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

Optimal. Leaf size=164 \[ -\frac{2 a \left (a^2 A b-2 a^3 B+3 a b^2 B-2 A b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{a^2 (A b-a B) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{(A b-2 a B) \tanh ^{-1}(\sin (c+d x))}{b^3 d}+\frac{B \tan (c+d x)}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.57818, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {4028, 4082, 3998, 3770, 3831, 2659, 208} \[ -\frac{2 a \left (a^2 A b-2 a^3 B+3 a b^2 B-2 A b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{a^2 (A b-a B) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{(A b-2 a B) \tanh ^{-1}(\sin (c+d x))}{b^3 d}+\frac{B \tan (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4028

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

Rule 4082

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

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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{\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx &=-\frac{a^2 (A b-a B) \tan (c+d x)}{b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\sec (c+d x) \left (-a b (A b-a B)-\left (a^2-b^2\right ) (A b-a B) \sec (c+d x)-b \left (a^2-b^2\right ) B \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{b^2 \left (a^2-b^2\right )}\\ &=\frac{B \tan (c+d x)}{b^2 d}-\frac{a^2 (A b-a B) \tan (c+d x)}{b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\sec (c+d x) \left (-a b^2 (A b-a B)-b \left (a^2-b^2\right ) (A b-2 a B) \sec (c+d x)\right )}{a+b \sec (c+d x)} \, dx}{b^3 \left (a^2-b^2\right )}\\ &=\frac{B \tan (c+d x)}{b^2 d}-\frac{a^2 (A b-a B) \tan (c+d x)}{b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{(A b-2 a B) \int \sec (c+d x) \, dx}{b^3}-\frac{\left (a \left (a^2 A b-2 A b^3-2 a^3 B+3 a b^2 B\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{b^3 \left (a^2-b^2\right )}\\ &=\frac{(A b-2 a B) \tanh ^{-1}(\sin (c+d x))}{b^3 d}+\frac{B \tan (c+d x)}{b^2 d}-\frac{a^2 (A b-a B) \tan (c+d x)}{b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (a \left (a^2 A b-2 A b^3-2 a^3 B+3 a b^2 B\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{b^4 \left (a^2-b^2\right )}\\ &=\frac{(A b-2 a B) \tanh ^{-1}(\sin (c+d x))}{b^3 d}+\frac{B \tan (c+d x)}{b^2 d}-\frac{a^2 (A b-a B) \tan (c+d x)}{b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (2 a \left (a^2 A b-2 A b^3-2 a^3 B+3 a b^2 B\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 )}{b^4 \left (a^2-b^2\right ) d}\\ &=\frac{(A b-2 a B) \tanh ^{-1}(\sin (c+d x))}{b^3 d}-\frac{2 a \left (a^2 A b-2 A b^3-2 a^3 B+3 a b^2 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{3/2} b^3 (a+b)^{3/2} d}+\frac{B \tan (c+d x)}{b^2 d}-\frac{a^2 (A b-a B) \tan (c+d x)}{b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 2.08354, size = 240, normalized size = 1.46 \[ \frac{-\frac{2 a \left (-a^2 A b+2 a^3 B-3 a b^2 B+2 A b^3\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{a^2 b (a B-A b) \sin (c+d x)}{(a-b) (a+b) (a \cos (c+d x)+b)}+2 a B \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-2 a B \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-A b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+A b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+b B \tan (c+d x)}{b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a*(-(a^2*A*b) + 2*A*b^3 + 2*a^3*B - 3*a*b^2*B)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2
- b^2)^(3/2) - A*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*a*B*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]
+ A*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 2*a*B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a^2*b*(-(A*
b) + a*B)*Sin[c + d*x])/((a - b)*(a + b)*(b + a*Cos[c + d*x])) + b*B*Tan[c + d*x])/(b^3*d)

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Maple [B]  time = 0.085, size = 510, normalized size = 3.1 \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)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,x)

[Out]

2/d*a^2/b/(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)*A-2/d*a^3/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)*B-2/d*a^3/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))*A+4/d*a/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arct
anh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A+4/d*a^4/b^3/(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*a^2/b/(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-1/d/b^2/(tan(1/2*d*x+1/2*c)+1)*B+1/d/b^2*ln(tan(1/2*d*x+1/2*c)+1)*A-2/d/b^3*ln(t
an(1/2*d*x+1/2*c)+1)*B*a-1/d/b^2/(tan(1/2*d*x+1/2*c)-1)*B-1/d/b^2*ln(tan(1/2*d*x+1/2*c)-1)*A+2/d/b^3*ln(tan(1/
2*d*x+1/2*c)-1)*B*a

