3.82 \(\int \sqrt{a+a \cos (c+d x)} (A+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)
Optimal. Leaf size=196 \[ \frac{a (35 A+48 C) \tan (c+d x)}{64 d \sqrt{a \cos (c+d x)+a}}+\frac{\sqrt{a} (35 A+48 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{64 d}+\frac{a (35 A+48 C) \tan (c+d x) \sec (c+d x)}{96 d \sqrt{a \cos (c+d x)+a}}+\frac{A \tan (c+d x) \sec ^3(c+d x) \sqrt{a \cos (c+d x)+a}}{4 d}+\frac{a A \tan (c+d x) \sec ^2(c+d x)}{24 d \sqrt{a \cos (c+d x)+a}} \]
[Out]
(Sqrt[a]*(35*A + 48*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(64*d) + (a*(35*A + 48*C)*Tan
[c + d*x])/(64*d*Sqrt[a + a*Cos[c + d*x]]) + (a*(35*A + 48*C)*Sec[c + d*x]*Tan[c + d*x])/(96*d*Sqrt[a + a*Cos[
c + d*x]]) + (a*A*Sec[c + d*x]^2*Tan[c + d*x])/(24*d*Sqrt[a + a*Cos[c + d*x]]) + (A*Sqrt[a + a*Cos[c + d*x]]*S
ec[c + d*x]^3*Tan[c + d*x])/(4*d)
________________________________________________________________________________________
Rubi [A] time = 0.471043, antiderivative size = 196, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used =
{3044, 2980, 2772, 2773, 206} \[ \frac{a (35 A+48 C) \tan (c+d x)}{64 d \sqrt{a \cos (c+d x)+a}}+\frac{\sqrt{a} (35 A+48 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{64 d}+\frac{a (35 A+48 C) \tan (c+d x) \sec (c+d x)}{96 d \sqrt{a \cos (c+d x)+a}}+\frac{A \tan (c+d x) \sec ^3(c+d x) \sqrt{a \cos (c+d x)+a}}{4 d}+\frac{a A \tan (c+d x) \sec ^2(c+d x)}{24 d \sqrt{a \cos (c+d x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[a + a*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]
[Out]
(Sqrt[a]*(35*A + 48*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(64*d) + (a*(35*A + 48*C)*Tan
[c + d*x])/(64*d*Sqrt[a + a*Cos[c + d*x]]) + (a*(35*A + 48*C)*Sec[c + d*x]*Tan[c + d*x])/(96*d*Sqrt[a + a*Cos[
c + d*x]]) + (a*A*Sec[c + d*x]^2*Tan[c + d*x])/(24*d*Sqrt[a + a*Cos[c + d*x]]) + (A*Sqrt[a + a*Cos[c + d*x]]*S
ec[c + d*x]^3*Tan[c + d*x])/(4*d)
Rule 3044
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^
m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n +
2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b
*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0
])
Rule 2980
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A
, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1]
Rule 2772
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[((b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x]
+ Dist[((2*n + 3)*(b*c - a*d))/(2*b*(n + 1)*(c^2 - d^2)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
Rule 2773
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*
b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^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 \sqrt{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\frac{A \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int \sqrt{a+a \cos (c+d x)} \left (\frac{a A}{2}+\frac{1}{2} a (5 A+8 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{4 a}\\ &=\frac{a A \sec ^2(c+d x) \tan (c+d x)}{24 d \sqrt{a+a \cos (c+d x)}}+\frac{A \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{48} (35 A+48 C) \int \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \, dx\\ &=\frac{a (35 A+48 C) \sec (c+d x) \tan (c+d x)}{96 d \sqrt{a+a \cos (c+d x)}}+\frac{a A \sec ^2(c+d x) \tan (c+d x)}{24 d \sqrt{a+a \cos (c+d x)}}+\frac{A \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{64} (35 A+48 C) \int \sqrt{a+a \cos (c+d x)} \sec ^2(c+d x) \, dx\\ &=\frac{a (35 A+48 C) \tan (c+d x)}{64 d \sqrt{a+a \cos (c+d x)}}+\frac{a (35 A+48 C) \sec (c+d x) \tan (c+d x)}{96 d \sqrt{a+a \cos (c+d x)}}+\frac{a A \sec ^2(c+d x) \tan (c+d x)}{24 d \sqrt{a+a \cos (c+d x)}}+\frac{A \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{128} (35 A+48 C) \int \sqrt{a+a \cos (c+d x)} \sec (c+d x) \, dx\\ &=\frac{a (35 A+48 C) \tan (c+d x)}{64 d \sqrt{a+a \cos (c+d x)}}+\frac{a (35 A+48 C) \sec (c+d x) \tan (c+d x)}{96 d \sqrt{a+a \cos (c+d x)}}+\frac{a A \sec ^2(c+d x) \tan (c+d x)}{24 d \sqrt{a+a \cos (c+d x)}}+\frac{A \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{(a (35 A+48 C)) \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{64 d}\\ &=\frac{\sqrt{a} (35 A+48 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{64 d}+\frac{a (35 A+48 C) \tan (c+d x)}{64 d \sqrt{a+a \cos (c+d x)}}+\frac{a (35 A+48 C) \sec (c+d x) \tan (c+d x)}{96 d \sqrt{a+a \cos (c+d x)}}+\frac{a A \sec ^2(c+d x) \tan (c+d x)}{24 d \sqrt{a+a \cos (c+d x)}}+\frac{A \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}
Mathematica [A] time = 1.83642, size = 145, normalized size = 0.74 \[ \frac{\sqrt{a (\cos (c+d x)+1)} \left (\frac{1}{2} \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^4(c+d x) ((539 A+432 C) \cos (c+d x)+4 (35 A+48 C) \cos (2 (c+d x))+105 A \cos (3 (c+d x))+332 A+144 C \cos (3 (c+d x))+192 C)+3 \sqrt{2} (35 A+48 C) \sec \left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{384 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[a + a*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]
[Out]
(Sqrt[a*(1 + Cos[c + d*x])]*(3*Sqrt[2]*(35*A + 48*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Sec[(c + d*x)/2] + ((33
2*A + 192*C + (539*A + 432*C)*Cos[c + d*x] + 4*(35*A + 48*C)*Cos[2*(c + d*x)] + 105*A*Cos[3*(c + d*x)] + 144*C
*Cos[3*(c + d*x)])*Sec[c + d*x]^4*Tan[(c + d*x)/2])/2))/(384*d)
________________________________________________________________________________________
Maple [B] time = 0.116, size = 1631, normalized size = 8.3 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^5*(a+a*cos(d*x+c))^(1/2),x)
[Out]
1/24*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(48*a*(35*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1
/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))+35*A*ln(4/(2*cos(1/2*d*x+1/2*c)+
