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

Optimal. Leaf size=209 \[ \frac{a^2 (75 A+88 B) \sin (c+d x)}{64 d \sqrt{a \sec (c+d x)+a}}+\frac{a^{3/2} (75 A+88 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{64 d}+\frac{a^2 (9 A+8 B) \sin (c+d x) \cos ^2(c+d x)}{24 d \sqrt{a \sec (c+d x)+a}}+\frac{a^2 (75 A+88 B) \sin (c+d x) \cos (c+d x)}{96 d \sqrt{a \sec (c+d x)+a}}+\frac{a A \sin (c+d x) \cos ^3(c+d x) \sqrt{a \sec (c+d x)+a}}{4 d} \]

[Out]

(a^(3/2)*(75*A + 88*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(64*d) + (a^2*(75*A + 88*B)*Si
n[c + d*x])/(64*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(75*A + 88*B)*Cos[c + d*x]*Sin[c + d*x])/(96*d*Sqrt[a + a*S
ec[c + d*x]]) + (a^2*(9*A + 8*B)*Cos[c + d*x]^2*Sin[c + d*x])/(24*d*Sqrt[a + a*Sec[c + d*x]]) + (a*A*Cos[c + d
*x]^3*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(4*d)

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Rubi [A]  time = 0.448674, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {4017, 4015, 3805, 3774, 203} \[ \frac{a^2 (75 A+88 B) \sin (c+d x)}{64 d \sqrt{a \sec (c+d x)+a}}+\frac{a^{3/2} (75 A+88 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{64 d}+\frac{a^2 (9 A+8 B) \sin (c+d x) \cos ^2(c+d x)}{24 d \sqrt{a \sec (c+d x)+a}}+\frac{a^2 (75 A+88 B) \sin (c+d x) \cos (c+d x)}{96 d \sqrt{a \sec (c+d x)+a}}+\frac{a A \sin (c+d x) \cos ^3(c+d x) \sqrt{a \sec (c+d x)+a}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(3/2)*(75*A + 88*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(64*d) + (a^2*(75*A + 88*B)*Si
n[c + d*x])/(64*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(75*A + 88*B)*Cos[c + d*x]*Sin[c + d*x])/(96*d*Sqrt[a + a*S
ec[c + d*x]]) + (a^2*(9*A + 8*B)*Cos[c + d*x]^2*Sin[c + d*x])/(24*d*Sqrt[a + a*Sec[c + d*x]]) + (a*A*Cos[c + d
*x]^3*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(4*d)

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} (A+B \sec (c+d x)) \, dx &=\frac{a A \cos ^3(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{1}{4} \int \cos ^3(c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{1}{2} a (9 A+8 B)+\frac{1}{2} a (5 A+8 B) \sec (c+d x)\right ) \, dx\\ &=\frac{a^2 (9 A+8 B) \cos ^2(c+d x) \sin (c+d x)}{24 d \sqrt{a+a \sec (c+d x)}}+\frac{a A \cos ^3(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{1}{48} (a (75 A+88 B)) \int \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^2 (75 A+88 B) \cos (c+d x) \sin (c+d x)}{96 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (9 A+8 B) \cos ^2(c+d x) \sin (c+d x)}{24 d \sqrt{a+a \sec (c+d x)}}+\frac{a A \cos ^3(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{1}{64} (a (75 A+88 B)) \int \cos (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^2 (75 A+88 B) \sin (c+d x)}{64 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (75 A+88 B) \cos (c+d x) \sin (c+d x)}{96 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (9 A+8 B) \cos ^2(c+d x) \sin (c+d x)}{24 d \sqrt{a+a \sec (c+d x)}}+\frac{a A \cos ^3(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{1}{128} (a (75 A+88 B)) \int \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^2 (75 A+88 B) \sin (c+d x)}{64 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (75 A+88 B) \cos (c+d x) \sin (c+d x)}{96 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (9 A+8 B) \cos ^2(c+d x) \sin (c+d x)}{24 d \sqrt{a+a \sec (c+d x)}}+\frac{a A \cos ^3(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{4 d}-\frac{\left (a^2 (75 A+88 B)\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 )}{64 d}\\ &=\frac{a^{3/2} (75 A+88 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{64 d}+\frac{a^2 (75 A+88 B) \sin (c+d x)}{64 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (75 A+88 B) \cos (c+d x) \sin (c+d x)}{96 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (9 A+8 B) \cos ^2(c+d x) \sin (c+d x)}{24 d \sqrt{a+a \sec (c+d x)}}+\frac{a A \cos ^3(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 1.3009, size = 154, normalized size = 0.74 \[ \frac{a \cos (c+d x) \sqrt{a (\sec (c+d x)+1)} \left (\sin (c+d x) \sqrt{1-\sec (c+d x)} (2 (93 A+88 B) \cos (c+d x)+4 (15 A+8 B) \cos (2 (c+d x))+12 A \cos (3 (c+d x))+285 A+296 B)+3 (75 A+88 B) \tan (c+d x) \tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )\right )}{192 d (\cos (c+d x)+1) \sqrt{1-\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Cos[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*((285*A + 296*B + 2*(93*A + 88*B)*Cos[c + d*x] + 4*(15*A + 8*B)*Cos
[2*(c + d*x)] + 12*A*Cos[3*(c + d*x)])*Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] + 3*(75*A + 88*B)*ArcTanh[Sqrt[1 -
Sec[c + d*x]]]*Tan[c + d*x]))/(192*d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])

