3.135 \(\int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=237 \[ \frac{2 a^3 (209 A+194 B) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (11 A+14 B) \tan (c+d x) \sec ^3(c+d x) \sqrt{a \sec (c+d x)+a}}{99 d}+\frac{2 a^3 (803 A+710 B) \tan (c+d x)}{495 d \sqrt{a \sec (c+d x)+a}}-\frac{4 a^2 (803 A+710 B) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3465 d}+\frac{2 a (803 A+710 B) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 d}+\frac{2 a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d} \]

[Out]

(2*a^3*(803*A + 710*B)*Tan[c + d*x])/(495*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(209*A + 194*B)*Sec[c + d*x]^3*
Tan[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) - (4*a^2*(803*A + 710*B)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])
/(3465*d) + (2*a^2*(11*A + 14*B)*Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(99*d) + (2*a*(803*A +
710*B)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(1155*d) + (2*a*B*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*Ta
n[c + d*x])/(11*d)

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Rubi [A]  time = 0.656926, antiderivative size = 237, 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 = {4018, 4016, 3800, 4001, 3792} \[ \frac{2 a^3 (209 A+194 B) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (11 A+14 B) \tan (c+d x) \sec ^3(c+d x) \sqrt{a \sec (c+d x)+a}}{99 d}+\frac{2 a^3 (803 A+710 B) \tan (c+d x)}{495 d \sqrt{a \sec (c+d x)+a}}-\frac{4 a^2 (803 A+710 B) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3465 d}+\frac{2 a (803 A+710 B) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 d}+\frac{2 a B \tan (c+d x) \sec ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]

[Out]

(2*a^3*(803*A + 710*B)*Tan[c + d*x])/(495*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(209*A + 194*B)*Sec[c + d*x]^3*
Tan[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) - (4*a^2*(803*A + 710*B)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])
/(3465*d) + (2*a^2*(11*A + 14*B)*Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(99*d) + (2*a*(803*A +
710*B)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(1155*d) + (2*a*B*Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*Ta
n[c + d*x])/(11*d)

Rule 4018

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

Rule 4016

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[(-2*b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]]
), x] + Dist[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1)), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^n, x], x]
/; FreeQ[{a, b, d, e, f, A, B, n}, 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 3800

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b
*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rule 4001

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

Rule 3792

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

Rubi steps

\begin{align*} \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx &=\frac{2 a B \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac{2}{11} \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac{1}{2} a (11 A+6 B)+\frac{1}{2} a (11 A+14 B) \sec (c+d x)\right ) \, dx\\ &=\frac{2 a^2 (11 A+14 B) \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac{2 a B \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac{4}{99} \int \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{3}{4} a^2 (55 A+46 B)+\frac{1}{4} a^2 (209 A+194 B) \sec (c+d x)\right ) \, dx\\ &=\frac{2 a^3 (209 A+194 B) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (11 A+14 B) \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac{2 a B \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac{1}{231} \left (a^2 (803 A+710 B)\right ) \int \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a^3 (209 A+194 B) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (11 A+14 B) \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac{2 a (803 A+710 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac{2 a B \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac{(2 a (803 A+710 B)) \int \sec (c+d x) \left (\frac{3 a}{2}-a \sec (c+d x)\right ) \sqrt{a+a \sec (c+d x)} \, dx}{1155}\\ &=\frac{2 a^3 (209 A+194 B) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}-\frac{4 a^2 (803 A+710 B) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac{2 a^2 (11 A+14 B) \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac{2 a (803 A+710 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac{2 a B \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}+\frac{1}{495} \left (a^2 (803 A+710 B)\right ) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a^3 (803 A+710 B) \tan (c+d x)}{495 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^3 (209 A+194 B) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}-\frac{4 a^2 (803 A+710 B) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac{2 a^2 (11 A+14 B) \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{99 d}+\frac{2 a (803 A+710 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 d}+\frac{2 a B \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{11 d}\\ \end{align*}

