3.372 \(\int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=164 \[ \frac{a^2 (11 B+14 C) \sin (c+d x)}{8 d \sqrt{a \sec (c+d x)+a}}+\frac{a^{3/2} (11 B+14 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{8 d}+\frac{a^2 (7 B+6 C) \sin (c+d x) \cos (c+d x)}{12 d \sqrt{a \sec (c+d x)+a}}+\frac{a B \sin (c+d x) \cos ^2(c+d x) \sqrt{a \sec (c+d x)+a}}{3 d} \]

[Out]

(a^(3/2)*(11*B + 14*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*d) + (a^2*(11*B + 14*C)*Sin
[c + d*x])/(8*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(7*B + 6*C)*Cos[c + d*x]*Sin[c + d*x])/(12*d*Sqrt[a + a*Sec[c
+ d*x]]) + (a*B*Cos[c + d*x]^2*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d)

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Rubi [A]  time = 0.470371, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4072, 4017, 4015, 3805, 3774, 203} \[ \frac{a^2 (11 B+14 C) \sin (c+d x)}{8 d \sqrt{a \sec (c+d x)+a}}+\frac{a^{3/2} (11 B+14 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{8 d}+\frac{a^2 (7 B+6 C) \sin (c+d x) \cos (c+d x)}{12 d \sqrt{a \sec (c+d x)+a}}+\frac{a B \sin (c+d x) \cos ^2(c+d x) \sqrt{a \sec (c+d x)+a}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(3/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^(3/2)*(11*B + 14*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*d) + (a^2*(11*B + 14*C)*Sin
[c + d*x])/(8*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(7*B + 6*C)*Cos[c + d*x]*Sin[c + d*x])/(12*d*Sqrt[a + a*Sec[c
+ d*x]]) + (a*B*Cos[c + d*x]^2*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d)

Rule 4072

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 4017

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

Rule 4015

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[(A*b^2*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(a*f*n*Sqrt[a + b*Csc[e + f*x]]), x] +
Dist[(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; Fr
eeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] &&
LtQ[n, 0]

Rule 3805

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(a*Cot[
e + f*x]*(d*Csc[e + f*x])^n)/(f*n*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[(a*(2*n + 1))/(2*b*d*n), Int[Sqrt[a + b
*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2
^(-1)] && IntegerQ[2*n]

Rule 3774

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

Rule 203

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

Rubi steps

\begin{align*} \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} (B+C \sec (c+d x)) \, dx\\ &=\frac{a B \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{1}{3} \int \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{1}{2} a (7 B+6 C)+\frac{3}{2} a (B+2 C) \sec (c+d x)\right ) \, dx\\ &=\frac{a^2 (7 B+6 C) \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}+\frac{a B \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{1}{8} (a (11 B+14 C)) \int \cos (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^2 (11 B+14 C) \sin (c+d x)}{8 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (7 B+6 C) \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}+\frac{a B \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{1}{16} (a (11 B+14 C)) \int \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^2 (11 B+14 C) \sin (c+d x)}{8 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (7 B+6 C) \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}+\frac{a B \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{3 d}-\frac{\left (a^2 (11 B+14 C)\right ) \operatorname{Subst}\left (\int \frac{1}{a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{8 d}\\ &=\frac{a^{3/2} (11 B+14 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{8 d}+\frac{a^2 (11 B+14 C) \sin (c+d x)}{8 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (7 B+6 C) \cos (c+d x) \sin (c+d x)}{12 d \sqrt{a+a \sec (c+d x)}}+\frac{a B \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [C]  time = 11.3871, size = 740, normalized size = 4.51 \[ a \left (\frac{B (\cos (c+d x)+1) \tan (c+d x) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{a (\sec (c+d x)+1)} \text{Hypergeometric2F1}\left (\frac{1}{2},3,\frac{3}{2},1-\sec (c+d x)\right )}{d (\sec (c+d x)+1)}+\frac{B (\cos (c+d x)+1) \tan (c+d x) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{a (\sec (c+d x)+1)} \left (\cos (c+d x) \sqrt{1-\sec (c+d x)}+\tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )\right )}{4 d \sqrt{-\tan ^2(c+d x)} \sqrt{\sec (c+d x)+1}}+\frac{B (\cos (c+d x)+1) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{a (\sec (c+d x)+1)} \left (\sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec (c+d x)+1} \left (7 \sin \left (\frac{1}{2} (c+d x)\right )-2 \sin \left (\frac{3}{2} (c+d x)\right )+3 \sin \left (\frac{5}{2} (c+d x)\right )+2 \sin \left (\frac{7}{2} (c+d x)\right )-3 \sqrt{2} \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \sqrt{\cos (c+d x)}\right )+\frac{12 \tan (c+d x) \left (\cos (c+d x) \sqrt{1-\sec (c+d x)}+\tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )\right )}{\sqrt{-\tan ^2(c+d x)}}\right )}{96 d \sqrt{\sec (c+d x)+1}}-\frac{C (\cos (c+d x)+1) \sec \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{a (\sec (c+d x)+1)} \left (\sin \left (\frac{1}{2} (c+d x)\right )-2 \sin \left (\frac{3}{2} (c+d x)\right )-\sin \left (\frac{5}{2} (c+d x)\right )+\sqrt{2} \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \sqrt{\cos (c+d x)}\right )}{16 d}+\frac{C \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \sqrt{\cos (c+d x)} (\cos (c+d x)+1) \sec \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{a (\sec (c+d x)+1)}}{2 \sqrt{2} d}+\frac{C (\cos (c+d x)+1) \tan (c+d x) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{a (\sec (c+d x)+1)} \left (\cos (c+d x) \sqrt{1-\sec (c+d x)}+\tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )\right )}{2 d \sqrt{-\tan ^2(c+d x)} \sqrt{\sec (c+d x)+1}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(3/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

