3.985 \(\int \frac{(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^5(c+d x)}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=285 \[ -\frac{2 b^3 \left (A b^2-a (b B-a C)\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d \sqrt{a-b} \sqrt{a+b}}-\frac{\tan (c+d x) \left (a^2 b (2 A+3 C)-2 a^3 B-3 a b^2 B+3 A b^3\right )}{3 a^4 d}+\frac{\left (4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)-4 a^3 b B-8 a b^3 B+8 A b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 a^5 d}+\frac{\tan (c+d x) \sec (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{8 a^3 d}-\frac{(A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d}+\frac{A \tan (c+d x) \sec ^3(c+d x)}{4 a d} \]

[Out]

(-2*b^3*(A*b^2 - a*(b*B - a*C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^5*Sqrt[a - b]*Sqrt[a +
b]*d) + ((8*A*b^4 - 4*a^3*b*B - 8*a*b^3*B + 4*a^2*b^2*(A + 2*C) + a^4*(3*A + 4*C))*ArcTanh[Sin[c + d*x]])/(8*a
^5*d) - ((3*A*b^3 - 2*a^3*B - 3*a*b^2*B + a^2*b*(2*A + 3*C))*Tan[c + d*x])/(3*a^4*d) + ((4*A*b^2 - 4*a*b*B + a
^2*(3*A + 4*C))*Sec[c + d*x]*Tan[c + d*x])/(8*a^3*d) - ((A*b - a*B)*Sec[c + d*x]^2*Tan[c + d*x])/(3*a^2*d) + (
A*Sec[c + d*x]^3*Tan[c + d*x])/(4*a*d)

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Rubi [A]  time = 1.28194, antiderivative size = 285, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {3055, 3001, 3770, 2659, 205} \[ -\frac{2 b^3 \left (A b^2-a (b B-a C)\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d \sqrt{a-b} \sqrt{a+b}}-\frac{\tan (c+d x) \left (a^2 b (2 A+3 C)-2 a^3 B-3 a b^2 B+3 A b^3\right )}{3 a^4 d}+\frac{\left (4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)-4 a^3 b B-8 a b^3 B+8 A b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 a^5 d}+\frac{\tan (c+d x) \sec (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )}{8 a^3 d}-\frac{(A b-a B) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d}+\frac{A \tan (c+d x) \sec ^3(c+d x)}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^3*(A*b^2 - a*(b*B - a*C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^5*Sqrt[a - b]*Sqrt[a +
b]*d) + ((8*A*b^4 - 4*a^3*b*B - 8*a*b^3*B + 4*a^2*b^2*(A + 2*C) + a^4*(3*A + 4*C))*ArcTanh[Sin[c + d*x]])/(8*a
^5*d) - ((3*A*b^3 - 2*a^3*B - 3*a*b^2*B + a^2*b*(2*A + 3*C))*Tan[c + d*x])/(3*a^4*d) + ((4*A*b^2 - 4*a*b*B + a
^2*(3*A + 4*C))*Sec[c + d*x]*Tan[c + d*x])/(8*a^3*d) - ((A*b - a*B)*Sec[c + d*x]^2*Tan[c + d*x])/(3*a^2*d) + (
A*Sec[c + d*x]^3*Tan[c + d*x])/(4*a*d)

