\(\int \frac {\cot ^3(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\) [354]

Optimal result

Integrand size = 23, antiderivative size = 111 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {b^3}{2 a^2 (a+b)^2 f \left (b+a \cos ^2(e+f x)\right )}-\frac {\csc ^2(e+f x)}{2 (a+b)^2 f}-\frac {b^2 (3 a+b) \log \left (b+a \cos ^2(e+f x)\right )}{2 a^2 (a+b)^3 f}-\frac {(a+3 b) \log (\sin (e+f x))}{(a+b)^3 f} \] Output:

-1/2*b^3/a^2/(a+b)^2/f/(b+a*cos(f*x+e)^2)-1/2*csc(f*x+e)^2/(a+b)^2/f-1/2*b 
^2*(3*a+b)*ln(b+a*cos(f*x+e)^2)/a^2/(a+b)^3/f-(a+3*b)*ln(sin(f*x+e))/(a+b) 
^3/f
 

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.17 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x)))^2 \left ((a+b) \csc ^2(e+f x)+2 (a+3 b) \log (\sin (e+f x))+\frac {b^2 \left (\frac {2 b (a+b)}{a+2 b+a \cos (2 (e+f x))}+(3 a+b) \log \left (a+b-a \sin ^2(e+f x)\right )\right )}{a^2}\right ) \sec ^4(e+f x)}{8 (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^2} \] Input:

Integrate[Cot[e + f*x]^3/(a + b*Sec[e + f*x]^2)^2,x]
 

Output:

-1/8*((a + 2*b + a*Cos[2*(e + f*x)])^2*((a + b)*Csc[e + f*x]^2 + 2*(a + 3* 
b)*Log[Sin[e + f*x]] + (b^2*((2*b*(a + b))/(a + 2*b + a*Cos[2*(e + f*x)]) 
+ (3*a + b)*Log[a + b - a*Sin[e + f*x]^2]))/a^2)*Sec[e + f*x]^4)/((a + b)^ 
3*f*(a + b*Sec[e + f*x]^2)^2)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.97, 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.

\(\displaystyle \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (e+f x)^3 \left (a+b \sec (e+f x)^2\right )^2}dx\)

\(\Big \downarrow \) 4626

\(\displaystyle -\frac {\int \frac {\cos ^7(e+f x)}{\left (1-\cos ^2(e+f x)\right )^2 \left (a \cos ^2(e+f x)+b\right )^2}d\cos (e+f x)}{f}\)

\(\Big \downarrow \) 354

\(\displaystyle -\frac {\int \frac {\cos ^6(e+f x)}{\left (1-\cos ^2(e+f x)\right )^2 \left (a \cos ^2(e+f x)+b\right )^2}d\cos ^2(e+f x)}{2 f}\)

\(\Big \downarrow \) 99

\(\displaystyle -\frac {\int \left (-\frac {b^3}{a (a+b)^2 \left (a \cos ^2(e+f x)+b\right )^2}+\frac {(3 a+b) b^2}{a (a+b)^3 \left (a \cos ^2(e+f x)+b\right )}+\frac {a+3 b}{(a+b)^3 \left (\cos ^2(e+f x)-1\right )}+\frac {1}{(a+b)^2 \left (\cos ^2(e+f x)-1\right )^2}\right )d\cos ^2(e+f x)}{2 f}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\frac {b^3}{a^2 (a+b)^2 \left (a \cos ^2(e+f x)+b\right )}+\frac {b^2 (3 a+b) \log \left (a \cos ^2(e+f x)+b\right )}{a^2 (a+b)^3}+\frac {1}{(a+b)^2 \left (1-\cos ^2(e+f x)\right )}+\frac {(a+3 b) \log \left (1-\cos ^2(e+f x)\right )}{(a+b)^3}}{2 f}\)

Input:

Int[Cot[e + f*x]^3/(a + b*Sec[e + f*x]^2)^2,x]
 

Output:

-1/2*(1/((a + b)^2*(1 - Cos[e + f*x]^2)) + b^3/(a^2*(a + b)^2*(b + a*Cos[e 
 + f*x]^2)) + ((a + 3*b)*Log[1 - Cos[e + f*x]^2])/(a + b)^3 + (b^2*(3*a + 
b)*Log[b + a*Cos[e + f*x]^2])/(a^2*(a + b)^3))/f
 

Defintions of rubi rules used

rule 99

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]))
 

rule 354

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4626

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]
 

Maple [A] (verified)

