3.2.50 \(\int \frac {\sec (c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^4} \, dx\) [150]

3.2.50.1 Optimal result

Integrand size = 31, antiderivative size = 142 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {2 (4 A-3 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(6 A+13 C) \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {(6 A+13 C) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )} \]

output

-1/7*(A+C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*sec(d*x+c))^4+2/35*(4*A-3*C)*tan(d 
*x+c)/a/d/(a+a*sec(d*x+c))^3+1/105*(6*A+13*C)*tan(d*x+c)/d/(a^2+a^2*sec(d* 
x+c))^2+1/105*(6*A+13*C)*tan(d*x+c)/d/(a^4+a^4*sec(d*x+c))
 

3.2.50.2 Mathematica [A] (verified)

Time = 3.91 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.20 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^7\left (\frac {1}{2} (c+d x)\right ) \left (70 (9 A+2 C) \sin \left (\frac {d x}{2}\right )-70 (9 A+2 C) \sin \left (c+\frac {d x}{2}\right )+441 A \sin \left (c+\frac {3 d x}{2}\right )+168 C \sin \left (c+\frac {3 d x}{2}\right )-315 A \sin \left (2 c+\frac {3 d x}{2}\right )+147 A \sin \left (2 c+\frac {5 d x}{2}\right )+56 C \sin \left (2 c+\frac {5 d x}{2}\right )-105 A \sin \left (3 c+\frac {5 d x}{2}\right )+36 A \sin \left (3 c+\frac {7 d x}{2}\right )+8 C \sin \left (3 c+\frac {7 d x}{2}\right )\right )}{6720 a^4 d} \]

input

Integrate[(Sec[c + d*x]*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^4,x]
 

output

(Sec[c/2]*Sec[(c + d*x)/2]^7*(70*(9*A + 2*C)*Sin[(d*x)/2] - 70*(9*A + 2*C) 
*Sin[c + (d*x)/2] + 441*A*Sin[c + (3*d*x)/2] + 168*C*Sin[c + (3*d*x)/2] - 
315*A*Sin[2*c + (3*d*x)/2] + 147*A*Sin[2*c + (5*d*x)/2] + 56*C*Sin[2*c + ( 
5*d*x)/2] - 105*A*Sin[3*c + (5*d*x)/2] + 36*A*Sin[3*c + (7*d*x)/2] + 8*C*S 
in[3*c + (7*d*x)/2]))/(6720*a^4*d)
 

3.2.50.3 Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {3042, 4567, 3042, 4488, 3042, 4283, 3042, 4281}

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.

\(\displaystyle \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\)

\(\Big \downarrow \) 4567

\(\displaystyle \frac {\int \frac {\sec (c+d x) (a (6 A-C)-a (2 A-5 C) \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a (6 A-C)-a (2 A-5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4488

\(\displaystyle \frac {\frac {1}{5} (6 A+13 C) \int \frac {\sec (c+d x)}{(\sec (c+d x) a+a)^2}dx+\frac {2 a (4 A-3 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} (6 A+13 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx+\frac {2 a (4 A-3 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4283

\(\displaystyle \frac {\frac {1}{5} (6 A+13 C) \left (\frac {\int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )+\frac {2 a (4 A-3 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} (6 A+13 C) \left (\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )+\frac {2 a (4 A-3 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4281

\(\displaystyle \frac {\frac {2 a (4 A-3 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {1}{5} (6 A+13 C) \left (\frac {\tan (c+d x)}{3 a d (a \sec (c+d x)+a)}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

input

Int[(Sec[c + d*x]*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^4,x]
 

output

-1/7*((A + C)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^4) + ((2* 
a*(4*A - 3*C)*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((6*A + 13*C)*( 
Tan[c + d*x]/(3*d*(a + a*Sec[c + d*x])^2) + Tan[c + d*x]/(3*a*d*(a + a*Sec 
[c + d*x]))))/5)/(7*a^2)
 

3.2.50.3.1 Defintions of rubi rules used

rule 3042

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

rule 4281

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo 
l] :> Simp[-Cot[e + f*x]/(f*(b + a*Csc[e + f*x])), x] /; FreeQ[{a, b, e, f} 
, x] && EqQ[a^2 - b^2, 0]
 

rule 4283

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ 
Symbol] :> Simp[b*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] 
 + Simp[(m + 1)/(a*(2*m + 1))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 
1), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1) 
] && IntegerQ[2*m]
 

rule 4488

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(A*b - a*B)*Cot[e + 
f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[(a*B*m + A*b*(m + 
1))/(a*b*(2*m + 1))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] 
 /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] 
&& NeQ[a*B*m + A*b*(m + 1), 0] && LtQ[m, -2^(-1)]
 

