\(\int \frac {\cot ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [196]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 21, antiderivative size = 145 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {3 b \csc (c+d x)}{a^4 d}-\frac {\csc ^2(c+d x)}{2 a^3 d}-\frac {\left (a^2-6 b^2\right ) \log (\sin (c+d x))}{a^5 d}+\frac {\left (a^2-6 b^2\right ) \log (a+b \sin (c+d x))}{a^5 d}-\frac {a^2-b^2}{2 a^3 d (a+b \sin (c+d x))^2}-\frac {a^2-3 b^2}{a^4 d (a+b \sin (c+d x))} \] Output:

3*b*csc(d*x+c)/a^4/d-1/2*csc(d*x+c)^2/a^3/d-(a^2-6*b^2)*ln(sin(d*x+c))/a^5 
/d+(a^2-6*b^2)*ln(a+b*sin(d*x+c))/a^5/d-1/2*(a^2-b^2)/a^3/d/(a+b*sin(d*x+c 
))^2-(a^2-3*b^2)/a^4/d/(a+b*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.83 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {-6 a b \csc (c+d x)+a^2 \csc ^2(c+d x)+2 \left (a^2-6 b^2\right ) \log (\sin (c+d x))-2 \left (a^2-6 b^2\right ) \log (a+b \sin (c+d x))+\frac {a^2 (a-b) (a+b)}{(a+b \sin (c+d x))^2}+\frac {2 a \left (a^2-3 b^2\right )}{a+b \sin (c+d x)}}{2 a^5 d} \] Input:

Integrate[Cot[c + d*x]^3/(a + b*Sin[c + d*x])^3,x]
 

Output:

-1/2*(-6*a*b*Csc[c + d*x] + a^2*Csc[c + d*x]^2 + 2*(a^2 - 6*b^2)*Log[Sin[c 
 + d*x]] - 2*(a^2 - 6*b^2)*Log[a + b*Sin[c + d*x]] + (a^2*(a - b)*(a + b)) 
/(a + b*Sin[c + d*x])^2 + (2*a*(a^2 - 3*b^2))/(a + b*Sin[c + d*x]))/(a^5*d 
)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3200, 522, 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.

\(\displaystyle \int \frac {\cot ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (c+d x)^3 (a+b \sin (c+d x))^3}dx\)

\(\Big \downarrow \) 3200

\(\displaystyle \frac {\int \frac {\csc ^3(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )}{b^3 (a+b \sin (c+d x))^3}d(b \sin (c+d x))}{d}\)

\(\Big \downarrow \) 522

\(\displaystyle \frac {\int \left (\frac {\csc ^3(c+d x)}{a^3 b}-\frac {3 \csc ^2(c+d x)}{a^4}+\frac {\left (6 b^2-a^2\right ) \csc (c+d x)}{a^5 b}+\frac {a^2-6 b^2}{a^5 (a+b \sin (c+d x))}+\frac {a^2-3 b^2}{a^4 (a+b \sin (c+d x))^2}+\frac {a^2-b^2}{a^3 (a+b \sin (c+d x))^3}\right )d(b \sin (c+d x))}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {3 b \csc (c+d x)}{a^4}-\frac {\csc ^2(c+d x)}{2 a^3}-\frac {\left (a^2-6 b^2\right ) \log (b \sin (c+d x))}{a^5}+\frac {\left (a^2-6 b^2\right ) \log (a+b \sin (c+d x))}{a^5}-\frac {a^2-3 b^2}{a^4 (a+b \sin (c+d x))}-\frac {a^2-b^2}{2 a^3 (a+b \sin (c+d x))^2}}{d}\)

Input:

Int[Cot[c + d*x]^3/(a + b*Sin[c + d*x])^3,x]
 

Output:

((3*b*Csc[c + d*x])/a^4 - Csc[c + d*x]^2/(2*a^3) - ((a^2 - 6*b^2)*Log[b*Si 
n[c + d*x]])/a^5 + ((a^2 - 6*b^2)*Log[a + b*Sin[c + d*x]])/a^5 - (a^2 - b^ 
2)/(2*a^3*(a + b*Sin[c + d*x])^2) - (a^2 - 3*b^2)/(a^4*(a + b*Sin[c + d*x] 
)))/d
 