Time = 4.21 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.27

method	result	size
derivativedivides	\(\frac {-\frac {b^{2} \left (\frac {\left (a +b \right ) b}{a^{2} \left (b +a \cos \left (f x +e \right )^{2}\right )}+\frac {\left (3 a +b \right ) \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{a^{2}}\right )}{2 \left (a +b \right )^{3}}-\frac {1}{4 \left (a +b \right )^{2} \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (-a -3 b \right ) \ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{3}}+\frac {1}{4 \left (a +b \right )^{2} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (-a -3 b \right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{3}}}{f}\)	\(141\)
default	\(\frac {-\frac {b^{2} \left (\frac {\left (a +b \right ) b}{a^{2} \left (b +a \cos \left (f x +e \right )^{2}\right )}+\frac {\left (3 a +b \right ) \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{a^{2}}\right )}{2 \left (a +b \right )^{3}}-\frac {1}{4 \left (a +b \right )^{2} \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (-a -3 b \right ) \ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{3}}+\frac {1}{4 \left (a +b \right )^{2} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (-a -3 b \right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{3}}}{f}\)	\(141\)
risch	\(-\frac {i x}{a^{2}}+\frac {2 i a x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}+\frac {2 i a e}{f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {6 i b x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}+\frac {6 i b e}{f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {6 i b^{2} x}{a \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {6 i b^{2} e}{a f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 i b^{3} x}{a^{2} \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 i b^{3} e}{a^{2} f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 a^{3} {\mathrm e}^{6 i \left (f x +e \right )}-2 b^{3} {\mathrm e}^{6 i \left (f x +e \right )}+4 a^{3} {\mathrm e}^{4 i \left (f x +e \right )}+8 a^{2} b \,{\mathrm e}^{4 i \left (f x +e \right )}+4 b^{3} {\mathrm e}^{4 i \left (f x +e \right )}+2 a^{3} {\mathrm e}^{2 i \left (f x +e \right )}-2 b^{3} {\mathrm e}^{2 i \left (f x +e \right )}}{a^{2} f \left (a +b \right )^{2} \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right ) \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) a}{f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) b}{f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {3 b^{2} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 a f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {b^{3} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 a^{2} f \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}\)	\(616\)

Input:

int(cot(f*x+e)^3/(a+b*sec(f*x+e)^2)^2,x,method=_RETURNVERBOSE)
 

Output:

1/f*(-1/2*b^2/(a+b)^3*((a+b)*b/a^2/(b+a*cos(f*x+e)^2)+(3*a+b)/a^2*ln(b+a*c 
os(f*x+e)^2))-1/4/(a+b)^2/(1+cos(f*x+e))+1/2*(-a-3*b)/(a+b)^3*ln(1+cos(f*x 
+e))+1/4/(a+b)^2/(-1+cos(f*x+e))+1/2*(-a-3*b)/(a+b)^3*ln(-1+cos(f*x+e)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 312 vs. \(2 (105) = 210\).

Time = 0.34 (sec) , antiderivative size = 312, normalized size of antiderivative = 2.81 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {a^{3} b + a^{2} b^{2} + a b^{3} + b^{4} + {\left (a^{4} + a^{3} b - a b^{3} - b^{4}\right )} \cos \left (f x + e\right )^{2} - {\left ({\left (3 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (f x + e\right )^{4} - 3 \, a b^{3} - b^{4} - {\left (3 \, a^{2} b^{2} - 2 \, a b^{3} - b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) - 2 \, {\left ({\left (a^{4} + 3 \, a^{3} b\right )} \cos \left (f x + e\right )^{4} - a^{3} b - 3 \, a^{2} b^{2} - {\left (a^{4} + 2 \, a^{3} b - 3 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (f x + e\right )\right )}{2 \, {\left ({\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} f \cos \left (f x + e\right )^{4} - {\left (a^{6} + 2 \, a^{5} b - 2 \, a^{3} b^{3} - a^{2} b^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} f\right )}} \] Input:

integrate(cot(f*x+e)^3/(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")
 

Output:

1/2*(a^3*b + a^2*b^2 + a*b^3 + b^4 + (a^4 + a^3*b - a*b^3 - b^4)*cos(f*x + 
 e)^2 - ((3*a^2*b^2 + a*b^3)*cos(f*x + e)^4 - 3*a*b^3 - b^4 - (3*a^2*b^2 - 
 2*a*b^3 - b^4)*cos(f*x + e)^2)*log(a*cos(f*x + e)^2 + b) - 2*((a^4 + 3*a^ 
3*b)*cos(f*x + e)^4 - a^3*b - 3*a^2*b^2 - (a^4 + 2*a^3*b - 3*a^2*b^2)*cos( 
f*x + e)^2)*log(1/2*sin(f*x + e)))/((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)* 
f*cos(f*x + e)^4 - (a^6 + 2*a^5*b - 2*a^3*b^3 - a^2*b^4)*f*cos(f*x + e)^2 
- (a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*f)
 

Sympy [F]

\[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\int \frac {\cot ^{3}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2}}\, dx \] Input:

integrate(cot(f*x+e)**3/(a+b*sec(f*x+e)**2)**2,x)
 

Output:

Integral(cot(e + f*x)**3/(a + b*sec(e + f*x)**2)**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.73 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {{\left (3 \, a b^{2} + b^{3}\right )} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}} + \frac {{\left (a + 3 \, b\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {a^{3} + a^{2} b - {\left (a^{3} - b^{3}\right )} \sin \left (f x + e\right )^{2}}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \sin \left (f x + e\right )^{4} - {\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} \sin \left (f x + e\right )^{2}}}{2 \, f} \] Input:

integrate(cot(f*x+e)^3/(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")
 

Output:

-1/2*((3*a*b^2 + b^3)*log(a*sin(f*x + e)^2 - a - b)/(a^5 + 3*a^4*b + 3*a^3 
*b^2 + a^2*b^3) + (a + 3*b)*log(sin(f*x + e)^2)/(a^3 + 3*a^2*b + 3*a*b^2 + 
 b^3) - (a^3 + a^2*b - (a^3 - b^3)*sin(f*x + e)^2)/((a^5 + 2*a^4*b + a^3*b 
^2)*sin(f*x + e)^4 - (a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*sin(f*x + e)^2) 
)/f
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.56 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {{\left (3 \, a b^{2} + b^{3}\right )} \log \left ({\left | a \cos \left (f x + e\right )^{2} + b \right |}\right )}{2 \, {\left (a^{5} f + 3 \, a^{4} b f + 3 \, a^{3} b^{2} f + a^{2} b^{3} f\right )}} - \frac {{\left (a + 3 \, b\right )} \log \left ({\left | -\cos \left (f x + e\right )^{2} + 1 \right |}\right )}{2 \, {\left (a^{3} f + 3 \, a^{2} b f + 3 \, a b^{2} f + b^{3} f\right )}} + \frac {a^{2} b + b^{3} + {\left (a^{3} - b^{3}\right )} \cos \left (f x + e\right )^{2}}{2 \, {\left (a \cos \left (f x + e\right )^{2} + b\right )} {\left (\cos \left (f x + e\right )^{2} - 1\right )} {\left (a + b\right )}^{2} a^{2} f} \] Input:

integrate(cot(f*x+e)^3/(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")
 

Output:

-1/2*(3*a*b^2 + b^3)*log(abs(a*cos(f*x + e)^2 + b))/(a^5*f + 3*a^4*b*f + 3 
*a^3*b^2*f + a^2*b^3*f) - 1/2*(a + 3*b)*log(abs(-cos(f*x + e)^2 + 1))/(a^3 
*f + 3*a^2*b*f + 3*a*b^2*f + b^3*f) + 1/2*(a^2*b + b^3 + (a^3 - b^3)*cos(f 
*x + e)^2)/((a*cos(f*x + e)^2 + b)*(cos(f*x + e)^2 - 1)*(a + b)^2*a^2*f)
 

Mupad [B] (verification not implemented)

Time = 15.42 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.44 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,a^2\,f}-\frac {\frac {1}{2\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (a\,b-b^2\right )}{2\,a\,{\left (a+b\right )}^2}}{f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^4+\left (a+b\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2\right )}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a+3\,b\right )}{f\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}-\frac {b^2\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )\,\left (3\,a+b\right )}{2\,a^2\,f\,{\left (a+b\right )}^3} \] Input:

int(cot(e + f*x)^3/(a + b/cos(e + f*x)^2)^2,x)
 

Output:

log(tan(e + f*x)^2 + 1)/(2*a^2*f) - (1/(2*(a + b)) + (tan(e + f*x)^2*(a*b 
- b^2))/(2*a*(a + b)^2))/(f*(tan(e + f*x)^2*(a + b) + b*tan(e + f*x)^4)) - 
 (log(tan(e + f*x))*(a + 3*b))/(f*(3*a*b^2 + 3*a^2*b + a^3 + b^3)) - (b^2* 
log(a + b + b*tan(e + f*x)^2)*(3*a + b))/(2*a^2*f*(a + b)^3)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 1057, normalized size of antiderivative = 9.52 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:

int(cot(f*x+e)^3/(a+b*sec(f*x+e)^2)^2,x)
 

Output:

(2*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**4*a**4 + 6*log(tan((e + f*x) 
/2)**2 + 1)*sin(e + f*x)**4*a**3*b + 6*log(tan((e + f*x)/2)**2 + 1)*sin(e 
+ f*x)**4*a**2*b**2 + 2*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**4*a*b** 
3 - 2*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**2*a**4 - 8*log(tan((e + f 
*x)/2)**2 + 1)*sin(e + f*x)**2*a**3*b - 12*log(tan((e + f*x)/2)**2 + 1)*si 
n(e + f*x)**2*a**2*b**2 - 8*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**2*a 
*b**3 - 2*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**2*b**4 - 3*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**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*b**3 + 3*log(sqrt(a + b)*t 
an((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x 
)**2*a**2*b**2 + 4*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*s 
qrt(a)*tan((e + f*x)/2))*sin(e + f*x)**2*a*b**3 + 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*b* 
*4 - 3*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**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*b**3 + 3* 
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**2 + 4*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*b**3 + log(...