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3770

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

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 205

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

Rubi steps

\begin{align*} \int \frac{\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{a+b \cos (c+d x)} \, dx &=\frac{A \sec ^3(c+d x) \tan (c+d x)}{4 a d}+\frac{\int \frac{\left (-4 (A b-a B)+a (3 A+4 C) \cos (c+d x)+3 A b \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx}{4 a}\\ &=-\frac{(A b-a B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d}+\frac{A \sec ^3(c+d x) \tan (c+d x)}{4 a d}+\frac{\int \frac{\left (3 \left (4 b (A b-a B)+a^2 (3 A+4 C)\right )+a (A b+8 a B) \cos (c+d x)-8 b (A b-a B) \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx}{12 a^2}\\ &=\frac{\left (4 A b^2-4 a b B+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 a^3 d}-\frac{(A b-a B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d}+\frac{A \sec ^3(c+d x) \tan (c+d x)}{4 a d}+\frac{\int \frac{\left (-8 \left (3 A b^3-2 a^3 B-3 a b^2 B+a^2 b (2 A+3 C)\right )-a \left (4 A b^2-4 a b B-3 a^2 (3 A+4 C)\right ) \cos (c+d x)+3 b \left (4 A b^2-4 a b B+a^2 (3 A+4 C)\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{24 a^3}\\ &=-\frac{\left (3 A b^3-2 a^3 B-3 a b^2 B+a^2 b (2 A+3 C)\right ) \tan (c+d x)}{3 a^4 d}+\frac{\left (4 A b^2-4 a b B+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 a^3 d}-\frac{(A b-a B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d}+\frac{A \sec ^3(c+d x) \tan (c+d x)}{4 a d}+\frac{\int \frac{\left (3 \left (8 A b^4-4 a^3 b B-8 a b^3 B+4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right )+3 a b \left (4 A b^2-4 a b B+a^2 (3 A+4 C)\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{24 a^4}\\ &=-\frac{\left (3 A b^3-2 a^3 B-3 a b^2 B+a^2 b (2 A+3 C)\right ) \tan (c+d x)}{3 a^4 d}+\frac{\left (4 A b^2-4 a b B+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 a^3 d}-\frac{(A b-a B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d}+\frac{A \sec ^3(c+d x) \tan (c+d x)}{4 a d}+\frac{\left (8 A b^4-4 a^3 b B-8 a b^3 B+4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right ) \int \sec (c+d x) \, dx}{8 a^5}-\frac{\left (b^3 \left (A b^2-a (b B-a C)\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{a^5}\\ &=\frac{\left (8 A b^4-4 a^3 b B-8 a b^3 B+4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 a^5 d}-\frac{\left (3 A b^3-2 a^3 B-3 a b^2 B+a^2 b (2 A+3 C)\right ) \tan (c+d x)}{3 a^4 d}+\frac{\left (4 A b^2-4 a b B+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 a^3 d}-\frac{(A b-a B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d}+\frac{A \sec ^3(c+d x) \tan (c+d x)}{4 a d}-\frac{\left (2 b^3 \left (A b^2-a (b B-a C)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^5 d}\\ &=-\frac{2 b^3 \left (A b^2-a (b B-a C)\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 \sqrt{a-b} \sqrt{a+b} d}+\frac{\left (8 A b^4-4 a^3 b B-8 a b^3 B+4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 a^5 d}-\frac{\left (3 A b^3-2 a^3 B-3 a b^2 B+a^2 b (2 A+3 C)\right ) \tan (c+d x)}{3 a^4 d}+\frac{\left (4 A b^2-4 a b B+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 a^3 d}-\frac{(A b-a B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d}+\frac{A \sec ^3(c+d x) \tan (c+d x)}{4 a d}\\ \end{align*}

Mathematica [A]  time = 1.42551, size = 406, normalized size = 1.42 \[ \frac{\frac{96 b^3 \left (a (a C-b B)+A b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}-6 \left (4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)-4 a^3 b B-8 a b^3 B+8 A b^4\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+6 \left (4 a^2 b^2 (A+2 C)+a^4 (3 A+4 C)-4 a^3 b B-8 a b^3 B+8 A b^4\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+a \tan (c+d x) \sec ^3(c+d x) \left (3 a \cos (2 (c+d x)) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )+4 \cos (c+d x) \left (-a^2 b (10 A+9 C)+10 a^3 B+9 a b^2 B-9 A b^3\right )-8 a^2 A b \cos (3 (c+d x))+21 a^3 A-12 a^2 b B-12 a^2 b C \cos (3 (c+d x))+8 a^3 B \cos (3 (c+d x))+12 a^3 C+12 a A b^2+12 a b^2 B \cos (3 (c+d x))-12 A b^3 \cos (3 (c+d x))\right )}{48 a^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((96*b^3*(A*b^2 + a*(-(b*B) + a*C))*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] - 6
*(8*A*b^4 - 4*a^3*b*B - 8*a*b^3*B + 4*a^2*b^2*(A + 2*C) + a^4*(3*A + 4*C))*Log[Cos[(c + d*x)/2] - Sin[(c + d*x
)/2]] + 6*(8*A*b^4 - 4*a^3*b*B - 8*a*b^3*B + 4*a^2*b^2*(A + 2*C) + a^4*(3*A + 4*C))*Log[Cos[(c + d*x)/2] + Sin
[(c + d*x)/2]] + a*(21*a^3*A + 12*a*A*b^2 - 12*a^2*b*B + 12*a^3*C + 4*(-9*A*b^3 + 10*a^3*B + 9*a*b^2*B - a^2*b
*(10*A + 9*C))*Cos[c + d*x] + 3*a*(4*A*b^2 - 4*a*b*B + a^2*(3*A + 4*C))*Cos[2*(c + d*x)] - 8*a^2*A*b*Cos[3*(c
+ d*x)] - 12*A*b^3*Cos[3*(c + d*x)] + 8*a^3*B*Cos[3*(c + d*x)] + 12*a*b^2*B*Cos[3*(c + d*x)] - 12*a^2*b*C*Cos[
3*(c + d*x)])*Sec[c + d*x]^3*Tan[c + d*x])/(48*a^5*d)