Defintions of rubi rules used

rule 522
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. 
), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], 
x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
 

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 3200
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p 
_.), x_Symbol] :> Simp[1/f   Subst[Int[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1) 
/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b 
^2, 0] && IntegerQ[(p + 1)/2]
 
Maple [A] (verified)

Time = 5.35 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.90

method result size
derivativedivides \(\frac {-\frac {1}{2 a^{3} \sin \left (d x +c \right )^{2}}+\frac {\left (-a^{2}+6 b^{2}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{5}}+\frac {3 b}{a^{4} \sin \left (d x +c \right )}+\frac {\left (a^{2}-6 b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{5}}-\frac {a^{2}-3 b^{2}}{a^{4} \left (a +b \sin \left (d x +c \right )\right )}-\frac {a^{2}-b^{2}}{2 a^{3} \left (a +b \sin \left (d x +c \right )\right )^{2}}}{d}\) \(131\)
default \(\frac {-\frac {1}{2 a^{3} \sin \left (d x +c \right )^{2}}+\frac {\left (-a^{2}+6 b^{2}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{5}}+\frac {3 b}{a^{4} \sin \left (d x +c \right )}+\frac {\left (a^{2}-6 b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{5}}-\frac {a^{2}-3 b^{2}}{a^{4} \left (a +b \sin \left (d x +c \right )\right )}-\frac {a^{2}-b^{2}}{2 a^{3} \left (a +b \sin \left (d x +c \right )\right )^{2}}}{d}\) \(131\)
risch \(-\frac {2 i \left (3 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-18 i {\mathrm e}^{6 i \left (d x +c \right )} a \,b^{2}+a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}-6 b^{3} {\mathrm e}^{7 i \left (d x +c \right )}-10 i {\mathrm e}^{4 i \left (d x +c \right )} a^{3}+36 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+5 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}+18 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+3 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-18 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-5 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}-18 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+6 b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )} b -b +2 i {\mathrm e}^{i \left (d x +c \right )} a \right )^{2} d \,a^{4}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{a^{3} d}-\frac {6 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) b^{2}}{a^{5} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{3} d}+\frac {6 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{2}}{a^{5} d}\) \(380\)

Input:

int(cot(d*x+c)^3/(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(-1/2/a^3/sin(d*x+c)^2+(-a^2+6*b^2)/a^5*ln(sin(d*x+c))+3/a^4*b/sin(d*x 
+c)+(a^2-6*b^2)/a^5*ln(a+b*sin(d*x+c))-(a^2-3*b^2)/a^4/(a+b*sin(d*x+c))-1/ 
2*(a^2-b^2)/a^3/(a+b*sin(d*x+c))^2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 404 vs. \(2 (141) = 282\).

Time = 0.12 (sec) , antiderivative size = 404, normalized size of antiderivative = 2.79 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {4 \, a^{4} - 18 \, a^{2} b^{2} - 3 \, {\left (a^{4} - 6 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left ({\left (a^{2} b^{2} - 6 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{4} - 5 \, a^{2} b^{2} - 6 \, b^{4} - {\left (a^{4} - 4 \, a^{2} b^{2} - 12 \, b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{3} b - 6 \, a b^{3} - {\left (a^{3} b - 6 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 2 \, {\left ({\left (a^{2} b^{2} - 6 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{4} - 5 \, a^{2} b^{2} - 6 \, b^{4} - {\left (a^{4} - 4 \, a^{2} b^{2} - 12 \, b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{3} b - 6 \, a b^{3} - {\left (a^{3} b - 6 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 2 \, {\left (a^{3} b + 6 \, a b^{3} + {\left (a^{3} b - 6 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{5} b^{2} d \cos \left (d x + c\right )^{4} - {\left (a^{7} + 2 \, a^{5} b^{2}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{7} + a^{5} b^{2}\right )} d - 2 \, {\left (a^{6} b d \cos \left (d x + c\right )^{2} - a^{6} b d\right )} \sin \left (d x + c\right )\right )}} \] Input:

integrate(cot(d*x+c)^3/(a+b*sin(d*x+c))^3,x, algorithm="fricas")
 

