\(\int \cot (c+d x) (a+b \tan (c+d x))^4 \, dx\) [449]

Optimal result

Integrand size = 19, antiderivative size = 92 \[ \int \cot (c+d x) (a+b \tan (c+d x))^4 \, dx=4 a b \left (a^2-b^2\right ) x-\frac {b^2 \left (6 a^2-b^2\right ) \log (\cos (c+d x))}{d}+\frac {a^4 \log (\sin (c+d x))}{d}+\frac {3 a b^3 \tan (c+d x)}{d}+\frac {b^2 (a+b \tan (c+d x))^2}{2 d} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.02 \[ \int \cot (c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {-(a+i b)^4 \log (i-\tan (c+d x))+2 a^4 \log (\tan (c+d x))-(a-i b)^4 \log (i+\tan (c+d x))+6 a b^3 \tan (c+d x)+b^2 (a+b \tan (c+d x))^2}{2 d} \] Input:

Integrate[Cot[c + d*x]*(a + b*Tan[c + d*x])^4,x]
 

Output:

(-((a + I*b)^4*Log[I - Tan[c + d*x]]) + 2*a^4*Log[Tan[c + d*x]] - (a - I*b 
)^4*Log[I + Tan[c + d*x]] + 6*a*b^3*Tan[c + d*x] + b^2*(a + b*Tan[c + d*x] 
)^2)/(2*d)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.579, Rules used = {3042, 4049, 27, 3042, 4120, 25, 3042, 4107, 3042, 25, 3956}

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.

\(\displaystyle \int \cot (c+d x) (a+b \tan (c+d x))^4 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (c+d x))^4}{\tan (c+d x)}dx\)

\(\Big \downarrow \) 4049

\(\displaystyle \frac {1}{2} \int 2 \cot (c+d x) (a+b \tan (c+d x)) \left (a^3+3 b^2 \tan ^2(c+d x) a+b \left (3 a^2-b^2\right ) \tan (c+d x)\right )dx+\frac {b^2 (a+b \tan (c+d x))^2}{2 d}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4120

\(\displaystyle -\int -\cot (c+d x) \left (a^4+4 b \left (a^2-b^2\right ) \tan (c+d x) a+b^2 \left (6 a^2-b^2\right ) \tan ^2(c+d x)\right )dx+\frac {3 a b^3 \tan (c+d x)}{d}+\frac {b^2 (a+b \tan (c+d x))^2}{2 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \int \cot (c+d x) \left (a^4+4 b \left (a^2-b^2\right ) \tan (c+d x) a+b^2 \left (6 a^2-b^2\right ) \tan ^2(c+d x)\right )dx+\frac {3 a b^3 \tan (c+d x)}{d}+\frac {b^2 (a+b \tan (c+d x))^2}{2 d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4107

\(\displaystyle a^4 \int \cot (c+d x)dx+b^2 \left (6 a^2-b^2\right ) \int \tan (c+d x)dx+4 a b x \left (a^2-b^2\right )+\frac {3 a b^3 \tan (c+d x)}{d}+\frac {b^2 (a+b \tan (c+d x))^2}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle a^4 \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx+b^2 \left (6 a^2-b^2\right ) \int \tan (c+d x)dx+4 a b x \left (a^2-b^2\right )+\frac {3 a b^3 \tan (c+d x)}{d}+\frac {b^2 (a+b \tan (c+d x))^2}{2 d}\)

\(\Big \downarrow \) 25

\(\displaystyle a^4 \left (-\int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx\right )+b^2 \left (6 a^2-b^2\right ) \int \tan (c+d x)dx+4 a b x \left (a^2-b^2\right )+\frac {3 a b^3 \tan (c+d x)}{d}+\frac {b^2 (a+b \tan (c+d x))^2}{2 d}\)

\(\Big \downarrow \) 3956

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

Input:

Int[Cot[c + d*x]*(a + b*Tan[c + d*x])^4,x]
 

Output:

4*a*b*(a^2 - b^2)*x - (b^2*(6*a^2 - b^2)*Log[Cos[c + d*x]])/d + (a^4*Log[- 
Sin[c + d*x]])/d + (3*a*b^3*Tan[c + d*x])/d + (b^2*(a + b*Tan[c + d*x])^2) 
/(2*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