a*((C*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x])*Sec[c/2 + (d*x)/2]^2*Sec[(c + d*x
)/2]*Sqrt[a*(1 + Sec[c + d*x])])/(2*Sqrt[2]*d) - (C*(1 + Cos[c + d*x])*Sec[c/2 + (d*x)/2]^2*Sec[(c + d*x)/2]*S
qrt[a*(1 + Sec[c + d*x])]*(Sqrt[2]*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Sqrt[Cos[c + d*x]] + Sin[(c + d*x)/2] - 2*
Sin[(3*(c + d*x))/2] - Sin[(5*(c + d*x))/2]))/(16*d) + (B*(1 + Cos[c + d*x])*Hypergeometric2F1[1/2, 3, 3/2, 1
- Sec[c + d*x]]*Sec[c/2 + (d*x)/2]^2*Sqrt[a*(1 + Sec[c + d*x])]*Tan[c + d*x])/(d*(1 + Sec[c + d*x])) + (B*(1 +
Cos[c + d*x])*Sec[c/2 + (d*x)/2]^2*(ArcTanh[Sqrt[1 - Sec[c + d*x]]] + Cos[c + d*x]*Sqrt[1 - Sec[c + d*x]])*Sq
rt[a*(1 + Sec[c + d*x])]*Tan[c + d*x])/(4*d*Sqrt[1 + Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2]) + (C*(1 + Cos[c + d*
x])*Sec[c/2 + (d*x)/2]^2*(ArcTanh[Sqrt[1 - Sec[c + d*x]]] + Cos[c + d*x]*Sqrt[1 - Sec[c + d*x]])*Sqrt[a*(1 + S
ec[c + d*x])]*Tan[c + d*x])/(2*d*Sqrt[1 + Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2]) + (B*(1 + Cos[c + d*x])*Sec[c/2
+ (d*x)/2]^2*Sqrt[a*(1 + Sec[c + d*x])]*(Sec[(c + d*x)/2]*Sqrt[1 + Sec[c + d*x]]*(-3*Sqrt[2]*ArcSin[Sqrt[2]*S
in[(c + d*x)/2]]*Sqrt[Cos[c + d*x]] + 7*Sin[(c + d*x)/2] - 2*Sin[(3*(c + d*x))/2] + 3*Sin[(5*(c + d*x))/2] + 2
*Sin[(7*(c + d*x))/2]) + (12*(ArcTanh[Sqrt[1 - Sec[c + d*x]]] + Cos[c + d*x]*Sqrt[1 - Sec[c + d*x]])*Tan[c + d
*x])/Sqrt[-Tan[c + d*x]^2]))/(96*d*Sqrt[1 + Sec[c + d*x]]))

