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

Optimal. Leaf size=208 \[ \frac{2 (35 A-7 B+31 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{105 a d}-\frac{4 (35 A-49 B+37 C) \sin (c+d x)}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{\sqrt{2} (A-B+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 (7 B-C) \sin (c+d x) \cos ^2(c+d x)}{35 d \sqrt{a \cos (c+d x)+a}}+\frac{2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt{a \cos (c+d x)+a}} \]

[Out]

(Sqrt[2]*(A - B + C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a]*d) - (4*(35*
A - 49*B + 37*C)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*(7*B - C)*Cos[c + d*x]^2*Sin[c + d*x])/(3
5*d*Sqrt[a + a*Cos[c + d*x]]) + (2*C*Cos[c + d*x]^3*Sin[c + d*x])/(7*d*Sqrt[a + a*Cos[c + d*x]]) + (2*(35*A -
7*B + 31*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(105*a*d)

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Rubi [A]  time = 0.611791, antiderivative size = 208, 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 = {3045, 2983, 2968, 3023, 2751, 2649, 206} \[ \frac{2 (35 A-7 B+31 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{105 a d}-\frac{4 (35 A-49 B+37 C) \sin (c+d x)}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{\sqrt{2} (A-B+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 (7 B-C) \sin (c+d x) \cos ^2(c+d x)}{35 d \sqrt{a \cos (c+d x)+a}}+\frac{2 C \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*(A - B + C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a]*d) - (4*(35*
A - 49*B + 37*C)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*(7*B - C)*Cos[c + d*x]^2*Sin[c + d*x])/(3
5*d*Sqrt[a + a*Cos[c + d*x]]) + (2*C*Cos[c + d*x]^3*Sin[c + d*x])/(7*d*Sqrt[a + a*Cos[c + d*x]]) + (2*(35*A -
7*B + 31*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(105*a*d)

Rule 3045

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

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 )}{\sqrt{a+a \cos (c+d x)}} \, dx &=\frac{2 C \cos ^3(c+d x) \sin (c+d x)}{7 d \sqrt{a+a \cos (c+d x)}}+\frac{2 \int \frac{\cos ^2(c+d x) \left (\frac{1}{2} a (7 A+6 C)+\frac{1}{2} a (7 B-C) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{7 a}\\ &=\frac{2 (7 B-C) \cos ^2(c+d x) \sin (c+d x)}{35 d \sqrt{a+a \cos (c+d x)}}+\frac{2 C \cos ^3(c+d x) \sin (c+d x)}{7 d \sqrt{a+a \cos (c+d x)}}+\frac{4 \int \frac{\cos (c+d x) \left (a^2 (7 B-C)+\frac{1}{4} a^2 (35 A-7 B+31 C) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{35 a^2}\\ &=\frac{2 (7 B-C) \cos ^2(c+d x) \sin (c+d x)}{35 d \sqrt{a+a \cos (c+d x)}}+\frac{2 C \cos ^3(c+d x) \sin (c+d x)}{7 d \sqrt{a+a \cos (c+d x)}}+\frac{4 \int \frac{a^2 (7 B-C) \cos (c+d x)+\frac{1}{4} a^2 (35 A-7 B+31 C) \cos ^2(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{35 a^2}\\ &=\frac{2 (7 B-C) \cos ^2(c+d x) \sin (c+d x)}{35 d \sqrt{a+a \cos (c+d x)}}+\frac{2 C \cos ^3(c+d x) \sin (c+d x)}{7 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (35 A-7 B+31 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{105 a d}+\frac{8 \int \frac{\frac{1}{8} a^3 (35 A-7 B+31 C)-\frac{1}{4} a^3 (35 A-49 B+37 C) \cos (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{105 a^3}\\ &=-\frac{4 (35 A-49 B+37 C) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (7 B-C) \cos ^2(c+d x) \sin (c+d x)}{35 d \sqrt{a+a \cos (c+d x)}}+\frac{2 C \cos ^3(c+d x) \sin (c+d x)}{7 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (35 A-7 B+31 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{105 a d}+(A-B+C) \int \frac{1}{\sqrt{a+a \cos (c+d x)}} \, dx\\ &=-\frac{4 (35 A-49 B+37 C) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (7 B-C) \cos ^2(c+d x) \sin (c+d x)}{35 d \sqrt{a+a \cos (c+d x)}}+\frac{2 C \cos ^3(c+d x) \sin (c+d x)}{7 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (35 A-7 B+31 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{105 a d}-\frac{(2 (A-B+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 )}{d}\\ &=\frac{\sqrt{2} (A-B+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \cos (c+d x)}}\right )}{\sqrt{a} d}-\frac{4 (35 A-49 B+37 C) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (7 B-C) \cos ^2(c+d x) \sin (c+d x)}{35 d \sqrt{a+a \cos (c+d x)}}+\frac{2 C \cos ^3(c+d x) \sin (c+d x)}{7 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (35 A-7 B+31 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{105 a d}\\ \end{align*}

