3.932 \(\int \frac{a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=140 \[ -\frac{2 b \left (2 a^2 b B-3 a^3 C+a b^2 C-b^3 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{b^2 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{x (b B-a C)}{a^2} \]

[Out]

((b*B - a*C)*x)/a^2 - (2*b*(2*a^2*b*B - b^3*B - 3*a^3*C + a*b^2*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt
[a + b]])/(a^2*(a - b)^(3/2)*(a + b)^(3/2)*d) + (b^2*(b*B - 2*a*C)*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c
+ d*x]))

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Rubi [A]  time = 0.389699, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 48, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {24, 3923, 3919, 3831, 2659, 208} \[ -\frac{2 b \left (2 a^2 b B-3 a^3 C+a b^2 C-b^3 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{b^2 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{x (b B-a C)}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*B - a*C)*x)/a^2 - (2*b*(2*a^2*b*B - b^3*B - 3*a^3*C + a*b^2*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt
[a + b]])/(a^2*(a - b)^(3/2)*(a + b)^(3/2)*d) + (b^2*(b*B - 2*a*C)*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c
+ d*x]))

Rule 24

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_Symbol] :> Dist[1/b^2, Int[u*(a + b*
v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] &&
LeQ[m, -1]

Rule 3923

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

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx &=\frac{\int \frac{b^2 (b B-a C)+b^3 C \sec (c+d x)}{(a+b \sec (c+d x))^2} \, dx}{b^2}\\ &=\frac{b^2 (b B-2 a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{-b^2 \left (a^2-b^2\right ) (b B-a C)+a b^3 (b B-2 a C) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a b^2 \left (a^2-b^2\right )}\\ &=\frac{(b B-a C) x}{a^2}+\frac{b^2 (b B-2 a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (b \left (2 a^2 b B-b^3 B-3 a^3 C+a b^2 C\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=\frac{(b B-a C) x}{a^2}+\frac{b^2 (b B-2 a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (2 a^2 b B-b^3 B-3 a^3 C+a b^2 C\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=\frac{(b B-a C) x}{a^2}+\frac{b^2 (b B-2 a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (2 \left (2 a^2 b B-b^3 B-3 a^3 C+a b^2 C\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^2 \left (a^2-b^2\right ) d}\\ &=\frac{(b B-a C) x}{a^2}-\frac{2 b \left (2 a^2 b B-b^3 B-3 a^3 C+a b^2 C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac{b^2 (b B-2 a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.827663, size = 211, normalized size = 1.51 \[ \frac{\sec (c+d x) (a \cos (c+d x)+b) (-a C+b B+b C \sec (c+d x)) \left (-\frac{2 b \left (-2 a^2 b B+3 a^3 C-a b^2 C+b^3 B\right ) (a \cos (c+d x)+b) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{a b^2 (b B-2 a C) \sin (c+d x)}{(a-b) (a+b)}+(c+d x) (b B-a C) (a \cos (c+d x)+b)\right )}{a^2 d (a+b \sec (c+d x))^2 ((b B-a C) \cos (c+d x)+b C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*Cos[c + d*x])*Sec[c + d*x]*(b*B - a*C + b*C*Sec[c + d*x])*((b*B - a*C)*(c + d*x)*(b + a*Cos[c + d*x])
- (2*b*(-2*a^2*b*B + b^3*B + 3*a^3*C - a*b^2*C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Co
s[c + d*x]))/(a^2 - b^2)^(3/2) + (a*b^2*(b*B - 2*a*C)*Sin[c + d*x])/((a - b)*(a + b))))/(a^2*d*(b*C + (b*B - a
*C)*Cos[c + d*x])*(a + b*Sec[c + d*x])^2)

