\(\int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx\) [123]

Optimal result

Integrand size = 35, antiderivative size = 122 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {(3 A+4 C) \text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{8 d \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{4 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {(3 A+4 C) \sin (c+d x)}{8 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \] Output:

1/8*(3*A+4*C)*arctanh(sin(d*x+c))*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)+ 
1/4*A*sin(d*x+c)/d/cos(d*x+c)^(7/2)/(b*cos(d*x+c))^(1/2)+1/8*(3*A+4*C)*sin 
(d*x+c)/d/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.66 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {(3 A+4 C) \text {arctanh}(\sin (c+d x)) \cos ^4(c+d x)+\left (2 A+(3 A+4 C) \cos ^2(c+d x)\right ) \sin (c+d x)}{8 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \] Input:

Integrate[(A + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(9/2)*Sqrt[b*Cos[c + d*x]]) 
,x]
 

Output:

((3*A + 4*C)*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^4 + (2*A + (3*A + 4*C)*Cos 
[c + d*x]^2)*Sin[c + d*x])/(8*d*Cos[c + d*x]^(7/2)*Sqrt[b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.75, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2032, 3042, 3491, 3042, 4255, 3042, 4257}

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.

\(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx\)

\(\Big \downarrow \) 2032

\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \left (C \cos ^2(c+d x)+A\right ) \sec ^5(c+d x)dx}{\sqrt {b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx}{\sqrt {b \cos (c+d x)}}\)

\(\Big \downarrow \) 3491

\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} (3 A+4 C) \int \sec ^3(c+d x)dx+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} (3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {b \cos (c+d x)}}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} (3 A+4 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} (3 A+4 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {b \cos (c+d x)}}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{4} (3 A+4 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {b \cos (c+d x)}}\)

Input:

Int[(A + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(9/2)*Sqrt[b*Cos[c + d*x]]),x]
 

Output:

(Sqrt[Cos[c + d*x]]*((A*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((3*A + 4*C)* 
(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4))/Sqr 
t[b*Cos[c + d*x]]
 

Defintions of rubi rules used

rule 2032

Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[a^(m - 1/ 
2)*b^(n + 1/2)*(Sqrt[a*v]/Sqrt[b*v])   Int[v^(m + n)*Fx, x], x] /; FreeQ[{a 
, b, m}, x] &&  !IntegerQ[m] && ILtQ[n - 1/2, 0] && IntegerQ[m + n]
 

rule 3042

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

rule 3491

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x 
_)]^2), x_Symbol] :> Simp[A*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m 
+ 1))), x] + Simp[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1))   Int[(b*Sin[e + f* 
x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]
 

rule 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4257

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

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.43

method	result	size
default	\(\frac {-3 A \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{4}-4 C \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{4}+3 A \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{4}+4 C \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{4}+\left (3 \cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right ) A +4 C \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )}{8 d \cos \left (d x +c \right )^{\frac {7}{2}} \sqrt {b \cos \left (d x +c \right )}}\)	\(174\)
parts	\(-\frac {A \left (3 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{4}-3 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{4}-3 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )-2 \sin \left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{\frac {7}{2}} \sqrt {b \cos \left (d x +c \right )}}+\frac {C \left (\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{2}-\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{2}+\sin \left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{\frac {3}{2}} \sqrt {b \cos \left (d x +c \right )}}\)	\(189\)
risch	\(-\frac {i \left (3 A \,{\mathrm e}^{6 i \left (d x +c \right )}+4 C \,{\mathrm e}^{6 i \left (d x +c \right )}+11 A \,{\mathrm e}^{4 i \left (d x +c \right )}+4 C \,{\mathrm e}^{4 i \left (d x +c \right )}-11 A \,{\mathrm e}^{2 i \left (d x +c \right )}-4 C \,{\mathrm e}^{2 i \left (d x +c \right )}-3 A -4 C \right )}{8 \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} d}+\frac {\sqrt {\cos \left (d x +c \right )}\, \left (3 A +4 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 \sqrt {b \cos \left (d x +c \right )}\, d}-\frac {\sqrt {\cos \left (d x +c \right )}\, \left (3 A +4 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 \sqrt {b \cos \left (d x +c \right )}\, d}\)	\(204\)

Input:

int((A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(b*cos(d*x+c))^(1/2),x,method=_RET 
URNVERBOSE)
 

Output:

1/8/d*(-3*A*ln(-cot(d*x+c)+csc(d*x+c)-1)*cos(d*x+c)^4-4*C*ln(-cot(d*x+c)+c 
sc(d*x+c)-1)*cos(d*x+c)^4+3*A*ln(-cot(d*x+c)+csc(d*x+c)+1)*cos(d*x+c)^4+4* 
C*ln(-cot(d*x+c)+csc(d*x+c)+1)*cos(d*x+c)^4+(3*cos(d*x+c)^2+2)*sin(d*x+c)* 
A+4*C*cos(d*x+c)^2*sin(d*x+c))/cos(d*x+c)^(7/2)/(b*cos(d*x+c))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.14 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\left [\frac {{\left (3 \, A + 4 \, C\right )} \sqrt {b} \cos \left (d x + c\right )^{5} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, {\left ({\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{16 \, b d \cos \left (d x + c\right )^{5}}, -\frac {{\left (3 \, A + 4 \, C\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{5} - {\left ({\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{8 \, b d \cos \left (d x + c\right )^{5}}\right ] \] Input:

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(b*cos(d*x+c))^(1/2),x, algo 
rithm="fricas")
 

Output:

[1/16*((3*A + 4*C)*sqrt(b)*cos(d*x + c)^5*log(-(b*cos(d*x + c)^3 - 2*sqrt( 
b*cos(d*x + c))*sqrt(b)*sqrt(cos(d*x + c))*sin(d*x + c) - 2*b*cos(d*x + c) 
)/cos(d*x + c)^3) + 2*((3*A + 4*C)*cos(d*x + c)^2 + 2*A)*sqrt(b*cos(d*x + 
c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b*d*cos(d*x + c)^5), -1/8*((3*A + 4* 
C)*sqrt(-b)*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(cos( 
d*x + c))))*cos(d*x + c)^5 - ((3*A + 4*C)*cos(d*x + c)^2 + 2*A)*sqrt(b*cos 
(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b*d*cos(d*x + c)^5)]
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate((A+C*cos(d*x+c)**2)/cos(d*x+c)**(9/2)/(b*cos(d*x+c))**(1/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2318 vs. \(2 (104) = 208\).

Time = 0.31 (sec) , antiderivative size = 2318, normalized size of antiderivative = 19.00 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(b*cos(d*x+c))^(1/2),x, algo 
rithm="maxima")
 

Output:

-1/16*((12*(sin(8*d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4 
*sin(2*d*x + 2*c))*cos(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 
44*(sin(8*d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*d 
*x + 2*c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 44*(sin( 
8*d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*d*x + 2*c 
))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 12*(sin(8*d*x + 
8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*d*x + 2*c))*cos(1 
/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 3*(2*(4*cos(6*d*x + 6*c) 
 + 6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(8*d*x + 8*c) + cos(8*d 
*x + 8*c)^2 + 8*(6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 
6*c) + 16*cos(6*d*x + 6*c)^2 + 12*(4*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c 
) + 36*cos(4*d*x + 4*c)^2 + 16*cos(2*d*x + 2*c)^2 + 4*(2*sin(6*d*x + 6*c) 
+ 3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + sin(8*d*x + 
8*c)^2 + 16*(3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 1 
6*sin(6*d*x + 6*c)^2 + 36*sin(4*d*x + 4*c)^2 + 48*sin(4*d*x + 4*c)*sin(2*d 
*x + 2*c) + 16*sin(2*d*x + 2*c)^2 + 8*cos(2*d*x + 2*c) + 1)*log(cos(1/2*ar 
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 
 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d* 
x + 2*c))) + 1) + 3*(2*(4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2* 
d*x + 2*c) + 1)*cos(8*d*x + 8*c) + cos(8*d*x + 8*c)^2 + 8*(6*cos(4*d*x ...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(b*cos(d*x+c))^(1/2),x, algo 
rithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\cos \left (c+d\,x\right )}^{9/2}\,\sqrt {b\,\cos \left (c+d\,x\right )}} \,d x \] Input:

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^(9/2)*(b*cos(c + d*x))^(1/2)),x)
 

Output:

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^(9/2)*(b*cos(c + d*x))^(1/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 317, normalized size of antiderivative = 2.60 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {\sqrt {b}\, \left (-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a -4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} c +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a +8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} c -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a -4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a +4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} c -6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a -8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} c +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a +4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c -3 \sin \left (d x +c \right )^{3} a -4 \sin \left (d x +c \right )^{3} c +5 \sin \left (d x +c \right ) a +4 \sin \left (d x +c \right ) c \right )}{8 b d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:

int((A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(b*cos(d*x+c))^(1/2),x)
 

Output:

(sqrt(b)*( - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a - 4*log(tan((c 
+ d*x)/2) - 1)*sin(c + d*x)**4*c + 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x 
)**2*a + 8*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*c - 3*log(tan((c + d* 
x)/2) - 1)*a - 4*log(tan((c + d*x)/2) - 1)*c + 3*log(tan((c + d*x)/2) + 1) 
*sin(c + d*x)**4*a + 4*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*c - 6*log 
(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a - 8*log(tan((c + d*x)/2) + 1)*sin 
(c + d*x)**2*c + 3*log(tan((c + d*x)/2) + 1)*a + 4*log(tan((c + d*x)/2) + 
1)*c - 3*sin(c + d*x)**3*a - 4*sin(c + d*x)**3*c + 5*sin(c + d*x)*a + 4*si 
n(c + d*x)*c))/(8*b*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))