Mathematica [A]  time = 0.637913, size = 118, normalized size = 0.57 \[ \frac{\cos \left (\frac{1}{2} (c+d x)\right ) \left (420 (A-B+C) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \sin \left (\frac{1}{2} (c+d x)\right ) ((140 A-28 B+169 C) \cos (c+d x)-140 A+6 (7 B-C) \cos (2 (c+d x))+406 B+15 C \cos (3 (c+d x))-178 C)\right )}{210 d \sqrt{a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*(420*(A - B + C)*ArcTanh[Sin[(c + d*x)/2]] + 2*(-140*A + 406*B - 178*C + (140*A - 28*B + 169
*C)*Cos[c + d*x] + 6*(7*B - C)*Cos[2*(c + d*x)] + 15*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(210*d*Sqrt[a*(1 +
Cos[c + d*x])])

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Maple [A]  time = 0.143, size = 324, normalized size = 1.6 \begin{align*}{\frac{1}{105\,d}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sqrt{a \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( -240\,C\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+168\,\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( B+2\,C \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-140\,\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( A+B+2\,C \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+105\,\sqrt{2}\ln \left ( 4\,{\frac{\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) aA-105\,\sqrt{2}\ln \left ( 4\,{\frac{\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) aB+105\,\sqrt{2}\ln \left ( 4\,{\frac{\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) aC+210\,B\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \right ){a}^{-{\frac{3}{2}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{a \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}}} \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))^(1/2),x)

[Out]

1/105*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-240*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+168*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*(B+2*C)*sin(1/2*d*x+1/2*c)^4-140*2^(1
/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*(A+B+2*C)*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-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*B+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*C+210*
B*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2))/a^(3/2)/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))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.01942, size = 524, normalized size = 2.52 \begin{align*} \frac{4 \,{\left (15 \, C \cos \left (d x + c\right )^{3} + 3 \,{\left (7 \, B - C\right )} \cos \left (d x + c\right )^{2} +{\left (35 \, A - 7 \, B + 31 \, C\right )} \cos \left (d x + c\right ) - 35 \, A + 91 \, B - 43 \, C\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right ) + \frac{105 \, \sqrt{2}{\left ({\left (A - B + C\right )} a \cos \left (d x + c\right ) +{\left (A - B + C\right )} a\right )} \log \left (-\frac{\cos \left (d x + c\right )^{2} - \frac{2 \, \sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt{a}}}{210 \,{\left (a d \cos \left (d x + c\right ) + a 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))^(1/2),x, algorithm="fricas")

[Out]

1/210*(4*(15*C*cos(d*x + c)^3 + 3*(7*B - C)*cos(d*x + c)^2 + (35*A - 7*B + 31*C)*cos(d*x + c) - 35*A + 91*B -
43*C)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c) + 105*sqrt(2)*((A - B + C)*a*cos(d*x + c) + (A - B + C)*a)*log(-(c
os(d*x + c)^2 - 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2
+ 2*cos(d*x + c) + 1))/sqrt(a))/(a*d*cos(d*x + c) + a*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))**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.87054, size = 271, normalized size = 1.3 \begin{align*} -\frac{\frac{105 \, \sqrt{2}{\left (A - B + 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 )}{\sqrt{a}} - \frac{2 \,{\left (105 \, \sqrt{2} B a^{3} -{\left ({\left (\sqrt{2}{\left (70 \, A a^{3} - 119 \, B a^{3} + 92 \, C a^{3}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 7 \, \sqrt{2}{\left (20 \, A a^{3} - 37 \, B a^{3} + 16 \, C a^{3}\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 35 \, \sqrt{2}{\left (2 \, A a^{3} - 7 \, B a^{3} + 4 \, C a^{3}\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{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{7}{2}}}}{105 \, 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))^(1/2),x, algorithm="giac")

[Out]

-1/105*(105*sqrt(2)*(A - B + 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)))/s
qrt(a) - 2*(105*sqrt(2)*B*a^3 - ((sqrt(2)*(70*A*a^3 - 119*B*a^3 + 92*C*a^3)*tan(1/2*d*x + 1/2*c)^2 + 7*sqrt(2)
*(20*A*a^3 - 37*B*a^3 + 16*C*a^3))*tan(1/2*d*x + 1/2*c)^2 + 35*sqrt(2)*(2*A*a^3 - 7*B*a^3 + 4*C*a^3))*tan(1/2*
d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(7/2))/d