3.3.51 \(\int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx\) [251]

Optimal. Leaf size=134 \[ -\frac {\left (2 a A b-2 a^2 B-b^2 B\right ) x}{2 b^3}+\frac {2 a^2 (A b-a B) \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^3 \sqrt {a+b} d}+\frac {(A b-a B) \sin (c+d x)}{b^2 d}+\frac {B \cos (c+d x) \sin (c+d x)}{2 b d} \]

[Out]

-1/2*(2*A*a*b-2*B*a^2-B*b^2)*x/b^3+(A*b-B*a)*sin(d*x+c)/b^2/d+1/2*B*cos(d*x+c)*sin(d*x+c)/b/d+2*a^2*(A*b-B*a)*
arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/b^3/d/(a-b)^(1/2)/(a+b)^(1/2)

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Rubi [A]
time = 0.20, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3069, 3102, 2814, 2738, 211} \begin {gather*} \frac {2 a^2 (A b-a B) \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d \sqrt {a-b} \sqrt {a+b}}-\frac {x \left (-2 a^2 B+2 a A b-b^2 B\right )}{2 b^3}+\frac {(A b-a B) \sin (c+d x)}{b^2 d}+\frac {B \sin (c+d x) \cos (c+d x)}{2 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((2*a*A*b - 2*a^2*B - b^2*B)*x)/b^3 + (2*a^2*(A*b - a*B)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b
]])/(Sqrt[a - b]*b^3*Sqrt[a + b]*d) + ((A*b - a*B)*Sin[c + d*x])/(b^2*d) + (B*Cos[c + d*x]*Sin[c + d*x])/(2*b*
d)

Rule 211

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

Rule 2738

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 2814

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

Rule 3069

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*
x])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f
*x])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*(m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c
- b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m
, 1] &&  !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3102

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

Rubi steps

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

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Mathematica [A]
time = 0.35, size = 121, normalized size = 0.90 \begin {gather*} \frac {2 \left (-2 a A b+2 a^2 B+b^2 B\right ) (c+d x)+\frac {8 a^2 (-A b+a B) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+4 b (A b-a B) \sin (c+d x)+b^2 B \sin (2 (c+d x))}{4 b^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.23, size = 169, normalized size = 1.26

method	result	size
derivativedivides	\(\frac {\frac {2 a^{2} \left (A b -a B \right ) \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \left (\frac {\left (-A \,b^{2}+B a b +\frac {1}{2} B \,b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-A \,b^{2}+B a b -\frac {1}{2} B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (2 A a b -2 B \,a^{2}-B \,b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{3}}}{d}\)	\(169\)
default	\(\frac {\frac {2 a^{2} \left (A b -a B \right ) \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \left (\frac {\left (-A \,b^{2}+B a b +\frac {1}{2} B \,b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-A \,b^{2}+B a b -\frac {1}{2} B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (2 A a b -2 B \,a^{2}-B \,b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{3}}}{d}\)	\(169\)
risch	\(-\frac {x A a}{b^{2}}+\frac {x B \,a^{2}}{b^{3}}+\frac {B x}{2 b}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A}{2 b d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} a B}{2 b^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A}{2 b d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} a B}{2 b^{2} d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}+\frac {B \sin \left (2 d x +2 c \right )}{4 b d}\)	\(420\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(2*a^2*(A*b-B*a)/b^3/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))-2/b^3*(((-A*
b^2+B*a*b+1/2*B*b^2)*tan(1/2*d*x+1/2*c)^3+(-A*b^2+B*a*b-1/2*B*b^2)*tan(1/2*d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^2
)^2+1/2*(2*A*a*b-2*B*a^2-B*b^2)*arctan(tan(1/2*d*x+1/2*c))))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

