\(\int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\) [857]

Optimal result

Integrand size = 33, antiderivative size = 149 \[ \int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {b B \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{2 d \sqrt {\cos (c+d x)}}+\frac {b B \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {A b \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}+\frac {A b \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x)} \] Output:

1/2*b*B*arctanh(sin(d*x+c))*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)+1/2*b* 
B*(b*cos(d*x+c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(5/2)+A*b*(b*cos(d*x+c))^(1 
/2)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+1/3*A*b*(b*cos(d*x+c))^(1/2)*sin(d*x+c)^ 
3/d/cos(d*x+c)^(7/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.52 \[ \int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {b \sqrt {b \cos (c+d x)} \left (3 B \text {arctanh}(\sin (c+d x)) \cos ^2(c+d x)+3 B \sin (c+d x)+2 A (2+\cos (2 (c+d x))) \tan (c+d x)\right )}{6 d \cos ^{\frac {5}{2}}(c+d x)} \] Input:

Integrate[((b*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(11/2 
),x]
 

Output:

(b*Sqrt[b*Cos[c + d*x]]*(3*B*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^2 + 3*B*Si 
n[c + d*x] + 2*A*(2 + Cos[2*(c + d*x)])*Tan[c + d*x]))/(6*d*Cos[c + d*x]^( 
5/2))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.59, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2031, 3042, 3227, 3042, 4254, 2009, 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.

\(\displaystyle \int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 2031

\(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \int (A+B \cos (c+d x)) \sec ^4(c+d x)dx}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

Input:

Int[((b*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(11/2),x]
 

Output:

(b*Sqrt[b*Cos[c + d*x]]*(B*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Ta 
n[c + d*x])/(2*d)) - (A*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/d))/Sqrt[Cos[c 
 + d*x]]
 

Defintions of rubi rules used

rule 2009

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

rule 2031

Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[a^(m + 1/ 
2)*b^(n - 1/2)*(Sqrt[b*v]/Sqrt[a*v])   Int[v^(m + n)*Fx, x], x] /; FreeQ[{a 
, b, m}, x] &&  !IntegerQ[m] && IGtQ[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 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 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 = 10.26 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.79

method	result	size
default	\(\frac {b \left (-3 B \cos \left (d x +c \right )^{3} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+3 B \cos \left (d x +c \right )^{3} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )+\left (4 \cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right ) A +3 B \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right ) b}}{6 d \cos \left (d x +c \right )^{\frac {7}{2}}}\)	\(117\)
parts	\(\frac {A \sin \left (d x +c \right ) \left (2 \cos \left (d x +c \right )^{2}+1\right ) \sqrt {\cos \left (d x +c \right ) b}\, b}{3 d \cos \left (d x +c \right )^{\frac {7}{2}}}+\frac {B \left (\cos \left (d x +c \right )^{2} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )-\cos \left (d x +c \right )^{2} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+\sin \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right ) b}\, b}{2 d \cos \left (d x +c \right )^{\frac {5}{2}}}\)	\(130\)
risch	\(-\frac {i b \sqrt {\cos \left (d x +c \right ) b}\, \left (3 B \,{\mathrm e}^{5 i \left (d x +c \right )}-12 A \,{\mathrm e}^{2 i \left (d x +c \right )}-3 B \,{\mathrm e}^{i \left (d x +c \right )}-4 A \right )}{3 \sqrt {\cos \left (d x +c \right )}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {b \sqrt {\cos \left (d x +c \right ) b}\, B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 \sqrt {\cos \left (d x +c \right )}\, d}-\frac {b \sqrt {\cos \left (d x +c \right ) b}\, B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 \sqrt {\cos \left (d x +c \right )}\, d}\)	\(156\)

Input:

int((cos(d*x+c)*b)^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(11/2),x,method=_RETU 
RNVERBOSE)
 

Output:

1/6*b/d*(-3*B*cos(d*x+c)^3*ln(-cot(d*x+c)+csc(d*x+c)-1)+3*B*cos(d*x+c)^3*l 
n(-cot(d*x+c)+csc(d*x+c)+1)+(4*cos(d*x+c)^2+2)*sin(d*x+c)*A+3*B*cos(d*x+c) 
*sin(d*x+c))*(cos(d*x+c)*b)^(1/2)/cos(d*x+c)^(7/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.74 \[ \int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\left [\frac {3 \, B b^{\frac {3}{2}} \cos \left (d x + c\right )^{4} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, {\left (4 \, A b \cos \left (d x + c\right )^{2} + 3 \, B b \cos \left (d x + c\right ) + 2 \, A b\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{4}}, -\frac {3 \, B \sqrt {-b} b \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{4} - {\left (4 \, A b \cos \left (d x + c\right )^{2} + 3 \, B b \cos \left (d x + c\right ) + 2 \, A b\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{4}}\right ] \] Input:

integrate((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(11/2),x, algor 
ithm="fricas")
 

Output:

[1/12*(3*B*b^(3/2)*cos(d*x + c)^4*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*(4*A*b*cos(d*x + c)^2 + 3*B*b*cos(d*x + c) + 2*A*b)*sqrt(b*c 
os(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4), -1/6*(3* 
B*sqrt(-b)*b*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(cos 
(d*x + c))))*cos(d*x + c)^4 - (4*A*b*cos(d*x + c)^2 + 3*B*b*cos(d*x + c) + 
 2*A*b)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + 
 c)^4)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((b*cos(d*x+c))**(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)**(11/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 992 vs. \(2 (127) = 254\).

Time = 0.34 (sec) , antiderivative size = 992, normalized size of antiderivative = 6.66 \[ \int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(11/2),x, algor 
ithm="maxima")
 

Output:

-1/12*(16*(3*b*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) + 9*b*cos(4*d*x + 4*c)*si 
n(2*d*x + 2*c) - (3*b*cos(2*d*x + 2*c) + b)*sin(6*d*x + 6*c) - 3*(3*b*cos( 
2*d*x + 2*c) + b)*sin(4*d*x + 4*c))*A*sqrt(b)/(2*(3*cos(4*d*x + 4*c) + 3*c 
os(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c) + cos(6*d*x + 6*c)^2 + 6*(3*cos(2*d* 
x + 2*c) + 1)*cos(4*d*x + 4*c) + 9*cos(4*d*x + 4*c)^2 + 9*cos(2*d*x + 2*c) 
^2 + 6*(sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + sin(6*d*x 
+ 6*c)^2 + 9*sin(4*d*x + 4*c)^2 + 18*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9 
*sin(2*d*x + 2*c)^2 + 6*cos(2*d*x + 2*c) + 1) + 3*(4*(b*sin(4*d*x + 4*c) + 
 2*b*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)) 
) - 4*(b*sin(4*d*x + 4*c) + 2*b*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d* 
x + 2*c), cos(2*d*x + 2*c))) - (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*s 
in(2*d*x + 2*c)^2 + 2*(2*b*cos(2*d*x + 2*c) + b)*cos(4*d*x + 4*c) + 4*b*co 
s(2*d*x + 2*c) + b)*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) + (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) + b)*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...
 

Giac [A] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00 \[ \int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {{\left (3 \, B \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 3 \, B \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) - \frac {2 \, {\left (6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}\right )} b^{\frac {3}{2}}}{6 \, d} \] Input:

integrate((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(11/2),x, algor 
ithm="giac")
 

Output:

1/6*(3*B*log(tan(1/2*d*x + 1/2*c) + 1) - 3*B*log(tan(1/2*d*x + 1/2*c) - 1) 
 - 2*(6*A*tan(1/2*d*x + 1/2*c)^5 - 3*B*tan(1/2*d*x + 1/2*c)^5 - 4*A*tan(1/ 
2*d*x + 1/2*c)^3 + 6*A*tan(1/2*d*x + 1/2*c) + 3*B*tan(1/2*d*x + 1/2*c))/(t 
an(1/2*d*x + 1/2*c)^6 - 3*tan(1/2*d*x + 1/2*c)^4 + 3*tan(1/2*d*x + 1/2*c)^ 
2 - 1))*b^(3/2)/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\cos \left (c+d\,x\right )}^{11/2}} \,d x \] Input:

int(((b*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x)))/cos(c + d*x)^(11/2),x)
 

Output:

int(((b*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x)))/cos(c + d*x)^(11/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.10 \[ \int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {\sqrt {b}\, b \left (-3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b -3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +4 \sin \left (d x +c \right )^{3} a -6 \sin \left (d x +c \right ) a \right )}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

int((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(11/2),x)
 

Output:

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