3.142 \(\int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=209 \[ \frac{a^3 (163 A+200 B) \sin (c+d x)}{64 d \sqrt{a \sec (c+d x)+a}}+\frac{a^{5/2} (163 A+200 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{64 d}+\frac{a^2 (11 A+8 B) \sin (c+d x) \cos ^2(c+d x) \sqrt{a \sec (c+d x)+a}}{24 d}+\frac{a^3 (95 A+104 B) \sin (c+d x) \cos (c+d x)}{96 d \sqrt{a \sec (c+d x)+a}}+\frac{a A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d} \]

[Out]

(a^(5/2)*(163*A + 200*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(64*d) + (a^3*(163*A + 200*B
)*Sin[c + d*x])/(64*d*Sqrt[a + a*Sec[c + d*x]]) + (a^3*(95*A + 104*B)*Cos[c + d*x]*Sin[c + d*x])/(96*d*Sqrt[a
+ a*Sec[c + d*x]]) + (a^2*(11*A + 8*B)*Cos[c + d*x]^2*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(24*d) + (a*A*Cos
[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(4*d)

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Rubi [A]  time = 0.580095, 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} \[ \frac{a^3 (163 A+200 B) \sin (c+d x)}{64 d \sqrt{a \sec (c+d x)+a}}+\frac{a^{5/2} (163 A+200 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{64 d}+\frac{a^2 (11 A+8 B) \sin (c+d x) \cos ^2(c+d x) \sqrt{a \sec (c+d x)+a}}{24 d}+\frac{a^3 (95 A+104 B) \sin (c+d x) \cos (c+d x)}{96 d \sqrt{a \sec (c+d x)+a}}+\frac{a A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]

[Out]

(a^(5/2)*(163*A + 200*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(64*d) + (a^3*(163*A + 200*B
)*Sin[c + d*x])/(64*d*Sqrt[a + a*Sec[c + d*x]]) + (a^3*(95*A + 104*B)*Cos[c + d*x]*Sin[c + d*x])/(96*d*Sqrt[a
+ a*Sec[c + d*x]]) + (a^2*(11*A + 8*B)*Cos[c + d*x]^2*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(24*d) + (a*A*Cos
[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*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))^{5/2} (A+B \sec (c+d x)) \, dx &=\frac{a A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d}+\frac{1}{4} \int \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac{1}{2} a (11 A+8 B)+\frac{1}{2} a (3 A+8 B) \sec (c+d x)\right ) \, dx\\ &=\frac{a^2 (11 A+8 B) \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{a A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d}+\frac{1}{12} \int \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{1}{4} a^2 (95 A+104 B)+\frac{3}{4} a^2 (17 A+24 B) \sec (c+d x)\right ) \, dx\\ &=\frac{a^3 (95 A+104 B) \cos (c+d x) \sin (c+d x)}{96 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (11 A+8 B) \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{a A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d}+\frac{1}{64} \left (a^2 (163 A+200 B)\right ) \int \cos (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^3 (163 A+200 B) \sin (c+d x)}{64 d \sqrt{a+a \sec (c+d x)}}+\frac{a^3 (95 A+104 B) \cos (c+d x) \sin (c+d x)}{96 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (11 A+8 B) \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{a A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d}+\frac{1}{128} \left (a^2 (163 A+200 B)\right ) \int \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a^3 (163 A+200 B) \sin (c+d x)}{64 d \sqrt{a+a \sec (c+d x)}}+\frac{a^3 (95 A+104 B) \cos (c+d x) \sin (c+d x)}{96 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (11 A+8 B) \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{a A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d}-\frac{\left (a^3 (163 A+200 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^{5/2} (163 A+200 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{64 d}+\frac{a^3 (163 A+200 B) \sin (c+d x)}{64 d \sqrt{a+a \sec (c+d x)}}+\frac{a^3 (95 A+104 B) \cos (c+d x) \sin (c+d x)}{96 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 (11 A+8 B) \cos ^2(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{a A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d}\\ \end{align*}

Mathematica [C]  time = 1.292, size = 366, normalized size = 1.75 \[ \frac{a^2 \sin (c+d x) \sqrt{a (\sec (c+d x)+1)} \left (4608 A \sqrt{1-\sec (c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{2},5,\frac{3}{2},1-\sec (c+d x)\right )+7680 B \sqrt{1-\sec (c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{2},4,\frac{3}{2},1-\sec (c+d x)\right )+2079 A \sqrt{1-\sec (c+d x)}+7641 A \cos (c+d x) \sqrt{1-\sec (c+d x)}+2097 A \cos (2 (c+d x)) \sqrt{1-\sec (c+d x)}+522 A \cos (3 (c+d x)) \sqrt{1-\sec (c+d x)}+18 A \cos (4 (c+d x)) \sqrt{1-\sec (c+d x)}+6075 A \tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )+1240 B \sqrt{1-\sec (c+d x)}+6360 B \cos (c+d x) \sqrt{1-\sec (c+d x)}+1240 B \cos (2 (c+d x)) \sqrt{1-\sec (c+d x)}-80 B \cos (3 (c+d x)) \sqrt{1-\sec (c+d x)}+6600 B \tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )\right )}{2880 d (\cos (c+d x)+1) \sqrt{1-\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]

