3.7.45 \(\int \cos ^5(c+d x) (a+b \sec (c+d x)) (A+C \sec ^2(c+d x)) \, dx\) [645]

Optimal. Leaf size=131 \[ \frac {1}{8} b (3 A+4 C) x+\frac {a (4 A+5 C) \sin (c+d x)}{5 d}+\frac {b (3 A+4 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {A b \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {a (4 A+5 C) \sin ^3(c+d x)}{15 d} \]

[Out]

1/8*b*(3*A+4*C)*x+1/5*a*(4*A+5*C)*sin(d*x+c)/d+1/8*b*(3*A+4*C)*cos(d*x+c)*sin(d*x+c)/d+1/4*A*b*cos(d*x+c)^3*si
n(d*x+c)/d+1/5*a*A*cos(d*x+c)^4*sin(d*x+c)/d-1/15*a*(4*A+5*C)*sin(d*x+c)^3/d

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Rubi [A]
time = 0.13, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4160, 4132, 2713, 4130, 2715, 8} \begin {gather*} -\frac {a (4 A+5 C) \sin ^3(c+d x)}{15 d}+\frac {a (4 A+5 C) \sin (c+d x)}{5 d}+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}+\frac {b (3 A+4 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {A b \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {1}{8} b x (3 A+4 C) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + b*Sec[c + d*x])*(A + C*Sec[c + d*x]^2),x]

[Out]

(b*(3*A + 4*C)*x)/8 + (a*(4*A + 5*C)*Sin[c + d*x])/(5*d) + (b*(3*A + 4*C)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (
A*b*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (a*A*Cos[c + d*x]^4*Sin[c + d*x])/(5*d) - (a*(4*A + 5*C)*Sin[c + d*x]
^3)/(15*d)

Rule 8

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

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2715

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

Rule 4130

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

Rule 4132

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 4160

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Dist[1/(d*n), Int[(d*Csc[e +
f*x])^(n + 1)*Simp[A*b*n + a*(C*n + A*(n + 1))*Csc[e + f*x] + b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b,
d, e, f, A, C}, x] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \cos ^5(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {1}{5} \int \cos ^4(c+d x) \left (-5 A b-a (4 A+5 C) \sec (c+d x)-5 b C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {1}{5} \int \cos ^4(c+d x) \left (-5 A b-5 b C \sec ^2(c+d x)\right ) \, dx+\frac {1}{5} (a (4 A+5 C)) \int \cos ^3(c+d x) \, dx\\ &=\frac {A b \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}+\frac {1}{4} (b (3 A+4 C)) \int \cos ^2(c+d x) \, dx-\frac {(a (4 A+5 C)) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac {a (4 A+5 C) \sin (c+d x)}{5 d}+\frac {b (3 A+4 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {A b \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {a (4 A+5 C) \sin ^3(c+d x)}{15 d}+\frac {1}{8} (b (3 A+4 C)) \int 1 \, dx\\ &=\frac {1}{8} b (3 A+4 C) x+\frac {a (4 A+5 C) \sin (c+d x)}{5 d}+\frac {b (3 A+4 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {A b \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {a (4 A+5 C) \sin ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]
time = 0.28, size = 108, normalized size = 0.82 \begin {gather*} \frac {180 A b c+240 b c C+180 A b d x+240 b C d x+60 a (5 A+8 C) \sin (c+d x)-160 a C \sin ^3(c+d x)+120 b (A+C) \sin (2 (c+d x))+50 a A \sin (3 (c+d x))+15 A b \sin (4 (c+d x))+6 a A \sin (5 (c+d x))}{480 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + b*Sec[c + d*x])*(A + C*Sec[c + d*x]^2),x]

[Out]

(180*A*b*c + 240*b*c*C + 180*A*b*d*x + 240*b*C*d*x + 60*a*(5*A + 8*C)*Sin[c + d*x] - 160*a*C*Sin[c + d*x]^3 +
120*b*(A + C)*Sin[2*(c + d*x)] + 50*a*A*Sin[3*(c + d*x)] + 15*A*b*Sin[4*(c + d*x)] + 6*a*A*Sin[5*(c + d*x)])/(
480*d)

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Maple [A]
time = 0.14, size = 117, normalized size = 0.89

method	result	size
derivativedivides	\(\frac {\frac {a A \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+A b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+C b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(117\)
default	\(\frac {\frac {a A \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+A b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+C b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(117\)
risch	\(\frac {3 A b x}{8}+\frac {b x C}{2}+\frac {5 a A \sin \left (d x +c \right )}{8 d}+\frac {3 \sin \left (d x +c \right ) a C}{4 d}+\frac {a A \sin \left (5 d x +5 c \right )}{80 d}+\frac {A b \sin \left (4 d x +4 c \right )}{32 d}+\frac {5 a A \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) a C}{12 d}+\frac {A b \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C b}{4 d}\)	\(134\)
norman	\(\frac {\left (\frac {3}{8} A b +\frac {1}{2} C b \right ) x +\left (-\frac {15}{8} A b -\frac {5}{2} C b \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {15}{8} A b -\frac {5}{2} C b \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {3}{8} A b +\frac {1}{2} C b \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {3}{8} A b +\frac {1}{2} C b \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {3}{8} A b +\frac {1}{2} C b \right ) x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {9}{8} A b +\frac {3}{2} C b \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {9}{8} A b +\frac {3}{2} C b \right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2 \left (2 a A -3 A b -2 a C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 \left (2 a A +3 A b -2 a C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (8 a A -5 A b +8 a C -4 C b \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (8 a A +5 A b +8 a C +4 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (88 a A -5 A b -40 a C +60 C b \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d}+\frac {\left (88 a A +5 A b -40 a C -60 C b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d}-\frac {8 a \left (19 A +5 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}\)	\(410\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+b*sec(d*x+c))*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/d*(1/5*a*A*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+A*b*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+
3/8*d*x+3/8*c)+1/3*a*C*(2+cos(d*x+c)^2)*sin(d*x+c)+C*b*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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Maxima [A]
time = 0.27, size = 113, normalized size = 0.86 \begin {gather*} \frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A b + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b}{480 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))
*C*a + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*b + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*C
*b)/d