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Fricas [A]
time = 0.41, size = 426, normalized size = 3.18 \begin {gather*} \left [\frac {{\left (2 \, B a^{4} - 2 \, A a^{3} b - B a^{2} b^{2} + 2 \, A a b^{3} - B b^{4}\right )} d x + {\left (B a^{3} - A a^{2} b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (2 \, B a^{3} b - 2 \, A a^{2} b^{2} - 2 \, B a b^{3} + 2 \, A b^{4} - {\left (B a^{2} b^{2} - B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )} d}, \frac {{\left (2 \, B a^{4} - 2 \, A a^{3} b - B a^{2} b^{2} + 2 \, A a b^{3} - B b^{4}\right )} d x - 2 \, {\left (B a^{3} - A a^{2} b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (2 \, B a^{3} b - 2 \, A a^{2} b^{2} - 2 \, B a b^{3} + 2 \, A b^{4} - {\left (B a^{2} b^{2} - B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 0.46, size = 227, normalized size = 1.69 \begin {gather*} \frac {\frac {{\left (2 \, B a^{2} - 2 \, A a b + B b^{2}\right )} {\left (d x + c\right )}}{b^{3}} + \frac {4 \, {\left (B a^{3} - A a^{2} b\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 )}}{\sqrt {a^{2} - b^{2}} b^{3}} - \frac {2 \, {\left (2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((2*B*a^2 - 2*A*a*b + B*b^2)*(d*x + c)/b^3 + 4*(B*a^3 - A*a^2*b)*(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)*b^3) -
2*(2*B*a*tan(1/2*d*x + 1/2*c)^3 - 2*A*b*tan(1/2*d*x + 1/2*c)^3 + B*b*tan(1/2*d*x + 1/2*c)^3 + 2*B*a*tan(1/2*d
*x + 1/2*c) - 2*A*b*tan(1/2*d*x + 1/2*c) - B*b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*b^2))/d

