3.62 \(\int \frac{(A+C \cos ^2(c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=129 \[ \frac{2 (36 A+C) \tan (c+d x)}{15 a^3 d}-\frac{3 A \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{3 A \tan (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{(9 A-C) \tan (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac{(A+C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

(-3*A*ArcTanh[Sin[c + d*x]])/(a^3*d) + (2*(36*A + C)*Tan[c + d*x])/(15*a^3*d) - ((A + C)*Tan[c + d*x])/(5*d*(a
+ a*Cos[c + d*x])^3) - ((9*A - C)*Tan[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - (3*A*Tan[c + d*x])/(d*(a^3
+ a^3*Cos[c + d*x]))

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Rubi [A]  time = 0.442715, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3042, 2978, 2748, 3767, 8, 3770} \[ \frac{2 (36 A+C) \tan (c+d x)}{15 a^3 d}-\frac{3 A \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{3 A \tan (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{(9 A-C) \tan (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac{(A+C) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*A*ArcTanh[Sin[c + d*x]])/(a^3*d) + (2*(36*A + C)*Tan[c + d*x])/(15*a^3*d) - ((A + C)*Tan[c + d*x])/(5*d*(a
+ a*Cos[c + d*x])^3) - ((9*A - C)*Tan[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - (3*A*Tan[c + d*x])/(d*(a^3
+ a^3*Cos[c + d*x]))

Rule 3042

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x
])^(n + 1))/(f*(b*c - a*d)*(2*m + 1)), x] + Dist[1/(b*(b*c - a*d)*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)
*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2
) + C*(b*c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 2748

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

Rule 3767

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

Rule 8

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx &=-\frac{(A+C) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{(a (6 A+C)-a (3 A-2 C) \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A+C) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(9 A-C) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{\int \frac{\left (a^2 (27 A+2 C)-2 a^2 (9 A-C) \cos (c+d x)\right ) \sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(A+C) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(9 A-C) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{3 A \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{\int \left (2 a^3 (36 A+C)-45 a^3 A \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{15 a^6}\\ &=-\frac{(A+C) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(9 A-C) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{3 A \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}-\frac{(3 A) \int \sec (c+d x) \, dx}{a^3}+\frac{(2 (36 A+C)) \int \sec ^2(c+d x) \, dx}{15 a^3}\\ &=-\frac{3 A \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{(A+C) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(9 A-C) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{3 A \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}-\frac{(2 (36 A+C)) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=-\frac{3 A \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{2 (36 A+C) \tan (c+d x)}{15 a^3 d}-\frac{(A+C) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(9 A-C) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{3 A \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 6.30674, size = 596, normalized size = 4.62 \[ \frac{\frac{\sec \left (\frac{c}{2}\right ) \sec (c) \cos (c+d x) \cos \left (\frac{c}{2}+\frac{d x}{2}\right ) \left (-600 A \sin \left (c-\frac{d x}{2}\right )+375 A \sin \left (c+\frac{d x}{2}\right )-480 A \sin \left (2 c+\frac{d x}{2}\right )-60 A \sin \left (c+\frac{3 d x}{2}\right )+402 A \sin \left (2 c+\frac{3 d x}{2}\right )-225 A \sin \left (3 c+\frac{3 d x}{2}\right )+315 A \sin \left (c+\frac{5 d x}{2}\right )+30 A \sin \left (2 c+\frac{5 d x}{2}\right )+240 A \sin \left (3 c+\frac{5 d x}{2}\right )-45 A \sin \left (4 c+\frac{5 d x}{2}\right )+72 A \sin \left (2 c+\frac{7 d x}{2}\right )+15 A \sin \left (3 c+\frac{7 d x}{2}\right )+57 A \sin \left (4 c+\frac{7 d x}{2}\right )-255 A \sin \left (\frac{d x}{2}\right )+567 A \sin \left (\frac{3 d x}{2}\right )-10 C \sin \left (c-\frac{d x}{2}\right )+10 C \sin \left (c+\frac{d x}{2}\right )-20 C \sin \left (2 c+\frac{d x}{2}\right )+22 C \sin \left (2 c+\frac{3 d x}{2}\right )+10 C \sin \left (c+\frac{5 d x}{2}\right )+10 C \sin \left (3 c+\frac{5 d x}{2}\right )+2 C \sin \left (2 c+\frac{7 d x}{2}\right )+2 C \sin \left (4 c+\frac{7 d x}{2}\right )-20 C \sin \left (\frac{d x}{2}\right )+22 C \sin \left (\frac{3 d x}{2}\right )\right ) \left (A \sec ^2(c+d x)+C\right )}{60 d (\cos (c+d x)+1)^3 (2 A+C \cos (2 c+2 d x)+C)}+\frac{48 A \cos ^2(c+d x) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (A \sec ^2(c+d x)+C\right ) \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{d (\cos (c+d x)+1)^3 (2 A+C \cos (2 c+2 d x)+C)}-\frac{48 A \cos ^2(c+d x) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (A \sec ^2(c+d x)+C\right ) \log \left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+\cos \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{d (\cos (c+d x)+1)^3 (2 A+C \cos (2 c+2 d x)+C)}}{a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + a*Cos[c + d*x])^3,x]

[Out]