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Maple [B]  time = 0.381, size = 581, normalized size = 3.5 \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)^4*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-1/192/d*a*(33*B*cos(d*x+c)^2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/
(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)+42*C*cos(d*x+c)^2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1
))^(5/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)+66*B*cos(d
*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*si
n(d*x+c)/cos(d*x+c))*sin(d*x+c)+84*C*cos(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*arctanh(1/2*2^(1/
2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)+33*B*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x
+c)+1))^(5/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)+42*C*
2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+
c)/cos(d*x+c))*sin(d*x+c)+64*B*cos(d*x+c)^6+112*B*cos(d*x+c)^5+96*C*cos(d*x+c)^5+88*B*cos(d*x+c)^4+240*C*cos(d
*x+c)^4-264*B*cos(d*x+c)^3-336*C*cos(d*x+c)^3)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)/cos(d*x+c)^2

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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)^4*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.679484, size = 949, normalized size = 5.79 \begin{align*} \left [\frac{3 \,{\left ({\left (11 \, B + 14 \, C\right )} a \cos \left (d x + c\right ) +{\left (11 \, B + 14 \, C\right )} a\right )} \sqrt{-a} \log \left (\frac{2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \,{\left (8 \, B a \cos \left (d x + c\right )^{3} + 2 \,{\left (11 \, B + 6 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \,{\left (11 \, B + 14 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{48 \,{\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac{3 \,{\left ({\left (11 \, B + 14 \, C\right )} a \cos \left (d x + c\right ) +{\left (11 \, B + 14 \, C\right )} a\right )} \sqrt{a} \arctan \left (\frac{\sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) -{\left (8 \, B a \cos \left (d x + c\right )^{3} + 2 \,{\left (11 \, B + 6 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \,{\left (11 \, B + 14 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{24 \,{\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/48*(3*((11*B + 14*C)*a*cos(d*x + c) + (11*B + 14*C)*a)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((
a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*(8*B
*a*cos(d*x + c)^3 + 2*(11*B + 6*C)*a*cos(d*x + c)^2 + 3*(11*B + 14*C)*a*cos(d*x + c))*sqrt((a*cos(d*x + c) + a
)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/24*(3*((11*B + 14*C)*a*cos(d*x + c) + (11*B + 14*C)*a)*
sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (8*B*a*cos(d*x +
c)^3 + 2*(11*B + 6*C)*a*cos(d*x + c)^2 + 3*(11*B + 14*C)*a*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*sin(d*x + c))/(d*cos(d*x + c) + 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)**4*(a+a*sec(d*x+c))**(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 7.01428, size = 1166, normalized size = 7.11 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

-1/48*(3*(11*B*sqrt(-a)*a*sgn(cos(d*x + c)) + 14*C*sqrt(-a)*a*sgn(cos(d*x + 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))^2 - a*(2*sqrt(2) + 3))) - 3*(11*B*sqrt(-a)*a*sgn(cos(d*x + c)
) + 14*C*sqrt(-a)*a*sgn(cos(d*x + 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))^2 + a*(2*sqrt(2) - 3))) + 4*sqrt(2)*(33*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2
+ a))^10*B*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 42*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^
2 + a))^10*C*sqrt(-a)*a^2*sgn(cos(d*x + c)) - 303*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c
)^2 + a))^8*B*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 822*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*
c)^2 + a))^8*C*sqrt(-a)*a^3*sgn(cos(d*x + c)) + 2394*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/
2*c)^2 + a))^6*B*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 3780*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x +
1/2*c)^2 + a))^6*C*sqrt(-a)*a^4*sgn(cos(d*x + c)) - 1806*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x
+ 1/2*c)^2 + a))^4*B*sqrt(-a)*a^5*sgn(cos(d*x + c)) - 2508*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*
x + 1/2*c)^2 + a))^4*C*sqrt(-a)*a^5*sgn(cos(d*x + c)) + 309*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d
*x + 1/2*c)^2 + a))^2*B*sqrt(-a)*a^6*sgn(cos(d*x + c)) + 498*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*
d*x + 1/2*c)^2 + a))^2*C*sqrt(-a)*a^6*sgn(cos(d*x + c)) - 19*B*sqrt(-a)*a^7*sgn(cos(d*x + c)) - 30*C*sqrt(-a)*
a^7*sgn(cos(d*x + c)))/((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(-a)*
tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^3)/d