\(\int \sec ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\) [996]

Optimal result

Integrand size = 31, antiderivative size = 69 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=a^3 B x+\frac {(A+B) \sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}-\frac {2 a^5 B \cos (c+d x)}{d \left (a^2-a^2 \sin (c+d x)\right )} \] Output:

a^3*B*x+1/3*(A+B)*sec(d*x+c)^3*(a+a*sin(d*x+c))^3/d-2*a^5*B*cos(d*x+c)/d/( 
a^2-a^2*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 7.29 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.75 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {a^3 \left (-3 (2 A+3 B (2+c+d x)) \cos \left (\frac {1}{2} (c+d x)\right )+(2 A+B (14+3 c+3 d x)) \cos \left (\frac {3}{2} (c+d x)\right )+6 B (2 (2+c+d x)+(c+d x) \cos (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3} \] Input:

Integrate[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]
 

Output:

-1/6*(a^3*(-3*(2*A + 3*B*(2 + c + d*x))*Cos[(c + d*x)/2] + (2*A + B*(14 + 
3*c + 3*d*x))*Cos[(3*(c + d*x))/2] + 6*B*(2*(2 + c + d*x) + (c + d*x)*Cos[ 
c + d*x])*Sin[(c + d*x)/2]))/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {3042, 3334, 3042, 3149, 3042, 3159, 24}

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.

\(\displaystyle \int \sec ^4(c+d x) (a \sin (c+d x)+a)^3 (A+B \sin (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (c+d x)+a)^3 (A+B \sin (c+d x))}{\cos (c+d x)^4}dx\)

\(\Big \downarrow \) 3334

\(\displaystyle \frac {(A+B) \sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d}-a B \int \sec ^2(c+d x) (\sin (c+d x) a+a)^2dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(A+B) \sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d}-a B \int \frac {(\sin (c+d x) a+a)^2}{\cos (c+d x)^2}dx\)

\(\Big \downarrow \) 3149

\(\displaystyle \frac {(A+B) \sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d}-a^5 B \int \frac {\cos ^2(c+d x)}{(a-a \sin (c+d x))^2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(A+B) \sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d}-a^5 B \int \frac {\cos (c+d x)^2}{(a-a \sin (c+d x))^2}dx\)

\(\Big \downarrow \) 3159

\(\displaystyle \frac {(A+B) \sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d}-a^5 B \left (\frac {2 \cos (c+d x)}{d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {\int 1dx}{a^2}\right )\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {(A+B) \sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d}-a^5 B \left (\frac {2 \cos (c+d x)}{d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {x}{a^2}\right )\)

Input:

Int[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]
 

Output:

((A + B)*Sec[c + d*x]^3*(a + a*Sin[c + d*x])^3)/(3*d) - a^5*B*(-(x/a^2) + 
(2*Cos[c + d*x])/(d*(a^2 - a^2*Sin[c + d*x])))
 

Defintions of rubi rules used

rule 24

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

rule 3042

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

rule 3149

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(a/g)^(2*m)   Int[(g*Cos[e + f*x])^(2*m + p)/( 
a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2 
, 0] && IntegerQ[m] && LtQ[p, -1] && GeQ[2*m + p, 0]
 

rule 3159

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[2*g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f 
*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Simp[g^2*((p - 1)/(b^2*(2*m + p + 1 
)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; 
FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] & 
& NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]
 

rule 3334

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b* 
c + a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(p + 1))) 
, x] + Simp[b*((a*d*m + b*c*(m + p + 1))/(a*g^2*(p + 1)))   Int[(g*Cos[e + 
f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, 
 f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.16 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.25

