}{f}\) | \(144\) |
| default | \(\frac {\frac {\ln \left (-1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{3}}+\frac {b \left (\frac {b \left (3 a^{2}+5 a b +2 b^{2}\right )}{a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )}+\frac {\left (3 a^{2}+3 a b +b^{2}\right ) \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{a^{3}}-\frac {b^{2} \left (a^{2}+2 a b +b^{2}\right )}{2 a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )^{2}}\right )}{2 \left (a +b \right )^{3}}+\frac {\ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{3}}}{f}\) | \(144\) |
| risch | \(\frac {i x}{a^{3}}-\frac {2 i x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}-\frac {2 i e}{f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {6 i b x}{a \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {6 i b e}{a f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {6 i b^{2} x}{a^{2} \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {6 i b^{2} e}{a^{2} f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {2 i b^{3} x}{a^{3} \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {2 i b^{3} e}{a^{3} f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 b^{2} \left (3 a^{2} {\mathrm e}^{6 i \left (f x +e \right )}+2 a b \,{\mathrm e}^{6 i \left (f x +e \right )}+6 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+14 a b \,{\mathrm e}^{4 i \left (f x +e \right )}+6 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+3 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+2 a b \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{a^{3} \left (a +b \right )^{2} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2}}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {3 b \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 a f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {3 b^{2} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 a^{2} f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {b^{3} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 a^{3} f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}\) | \(628\) |
Input:
int(cot(f*x+e)/(a+b*sec(f*x+e)^2)^3,x,method=_RETURNVERBOSE)
Output:
1/f*(1/2/(a+b)^3*ln(-1+cos(f*x+e))+1/2*b/(a+b)^3*(1/a^3*b*(3*a^2+5*a*b+2*b ^2)/(b+a*cos(f*x+e)^2)+(3*a^2+3*a*b+b^2)/a^3*ln(b+a*cos(f*x+e)^2)-1/2*b^2* (a^2+2*a*b+b^2)/a^3/(b+a*cos(f*x+e)^2)^2)+1/2/(a+b)^3*ln(1+cos(f*x+e)))
Leaf count of result is larger than twice the leaf count of optimal. 307 vs. \(2 (124) = 248\).
Time = 0.41 (sec) , antiderivative size = 307, normalized size of antiderivative = 2.36 \[ \int \frac {\cot (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {5 \, a^{2} b^{3} + 8 \, a b^{4} + 3 \, b^{5} + 2 \, {\left (3 \, a^{3} b^{2} + 5 \, a^{2} b^{3} + 2 \, a b^{4}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5} + {\left (3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) + 4 \, {\left (a^{5} \cos \left (f x + e\right )^{4} + 2 \, a^{4} b \cos \left (f x + e\right )^{2} + a^{3} b^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (f x + e\right )\right )}{4 \, {\left ({\left (a^{8} + 3 \, a^{7} b + 3 \, a^{6} b^{2} + a^{5} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{7} b + 3 \, a^{6} b^{2} + 3 \, a^{5} b^{3} + a^{4} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{6} b^{2} + 3 \, a^{5} b^{3} + 3 \, a^{4} b^{4} + a^{3} b^{5}\right )} f\right )}} \] Input:
integrate(cot(f*x+e)/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")
Output:
1/4*(5*a^2*b^3 + 8*a*b^4 + 3*b^5 + 2*(3*a^3*b^2 + 5*a^2*b^3 + 2*a*b^4)*cos (f*x + e)^2 + 2*(3*a^2*b^3 + 3*a*b^4 + b^5 + (3*a^4*b + 3*a^3*b^2 + a^2*b^ 3)*cos(f*x + e)^4 + 2*(3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*cos(f*x + e)^2)*log( a*cos(f*x + e)^2 + b) + 4*(a^5*cos(f*x + e)^4 + 2*a^4*b*cos(f*x + e)^2 + a ^3*b^2)*log(1/2*sin(f*x + e)))/((a^8 + 3*a^7*b + 3*a^6*b^2 + a^5*b^3)*f*co s(f*x + e)^4 + 2*(a^7*b + 3*a^6*b^2 + 3*a^5*b^3 + a^4*b^4)*f*cos(f*x + e)^ 2 + (a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*f)
Timed out. \[ \int \frac {\cot (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Timed out} \] Input:
integrate(cot(f*x+e)/(a+b*sec(f*x+e)**2)**3,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.87 \[ \int \frac {\cot (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {2 \, {\left (3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}} + \frac {6 \, a^{2} b^{2} + 9 \, a b^{3} + 3 \, b^{4} - 2 \, {\left (3 \, a^{2} b^{2} + 2 \, a b^{3}\right )} \sin \left (f x + e\right )^{2}}{a^{7} + 4 \, a^{6} b + 6 \, a^{5} b^{2} + 4 \, a^{4} b^{3} + a^{3} b^{4} + {\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2}\right )} \sin \left (f x + e\right )^{4} - 2 \, {\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} \sin \left (f x + e\right )^{2}} + \frac {2 \, \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}}{4 \, f} \] Input:
integrate(cot(f*x+e)/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")
Output:
1/4*(2*(3*a^2*b + 3*a*b^2 + b^3)*log(a*sin(f*x + e)^2 - a - b)/(a^6 + 3*a^ 5*b + 3*a^4*b^2 + a^3*b^3) + (6*a^2*b^2 + 9*a*b^3 + 3*b^4 - 2*(3*a^2*b^2 + 2*a*b^3)*sin(f*x + e)^2)/(a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4 + (a^7 + 2*a^6*b + a^5*b^2)*sin(f*x + e)^4 - 2*(a^7 + 3*a^6*b + 3*a^5*b^2 + a^4*b^3)*sin(f*x + e)^2) + 2*log(sin(f*x + e)^2)/(a^3 + 3*a^2*b + 3*a*b ^2 + b^3))/f
Time = 0.16 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.46 \[ \int \frac {\cot (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {{\left (3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left ({\left | a \cos \left (f x + e\right )^{2} + b \right |}\right )}{2 \, {\left (a^{6} f + 3 \, a^{5} b f + 3 \, a^{4} b^{2} f + a^{3} b^{3} f\right )}} + \frac {\log \left ({\left | -\cos \left (f x + e\right )^{2} + 1 \right |}\right )}{2 \, {\left (a^{3} f + 3 \, a^{2} b f + 3 \, a b^{2} f + b^{3} f\right )}} + \frac {2 \, {\left (3 \, a^{2} b^{2} + 5 \, a b^{3} + 2 \, b^{4}\right )} \cos \left (f x + e\right )^{2} + \frac {5 \, a^{2} b^{3} + 8 \, a b^{4} + 3 \, b^{5}}{a}}{4 \, {\left (a \cos \left (f x + e\right )^{2} + b\right )}^{2} {\left (a + b\right )}^{3} a^{2} f} \] Input:
integrate(cot(f*x+e)/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")
Output:
1/2*(3*a^2*b + 3*a*b^2 + b^3)*log(abs(a*cos(f*x + e)^2 + b))/(a^6*f + 3*a^ 5*b*f + 3*a^4*b^2*f + a^3*b^3*f) + 1/2*log(abs(-cos(f*x + e)^2 + 1))/(a^3* f + 3*a^2*b*f + 3*a*b^2*f + b^3*f) + 1/4*(2*(3*a^2*b^2 + 5*a*b^3 + 2*b^4)* cos(f*x + e)^2 + (5*a^2*b^3 + 8*a*b^4 + 3*b^5)/a)/((a*cos(f*x + e)^2 + b)^ 2*(a + b)^3*a^2*f)
Time = 18.95 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.46 \[ \int \frac {\cot (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )}{f\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}-\frac {\frac {2\,b^2+5\,a\,b}{4\,a^2\,\left (a+b\right )}+\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (b^2+2\,a\,b\right )}{2\,a^2\,{\left (a+b\right )}^2}}{f\,\left (2\,a\,b+a^2+b^2+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}-\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,a^3\,f}+\frac {b\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )\,\left (3\,a^2+3\,a\,b+b^2\right )}{2\,a^3\,f\,{\left (a+b\right )}^3} \] Input:
int(cot(e + f*x)/(a + b/cos(e + f*x)^2)^3,x)
Output:
log(tan(e + f*x))/(f*(3*a*b^2 + 3*a^2*b + a^3 + b^3)) - ((5*a*b + 2*b^2)/( 4*a^2*(a + b)) + (b*tan(e + f*x)^2*(2*a*b + b^2))/(2*a^2*(a + b)^2))/(f*(2 *a*b + a^2 + b^2 + tan(e + f*x)^2*(2*a*b + 2*b^2) + b^2*tan(e + f*x)^4)) - log(tan(e + f*x)^2 + 1)/(2*a^3*f) + (b*log(a + b + b*tan(e + f*x)^2)*(3*a *b + 3*a^2 + b^2))/(2*a^3*f*(a + b)^3)
Time = 0.23 (sec) , antiderivative size = 1871, normalized size of antiderivative = 14.39 \[ \int \frac {\cot (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:
int(cot(f*x+e)/(a+b*sec(f*x+e)^2)^3,x)
Output:
( - 4*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**4*a**5 - 12*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**4*a**4*b - 12*log(tan((e + f*x)/2)**2 + 1)*s in(e + f*x)**4*a**3*b**2 - 4*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**4* a**2*b**3 + 8*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**2*a**5 + 32*log(t an((e + f*x)/2)**2 + 1)*sin(e + f*x)**2*a**4*b + 48*log(tan((e + f*x)/2)** 2 + 1)*sin(e + f*x)**2*a**3*b**2 + 32*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**2*a**2*b**3 + 8*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**2*a*b**4 - 4*log(tan((e + f*x)/2)**2 + 1)*a**5 - 20*log(tan((e + f*x)/2)**2 + 1)*a **4*b - 40*log(tan((e + f*x)/2)**2 + 1)*a**3*b**2 - 40*log(tan((e + f*x)/2 )**2 + 1)*a**2*b**3 - 20*log(tan((e + f*x)/2)**2 + 1)*a*b**4 - 4*log(tan(( e + f*x)/2)**2 + 1)*b**5 + 6*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*a**4*b + 6*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*a**3*b**2 + 2*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*a**2*b**3 - 12*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)**2*a**4*b - 24*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)**2*a**3*b**2 - 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)**2*a**2*b**3 - 4*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) -...