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Maple [B]  time = 0.098, size = 1335, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a/d*A/(tan(1/2*d*x+1/2*c)-1)^3+1/2/a/d*ln(tan(1/2*d*x+1/2*c)+1)*C+7/8/a/d*A/(tan(1/2*d*x+1/2*c)-1)^2-1/2/a
/d*ln(tan(1/2*d*x+1/2*c)-1)*C-7/8/a/d*A/(tan(1/2*d*x+1/2*c)+1)^2+1/4/d*A/a/(tan(1/2*d*x+1/2*c)-1)^4-1/4/d*A/a/
(tan(1/2*d*x+1/2*c)+1)^4-1/2/a/d*B/(tan(1/2*d*x+1/2*c)-1)^2+1/2/a/d*B/(tan(1/2*d*x+1/2*c)+1)^2-1/3/a/d*B/(tan(
1/2*d*x+1/2*c)-1)^3+1/2/a/d/(tan(1/2*d*x+1/2*c)-1)^2*C-1/3/a/d*B/(tan(1/2*d*x+1/2*c)+1)^3-1/2/a/d/(tan(1/2*d*x
+1/2*c)+1)^2*C-1/a/d*B/(tan(1/2*d*x+1/2*c)+1)-1/a/d*B/(tan(1/2*d*x+1/2*c)-1)+1/2/a/d/(tan(1/2*d*x+1/2*c)-1)*C+
1/2/a/d*A/(tan(1/2*d*x+1/2*c)+1)^3+1/2/a/d/(tan(1/2*d*x+1/2*c)+1)*C+5/8/a/d*A/(tan(1/2*d*x+1/2*c)+1)+5/8/a/d*A
/(tan(1/2*d*x+1/2*c)-1)-3/8/a/d*A*ln(tan(1/2*d*x+1/2*c)-1)+3/8/a/d*A*ln(tan(1/2*d*x+1/2*c)+1)+1/2/d/a^2*ln(tan
(1/2*d*x+1/2*c)-1)*b*B-1/2/d/a^2/(tan(1/2*d*x+1/2*c)+1)*b*B-1/2/d/a^2/(tan(1/2*d*x+1/2*c)-1)*b*B+1/2/d/a^3/(ta
n(1/2*d*x+1/2*c)-1)*A*b^2+1/2/d*A/a^2/(tan(1/2*d*x+1/2*c)-1)^2*b+1/2/d/a^3/(tan(1/2*d*x+1/2*c)+1)*A*b^2-1/2/d*
A/a^2/(tan(1/2*d*x+1/2*c)+1)^2*b+1/d*A/a^2/(tan(1/2*d*x+1/2*c)-1)*b+1/2/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*A*b^2+1
/d*A/a^2/(tan(1/2*d*x+1/2*c)+1)*b-1/2/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)*A*b^2+1/d/a^4/(tan(1/2*d*x+1/2*c)+1)*A*b^
3-1/d/a^3/(tan(1/2*d*x+1/2*c)+1)*b^2*B+1/d/a^2/(tan(1/2*d*x+1/2*c)+1)*b*C+1/d/a^4/(tan(1/2*d*x+1/2*c)-1)*A*b^3
-1/d/a^3/(tan(1/2*d*x+1/2*c)-1)*b^2*B+1/d/a^2/(tan(1/2*d*x+1/2*c)-1)*b*C+1/3/d/a^2/(tan(1/2*d*x+1/2*c)+1)^3*A*
b-1/2/d/a^3/(tan(1/2*d*x+1/2*c)+1)^2*A*b^2+1/2/d/a^2/(tan(1/2*d*x+1/2*c)+1)^2*b*B+1/d/a^5*ln(tan(1/2*d*x+1/2*c
)+1)*A*b^4-1/d/a^4*ln(tan(1/2*d*x+1/2*c)+1)*b^3*B+1/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*b^2*C+1/3/d/a^2/(tan(1/2*d*
x+1/2*c)-1)^3*A*b+1/2/d/a^3/(tan(1/2*d*x+1/2*c)-1)^2*A*b^2-1/2/d/a^2/(tan(1/2*d*x+1/2*c)-1)^2*b*B-1/d/a^5*ln(t
an(1/2*d*x+1/2*c)-1)*A*b^4+1/d/a^4*ln(tan(1/2*d*x+1/2*c)-1)*b^3*B-1/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)*b^2*C-2/d*b
^5/a^5/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A+2/d*b^4/a^4/((a+b)*(a-b))^(1
/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B-2/d*b^3/a^3/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/
2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C-1/2/d/a^2*ln(tan(1/2*d*x+1/2*c)+1)*b*B