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 31.3625, size = 2433, normalized size = 14.84 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(((2*B*a^5 - A*a^4*b - 3*B*a^3*b^2 + 2*A*a^2*b^3)*cos(d*x + c)^2 + (2*B*a^4*b - A*a^3*b^2 - 3*B*a^2*b^3 +
2*A*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
*B*a^6 - A*a^5*b - 4*B*a^4*b^2 + 2*A*a^3*b^3 + 2*B*a^2*b^4 - A*a*b^5)*cos(d*x + c)^2 + (2*B*a^5*b - A*a^4*b^2
- 4*B*a^3*b^3 + 2*A*a^2*b^4 + 2*B*a*b^5 - A*b^6)*cos(d*x + c))*log(sin(d*x + c) + 1) + ((2*B*a^6 - A*a^5*b - 4
*B*a^4*b^2 + 2*A*a^3*b^3 + 2*B*a^2*b^4 - A*a*b^5)*cos(d*x + c)^2 + (2*B*a^5*b - A*a^4*b^2 - 4*B*a^3*b^3 + 2*A*
a^2*b^4 + 2*B*a*b^5 - A*b^6)*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*(B*a^4*b^2 - 2*B*a^2*b^4 + B*b^6 + (2*B*
a^5*b - A*a^4*b^2 - 3*B*a^3*b^3 + A*a^2*b^4 + B*a*b^5)*cos(d*x + c))*sin(d*x + c))/((a^5*b^3 - 2*a^3*b^5 + a*b
^7)*d*cos(d*x + c)^2 + (a^4*b^4 - 2*a^2*b^6 + b^8)*d*cos(d*x + c)), 1/2*(2*((2*B*a^5 - A*a^4*b - 3*B*a^3*b^2 +
2*A*a^2*b^3)*cos(d*x + c)^2 + (2*B*a^4*b - A*a^3*b^2 - 3*B*a^2*b^3 + 2*A*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*B*a^6 - A*a^5*b - 4*B*a^4*b^
2 + 2*A*a^3*b^3 + 2*B*a^2*b^4 - A*a*b^5)*cos(d*x + c)^2 + (2*B*a^5*b - A*a^4*b^2 - 4*B*a^3*b^3 + 2*A*a^2*b^4 +
2*B*a*b^5 - A*b^6)*cos(d*x + c))*log(sin(d*x + c) + 1) + ((2*B*a^6 - A*a^5*b - 4*B*a^4*b^2 + 2*A*a^3*b^3 + 2*
B*a^2*b^4 - A*a*b^5)*cos(d*x + c)^2 + (2*B*a^5*b - A*a^4*b^2 - 4*B*a^3*b^3 + 2*A*a^2*b^4 + 2*B*a*b^5 - A*b^6)*
cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*(B*a^4*b^2 - 2*B*a^2*b^4 + B*b^6 + (2*B*a^5*b - A*a^4*b^2 - 3*B*a^3*b
^3 + A*a^2*b^4 + B*a*b^5)*cos(d*x + c))*sin(d*x + c))/((a^5*b^3 - 2*a^3*b^5 + a*b^7)*d*cos(d*x + c)^2 + (a^4*b
^4 - 2*a^2*b^6 + b^8)*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 )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.26911, size = 545, normalized size = 3.32 \begin{align*} \frac{\frac{2 \,{\left (2 \, B a^{4} - A a^{3} b - 3 \, B a^{2} b^{2} + 2 \, A a b^{3}\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^{2} b^{3} - b^{5}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{2 \,{\left (2 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + B b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + B 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 \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a + b\right )}{\left (a^{2} b^{2} - b^{4}\right )}} - \frac{{\left (2 \, B a - A b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{b^{3}} + \frac{{\left (2 \, B a - A b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{b^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(2*B*a^4 - A*a^3*b - 3*B*a^2*b^2 + 2*A*a*b^3)*(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^2*b^3 - b^5)*sqrt(-a^2 + b^2)) - 2*(2*
B*a^3*tan(1/2*d*x + 1/2*c)^3 - A*a^2*b*tan(1/2*d*x + 1/2*c)^3 - B*a^2*b*tan(1/2*d*x + 1/2*c)^3 - B*a*b^2*tan(1
/2*d*x + 1/2*c)^3 + B*b^3*tan(1/2*d*x + 1/2*c)^3 - 2*B*a^3*tan(1/2*d*x + 1/2*c) + A*a^2*b*tan(1/2*d*x + 1/2*c)
- B*a^2*b*tan(1/2*d*x + 1/2*c) + B*a*b^2*tan(1/2*d*x + 1/2*c) + B*b^3*tan(1/2*d*x + 1/2*c))/((a*tan(1/2*d*x +
1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b)*(a^2*b^2 - b^4)) - (2*B*a - A*b)*lo
g(abs(tan(1/2*d*x + 1/2*c) + 1))/b^3 + (2*B*a - A*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^3)/d