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Mupad [B]
time = 4.00, size = 2500, normalized size = 18.66 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((tan(c/2 + (d*x)/2)*(2*A*b - 2*B*a + B*b))/b^2 - (tan(c/2 + (d*x)/2)^3*(2*B*a - 2*A*b + B*b))/b^2)/(d*(2*tan(
c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 + 1)) - (atan(((((((8*(2*B*b^10 + 8*A*a^2*b^8 - 4*A*a^3*b^7 + 2*B*a^2*
b^8 - 6*B*a^3*b^7 + 4*B*a^4*b^6 - 4*A*a*b^9 - 2*B*a*b^9))/b^6 - (4*tan(c/2 + (d*x)/2)*(B*a^2*2i + B*b^2*1i - A
*a*b*2i)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/b^7)*(B*a^2*2i + B*b^2*1i - A*a*b*2i))/(2*b^3) - (8*tan(c/2 + (d*
x)/2)*(8*B^2*a^7 - B^2*b^7 + 3*B^2*a*b^6 - 16*B^2*a^6*b - 4*A^2*a^2*b^5 + 12*A^2*a^3*b^4 - 16*A^2*a^4*b^3 + 8*
A^2*a^5*b^2 - 7*B^2*a^2*b^5 + 13*B^2*a^3*b^4 - 16*B^2*a^4*b^3 + 16*B^2*a^5*b^2 + 4*A*B*a*b^6 - 16*A*B*a^6*b -
12*A*B*a^2*b^5 + 20*A*B*a^3*b^4 - 28*A*B*a^4*b^3 + 32*A*B*a^5*b^2))/b^4)*(B*a^2*2i + B*b^2*1i - A*a*b*2i)*1i)/
(2*b^3) - (((((8*(2*B*b^10 + 8*A*a^2*b^8 - 4*A*a^3*b^7 + 2*B*a^2*b^8 - 6*B*a^3*b^7 + 4*B*a^4*b^6 - 4*A*a*b^9 -
2*B*a*b^9))/b^6 + (4*tan(c/2 + (d*x)/2)*(B*a^2*2i + B*b^2*1i - A*a*b*2i)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/
b^7)*(B*a^2*2i + B*b^2*1i - A*a*b*2i))/(2*b^3) + (8*tan(c/2 + (d*x)/2)*(8*B^2*a^7 - B^2*b^7 + 3*B^2*a*b^6 - 16
*B^2*a^6*b - 4*A^2*a^2*b^5 + 12*A^2*a^3*b^4 - 16*A^2*a^4*b^3 + 8*A^2*a^5*b^2 - 7*B^2*a^2*b^5 + 13*B^2*a^3*b^4
- 16*B^2*a^4*b^3 + 16*B^2*a^5*b^2 + 4*A*B*a*b^6 - 16*A*B*a^6*b - 12*A*B*a^2*b^5 + 20*A*B*a^3*b^4 - 28*A*B*a^4*
b^3 + 32*A*B*a^5*b^2))/b^4)*(B*a^2*2i + B*b^2*1i - A*a*b*2i)*1i)/(2*b^3))/((16*(4*B^3*a^8 - 6*B^3*a^7*b + 4*A^
3*a^4*b^4 - 4*A^3*a^5*b^3 - B^3*a^3*b^5 + 2*B^3*a^4*b^4 - 5*B^3*a^5*b^3 + 6*B^3*a^6*b^2 - 12*A*B^2*a^7*b + A*B
^2*a^2*b^6 - 2*A*B^2*a^3*b^5 + 9*A*B^2*a^4*b^4 - 12*A*B^2*a^5*b^3 + 16*A*B^2*a^6*b^2 - 4*A^2*B*a^3*b^5 + 6*A^2
*B*a^4*b^4 - 14*A^2*B*a^5*b^3 + 12*A^2*B*a^6*b^2))/b^6 + (((((8*(2*B*b^10 + 8*A*a^2*b^8 - 4*A*a^3*b^7 + 2*B*a^
2*b^8 - 6*B*a^3*b^7 + 4*B*a^4*b^6 - 4*A*a*b^9 - 2*B*a*b^9))/b^6 - (4*tan(c/2 + (d*x)/2)*(B*a^2*2i + B*b^2*1i -
A*a*b*2i)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/b^7)*(B*a^2*2i + B*b^2*1i - A*a*b*2i))/(2*b^3) - (8*tan(c/2 + (
d*x)/2)*(8*B^2*a^7 - B^2*b^7 + 3*B^2*a*b^6 - 16*B^2*a^6*b - 4*A^2*a^2*b^5 + 12*A^2*a^3*b^4 - 16*A^2*a^4*b^3 +
8*A^2*a^5*b^2 - 7*B^2*a^2*b^5 + 13*B^2*a^3*b^4 - 16*B^2*a^4*b^3 + 16*B^2*a^5*b^2 + 4*A*B*a*b^6 - 16*A*B*a^6*b
- 12*A*B*a^2*b^5 + 20*A*B*a^3*b^4 - 28*A*B*a^4*b^3 + 32*A*B*a^5*b^2))/b^4)*(B*a^2*2i + B*b^2*1i - A*a*b*2i))/(
2*b^3) + (((((8*(2*B*b^10 + 8*A*a^2*b^8 - 4*A*a^3*b^7 + 2*B*a^2*b^8 - 6*B*a^3*b^7 + 4*B*a^4*b^6 - 4*A*a*b^9 -
