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

Optimal. Leaf size=229 \[ -\frac{(15 A-35 B+39 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{30 a^2 d}-\frac{(7 A-11 B+15 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}+\frac{(5 A-5 B+9 C) \sin (c+d x) \cos ^2(c+d x)}{10 a d \sqrt{a \cos (c+d x)+a}}+\frac{(45 A-65 B+93 C) \sin (c+d x)}{15 a d \sqrt{a \cos (c+d x)+a}} \]

[Out]

-((7*A - 11*B + 15*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d
) - ((A - B + C)*Cos[c + d*x]^3*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((45*A - 65*B + 93*C)*Sin[c +
d*x])/(15*a*d*Sqrt[a + a*Cos[c + d*x]]) + ((5*A - 5*B + 9*C)*Cos[c + d*x]^2*Sin[c + d*x])/(10*a*d*Sqrt[a + a*
Cos[c + d*x]]) - ((15*A - 35*B + 39*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(30*a^2*d)

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Rubi [A]  time = 0.669542, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {3041, 2983, 2968, 3023, 2751, 2649, 206} \[ -\frac{(15 A-35 B+39 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{30 a^2 d}-\frac{(7 A-11 B+15 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}+\frac{(5 A-5 B+9 C) \sin (c+d x) \cos ^2(c+d x)}{10 a d \sqrt{a \cos (c+d x)+a}}+\frac{(45 A-65 B+93 C) \sin (c+d x)}{15 a d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((7*A - 11*B + 15*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d
) - ((A - B + C)*Cos[c + d*x]^3*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((45*A - 65*B + 93*C)*Sin[c +
d*x])/(15*a*d*Sqrt[a + a*Cos[c + d*x]]) + ((5*A - 5*B + 9*C)*Cos[c + d*x]^2*Sin[c + d*x])/(10*a*d*Sqrt[a + a*
Cos[c + d*x]]) - ((15*A - 35*B + 39*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(30*a^2*d)

Rule 3041

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((a*A - b*B + 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)) + B*(
b*c*m + a*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(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, B, 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 2983

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*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(f*(
m + n + 1)), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*b*c*(
m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && (I
ntegerQ[n] || EqQ[m + 1/2, 0])

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2751

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

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx &=-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\cos ^2(c+d x) \left (-a (A-3 B+3 C)+\frac{1}{2} a (5 A-5 B+9 C) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(5 A-5 B+9 C) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{\cos (c+d x) \left (a^2 (5 A-5 B+9 C)-\frac{1}{4} a^2 (15 A-35 B+39 C) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{5 a^3}\\ &=-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(5 A-5 B+9 C) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{a^2 (5 A-5 B+9 C) \cos (c+d x)-\frac{1}{4} a^2 (15 A-35 B+39 C) \cos ^2(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{5 a^3}\\ &=-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(5 A-5 B+9 C) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}-\frac{(15 A-35 B+39 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{30 a^2 d}+\frac{2 \int \frac{-\frac{1}{8} a^3 (15 A-35 B+39 C)+\frac{1}{4} a^3 (45 A-65 B+93 C) \cos (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{15 a^4}\\ &=-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(45 A-65 B+93 C) \sin (c+d x)}{15 a d \sqrt{a+a \cos (c+d x)}}+\frac{(5 A-5 B+9 C) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}-\frac{(15 A-35 B+39 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{30 a^2 d}-\frac{(7 A-11 B+15 C) \int \frac{1}{\sqrt{a+a \cos (c+d x)}} \, dx}{4 a}\\ &=-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(45 A-65 B+93 C) \sin (c+d x)}{15 a d \sqrt{a+a \cos (c+d x)}}+\frac{(5 A-5 B+9 C) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}-\frac{(15 A-35 B+39 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{30 a^2 d}+\frac{(7 A-11 B+15 C) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{2 a d}\\ &=-\frac{(7 A-11 B+15 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \cos (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(45 A-65 B+93 C) \sin (c+d x)}{15 a d \sqrt{a+a \cos (c+d x)}}+\frac{(5 A-5 B+9 C) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}-\frac{(15 A-35 B+39 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{30 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.722273, size = 153, normalized size = 0.67 \[ \frac{15 (7 A-11 B+15 C) \cos ^5\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-\sin \left (\frac{1}{2} (c+d x)\right ) \cos ^3\left (\frac{1}{2} (c+d x)\right ) (3 (20 A-20 B+39 C) \cos (c+d x)+75 A+2 (5 B-3 C) \cos (2 (c+d x))-85 B+3 C \cos (3 (c+d x))+141 C)}{15 d \left (\sin ^2\left (\frac{1}{2} (c+d x)\right )-1\right ) (a (\cos (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*(7*A - 11*B + 15*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^5 - Cos[(c + d*x)/2]^3*(75*A - 85*B + 141*C
+ 3*(20*A - 20*B + 39*C)*Cos[c + d*x] + 2*(5*B - 3*C)*Cos[2*(c + d*x)] + 3*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/
2])/(15*d*(a*(1 + Cos[c + d*x]))^(3/2)*(-1 + Sin[(c + d*x)/2]^2))