Output:

-1/2*(4*a^4 - 18*a^2*b^2 - 3*(a^4 - 6*a^2*b^2)*cos(d*x + c)^2 - 2*((a^2*b^ 
2 - 6*b^4)*cos(d*x + c)^4 + a^4 - 5*a^2*b^2 - 6*b^4 - (a^4 - 4*a^2*b^2 - 1 
2*b^4)*cos(d*x + c)^2 + 2*(a^3*b - 6*a*b^3 - (a^3*b - 6*a*b^3)*cos(d*x + c 
)^2)*sin(d*x + c))*log(b*sin(d*x + c) + a) + 2*((a^2*b^2 - 6*b^4)*cos(d*x 
+ c)^4 + a^4 - 5*a^2*b^2 - 6*b^4 - (a^4 - 4*a^2*b^2 - 12*b^4)*cos(d*x + c) 
^2 + 2*(a^3*b - 6*a*b^3 - (a^3*b - 6*a*b^3)*cos(d*x + c)^2)*sin(d*x + c))* 
log(-1/2*sin(d*x + c)) - 2*(a^3*b + 6*a*b^3 + (a^3*b - 6*a*b^3)*cos(d*x + 
c)^2)*sin(d*x + c))/(a^5*b^2*d*cos(d*x + c)^4 - (a^7 + 2*a^5*b^2)*d*cos(d* 
x + c)^2 + (a^7 + a^5*b^2)*d - 2*(a^6*b*d*cos(d*x + c)^2 - a^6*b*d)*sin(d* 
x + c))
 

Sympy [F]

\[ \int \frac {\cot ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int \frac {\cot ^{3}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \] Input:

integrate(cot(d*x+c)**3/(a+b*sin(d*x+c))**3,x)
 

Output:

Integral(cot(c + d*x)**3/(a + b*sin(c + d*x))**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.08 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {4 \, a^{2} b \sin \left (d x + c\right ) - 2 \, {\left (a^{2} b - 6 \, b^{3}\right )} \sin \left (d x + c\right )^{3} - a^{3} - 3 \, {\left (a^{3} - 6 \, a b^{2}\right )} \sin \left (d x + c\right )^{2}}{a^{4} b^{2} \sin \left (d x + c\right )^{4} + 2 \, a^{5} b \sin \left (d x + c\right )^{3} + a^{6} \sin \left (d x + c\right )^{2}} + \frac {2 \, {\left (a^{2} - 6 \, b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{5}} - \frac {2 \, {\left (a^{2} - 6 \, b^{2}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{5}}}{2 \, d} \] Input:

integrate(cot(d*x+c)^3/(a+b*sin(d*x+c))^3,x, algorithm="maxima")
 

Output:

1/2*((4*a^2*b*sin(d*x + c) - 2*(a^2*b - 6*b^3)*sin(d*x + c)^3 - a^3 - 3*(a 
^3 - 6*a*b^2)*sin(d*x + c)^2)/(a^4*b^2*sin(d*x + c)^4 + 2*a^5*b*sin(d*x + 
c)^3 + a^6*sin(d*x + c)^2) + 2*(a^2 - 6*b^2)*log(b*sin(d*x + c) + a)/a^5 - 
 2*(a^2 - 6*b^2)*log(sin(d*x + c))/a^5)/d
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.09 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {{\left (a^{2} - 6 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{5} d} + \frac {{\left (a^{2} b - 6 \, b^{3}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{5} b d} - \frac {2 \, a^{2} b \sin \left (d x + c\right )^{3} - 12 \, b^{3} \sin \left (d x + c\right )^{3} + 3 \, a^{3} \sin \left (d x + c\right )^{2} - 18 \, a b^{2} \sin \left (d x + c\right )^{2} - 4 \, a^{2} b \sin \left (d x + c\right ) + a^{3}}{2 \, {\left (b \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right )\right )}^{2} a^{4} d} \] Input:

integrate(cot(d*x+c)^3/(a+b*sin(d*x+c))^3,x, algorithm="giac")
 

Output:

-(a^2 - 6*b^2)*log(abs(sin(d*x + c)))/(a^5*d) + (a^2*b - 6*b^3)*log(abs(b* 
sin(d*x + c) + a))/(a^5*b*d) - 1/2*(2*a^2*b*sin(d*x + c)^3 - 12*b^3*sin(d* 
x + c)^3 + 3*a^3*sin(d*x + c)^2 - 18*a*b^2*sin(d*x + c)^2 - 4*a^2*b*sin(d* 
x + c) + a^3)/((b*sin(d*x + c)^2 + a*sin(d*x + c))^2*a^4*d)
 

Mupad [B] (verification not implemented)

Time = 17.59 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.30 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (22\,a\,b^2-a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (26\,a^2\,b-8\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (22\,a^2\,b-32\,b^3\right )-\frac {a^3}{2}+4\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^4-96\,a^2\,b^2+112\,b^4\right )}{2\,a}}{d\,\left (4\,a^6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\,a^6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (8\,a^6+16\,a^4\,b^2\right )+16\,a^5\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+16\,a^5\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}+\frac {3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^4\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2-6\,b^2\right )}{a^5\,d}+\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (a^2-6\,b^2\right )}{a^5\,d} \] Input:

int(cot(c + d*x)^3/(a + b*sin(c + d*x))^3,x)
 

Output:

(tan(c/2 + (d*x)/2)^2*(22*a*b^2 - a^3) + tan(c/2 + (d*x)/2)^3*(26*a^2*b - 
8*b^3) + tan(c/2 + (d*x)/2)^5*(22*a^2*b - 32*b^3) - a^3/2 + 4*a^2*b*tan(c/ 
2 + (d*x)/2) - (tan(c/2 + (d*x)/2)^4*(a^4 + 112*b^4 - 96*a^2*b^2))/(2*a))/ 
(d*(4*a^6*tan(c/2 + (d*x)/2)^2 + 4*a^6*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d 
*x)/2)^4*(8*a^6 + 16*a^4*b^2) + 16*a^5*b*tan(c/2 + (d*x)/2)^3 + 16*a^5*b*t 
an(c/2 + (d*x)/2)^5)) - tan(c/2 + (d*x)/2)^2/(8*a^3*d) + (3*b*tan(c/2 + (d 
*x)/2))/(2*a^4*d) - (log(tan(c/2 + (d*x)/2))*(a^2 - 6*b^2))/(a^5*d) + (log 
(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2)*(a^2 - 6*b^2))/(a^5* 
d)
 

Reduce [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 534, normalized size of antiderivative = 3.68 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

int(cot(d*x+c)^3/(a+b*sin(d*x+c))^3,x)
 

Output:

(4*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**4*a 
**2*b**2 - 24*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c 
+ d*x)**4*b**4 + 8*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*s 
in(c + d*x)**3*a**3*b - 48*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)* 
b + a)*sin(c + d*x)**3*a*b**3 + 4*log(tan((c + d*x)/2)**2*a + 2*tan((c + d 
*x)/2)*b + a)*sin(c + d*x)**2*a**4 - 24*log(tan((c + d*x)/2)**2*a + 2*tan( 
(c + d*x)/2)*b + a)*sin(c + d*x)**2*a**2*b**2 - 4*log(tan((c + d*x)/2))*si 
n(c + d*x)**4*a**2*b**2 + 24*log(tan((c + d*x)/2))*sin(c + d*x)**4*b**4 - 
8*log(tan((c + d*x)/2))*sin(c + d*x)**3*a**3*b + 48*log(tan((c + d*x)/2))* 
sin(c + d*x)**3*a*b**3 - 4*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**4 + 24 
*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**2*b**2 - sin(c + d*x)**4*a**2*b* 
*2 - 12*sin(c + d*x)**4*b**4 - 6*sin(c + d*x)**3*a**3*b - 7*sin(c + d*x)** 
2*a**4 + 24*sin(c + d*x)**2*a**2*b**2 + 8*sin(c + d*x)*a**3*b - 2*a**4)/(4 
*sin(c + d*x)**2*a**5*d*(sin(c + d*x)**2*b**2 + 2*sin(c + d*x)*a*b + a**2) 
)