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Maple [B]  time = 0.105, size = 415, normalized size = 3. \begin{align*} 2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) Bb}{d{a}^{2}}}-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{ad}}-2\,{\frac{{b}^{3}\tan \left ( 1/2\,dx+c/2 \right ) B}{ad \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) }}+4\,{\frac{{b}^{2}\tan \left ( 1/2\,dx+c/2 \right ) C}{d \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) }}-4\,{\frac{B{b}^{2}}{d \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+2\,{\frac{B{b}^{4}}{d{a}^{2} \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+6\,{\frac{abC}{d \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-2\,{\frac{C{b}^{3}}{ad \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a^2*arctan(tan(1/2*d*x+1/2*c))*B*b-2/a/d*arctan(tan(1/2*d*x+1/2*c))*C-2/d*b^3/a/(a^2-b^2)*tan(1/2*d*x+1/2*
c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)*B+4/d*b^2/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2
*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)*C-4/d*b^2/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)
/((a+b)*(a-b))^(1/2))*B+2/d*b^4/a^2/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a
-b))^(1/2))*B+6/d*b*a/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C-
2/d*b^3/a/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.64219, size = 1526, normalized size = 10.9 \begin{align*} \left [-\frac{2 \,{\left (C a^{6} - B a^{5} b - 2 \, C a^{4} b^{2} + 2 \, B a^{3} b^{3} + C a^{2} b^{4} - B a b^{5}\right )} d x \cos \left (d x + c\right ) + 2 \,{\left (C a^{5} b - B a^{4} b^{2} - 2 \, C a^{3} b^{3} + 2 \, B a^{2} b^{4} + C a b^{5} - B b^{6}\right )} d x -{\left (3 \, C a^{3} b^{2} - 2 \, B a^{2} b^{3} - C a b^{4} + B b^{5} +{\left (3 \, C a^{4} b - 2 \, B a^{3} b^{2} - C a^{2} b^{3} + B a b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt{a^{2} - b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + 2 \,{\left (2 \, C a^{4} b^{2} - B a^{3} b^{3} - 2 \, C a^{2} b^{4} + B a b^{5}\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d\right )}}, -\frac{{\left (C a^{6} - B a^{5} b - 2 \, C a^{4} b^{2} + 2 \, B a^{3} b^{3} + C a^{2} b^{4} - B a b^{5}\right )} d x \cos \left (d x + c\right ) +{\left (C a^{5} b - B a^{4} b^{2} - 2 \, C a^{3} b^{3} + 2 \, B a^{2} b^{4} + C a b^{5} - B b^{6}\right )} d x -{\left (3 \, C a^{3} b^{2} - 2 \, B a^{2} b^{3} - C a b^{4} + B b^{5} +{\left (3 \, C a^{4} b - 2 \, B a^{3} b^{2} - C a^{2} b^{3} + B a b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) +{\left (2 \, C a^{4} b^{2} - B a^{3} b^{3} - 2 \, C a^{2} b^{4} + B a b^{5}\right )} \sin \left (d x + c\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*(C*a^6 - B*a^5*b - 2*C*a^4*b^2 + 2*B*a^3*b^3 + C*a^2*b^4 - B*a*b^5)*d*x*cos(d*x + c) + 2*(C*a^5*b - B
*a^4*b^2 - 2*C*a^3*b^3 + 2*B*a^2*b^4 + C*a*b^5 - B*b^6)*d*x - (3*C*a^3*b^2 - 2*B*a^2*b^3 - C*a*b^4 + B*b^5 + (
3*C*a^4*b - 2*B*a^3*b^2 - C*a^2*b^3 + B*a*b^4)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 -
2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2
+ 2*a*b*cos(d*x + c) + b^2)) + 2*(2*C*a^4*b^2 - B*a^3*b^3 - 2*C*a^2*b^4 + B*a*b^5)*sin(d*x + c))/((a^7 - 2*a^
5*b^2 + a^3*b^4)*d*cos(d*x + c) + (a^6*b - 2*a^4*b^3 + a^2*b^5)*d), -((C*a^6 - B*a^5*b - 2*C*a^4*b^2 + 2*B*a^3
*b^3 + C*a^2*b^4 - B*a*b^5)*d*x*cos(d*x + c) + (C*a^5*b - B*a^4*b^2 - 2*C*a^3*b^3 + 2*B*a^2*b^4 + C*a*b^5 - B*
b^6)*d*x - (3*C*a^3*b^2 - 2*B*a^2*b^3 - C*a*b^4 + B*b^5 + (3*C*a^4*b - 2*B*a^3*b^2 - C*a^2*b^3 + B*a*b^4)*cos(
d*x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) + (2*C*a^
4*b^2 - B*a^3*b^3 - 2*C*a^2*b^4 + B*a*b^5)*sin(d*x + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cos(d*x + c) + (a^6*b
- 2*a^4*b^3 + a^2*b^5)*d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(-B*b/(a**2 + 2*a*b*sec(c + d*x) + b**2*sec(c + d*x)**2), x) - Integral(C*a/(a**2 + 2*a*b*sec(c + d*x
) + b**2*sec(c + d*x)**2), x) - Integral(-C*b*sec(c + d*x)/(a**2 + 2*a*b*sec(c + d*x) + b**2*sec(c + d*x)**2),
x)

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Giac [A]  time = 1.35741, size = 301, normalized size = 2.15 \begin{align*} \frac{\frac{2 \,{\left (3 \, C a^{3} b - 2 \, B a^{2} b^{2} - C a b^{3} + B b^{4}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{{\left (C a - B b\right )}{\left (d x + c\right )}}{a^{2}} + \frac{2 \,{\left (2 \, C a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a^{3} - a b^{2}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b\right )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(3*C*a^3*b - 2*B*a^2*b^2 - C*a*b^3 + B*b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*
tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^4 - a^2*b^2)*sqrt(-a^2 + b^2)) - (C*a -
B*b)*(d*x + c)/a^2 + 2*(2*C*a*b^2*tan(1/2*d*x + 1/2*c) - B*b^3*tan(1/2*d*x + 1/2*c))/((a^3 - a*b^2)*(a*tan(1/2
*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)))/d