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Fricas [A]
time = 2.83, size = 94, normalized size = 0.72 \begin {gather*} \frac {15 \, {\left (3 \, A + 4 \, C\right )} b d x + {\left (24 \, A a \cos \left (d x + c\right )^{4} + 30 \, A b \cos \left (d x + c\right )^{3} + 8 \, {\left (4 \, A + 5 \, C\right )} a \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, A + 4 \, C\right )} b \cos \left (d x + c\right ) + 16 \, {\left (4 \, A + 5 \, C\right )} a\right )} \sin \left (d x + c\right )}{120 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/120*(15*(3*A + 4*C)*b*d*x + (24*A*a*cos(d*x + c)^4 + 30*A*b*cos(d*x + c)^3 + 8*(4*A + 5*C)*a*cos(d*x + c)^2
+ 15*(3*A + 4*C)*b*cos(d*x + c) + 16*(4*A + 5*C)*a)*sin(d*x + c))/d

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(a+b*sec(d*x+c))*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 302 vs. \(2 (119) = 238\).
time = 0.47, size = 302, normalized size = 2.31 \begin {gather*} \frac {15 \, {\left (3 \, A b + 4 \, C b\right )} {\left (d x + c\right )} + \frac {2 \, {\left (120 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 160 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 320 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 320 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/120*(15*(3*A*b + 4*C*b)*(d*x + c) + 2*(120*A*a*tan(1/2*d*x + 1/2*c)^9 + 120*C*a*tan(1/2*d*x + 1/2*c)^9 - 75*
A*b*tan(1/2*d*x + 1/2*c)^9 - 60*C*b*tan(1/2*d*x + 1/2*c)^9 + 160*A*a*tan(1/2*d*x + 1/2*c)^7 + 320*C*a*tan(1/2*
d*x + 1/2*c)^7 - 30*A*b*tan(1/2*d*x + 1/2*c)^7 - 120*C*b*tan(1/2*d*x + 1/2*c)^7 + 464*A*a*tan(1/2*d*x + 1/2*c)
^5 + 400*C*a*tan(1/2*d*x + 1/2*c)^5 + 160*A*a*tan(1/2*d*x + 1/2*c)^3 + 320*C*a*tan(1/2*d*x + 1/2*c)^3 + 30*A*b
*tan(1/2*d*x + 1/2*c)^3 + 120*C*b*tan(1/2*d*x + 1/2*c)^3 + 120*A*a*tan(1/2*d*x + 1/2*c) + 120*C*a*tan(1/2*d*x
+ 1/2*c) + 75*A*b*tan(1/2*d*x + 1/2*c) + 60*C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d

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Mupad [B]
time = 6.50, size = 250, normalized size = 1.91 \begin {gather*} \frac {\left (2\,A\,a-\frac {5\,A\,b}{4}+2\,C\,a-C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {8\,A\,a}{3}-\frac {A\,b}{2}+\frac {16\,C\,a}{3}-2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a}{15}+\frac {20\,C\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {8\,A\,a}{3}+\frac {A\,b}{2}+\frac {16\,C\,a}{3}+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a+\frac {5\,A\,b}{4}+2\,C\,a+C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {b\,\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,A+4\,C\right )}{4\,\left (\frac {3\,A\,b}{4}+C\,b\right )}\right )\,\left (3\,A+4\,C\right )}{4\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x)),x)

[Out]

(tan(c/2 + (d*x)/2)*(2*A*a + (5*A*b)/4 + 2*C*a + C*b) + tan(c/2 + (d*x)/2)^5*((116*A*a)/15 + (20*C*a)/3) + tan
(c/2 + (d*x)/2)^9*(2*A*a - (5*A*b)/4 + 2*C*a - C*b) + tan(c/2 + (d*x)/2)^3*((8*A*a)/3 + (A*b)/2 + (16*C*a)/3 +
2*C*b) + tan(c/2 + (d*x)/2)^7*((8*A*a)/3 - (A*b)/2 + (16*C*a)/3 - 2*C*b))/(d*(5*tan(c/2 + (d*x)/2)^2 + 10*tan
(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) + (b*atan((
b*tan(c/2 + (d*x)/2)*(3*A + 4*C))/(4*((3*A*b)/4 + C*b)))*(3*A + 4*C))/(4*d)

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