method	result	size
risch	\(a^{3} B x -\frac {2 \left (3 A \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-12 i B \,a^{3} {\mathrm e}^{i \left (d x +c \right )}+9 B \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-a^{3} A -7 a^{3} B \right )}{3 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d}\)	\(86\)
parallelrisch	\(-\frac {2 a^{3} \left (-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} x d B}{2}+\left (\frac {3}{2} d x B +A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+B \left (-\frac {3 d x}{2}+4\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {d x B}{2}+\frac {A}{3}-\frac {5 B}{3}\right )}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(89\)
derivativedivides	\(\frac {a^{3} A \left (\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}\right )+a^{3} B \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+d x +c \right )+\frac {a^{3} A \sin \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{3}}+3 a^{3} B \left (\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}\right )+\frac {a^{3} A}{\cos \left (d x +c \right )^{3}}+\frac {a^{3} B \sin \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{3}}-a^{3} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+\frac {a^{3} B}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(248\)
default	\(\frac {a^{3} A \left (\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}\right )+a^{3} B \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+d x +c \right )+\frac {a^{3} A \sin \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{3}}+3 a^{3} B \left (\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}\right )+\frac {a^{3} A}{\cos \left (d x +c \right )^{3}}+\frac {a^{3} B \sin \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{3}}-a^{3} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+\frac {a^{3} B}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(248\)
norman	\(\frac {\frac {\left (6 a^{3} A +2 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{d}+a^{3} B x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+a^{3} B x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}-\frac {20 a^{3} A -4 a^{3} B}{3 d}-a^{3} B x -\frac {\left (8 a^{3} A +24 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}-\frac {\left (38 a^{3} A +2 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d}-\frac {\left (46 a^{3} A +26 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}-\frac {\left (98 a^{3} A +182 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}-\frac {\left (212 a^{3} A +188 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3 d}-a^{3} B x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 a^{3} B x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+3 a^{3} B x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-3 a^{3} B x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-3 a^{3} B x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-\frac {2 a^{3} \left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a^{3} \left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}-\frac {8 a^{3} \left (7 A +9 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {4 a^{3} \left (11 A +5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {4 a^{3} \left (11 A +5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}-\frac {2 a^{3} \left (61 A +67 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}-\frac {2 a^{3} \left (61 A +67 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\)	\(531\)

Input:

int(sec(d*x+c)^4*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x,method=_RETURNVERBO 
SE)
 

Output:

a^3*B*x-2/3*(3*A*a^3*exp(2*I*(d*x+c))-12*I*B*a^3*exp(I*(d*x+c))+9*B*a^3*ex 
p(2*I*(d*x+c))-a^3*A-7*a^3*B)/(exp(I*(d*x+c))-I)^3/d
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 167 vs. \(2 (68) = 136\).

Time = 0.09 (sec) , antiderivative size = 167, normalized size of antiderivative = 2.42 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {6 \, B a^{3} d x + 2 \, {\left (A + B\right )} a^{3} - {\left (3 \, B a^{3} d x + {\left (A + 7 \, B\right )} a^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, B a^{3} d x + {\left (A - 5 \, B\right )} a^{3}\right )} \cos \left (d x + c\right ) - {\left (6 \, B a^{3} d x - 2 \, {\left (A + B\right )} a^{3} + {\left (3 \, B a^{3} d x - {\left (A + 7 \, B\right )} a^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \] Input:

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="f 
ricas")
 

Output:

-1/3*(6*B*a^3*d*x + 2*(A + B)*a^3 - (3*B*a^3*d*x + (A + 7*B)*a^3)*cos(d*x 
+ c)^2 + (3*B*a^3*d*x + (A - 5*B)*a^3)*cos(d*x + c) - (6*B*a^3*d*x - 2*(A 
+ B)*a^3 + (3*B*a^3*d*x - (A + 7*B)*a^3)*cos(d*x + c))*sin(d*x + c))/(d*co 
s(d*x + c)^2 - d*cos(d*x + c) + (d*cos(d*x + c) + 2*d)*sin(d*x + c) - 2*d)
 

Sympy [F(-1)]

Timed out. \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)**4*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (68) = 136\).

