\(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx\) [328]

Optimal result

Integrand size = 43, antiderivative size = 120 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\frac {(A+2 C) \text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{2 b d \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{2 b d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x)}{b d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.69 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\frac {2 C \coth ^{-1}(\sin (c+d x)) \cos ^2(c+d x)+A \text {arctanh}(\sin (c+d x)) \cos ^2(c+d x)+(A+2 B \cos (c+d x)) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2}} \] Input:

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(b*C 
os[c + d*x])^(3/2)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.67, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.209, Rules used = {2032, 3042, 3500, 3042, 3227, 3042, 4254, 24, 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.

\(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 2032

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx}{b \sqrt {b \cos (c+d x)}}\)

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4257

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

Input:

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(b*Cos[c + 
 d*x])^(3/2)),x]
 

Output:

(Sqrt[Cos[c + d*x]]*((A*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (((A + 2*C)*Arc 
Tanh[Sin[c + d*x]])/d + (2*B*Tan[c + d*x])/d)/2))/(b*Sqrt[b*Cos[c + d*x]])
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

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 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3500

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 
 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 

rule 4254

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

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 = 0.41 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.12

method	result	size
default	\(-\frac {A \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{2}-A \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{2}+4 C \,\operatorname {arctanh}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \cos \left (d x +c \right )^{2}-2 B \sin \left (d x +c \right ) \cos \left (d x +c \right )-A \sin \left (d x +c \right )}{2 b d \cos \left (d x +c \right )^{\frac {3}{2}} \sqrt {b \cos \left (d x +c \right )}}\)	\(134\)
risch	\(-\frac {i \left (A \,{\mathrm e}^{2 i \left (d x +c \right )}-A -4 B \cos \left (d x +c \right )\right )}{2 b \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {\sqrt {\cos \left (d x +c \right )}\, \left (A +2 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 b \sqrt {b \cos \left (d x +c \right )}\, d}+\frac {\sqrt {\cos \left (d x +c \right )}\, \left (A +2 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 b \sqrt {b \cos \left (d x +c \right )}\, d}\)	\(154\)
parts	\(\frac {A \left (\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{2}-\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{2}+\sin \left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{\frac {3}{2}} b \sqrt {b \cos \left (d x +c \right )}}+\frac {B \sin \left (d x +c \right )}{b d \sqrt {\cos \left (d x +c \right )}\, \sqrt {b \cos \left (d x +c \right )}}-\frac {2 C \,\operatorname {arctanh}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right )}}{d b \sqrt {b \cos \left (d x +c \right )}}\)	\(164\)

Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(3/2), 
x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.99 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\left [\frac {{\left (A + 2 \, C\right )} \sqrt {b} \cos \left (d x + c\right )^{3} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, {\left (2 \, B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, b^{2} d \cos \left (d x + c\right )^{3}}, -\frac {{\left (A + 2 \, C\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{3} - {\left (2 \, B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b^{2} d \cos \left (d x + c\right )^{3}}\right ] \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^ 
(3/2),x, algorithm="fricas")
 

Output:

[1/4*((A + 2*C)*sqrt(b)*cos(d*x + c)^3*log(-(b*cos(d*x + c)^3 - 2*sqrt(b*c 
os(d*x + c))*sqrt(b)*sqrt(cos(d*x + c))*sin(d*x + c) - 2*b*cos(d*x + c))/c 
os(d*x + c)^3) + 2*(2*B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))*sqrt(cos(d* 
x + c))*sin(d*x + c))/(b^2*d*cos(d*x + c)^3), -1/2*((A + 2*C)*sqrt(-b)*arc 
tan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(cos(d*x + c))))*cos 
(d*x + c)^3 - (2*B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c 
))*sin(d*x + c))/(b^2*d*cos(d*x + c)^3)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(3/2)/(b*cos(d*x+c) 
)**(3/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 802 vs. \(2 (104) = 208\).

Time = 0.38 (sec) , antiderivative size = 802, normalized size of antiderivative = 6.68 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^ 
(3/2),x, algorithm="maxima")
 

Output:

1/4*(8*B*sqrt(b)*sin(2*d*x + 2*c)/(b^2*cos(2*d*x + 2*c)^2 + b^2*sin(2*d*x 
+ 2*c)^2 + 2*b^2*cos(2*d*x + 2*c) + b^2) - (4*(sin(4*d*x + 4*c) + 2*sin(2* 
d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 4*(sin( 
4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2 
*d*x + 2*c))) - (2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 
 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*s 
in(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(cos(1 
/2*arctan2(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) + (2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos( 
4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 
 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*lo 
g(cos(1/2*arctan2(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) - 4*(cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c) + 1 
)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*(cos(4*d*x + 4* 
c) + 2*cos(2*d*x + 2*c) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 
 2*c))))*A/((b*cos(4*d*x + 4*c)^2 + 4*b*cos(2*d*x + 2*c)^2 + b*sin(4*d*x + 
 4*c)^2 + 4*b*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*b*sin(2*d*x + 2*c)^2 + 
 2*(2*b*cos(2*d*x + 2*c) + b)*cos(4*d*x + 4*c) + 4*b*cos(2*d*x + 2*c) +...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^ 
(3/2),x, algorithm="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. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(3/2)*(b*cos(c + 
 d*x))^(3/2)),x)
 

Output:

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(3/2)*(b*cos(c + 
 d*x))^(3/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.64 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {b}\, \left (-2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} c +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} c -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c -\sin \left (d x +c \right ) a \right )}{2 b^{2} d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(3/2), 
x)
 

Output:

(sqrt(b)*( - 2*cos(c + d*x)*sin(c + d*x)*b - log(tan((c + d*x)/2) - 1)*sin 
(c + d*x)**2*a - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*c + log(tan(( 
c + d*x)/2) - 1)*a + 2*log(tan((c + d*x)/2) - 1)*c + log(tan((c + d*x)/2) 
+ 1)*sin(c + d*x)**2*a + 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*c - l 
og(tan((c + d*x)/2) + 1)*a - 2*log(tan((c + d*x)/2) + 1)*c - sin(c + d*x)* 
a))/(2*b**2*d*(sin(c + d*x)**2 - 1))