[Out]

(a^2*(6075*A*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + 6600*B*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + 2079*A*Sqrt[1 - Sec[c
+ d*x]] + 1240*B*Sqrt[1 - Sec[c + d*x]] + 7641*A*Cos[c + d*x]*Sqrt[1 - Sec[c + d*x]] + 6360*B*Cos[c + d*x]*Sqr
t[1 - Sec[c + d*x]] + 2097*A*Cos[2*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 1240*B*Cos[2*(c + d*x)]*Sqrt[1 - Sec[c
+ d*x]] + 522*A*Cos[3*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] - 80*B*Cos[3*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 18*A*
Cos[4*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 7680*B*Hypergeometric2F1[1/2, 4, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec
[c + d*x]] + 4608*A*Hypergeometric2F1[1/2, 5, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]])*Sqrt[a*(1 + Sec[c
+ d*x])]*Sin[c + d*x])/(2880*d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])

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Maple [B]  time = 0.285, size = 765, 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))^(5/2)*(A+B*sec(d*x+c)),x)

[Out]

1/3072/d*a^2*(489*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)+600*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)+1467*A
*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*c
os(d*x+c)/(cos(d*x+c)+1))^(7/2)*2^(1/2)+1800*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)+1467*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)+1800*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)+489*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)+600*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)^8-2176*A*cos(d*x+c)^7-1024*B*cos(d*x+c)^7-2272*A*cos(d*x+c)^6-3328*B*cos(d*
x+c)^6-2608*A*cos(d*x+c)^5-5248*B*cos(d*x+c)^5+7824*A*cos(d*x+c)^4+9600*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))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.719529, size = 1103, normalized size = 5.28 \begin{align*} \left [\frac{3 \,{\left ({\left (163 \, A + 200 \, B\right )} a^{2} \cos \left (d x + c\right ) +{\left (163 \, A + 200 \, B\right )} a^{2}\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^{2} \cos \left (d x + c\right )^{4} + 8 \,{\left (23 \, A + 8 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \,{\left (163 \, A + 136 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (163 \, A + 200 \, B\right )} a^{2} \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 (163 \, A + 200 \, B\right )} a^{2} \cos \left (d x + c\right ) +{\left (163 \, A + 200 \, B\right )} a^{2}\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^{2} \cos \left (d x + c\right )^{4} + 8 \,{\left (23 \, A + 8 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \,{\left (163 \, A + 136 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (163 \, A + 200 \, B\right )} a^{2} \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))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

[1/384*(3*((163*A + 200*B)*a^2*cos(d*x + c) + (163*A + 200*B)*a^2)*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^2*cos(d*x + c)^4 + 8*(23*A + 8*B)*a^2*cos(d*x + c)^3 + 2*(163*A + 136*B)*a^2*cos(d*x + c)^2 + 3*(
163*A + 200*B)*a^2*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), -
1/192*(3*((163*A + 200*B)*a^2*cos(d*x + c) + (163*A + 200*B)*a^2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos
(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (48*A*a^2*cos(d*x + c)^4 + 8*(23*A + 8*B)*a^2*cos(d*x + c)^3
+ 2*(163*A + 136*B)*a^2*cos(d*x + c)^2 + 3*(163*A + 200*B)*a^2*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))**(5/2)*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 8.43719, size = 1480, normalized size = 7.08 \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))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/384*(3*(163*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 200*B*sqrt(-a)*a^2*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*(163*A*sqrt(-a)*a^2*sgn(co
s(d*x + c)) + 200*B*sqrt(-a)*a^2*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)*(489*(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^3*sgn(cos(d*x + c)) + 600*(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^3*sgn(cos(d*x + c)) - 10269*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*ta
n(1/2*d*x + 1/2*c)^2 + a))^12*A*sqrt(-a)*a^4*sgn(cos(d*x + c)) - 12600*(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^4*sgn(cos(d*x + c)) + 69885*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sq
rt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*A*sqrt(-a)*a^5*sgn(cos(d*x + c)) + 103992*(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^5*sgn(cos(d*x + c)) - 259233*(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^6*sgn(cos(d*x + c)) - 339864*(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^6*sgn(cos(d*x + c)) + 209979*(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^7*sgn(cos(d*x + c)) + 262920*(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^7*sgn(cos(d*x + c)) - 55511*(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^8*sgn(cos(d*x + c)) - 73640*(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^8*sgn(cos(d*x + c)) + 6687*(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^9*sgn(cos(d*x + c)) + 8808*(sq
rt(-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^9*sgn(cos(d*x + c)) - 299*A*
sqrt(-a)*a^10*sgn(cos(d*x + c)) - 392*B*sqrt(-a)*a^10*sgn(cos(d*x + c)))/((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqr
t(-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