Mathematica [B]  time = 6.16646, size = 487, normalized size = 2.05 \[ \frac{2 A \tan (c+d x) \sec ^3(c+d x) (a (\sec (c+d x)+1))^{5/2}}{9 d (\sec (c+d x)+1)^2}+\frac{38 A \tan (c+d x) \sec ^3(c+d x) (a (\sec (c+d x)+1))^{5/2}}{63 d (\sec (c+d x)+1)^3}+\frac{146 A \tan (c+d x) \sec ^2(c+d x) (a (\sec (c+d x)+1))^{5/2}}{105 d (\sec (c+d x)+1)^3}+\frac{584 A \tan (c+d x) \sec (c+d x) (a (\sec (c+d x)+1))^{5/2}}{315 d (\sec (c+d x)+1)^3}+\frac{1168 A \tan (c+d x) (a (\sec (c+d x)+1))^{5/2}}{315 d (\sec (c+d x)+1)^3}+\frac{2 B \tan (c+d x) \sec ^4(c+d x) (a (\sec (c+d x)+1))^{5/2}}{11 d (\sec (c+d x)+1)^2}+\frac{46 B \tan (c+d x) \sec ^4(c+d x) (a (\sec (c+d x)+1))^{5/2}}{99 d (\sec (c+d x)+1)^3}+\frac{710 B \tan (c+d x) \sec ^3(c+d x) (a (\sec (c+d x)+1))^{5/2}}{693 d (\sec (c+d x)+1)^3}+\frac{284 B \tan (c+d x) \sec ^2(c+d x) (a (\sec (c+d x)+1))^{5/2}}{231 d (\sec (c+d x)+1)^3}+\frac{1136 B \tan (c+d x) \sec (c+d x) (a (\sec (c+d x)+1))^{5/2}}{693 d (\sec (c+d x)+1)^3}+\frac{2272 B \tan (c+d x) (a (\sec (c+d x)+1))^{5/2}}{693 d (\sec (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]

[Out]

(1168*A*(a*(1 + Sec[c + d*x]))^(5/2)*Tan[c + d*x])/(315*d*(1 + Sec[c + d*x])^3) + (2272*B*(a*(1 + Sec[c + d*x]
))^(5/2)*Tan[c + d*x])/(693*d*(1 + Sec[c + d*x])^3) + (584*A*Sec[c + d*x]*(a*(1 + Sec[c + d*x]))^(5/2)*Tan[c +
d*x])/(315*d*(1 + Sec[c + d*x])^3) + (1136*B*Sec[c + d*x]*(a*(1 + Sec[c + d*x]))^(5/2)*Tan[c + d*x])/(693*d*(
1 + Sec[c + d*x])^3) + (146*A*Sec[c + d*x]^2*(a*(1 + Sec[c + d*x]))^(5/2)*Tan[c + d*x])/(105*d*(1 + Sec[c + d*
x])^3) + (284*B*Sec[c + d*x]^2*(a*(1 + Sec[c + d*x]))^(5/2)*Tan[c + d*x])/(231*d*(1 + Sec[c + d*x])^3) + (38*A
*Sec[c + d*x]^3*(a*(1 + Sec[c + d*x]))^(5/2)*Tan[c + d*x])/(63*d*(1 + Sec[c + d*x])^3) + (710*B*Sec[c + d*x]^3
*(a*(1 + Sec[c + d*x]))^(5/2)*Tan[c + d*x])/(693*d*(1 + Sec[c + d*x])^3) + (46*B*Sec[c + d*x]^4*(a*(1 + Sec[c
+ d*x]))^(5/2)*Tan[c + d*x])/(99*d*(1 + Sec[c + d*x])^3) + (2*A*Sec[c + d*x]^3*(a*(1 + Sec[c + d*x]))^(5/2)*Ta
n[c + d*x])/(9*d*(1 + Sec[c + d*x])^2) + (2*B*Sec[c + d*x]^4*(a*(1 + Sec[c + d*x]))^(5/2)*Tan[c + d*x])/(11*d*
(1 + Sec[c + d*x])^2)