rule 4567

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

3.2.50.4 Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.54

method	result	size
parallelrisch	\(-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {7 \left (-3 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{5}+7 \left (A -\frac {C}{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-7 A -7 C \right )}{56 a^{4} d}\)	\(76\)
derivativedivides	\(\frac {\frac {\left (-A -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {\left (3 A -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {\left (-3 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{8 d \,a^{4}}\)	\(90\)
default	\(\frac {\frac {\left (-A -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {\left (3 A -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {\left (-3 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{8 d \,a^{4}}\)	\(90\)
risch	\(\frac {2 i \left (105 A \,{\mathrm e}^{6 i \left (d x +c \right )}+315 A \,{\mathrm e}^{5 i \left (d x +c \right )}+630 A \,{\mathrm e}^{4 i \left (d x +c \right )}+140 C \,{\mathrm e}^{4 i \left (d x +c \right )}+630 A \,{\mathrm e}^{3 i \left (d x +c \right )}+140 C \,{\mathrm e}^{3 i \left (d x +c \right )}+441 A \,{\mathrm e}^{2 i \left (d x +c \right )}+168 C \,{\mathrm e}^{2 i \left (d x +c \right )}+147 A \,{\mathrm e}^{i \left (d x +c \right )}+56 C \,{\mathrm e}^{i \left (d x +c \right )}+36 A +8 C \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\)	\(150\)
norman	\(\frac {\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{56 a d}-\frac {\left (9 A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 a d}+\frac {\left (27 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}+\frac {\left (31 A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{280 a d}-\frac {\left (123 A -31 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{420 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2} a^{3}}\)	\(165\)

input

int(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x,method=_RETURNVERBO 
SE)
 

output

-1/56*tan(1/2*d*x+1/2*c)*((A+C)*tan(1/2*d*x+1/2*c)^6+7/5*(-3*A+C)*tan(1/2* 
d*x+1/2*c)^4+7*(A-1/3*C)*tan(1/2*d*x+1/2*c)^2-7*A-7*C)/a^4/d
 

3.2.50.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.87 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {{\left (4 \, {\left (9 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (39 \, A + 32 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (6 \, A + 13 \, C\right )} \cos \left (d x + c\right ) + 6 \, A + 13 \, C\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

input

integrate(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="f 
ricas")
 

output

1/105*(4*(9*A + 2*C)*cos(d*x + c)^3 + (39*A + 32*C)*cos(d*x + c)^2 + 4*(6* 
A + 13*C)*cos(d*x + c) + 6*A + 13*C)*sin(d*x + c)/(a^4*d*cos(d*x + c)^4 + 
4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a 
^4*d)
 

3.2.50.6 Sympy [F]

\[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\int \frac {A \sec {\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{3}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

input

integrate(sec(d*x+c)*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**4,x)
 

output

(Integral(A*sec(c + d*x)/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + 
d*x)**2 + 4*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**3/(sec(c + d* 
x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x))/a 
**4
 

3.2.50.7 Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.23 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\frac {C {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac {3 \, A {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]

input

integrate(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="m 
axima")
 

output

1/840*(C*(105*sin(d*x + c)/(cos(d*x + c) + 1) + 35*sin(d*x + c)^3/(cos(d*x 
 + c) + 1)^3 - 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 15*sin(d*x + c)^7/ 
(cos(d*x + c) + 1)^7)/a^4 + 3*A*(35*sin(d*x + c)/(cos(d*x + c) + 1) - 35*s 
in(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^ 
5 - 5*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4)/d
 

3.2.50.8 Giac [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.82 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=-\frac {15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 63 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 35 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{840 \, a^{4} d} \]

input

integrate(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="g 
iac")
 

output

-1/840*(15*A*tan(1/2*d*x + 1/2*c)^7 + 15*C*tan(1/2*d*x + 1/2*c)^7 - 63*A*t 
an(1/2*d*x + 1/2*c)^5 + 21*C*tan(1/2*d*x + 1/2*c)^5 + 105*A*tan(1/2*d*x + 
1/2*c)^3 - 35*C*tan(1/2*d*x + 1/2*c)^3 - 105*A*tan(1/2*d*x + 1/2*c) - 105* 
C*tan(1/2*d*x + 1/2*c))/(a^4*d)
 

3.2.50.9 Mupad [B] (verification not implemented)

Time = 16.19 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.62 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A+C\right )}{56\,a^4}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+C\right )}{8\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,A-C\right )}{24\,a^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (3\,A-C\right )}{40\,a^4}}{d} \]

input

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)*(a + a/cos(c + d*x))^4),x)
 

output

-((tan(c/2 + (d*x)/2)^7*(A + C))/(56*a^4) - (tan(c/2 + (d*x)/2)*(A + C))/( 
8*a^4) + (tan(c/2 + (d*x)/2)^3*(3*A - C))/(24*a^4) - (tan(c/2 + (d*x)/2)^5 
*(3*A - C))/(40*a^4))/d