((48*A*Cos[c/2 + (d*x)/2]^6*Cos[c + d*x]^2*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*(C + A*Sec[c + d*x]^2)
)/(d*(1 + Cos[c + d*x])^3*(2*A + C + C*Cos[2*c + 2*d*x])) - (48*A*Cos[c/2 + (d*x)/2]^6*Cos[c + d*x]^2*Log[Cos[
c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*(C + A*Sec[c + d*x]^2))/(d*(1 + Cos[c + d*x])^3*(2*A + C + C*Cos[2*c + 2*
d*x])) + (Cos[c/2 + (d*x)/2]*Cos[c + d*x]*Sec[c/2]*Sec[c]*(C + A*Sec[c + d*x]^2)*(-255*A*Sin[(d*x)/2] - 20*C*S
in[(d*x)/2] + 567*A*Sin[(3*d*x)/2] + 22*C*Sin[(3*d*x)/2] - 600*A*Sin[c - (d*x)/2] - 10*C*Sin[c - (d*x)/2] + 37
5*A*Sin[c + (d*x)/2] + 10*C*Sin[c + (d*x)/2] - 480*A*Sin[2*c + (d*x)/2] - 20*C*Sin[2*c + (d*x)/2] - 60*A*Sin[c
+ (3*d*x)/2] + 402*A*Sin[2*c + (3*d*x)/2] + 22*C*Sin[2*c + (3*d*x)/2] - 225*A*Sin[3*c + (3*d*x)/2] + 315*A*Si
n[c + (5*d*x)/2] + 10*C*Sin[c + (5*d*x)/2] + 30*A*Sin[2*c + (5*d*x)/2] + 240*A*Sin[3*c + (5*d*x)/2] + 10*C*Sin
[3*c + (5*d*x)/2] - 45*A*Sin[4*c + (5*d*x)/2] + 72*A*Sin[2*c + (7*d*x)/2] + 2*C*Sin[2*c + (7*d*x)/2] + 15*A*Si
n[3*c + (7*d*x)/2] + 57*A*Sin[4*c + (7*d*x)/2] + 2*C*Sin[4*c + (7*d*x)/2]))/(60*d*(1 + Cos[c + d*x])^3*(2*A +
C + C*Cos[2*c + 2*d*x])))/a^3

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Maple [A]  time = 0.061, size = 204, normalized size = 1.6 \begin{align*}{\frac{A}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{C}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{A}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{C}{6\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{17\,A}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{C}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{A}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+3\,{\frac{A\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }{d{a}^{3}}}-{\frac{A}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-3\,{\frac{A\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20/d/a^3*A*tan(1/2*d*x+1/2*c)^5+1/20/d/a^3*C*tan(1/2*d*x+1/2*c)^5+1/2/d/a^3*tan(1/2*d*x+1/2*c)^3*A+1/6/d/a^3
*C*tan(1/2*d*x+1/2*c)^3+17/4/d/a^3*A*tan(1/2*d*x+1/2*c)+1/4/d/a^3*C*tan(1/2*d*x+1/2*c)-1/d/a^3*A/(tan(1/2*d*x+
1/2*c)-1)+3/d/a^3*A*ln(tan(1/2*d*x+1/2*c)-1)-1/d/a^3*A/(tan(1/2*d*x+1/2*c)+1)-3/d/a^3*A*ln(tan(1/2*d*x+1/2*c)+
1)

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Maxima [A]  time = 1.03376, size = 315, normalized size = 2.44 \begin{align*} \frac{3 \, A{\left (\frac{40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac{a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}} + \frac{\frac{85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} + \frac{C{\left (\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(3*A*(40*sin(d*x + c)/((a^3 - a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x
+ c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 -
60*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) + C*(15*sin
(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)
/a^3)/d

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Fricas [A]  time = 1.46351, size = 576, normalized size = 4.47 \begin{align*} -\frac{45 \,{\left (A \cos \left (d x + c\right )^{4} + 3 \, A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \,{\left (A \cos \left (d x + c\right )^{4} + 3 \, A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (2 \,{\left (36 \, A + C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (57 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (117 \, A + 7 \, C\right )} \cos \left (d x + c\right ) + 15 \, A\right )} \sin \left (d x + c\right )}{30 \,{\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(45*(A*cos(d*x + c)^4 + 3*A*cos(d*x + c)^3 + 3*A*cos(d*x + c)^2 + A*cos(d*x + c))*log(sin(d*x + c) + 1)
- 45*(A*cos(d*x + c)^4 + 3*A*cos(d*x + c)^3 + 3*A*cos(d*x + c)^2 + A*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*
(2*(36*A + C)*cos(d*x + c)^3 + 3*(57*A + 2*C)*cos(d*x + c)^2 + (117*A + 7*C)*cos(d*x + c) + 15*A)*sin(d*x + c)
)/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**2/(a+a*cos(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.33288, size = 240, normalized size = 1.86 \begin{align*} -\frac{\frac{180 \, A \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{180 \, A \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac{120 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} a^{3}} - \frac{3 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 30 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 10 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 255 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

-1/60*(180*A*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 180*A*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 120*A*tan
(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^3) - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x
+ 1/2*c)^5 + 30*A*a^12*tan(1/2*d*x + 1/2*c)^3 + 10*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 255*A*a^12*tan(1/2*d*x + 1
/2*c) + 15*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d