2*B*a*b^9))/b^6 + (4*tan(c/2 + (d*x)/2)*(B*a^2*2i + B*b^2*1i - A*a*b*2i)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/b
^7)*(B*a^2*2i + B*b^2*1i - A*a*b*2i))/(2*b^3) + (8*tan(c/2 + (d*x)/2)*(8*B^2*a^7 - B^2*b^7 + 3*B^2*a*b^6 - 16*
B^2*a^6*b - 4*A^2*a^2*b^5 + 12*A^2*a^3*b^4 - 16*A^2*a^4*b^3 + 8*A^2*a^5*b^2 - 7*B^2*a^2*b^5 + 13*B^2*a^3*b^4 -
16*B^2*a^4*b^3 + 16*B^2*a^5*b^2 + 4*A*B*a*b^6 - 16*A*B*a^6*b - 12*A*B*a^2*b^5 + 20*A*B*a^3*b^4 - 28*A*B*a^4*b
^3 + 32*A*B*a^5*b^2))/b^4)*(B*a^2*2i + B*b^2*1i - A*a*b*2i))/(2*b^3)))*(B*a^2*2i + B*b^2*1i - A*a*b*2i)*1i)/(b
^3*d) + (a^2*atan(((a^2*(-(a + b)*(a - b))^(1/2)*(A*b - B*a)*((8*tan(c/2 + (d*x)/2)*(8*B^2*a^7 - B^2*b^7 + 3*B
^2*a*b^6 - 16*B^2*a^6*b - 4*A^2*a^2*b^5 + 12*A^2*a^3*b^4 - 16*A^2*a^4*b^3 + 8*A^2*a^5*b^2 - 7*B^2*a^2*b^5 + 13
*B^2*a^3*b^4 - 16*B^2*a^4*b^3 + 16*B^2*a^5*b^2 + 4*A*B*a*b^6 - 16*A*B*a^6*b - 12*A*B*a^2*b^5 + 20*A*B*a^3*b^4
- 28*A*B*a^4*b^3 + 32*A*B*a^5*b^2))/b^4 + (a^2*(-(a + b)*(a - b))^(1/2)*(A*b - B*a)*((8*(2*B*b^10 + 8*A*a^2*b^
8 - 4*A*a^3*b^7 + 2*B*a^2*b^8 - 6*B*a^3*b^7 + 4*B*a^4*b^6 - 4*A*a*b^9 - 2*B*a*b^9))/b^6 + (8*a^2*tan(c/2 + (d*
x)/2)*(-(a + b)*(a - b))^(1/2)*(A*b - B*a)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/(b^4*(b^5 - a^2*b^3))))/(b^5 -
a^2*b^3))*1i)/(b^5 - a^2*b^3) + (a^2*(-(a + b)*(a - b))^(1/2)*(A*b - B*a)*((8*tan(c/2 + (d*x)/2)*(8*B^2*a^7 -
B^2*b^7 + 3*B^2*a*b^6 - 16*B^2*a^6*b - 4*A^2*a^2*b^5 + 12*A^2*a^3*b^4 - 16*A^2*a^4*b^3 + 8*A^2*a^5*b^2 - 7*B^2
*a^2*b^5 + 13*B^2*a^3*b^4 - 16*B^2*a^4*b^3 + 16*B^2*a^5*b^2 + 4*A*B*a*b^6 - 16*A*B*a^6*b - 12*A*B*a^2*b^5 + 20
*A*B*a^3*b^4 - 28*A*B*a^4*b^3 + 32*A*B*a^5*b^2))/b^4 - (a^2*(-(a + b)*(a - b))^(1/2)*(A*b - B*a)*((8*(2*B*b^10
+ 8*A*a^2*b^8 - 4*A*a^3*b^7 + 2*B*a^2*b^8 - 6*B*a^3*b^7 + 4*B*a^4*b^6 - 4*A*a*b^9 - 2*B*a*b^9))/b^6 - (8*a^2*
tan(c/2 + (d*x)/2)*(-(a + b)*(a - b))^(1/2)*(A*b - B*a)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/(b^4*(b^5 - a^2*b^
3))))/(b^5 - a^2*b^3))*1i)/(b^5 - a^2*b^3))/((16*(4*B^3*a^8 - 6*B^3*a^7*b + 4*A^3*a^4*b^4 - 4*A^3*a^5*b^3 - B^
3*a^3*b^5 + 2*B^3*a^4*b^4 - 5*B^3*a^5*b^3 + 6*B^3*a^6*b^2 - 12*A*B^2*a^7*b + A*B^2*a^2*b^6 - 2*A*B^2*a^3*b^5 +
9*A*B^2*a^4*b^4 - 12*A*B^2*a^5*b^3 + 16*A*B^2*a^6*b^2 - 4*A^2*B*a^3*b^5 + 6*A^2*B*a^4*b^4 - 14*A^2*B*a^5*b^3
+ 12*A^2*B*a^6*b^2))/b^6 + (a^2*(-(a + b)*(a - b))^(1/2)*(A*b - B*a)*((8*tan(c/2 + (d*x)/2)*(8*B^2*a^7 - B^2*b
^7 + 3*B^2*a*b^6 - 16*B^2*a^6*b - 4*A^2*a^2*b^5 + 12*A^2*a^3*b^4 - 16*A^2*a^4*b^3 + 8*A^2*a^5*b^2 - 7*B^2*a^2*
b^5 + 13*B^2*a^3*b^4 - 16*B^2*a^4*b^3 + 16*B^2*a^5*b^2 + 4*A*B*a*b^6 - 16*A*B*a^6*b - 12*A*B*a^2*b^5 + 20*A*B*
a^3*b^4 - 28*A*B*a^4*b^3 + 32*A*B*a^5*b^2))/b^4...

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