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Maple [B]  time = 0.159, size = 533, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(96*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*sin(1/2*d*x+1/2*c)^6
-16*2^(1/2)*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(5*B+6*C)*sin(1/2*d*x+1/2*c)^4+5*2^(1/2)*(24*A*a^(1/2)*(a*s
in(1/2*d*x+1/2*c)^2)^(1/2)-21*A*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a))*a-8*B*(a*s
in(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+33*B*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a))*a+
48*C*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-45*C*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/
2)+a))*a)*sin(1/2*d*x+1/2*c)^2+105*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a))
*a*A-165*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a))*a*B+225*2^(1/2)*ln(4/cos(
1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a))*a*C-135*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^
(1/2)+135*B*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-255*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2
))/cos(1/2*d*x+1/2*c)/a^(5/2)/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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

[Out]

Timed out

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Fricas [A]  time = 1.98861, size = 641, normalized size = 2.8 \begin{align*} \frac{15 \, \sqrt{2}{\left ({\left (7 \, A - 11 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (7 \, A - 11 \, B + 15 \, C\right )} \cos \left (d x + c\right ) + 7 \, A - 11 \, B + 15 \, C\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} + 2 \, \sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \,{\left (12 \, C \cos \left (d x + c\right )^{3} + 4 \,{\left (5 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{2} + 12 \,{\left (5 \, A - 5 \, B + 9 \, C\right )} \cos \left (d x + c\right ) + 75 \, A - 95 \, B + 147 \, C\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{120 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*sqrt(2)*((7*A - 11*B + 15*C)*cos(d*x + c)^2 + 2*(7*A - 11*B + 15*C)*cos(d*x + c) + 7*A - 11*B + 15*C
)*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) - 2*a*cos(d*x + c)
- 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(12*C*cos(d*x + c)^3 + 4*(5*B - 3*C)*cos(d*x + c)^2 + 12*(5*
A - 5*B + 9*C)*cos(d*x + c) + 75*A - 95*B + 147*C)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^2*d*cos(d*x + c)^
2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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

[Out]

Timed out

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Giac [A]  time = 2.59542, size = 308, normalized size = 1.34 \begin{align*} \frac{\frac{15 \, \sqrt{2}{\left (7 \, A - 11 \, B + 15 \, C\right )} \log \left ({\left | -\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac{3}{2}}} + \frac{{\left ({\left ({\left (\frac{15 \, \sqrt{2}{\left (A a^{3} - B a^{3} + C a^{3}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{2}} + \frac{\sqrt{2}{\left (165 \, A a^{3} - 245 \, B a^{3} + 381 \, C a^{3}\right )}}{a^{2}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \frac{5 \, \sqrt{2}{\left (57 \, A a^{3} - 73 \, B a^{3} + 105 \, C a^{3}\right )}}{a^{2}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \frac{15 \, \sqrt{2}{\left (9 \, A a^{3} - 9 \, B a^{3} + 17 \, C a^{3}\right )}}{a^{2}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{5}{2}}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(15*sqrt(2)*(7*A - 11*B + 15*C)*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a
)))/a^(3/2) + (((15*sqrt(2)*(A*a^3 - B*a^3 + C*a^3)*tan(1/2*d*x + 1/2*c)^2/a^2 + sqrt(2)*(165*A*a^3 - 245*B*a^
3 + 381*C*a^3)/a^2)*tan(1/2*d*x + 1/2*c)^2 + 5*sqrt(2)*(57*A*a^3 - 73*B*a^3 + 105*C*a^3)/a^2)*tan(1/2*d*x + 1/
2*c)^2 + 15*sqrt(2)*(9*A*a^3 - 9*B*a^3 + 17*C*a^3)/a^2)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(5
/2))/d