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

rule 3956

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d 
*x], x]]/d, x] /; FreeQ[{c, d}, x]
 

rule 4049

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c 
+ d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) 
 Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n 
- 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ 
e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], 
x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 
, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I 
ntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) 
)
 

rule 4107

Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2 
)/tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[B*x, x] + (Simp[A   Int[1/Tan[ 
e + f*x], x], x] + Simp[C   Int[Tan[e + f*x], x], x]) /; FreeQ[{e, f, A, B, 
 C}, x] && NeQ[A, C]
 

rule 4120

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.) 
*(x_)])^(n_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[b*C*Tan[e + f*x]*((c + d*Tan[e + f*x])^(n + 
 1)/(d*f*(n + 2))), x] - Simp[1/(d*(n + 2))   Int[(c + d*Tan[e + f*x])^n*Si 
mp[b*c*C - a*A*d*(n + 2) - (A*b + a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C* 
d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, 
b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] && 
  !LtQ[n, -1]
 

Maple [A] (verified)

Time = 0.93 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.01

method	result	size
norman	\(\left (4 a^{3} b -4 a \,b^{3}\right ) x +\frac {b^{4} \tan \left (d x +c \right )^{2}}{2 d}+\frac {4 a \,b^{3} \tan \left (d x +c \right )}{d}+\frac {a^{4} \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {\left (a^{4}-6 b^{2} a^{2}+b^{4}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\)	\(93\)
derivativedivides	\(\frac {\frac {b^{4} \tan \left (d x +c \right )^{2}}{2}+4 a \,b^{3} \tan \left (d x +c \right )+a^{4} \ln \left (\tan \left (d x +c \right )\right )+\frac {\left (-a^{4}+6 b^{2} a^{2}-b^{4}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (4 a^{3} b -4 a \,b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(95\)
default	\(\frac {\frac {b^{4} \tan \left (d x +c \right )^{2}}{2}+4 a \,b^{3} \tan \left (d x +c \right )+a^{4} \ln \left (\tan \left (d x +c \right )\right )+\frac {\left (-a^{4}+6 b^{2} a^{2}-b^{4}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (4 a^{3} b -4 a \,b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\)	\(95\)
parallelrisch	\(\frac {a^{4} \left (2 \ln \left (\tan \left (d x +c \right )\right )-\ln \left (\sec \left (d x +c \right )^{2}\right )\right )+6 \ln \left (\sec \left (d x +c \right )^{2}\right ) a^{2} b^{2}-\ln \left (\sec \left (d x +c \right )^{2}\right ) b^{4}+b^{4} \tan \left (d x +c \right )^{2}+8 a^{3} b d x -8 a \,b^{3} d x +8 a \,b^{3} \tan \left (d x +c \right )}{2 d}\)	\(103\)
risch	\(4 a^{3} b x -4 a \,b^{3} x -i a^{4} x +6 i a^{2} b^{2} x -i b^{4} x +\frac {12 i a^{2} b^{2} c}{d}-\frac {2 i b^{4} c}{d}-\frac {2 i a^{4} c}{d}+\frac {2 i b^{3} \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+4 a \,{\mathrm e}^{2 i \left (d x +c \right )}+4 a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {6 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) b^{2} a^{2}}{d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) b^{4}}{d}+\frac {a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(185\)

Input:

int(cot(d*x+c)*(a+b*tan(d*x+c))^4,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.10 \[ \int \cot (c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {b^{4} \tan \left (d x + c\right )^{2} + a^{4} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) + 8 \, a b^{3} \tan \left (d x + c\right ) + 8 \, {\left (a^{3} b - a b^{3}\right )} d x - {\left (6 \, a^{2} b^{2} - b^{4}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, d} \] Input:

integrate(cot(d*x+c)*(a+b*tan(d*x+c))^4,x, algorithm="fricas")
 

Output:

1/2*(b^4*tan(d*x + c)^2 + a^4*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1)) + 8 
*a*b^3*tan(d*x + c) + 8*(a^3*b - a*b^3)*d*x - (6*a^2*b^2 - b^4)*log(1/(tan 
(d*x + c)^2 + 1)))/d
 

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.45 \[ \int \cot (c+d x) (a+b \tan (c+d x))^4 \, dx=\begin {cases} - \frac {a^{4} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a^{4} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 4 a^{3} b x + \frac {3 a^{2} b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - 4 a b^{3} x + \frac {4 a b^{3} \tan {\left (c + d x \right )}}{d} - \frac {b^{4} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {b^{4} \tan ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\left (c \right )}\right )^{4} \cot {\left (c \right )} & \text {otherwise} \end {cases} \] Input:

integrate(cot(d*x+c)*(a+b*tan(d*x+c))**4,x)
 

Output:

Piecewise((-a**4*log(tan(c + d*x)**2 + 1)/(2*d) + a**4*log(tan(c + d*x))/d 
 + 4*a**3*b*x + 3*a**2*b**2*log(tan(c + d*x)**2 + 1)/d - 4*a*b**3*x + 4*a* 
b**3*tan(c + d*x)/d - b**4*log(tan(c + d*x)**2 + 1)/(2*d) + b**4*tan(c + d 
*x)**2/(2*d), Ne(d, 0)), (x*(a + b*tan(c))**4*cot(c), True))
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.97 \[ \int \cot (c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {b^{4} \tan \left (d x + c\right )^{2} + 2 \, a^{4} \log \left (\tan \left (d x + c\right )\right ) + 8 \, a b^{3} \tan \left (d x + c\right ) + 8 \, {\left (a^{3} b - a b^{3}\right )} {\left (d x + c\right )} - {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} \] Input:

integrate(cot(d*x+c)*(a+b*tan(d*x+c))^4,x, algorithm="maxima")
 

Output:

1/2*(b^4*tan(d*x + c)^2 + 2*a^4*log(tan(d*x + c)) + 8*a*b^3*tan(d*x + c) + 
 8*(a^3*b - a*b^3)*(d*x + c) - (a^4 - 6*a^2*b^2 + b^4)*log(tan(d*x + c)^2 
+ 1))/d
 

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.10 \[ \int \cot (c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {a^{4} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} + \frac {4 \, {\left (a^{3} b - a b^{3}\right )} {\left (d x + c\right )}}{d} - \frac {{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} + \frac {b^{4} d \tan \left (d x + c\right )^{2} + 8 \, a b^{3} d \tan \left (d x + c\right )}{2 \, d^{2}} \] Input:

integrate(cot(d*x+c)*(a+b*tan(d*x+c))^4,x, algorithm="giac")
 

Output:

a^4*log(abs(tan(d*x + c)))/d + 4*(a^3*b - a*b^3)*(d*x + c)/d - 1/2*(a^4 - 
6*a^2*b^2 + b^4)*log(tan(d*x + c)^2 + 1)/d + 1/2*(b^4*d*tan(d*x + c)^2 + 8 
*a*b^3*d*tan(d*x + c))/d^2
 

Mupad [B] (verification not implemented)

Time = 1.16 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00 \[ \int \cot (c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {b^4\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^4}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,{\left (a+b\,1{}\mathrm {i}\right )}^4}{2\,d}+\frac {a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}+\frac {4\,a\,b^3\,\mathrm {tan}\left (c+d\,x\right )}{d} \] Input:

int(cot(c + d*x)*(a + b*tan(c + d*x))^4,x)
 

Output:

(b^4*tan(c + d*x)^2)/(2*d) - (log(tan(c + d*x) + 1i)*(a*1i + b)^4)/(2*d) - 
 (log(tan(c + d*x) - 1i)*(a + b*1i)^4)/(2*d) + (a^4*log(tan(c + d*x)))/d + 
 (4*a*b^3*tan(c + d*x))/d
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 459, normalized size of antiderivative = 4.99 \[ \int \cot (c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {-8 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{3}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} a^{4}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} a^{2} b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} b^{4}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{4}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{2} b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b^{4}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2} b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{4}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{4}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2} b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{4}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{4}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a^{4}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{4}+8 \sin \left (d x +c \right )^{2} a^{3} b d x -8 \sin \left (d x +c \right )^{2} a \,b^{3} d x -\sin \left (d x +c \right )^{2} b^{4}-8 a^{3} b d x +8 a \,b^{3} d x}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

int(cot(d*x+c)*(a+b*tan(d*x+c))^4,x)
 

Output:

( - 8*cos(c + d*x)*sin(c + d*x)*a*b**3 - 2*log(tan((c + d*x)/2)**2 + 1)*si 
n(c + d*x)**2*a**4 + 12*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a**2* 
b**2 - 2*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*b**4 + 2*log(tan((c 
+ d*x)/2)**2 + 1)*a**4 - 12*log(tan((c + d*x)/2)**2 + 1)*a**2*b**2 + 2*log 
(tan((c + d*x)/2)**2 + 1)*b**4 - 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x) 
**2*a**2*b**2 + 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**4 + 12*log( 
tan((c + d*x)/2) - 1)*a**2*b**2 - 2*log(tan((c + d*x)/2) - 1)*b**4 - 12*lo 
g(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b**2 + 2*log(tan((c + d*x)/2) 
 + 1)*sin(c + d*x)**2*b**4 + 12*log(tan((c + d*x)/2) + 1)*a**2*b**2 - 2*lo 
g(tan((c + d*x)/2) + 1)*b**4 + 2*log(tan((c + d*x)/2))*sin(c + d*x)**2*a** 
4 - 2*log(tan((c + d*x)/2))*a**4 + 8*sin(c + d*x)**2*a**3*b*d*x - 8*sin(c 
+ d*x)**2*a*b**3*d*x - sin(c + d*x)**2*b**4 - 8*a**3*b*d*x + 8*a*b**3*d*x) 
/(2*d*(sin(c + d*x)**2 - 1))