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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*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5/(a+b*cos(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [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+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

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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((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**5/(a+b*cos(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.30812, size = 1185, normalized size = 4.16 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(3*A*a^4 + 4*C*a^4 - 4*B*a^3*b + 4*A*a^2*b^2 + 8*C*a^2*b^2 - 8*B*a*b^3 + 8*A*b^4)*log(abs(tan(1/2*d*x
+ 1/2*c) + 1))/a^5 - 3*(3*A*a^4 + 4*C*a^4 - 4*B*a^3*b + 4*A*a^2*b^2 + 8*C*a^2*b^2 - 8*B*a*b^3 + 8*A*b^4)*log(a
bs(tan(1/2*d*x + 1/2*c) - 1))/a^5 + 48*(C*a^2*b^3 - B*a*b^4 + A*b^5)*(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)))/(sqrt(a^2 - b^2)*a^5) +
2*(15*A*a^3*tan(1/2*d*x + 1/2*c)^7 - 24*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 12*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 24*A
*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 12*B*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 24*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 12*A*a
*b^2*tan(1/2*d*x + 1/2*c)^7 - 24*B*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 24*A*b^3*tan(1/2*d*x + 1/2*c)^7 + 9*A*a^3*ta
n(1/2*d*x + 1/2*c)^5 + 40*B*a^3*tan(1/2*d*x + 1/2*c)^5 - 12*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 40*A*a^2*b*tan(1/2*
d*x + 1/2*c)^5 + 12*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 72*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 12*A*a*b^2*tan(1/2*d*
x + 1/2*c)^5 + 72*B*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 72*A*b^3*tan(1/2*d*x + 1/2*c)^5 + 9*A*a^3*tan(1/2*d*x + 1/2
*c)^3 - 40*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 12*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 40*A*a^2*b*tan(1/2*d*x + 1/2*c)^3
+ 12*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 72*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 12*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 -
72*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 72*A*b^3*tan(1/2*d*x + 1/2*c)^3 + 15*A*a^3*tan(1/2*d*x + 1/2*c) + 24*B*a^3
*tan(1/2*d*x + 1/2*c) + 12*C*a^3*tan(1/2*d*x + 1/2*c) - 24*A*a^2*b*tan(1/2*d*x + 1/2*c) - 12*B*a^2*b*tan(1/2*d
*x + 1/2*c) - 24*C*a^2*b*tan(1/2*d*x + 1/2*c) + 12*A*a*b^2*tan(1/2*d*x + 1/2*c) + 24*B*a*b^2*tan(1/2*d*x + 1/2
*c) - 24*A*b^3*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^4*a^4))/d