2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))+48*C*ln(-4/(-2*cos
(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))+48
*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^
(1/2)+2*a)))*sin(1/2*d*x+1/2*c)^8-48*(35*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+48*C*2^(1/2)*(a*sin(
1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+70*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2
*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+70*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*
d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*a+96*C*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(
a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+96*C*ln(4/(2*cos(1/2*d*x+1
/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*a)*sin(1/2*d
*x+1/2*c)^6+8*(385*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+528*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/
2)*a^(1/2)+315*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/
2)*cos(1/2*d*x+1/2*c)+2*a))*a+315*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*
2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*a+432*C*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a
*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+432*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a
*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*a)*sin(1/2*d*x+1/2*c)^4-4*(51
1*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+624*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+210*A*
ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/
2*c)+2*a))*a+210*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/
2*d*x+1/2*c)^2)^(1/2)+2*a))*a+288*C*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*
c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+288*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*
d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*a)*sin(1/2*d*x+1/2*c)^2+558*A*2^(1/2)*(a*sin(1
/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+105*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2
*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+105*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2
*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*a+480*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2
)*a^(1/2)+144*C*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2
)*cos(1/2*d*x+1/2*c)+2*a))*a+144*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2
^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*a)/a^(1/2)/(2*cos(1/2*d*x+1/2*c)+2^(1/2))^4/(2*cos(1/2*d*x+1/2*c)-
2^(1/2))^4/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((A+C*cos(d*x+c)^2)*sec(d*x+c)^5*(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
Timed out
________________________________________________________________________________________
Fricas [A] time = 1.83699, size = 551, normalized size = 2.81 \begin{align*} \frac{3 \,{\left ({\left (35 \, A + 48 \, C\right )} \cos \left (d x + c\right )^{5} +{\left (35 \, A + 48 \, C\right )} \cos \left (d x + c\right )^{4}\right )} \sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a}{\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \,{\left (3 \,{\left (35 \, A + 48 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \,{\left (35 \, A + 48 \, C\right )} \cos \left (d x + c\right )^{2} + 56 \, A \cos \left (d x + c\right ) + 48 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{768 \,{\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^5*(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
1/768*(3*((35*A + 48*C)*cos(d*x + c)^5 + (35*A + 48*C)*cos(d*x + c)^4)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos
(d*x + c)^2 - 4*sqrt(a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(
d*x + c)^2)) + 4*(3*(35*A + 48*C)*cos(d*x + c)^3 + 2*(35*A + 48*C)*cos(d*x + c)^2 + 56*A*cos(d*x + c) + 48*A)*
sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4)
________________________________________________________________________________________
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((A+C*cos(d*x+c)**2)*sec(d*x+c)**5*(a+a*cos(d*x+c))**(1/2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [B] time = 2.77409, size = 1083, normalized size = 5.53 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^5*(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")
[Out]
1/384*(3*(35*A*sqrt(a) + 48*C*sqrt(a))*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*(35*A*sqrt(a) + 48*C*sqrt(a))*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)*(279*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*t
an(1/2*d*x + 1/2*c)^2 + a))^14*A*a^(3/2) + 240*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 +
a))^14*C*a^(3/2) + 285*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^12*A*a^(5/2) - 196
8*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^12*C*a^(5/2) - 4605*(sqrt(a)*tan(1/2*d*x
+ 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^10*A*a^(7/2) - 2640*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan
(1/2*d*x + 1/2*c)^2 + a))^10*C*a^(7/2) + 37281*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 +
a))^8*A*a^(9/2) + 41616*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^8*C*a^(9/2) - 356
43*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*a^(11/2) - 42288*(sqrt(a)*tan(1/2*d
*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*a^(11/2) + 9175*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*t
an(1/2*d*x + 1/2*c)^2 + a))^4*A*a^(13/2) + 12528*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2
+ a))^4*C*a^(13/2) - 1311*(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^(15/2) -
1392*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*C*a^(15/2) + 43*A*a^(17/2) + 48*C*a
^(17/2))/((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