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Maple [B]  time = 0.294, size = 763, normalized size = 3.7 \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)*(A+B*sec(d*x+c)),x)

[Out]

1/3072/d*a*(225*A*sin(d*x+c)*cos(d*x+c)^3*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))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*2^(1/2)+264*B*sin(d*x+c)*cos(d*x+c)^3*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))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*2^(1/2)+675*A*si
n(d*x+c)*cos(d*x+c)^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))*(-2*cos(
d*x+c)/(cos(d*x+c)+1))^(7/2)*2^(1/2)+792*B*sin(d*x+c)*cos(d*x+c)^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))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*2^(1/2)+675*A*sin(d*x+c)*cos(d*x+c)*
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))*(-2*cos(d*x+c)/(cos(d*x+c)+1))
^(7/2)*2^(1/2)+792*B*sin(d*x+c)*cos(d*x+c)*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))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*2^(1/2)+225*A*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))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*2^(1/2)*sin(d*x+c)+264*B*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))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*2^(1/2)
*sin(d*x+c)-768*A*cos(d*x+c)^8-1152*A*cos(d*x+c)^7-1024*B*cos(d*x+c)^7-480*A*cos(d*x+c)^6-1792*B*cos(d*x+c)^6-
1200*A*cos(d*x+c)^5-1408*B*cos(d*x+c)^5+3600*A*cos(d*x+c)^4+4224*B*cos(d*x+c)^4)*(a*(cos(d*x+c)+1)/cos(d*x+c))
^(1/2)/sin(d*x+c)/cos(d*x+c)^3

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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)*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.717182, size = 1049, normalized size = 5.02 \begin{align*} \left [\frac{3 \,{\left ({\left (75 \, A + 88 \, B\right )} a \cos \left (d x + c\right ) +{\left (75 \, A + 88 \, B\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 (48 \, A a \cos \left (d x + c\right )^{4} + 8 \,{\left (15 \, A + 8 \, B\right )} a \cos \left (d x + c\right )^{3} + 2 \,{\left (75 \, A + 88 \, B\right )} a \cos \left (d x + c\right )^{2} + 3 \,{\left (75 \, A + 88 \, B\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 )}{384 \,{\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac{3 \,{\left ({\left (75 \, A + 88 \, B\right )} a \cos \left (d x + c\right ) +{\left (75 \, A + 88 \, B\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 (48 \, A a \cos \left (d x + c\right )^{4} + 8 \,{\left (15 \, A + 8 \, B\right )} a \cos \left (d x + c\right )^{3} + 2 \,{\left (75 \, A + 88 \, B\right )} a \cos \left (d x + c\right )^{2} + 3 \,{\left (75 \, A + 88 \, B\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 )}{192 \,{\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)*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

[1/384*(3*((75*A + 88*B)*a*cos(d*x + c) + (75*A + 88*B)*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*(48
*A*a*cos(d*x + c)^4 + 8*(15*A + 8*B)*a*cos(d*x + c)^3 + 2*(75*A + 88*B)*a*cos(d*x + c)^2 + 3*(75*A + 88*B)*a*c
os(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/192*(3*((75*A + 88
*B)*a*cos(d*x + c) + (75*A + 88*B)*a)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqr
t(a)*sin(d*x + c))) - (48*A*a*cos(d*x + c)^4 + 8*(15*A + 8*B)*a*cos(d*x + c)^3 + 2*(75*A + 88*B)*a*cos(d*x + c
)^2 + 3*(75*A + 88*B)*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)*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 7.80016, size = 1469, normalized size = 7.03 \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)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/384*(3*(75*A*sqrt(-a)*a*sgn(cos(d*x + c)) + 88*B*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*(75*A*sqrt(-a)*a*sgn(cos(d*x + c
)) + 88*B*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)*(225*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)
^2 + a))^14*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 264*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*
c)^2 + a))^14*B*sqrt(-a)*a^2*sgn(cos(d*x + c)) - 6261*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1
/2*c)^2 + a))^12*A*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 4008*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x
+ 1/2*c)^2 + a))^12*B*sqrt(-a)*a^3*sgn(cos(d*x + c)) + 35925*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*
d*x + 1/2*c)^2 + a))^10*A*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 33960*(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^4*sgn(cos(d*x + c)) - 127449*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a
*tan(1/2*d*x + 1/2*c)^2 + a))^8*A*sqrt(-a)*a^5*sgn(cos(d*x + c)) - 131784*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqr
t(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*B*sqrt(-a)*a^5*sgn(cos(d*x + c)) + 101667*(sqrt(-a)*tan(1/2*d*x + 1/2*c) -
sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*sqrt(-a)*a^6*sgn(cos(d*x + c)) + 108312*(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^6*sgn(cos(d*x + c)) - 26079*(sqrt(-a)*tan(1/2*d*x + 1
/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*sqrt(-a)*a^7*sgn(cos(d*x + c)) - 29432*(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^7*sgn(cos(d*x + c)) + 3303*(sqrt(-a)*tan(1/2*d*
x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A*sqrt(-a)*a^8*sgn(cos(d*x + c)) + 3384*(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^8*sgn(cos(d*x + c)) - 147*A*sqrt(-a)*a^9*sg
n(cos(d*x + c)) - 152*B*sqrt(-a)*a^9*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)^4)/d