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

Optimal. Leaf size=181 \[ \frac{(3 A-7 B+11 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(3 A-3 B+7 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{6 a^2 d}-\frac{(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac{(3 A-9 B+13 C) \tan (c+d x)}{3 a d \sqrt{a \sec (c+d x)+a}} \]

[Out]

((3*A - 7*B + 11*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) -
((A - B + C)*Sec[c + d*x]^2*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) - ((3*A - 9*B + 13*C)*Tan[c + d*x]
)/(3*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((3*A - 3*B + 7*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(6*a^2*d)

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Rubi [A]  time = 0.476456, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.116, Rules used = {4084, 4010, 4001, 3795, 203} \[ \frac{(3 A-7 B+11 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(3 A-3 B+7 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{6 a^2 d}-\frac{(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac{(3 A-9 B+13 C) \tan (c+d x)}{3 a d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*A - 7*B + 11*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) -
((A - B + C)*Sec[c + d*x]^2*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) - ((3*A - 9*B + 13*C)*Tan[c + d*x]
)/(3*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((3*A - 3*B + 7*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(6*a^2*d)

Rule 4084

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

Rule 4010

Int[csc[(e_.) + (f_.)*(x_)]^2*(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 + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), I
nt[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -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 3795

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, 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 \frac{\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{\int \frac{\sec ^2(c+d x) \left (2 a (B-C)+\frac{1}{2} a (3 A-3 B+7 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(3 A-3 B+7 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{6 a^2 d}+\frac{\int \frac{\sec (c+d x) \left (\frac{1}{4} a^2 (3 A-3 B+7 C)-\frac{1}{2} a^2 (3 A-9 B+13 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{3 a^3}\\ &=-\frac{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(3 A-9 B+13 C) \tan (c+d x)}{3 a d \sqrt{a+a \sec (c+d x)}}+\frac{(3 A-3 B+7 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{6 a^2 d}+\frac{(3 A-7 B+11 C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(3 A-9 B+13 C) \tan (c+d x)}{3 a d \sqrt{a+a \sec (c+d x)}}+\frac{(3 A-3 B+7 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{6 a^2 d}-\frac{(3 A-7 B+11 C) \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{2 a d}\\ &=\frac{(3 A-7 B+11 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(3 A-9 B+13 C) \tan (c+d x)}{3 a d \sqrt{a+a \sec (c+d x)}}+\frac{(3 A-3 B+7 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{6 a^2 d}\\ \end{align*}

Mathematica [C]  time = 25.2684, size = 7119, normalized size = 39.33 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [B]  time = 0.321, size = 867, normalized size = 4.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24/d/a^2*(-1+cos(d*x+c))*(9*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2
)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)-21*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(-(
-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)+33*C*(-2*co
s(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))
*cos(d*x+c)^2*sin(d*x+c)+18*A*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(-(-(-2*cos(d*x+c)/
(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))-42*B*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+
c)+1))^(3/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))+66*C*cos(d*x+c)*s
in(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c
)-1)/sin(d*x+c))+9*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c
)+cos(d*x+c)-1)/sin(d*x+c))*sin(d*x+c)-21*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x
+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*sin(d*x+c)+33*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(-(-
(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*sin(d*x+c)-12*A*cos(d*x+c)^3+60*B*co
s(d*x+c)^3-76*C*cos(d*x+c)^3+12*A*cos(d*x+c)^2-12*B*cos(d*x+c)^2+28*C*cos(d*x+c)^2-48*B*cos(d*x+c)+64*C*cos(d*
x+c)-16*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)^3/cos(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)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.637374, size = 1257, normalized size = 6.94 \begin{align*} \left [-\frac{3 \, \sqrt{2}{\left ({\left (3 \, A - 7 \, B + 11 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \,{\left (3 \, A - 7 \, B + 11 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (3 \, A - 7 \, B + 11 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt{-a} \log \left (\frac{2 \, \sqrt{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 ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \,{\left ({\left (3 \, A - 15 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{2} - 12 \,{\left (B - C\right )} \cos \left (d x + c\right ) - 4 \, C\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 (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}}, -\frac{3 \, \sqrt{2}{\left ({\left (3 \, A - 7 \, B + 11 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \,{\left (3 \, A - 7 \, B + 11 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (3 \, A - 7 \, B + 11 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \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 ) + 2 \,{\left ({\left (3 \, A - 15 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{2} - 12 \,{\left (B - C\right )} \cos \left (d x + c\right ) - 4 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{12 \,{\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/24*(3*sqrt(2)*((3*A - 7*B + 11*C)*cos(d*x + c)^3 + 2*(3*A - 7*B + 11*C)*cos(d*x + c)^2 + (3*A - 7*B + 11*C
)*cos(d*x + c))*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x
+ c) + 3*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((3*A - 15*B + 19
*C)*cos(d*x + c)^2 - 12*(B - C)*cos(d*x + c) - 4*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2
*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c)), -1/12*(3*sqrt(2)*((3*A - 7*B + 11*C)*cos(d*x
+ c)^3 + 2*(3*A - 7*B + 11*C)*cos(d*x + c)^2 + (3*A - 7*B + 11*C)*cos(d*x + c))*sqrt(a)*arctan(sqrt(2)*sqrt((
a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + 2*((3*A - 15*B + 19*C)*cos(d*x + c)^2
- 12*(B - C)*cos(d*x + c) - 4*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^3
+ 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**2/(a*(sec(c + d*x) + 1))**(3/2), x)

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Giac [B]  time = 9.28646, size = 491, normalized size = 2.71 \begin{align*} -\frac{\frac{{\left ({\left (\frac{3 \,{\left (\sqrt{2} A a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) - \sqrt{2} B a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) + \sqrt{2} C a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a} - \frac{2 \,{\left (3 \, \sqrt{2} A a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) - 15 \, \sqrt{2} B a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) + 23 \, \sqrt{2} C a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )\right )}}{a}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \frac{3 \,{\left (\sqrt{2} A a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) - 9 \, \sqrt{2} B a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) + 9 \, \sqrt{2} C a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )\right )}}{a}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}} - \frac{3 \,{\left (3 \, \sqrt{2} A - 7 \, \sqrt{2} B + 11 \, \sqrt{2} C\right )} \log \left ({\left | -\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt{-a} a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(((3*(sqrt(2)*A*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - sqrt(2)*B*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + sqrt(2
)*C*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*tan(1/2*d*x + 1/2*c)^2/a - 2*(3*sqrt(2)*A*a*sgn(tan(1/2*d*x + 1/2*c)^2
- 1) - 15*sqrt(2)*B*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 23*sqrt(2)*C*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))/a)*tan
(1/2*d*x + 1/2*c)^2 + 3*(sqrt(2)*A*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 9*sqrt(2)*B*a*sgn(tan(1/2*d*x + 1/2*c)^
2 - 1) + 9*sqrt(2)*C*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))/a)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a
)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)) - 3*(3*sqrt(2)*A - 7*sqrt(2)*B + 11*sqrt(2)*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)))/(sqrt(-a)*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d