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Maple [A]  time = 0.276, size = 163, normalized size = 0.7 \begin{align*} -{\frac{2\,{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 6424\,A \left ( \cos \left ( dx+c \right ) \right ) ^{5}+5680\,B \left ( \cos \left ( dx+c \right ) \right ) ^{5}+3212\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+2840\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4}+2409\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+2130\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+1430\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+1775\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+385\,A\cos \left ( dx+c \right ) +1120\,B\cos \left ( dx+c \right ) +315\,B \right ) }{3465\,d \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x)

[Out]

-2/3465/d*a^2*(-1+cos(d*x+c))*(6424*A*cos(d*x+c)^5+5680*B*cos(d*x+c)^5+3212*A*cos(d*x+c)^4+2840*B*cos(d*x+c)^4
+2409*A*cos(d*x+c)^3+2130*B*cos(d*x+c)^3+1430*A*cos(d*x+c)^2+1775*B*cos(d*x+c)^2+385*A*cos(d*x+c)+1120*B*cos(d
*x+c)+315*B)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*x+c)^5/sin(d*x+c)

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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(sec(d*x+c)^3*(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 = 0.49293, size = 409, normalized size = 1.73 \begin{align*} \frac{2 \,{\left (8 \,{\left (803 \, A + 710 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} + 4 \,{\left (803 \, A + 710 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \,{\left (803 \, A + 710 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 5 \,{\left (286 \, A + 355 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 35 \,{\left (11 \, A + 32 \, B\right )} a^{2} \cos \left (d x + c\right ) + 315 \, B a^{2}\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \,{\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

2/3465*(8*(803*A + 710*B)*a^2*cos(d*x + c)^5 + 4*(803*A + 710*B)*a^2*cos(d*x + c)^4 + 3*(803*A + 710*B)*a^2*co
s(d*x + c)^3 + 5*(286*A + 355*B)*a^2*cos(d*x + c)^2 + 35*(11*A + 32*B)*a^2*cos(d*x + c) + 315*B*a^2)*sqrt((a*c
os(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5)

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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(sec(d*x+c)**3*(a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 5.7378, size = 424, normalized size = 1.79 \begin{align*} -\frac{8 \,{\left (3465 \, \sqrt{2} A a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 3465 \, \sqrt{2} B a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (10395 \, \sqrt{2} A a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 8085 \, \sqrt{2} B a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (15939 \, \sqrt{2} A a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 15015 \, \sqrt{2} B a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) -{\left (14157 \, \sqrt{2} A a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 12375 \, \sqrt{2} B a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 4 \,{\left (1573 \, \sqrt{2} A a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 1375 \, \sqrt{2} B a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 2 \,{\left (143 \, \sqrt{2} A a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 125 \, \sqrt{2} B a^{8} \mathrm{sgn}\left (\cos \left (d x + c\right )\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 )^{2}\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 )^{2}\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 )}{3465 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{5} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

-8/3465*(3465*sqrt(2)*A*a^8*sgn(cos(d*x + c)) + 3465*sqrt(2)*B*a^8*sgn(cos(d*x + c)) - (10395*sqrt(2)*A*a^8*sg
n(cos(d*x + c)) + 8085*sqrt(2)*B*a^8*sgn(cos(d*x + c)) - (15939*sqrt(2)*A*a^8*sgn(cos(d*x + c)) + 15015*sqrt(2
)*B*a^8*sgn(cos(d*x + c)) - (14157*sqrt(2)*A*a^8*sgn(cos(d*x + c)) + 12375*sqrt(2)*B*a^8*sgn(cos(d*x + c)) - 4
*(1573*sqrt(2)*A*a^8*sgn(cos(d*x + c)) + 1375*sqrt(2)*B*a^8*sgn(cos(d*x + c)) - 2*(143*sqrt(2)*A*a^8*sgn(cos(d
*x + c)) + 125*sqrt(2)*B*a^8*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x +
1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*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)
^5*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)