Time = 0.12 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.38 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {3 \, A a^{3} \tan \left (d x + c\right )^{3} + 3 \, B a^{3} \tan \left (d x + c\right )^{3} + {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} B a^{3} - \frac {{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} A a^{3}}{\cos \left (d x + c\right )^{3}} - \frac {3 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} B a^{3}}{\cos \left (d x + c\right )^{3}} + \frac {3 \, A a^{3}}{\cos \left (d x + c\right )^{3}} + \frac {B a^{3}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \] Input:

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="m 
axima")
 

Output:

1/3*(3*A*a^3*tan(d*x + c)^3 + 3*B*a^3*tan(d*x + c)^3 + (tan(d*x + c)^3 + 3 
*tan(d*x + c))*A*a^3 + (tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*B*a 
^3 - (3*cos(d*x + c)^2 - 1)*A*a^3/cos(d*x + c)^3 - 3*(3*cos(d*x + c)^2 - 1 
)*B*a^3/cos(d*x + c)^3 + 3*A*a^3/cos(d*x + c)^3 + B*a^3/cos(d*x + c)^3)/d
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.35 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (d x + c\right )} B a^{3} - \frac {2 \, {\left (3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A a^{3} - 5 \, B a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{3 \, d} \] Input:

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="g 
iac")
 

Output:

1/3*(3*(d*x + c)*B*a^3 - 2*(3*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 3*B*a^3*tan(1 
/2*d*x + 1/2*c)^2 + 12*B*a^3*tan(1/2*d*x + 1/2*c) + A*a^3 - 5*B*a^3)/(tan( 
1/2*d*x + 1/2*c) - 1)^3)/d
 

Mupad [B] (verification not implemented)

Time = 33.90 (sec) , antiderivative size = 140, normalized size of antiderivative = 2.03 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=B\,a^3\,x-\frac {\frac {a^3\,\left (2\,A-10\,B+3\,B\,\left (c+d\,x\right )\right )}{3}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^3\,\left (6\,A-6\,B+9\,B\,\left (c+d\,x\right )\right )}{3}-3\,B\,a^3\,\left (c+d\,x\right )\right )+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^3\,\left (24\,B-9\,B\,\left (c+d\,x\right )\right )}{3}+3\,B\,a^3\,\left (c+d\,x\right )\right )-B\,a^3\,\left (c+d\,x\right )}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3} \] Input:

int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^3)/cos(c + d*x)^4,x)
 

Output:

B*a^3*x - ((a^3*(2*A - 10*B + 3*B*(c + d*x)))/3 + tan(c/2 + (d*x)/2)^2*((a 
^3*(6*A - 6*B + 9*B*(c + d*x)))/3 - 3*B*a^3*(c + d*x)) + tan(c/2 + (d*x)/2 
)*((a^3*(24*B - 9*B*(c + d*x)))/3 + 3*B*a^3*(c + d*x)) - B*a^3*(c + d*x))/ 
(d*(tan(c/2 + (d*x)/2) - 1)^3)
                                                                                    
                                                                                    
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.23 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^{3} \left (-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a +3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b d x +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b -9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b d x -6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b d x -18 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -3 b d x +8 b \right )}{3 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )} \] Input:

int(sec(d*x+c)^4*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)
 

Output:

(a**3*( - 2*tan((c + d*x)/2)**3*a + 3*tan((c + d*x)/2)**3*b*d*x + 2*tan((c 
 + d*x)/2)**3*b - 9*tan((c + d*x)/2)**2*b*d*x - 6*tan((c + d*x)/2)*a + 9*t 
an((c + d*x)/2)*b*d*x - 18*tan((c + d*x)/2)*b - 3*b*d*x + 8*b))/(3*d*(tan( 
(c + d*x)/2)**3 - 3*tan((c + d*x)/2)**2 + 3*tan((c + d*x)/2) - 1))