3.262 \(\int \frac {\cos ^2(c+d x) (B \cos (c+d x)+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=147 \[ \frac {2 (5 B-8 C) \sin (c+d x)}{3 a^2 d}+\frac {(5 B-8 C) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac {(4 B-7 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac {x (4 B-7 C)}{2 a^2}+\frac {(B-C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

-1/2*(4*B-7*C)*x/a^2+2/3*(5*B-8*C)*sin(d*x+c)/a^2/d-1/2*(4*B-7*C)*cos(d*x+c)*sin(d*x+c)/a^2/d+1/3*(5*B-8*C)*co
s(d*x+c)^2*sin(d*x+c)/a^2/d/(1+cos(d*x+c))+1/3*(B-C)*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^2

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Rubi [A]  time = 0.36, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {3029, 2977, 2734} \[ \frac {2 (5 B-8 C) \sin (c+d x)}{3 a^2 d}+\frac {(5 B-8 C) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac {(4 B-7 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac {x (4 B-7 C)}{2 a^2}+\frac {(B-C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((4*B - 7*C)*x)/(2*a^2) + (2*(5*B - 8*C)*Sin[c + d*x])/(3*a^2*d) - ((4*B - 7*C)*Cos[c + d*x]*Sin[c + d*x])/(2
*a^2*d) + ((5*B - 8*C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) + ((B - C)*Cos[c + d*x]^3*Sin
[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2)

Rule 2734

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

Rule 2977

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

Rule 3029

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

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx &=\int \frac {\cos ^3(c+d x) (B+C \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx\\ &=\frac {(B-C) \cos ^3(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \frac {\cos ^2(c+d x) (3 a (B-C)-a (2 B-5 C) \cos (c+d x))}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=\frac {(5 B-8 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac {(B-C) \cos ^3(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \cos (c+d x) \left (2 a^2 (5 B-8 C)-3 a^2 (4 B-7 C) \cos (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {(4 B-7 C) x}{2 a^2}+\frac {2 (5 B-8 C) \sin (c+d x)}{3 a^2 d}-\frac {(4 B-7 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {(5 B-8 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac {(B-C) \cos ^3(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}\\ \end {align*}

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Mathematica [B]  time = 0.87, size = 315, normalized size = 2.14 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (-36 d x (4 B-7 C) \cos \left (c+\frac {d x}{2}\right )-120 B \sin \left (c+\frac {d x}{2}\right )+164 B \sin \left (c+\frac {3 d x}{2}\right )+36 B \sin \left (2 c+\frac {3 d x}{2}\right )+12 B \sin \left (2 c+\frac {5 d x}{2}\right )+12 B \sin \left (3 c+\frac {5 d x}{2}\right )-48 B d x \cos \left (c+\frac {3 d x}{2}\right )-48 B d x \cos \left (2 c+\frac {3 d x}{2}\right )-36 d x (4 B-7 C) \cos \left (\frac {d x}{2}\right )+264 B \sin \left (\frac {d x}{2}\right )+147 C \sin \left (c+\frac {d x}{2}\right )-239 C \sin \left (c+\frac {3 d x}{2}\right )-63 C \sin \left (2 c+\frac {3 d x}{2}\right )-15 C \sin \left (2 c+\frac {5 d x}{2}\right )-15 C \sin \left (3 c+\frac {5 d x}{2}\right )+3 C \sin \left (3 c+\frac {7 d x}{2}\right )+3 C \sin \left (4 c+\frac {7 d x}{2}\right )+84 C d x \cos \left (c+\frac {3 d x}{2}\right )+84 C d x \cos \left (2 c+\frac {3 d x}{2}\right )-381 C \sin \left (\frac {d x}{2}\right )\right )}{48 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(-36*(4*B - 7*C)*d*x*Cos[(d*x)/2] - 36*(4*B - 7*C)*d*x*Cos[c + (d*x)/2] - 48*B*d*x*
Cos[c + (3*d*x)/2] + 84*C*d*x*Cos[c + (3*d*x)/2] - 48*B*d*x*Cos[2*c + (3*d*x)/2] + 84*C*d*x*Cos[2*c + (3*d*x)/
2] + 264*B*Sin[(d*x)/2] - 381*C*Sin[(d*x)/2] - 120*B*Sin[c + (d*x)/2] + 147*C*Sin[c + (d*x)/2] + 164*B*Sin[c +
(3*d*x)/2] - 239*C*Sin[c + (3*d*x)/2] + 36*B*Sin[2*c + (3*d*x)/2] - 63*C*Sin[2*c + (3*d*x)/2] + 12*B*Sin[2*c
+ (5*d*x)/2] - 15*C*Sin[2*c + (5*d*x)/2] + 12*B*Sin[3*c + (5*d*x)/2] - 15*C*Sin[3*c + (5*d*x)/2] + 3*C*Sin[3*c
+ (7*d*x)/2] + 3*C*Sin[4*c + (7*d*x)/2]))/(48*a^2*d*(1 + Cos[c + d*x])^2)

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fricas [A]  time = 0.43, size = 138, normalized size = 0.94 \[ -\frac {3 \, {\left (4 \, B - 7 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (4 \, B - 7 \, C\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (4 \, B - 7 \, C\right )} d x - {\left (3 \, C \cos \left (d x + c\right )^{3} + 6 \, {\left (B - C\right )} \cos \left (d x + c\right )^{2} + {\left (28 \, B - 43 \, C\right )} \cos \left (d x + c\right ) + 20 \, B - 32 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(4*B - 7*C)*d*x*cos(d*x + c)^2 + 6*(4*B - 7*C)*d*x*cos(d*x + c) + 3*(4*B - 7*C)*d*x - (3*C*cos(d*x + c
)^3 + 6*(B - C)*cos(d*x + c)^2 + (28*B - 43*C)*cos(d*x + c) + 20*B - 32*C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2
+ 2*a^2*d*cos(d*x + c) + a^2*d)

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giac [A]  time = 0.41, size = 164, normalized size = 1.12 \[ -\frac {\frac {3 \, {\left (d x + c\right )} {\left (4 \, B - 7 \, C\right )}}{a^{2}} - \frac {6 \, {\left (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}} + \frac {B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(d*x + c)*(4*B - 7*C)/a^2 - 6*(2*B*tan(1/2*d*x + 1/2*c)^3 - 5*C*tan(1/2*d*x + 1/2*c)^3 + 2*B*tan(1/2*d
*x + 1/2*c) - 3*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^2) + (B*a^4*tan(1/2*d*x + 1/2*c)^3 -
C*a^4*tan(1/2*d*x + 1/2*c)^3 - 15*B*a^4*tan(1/2*d*x + 1/2*c) + 21*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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maple [A]  time = 0.14, size = 252, normalized size = 1.71 \[ -\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {5 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {3 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{2}}+\frac {7 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/d/a^2*B*tan(1/2*d*x+1/2*c)^3+1/6/d/a^2*C*tan(1/2*d*x+1/2*c)^3+5/2/d/a^2*B*tan(1/2*d*x+1/2*c)-7/2/d/a^2*C*
tan(1/2*d*x+1/2*c)-5/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*C*tan(1/2*d*x+1/2*c)^3+2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^
2*B*tan(1/2*d*x+1/2*c)^3-3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*C*tan(1/2*d*x+1/2*c)+2/d/a^2/(1+tan(1/2*d*x+1/2*c)
^2)^2*B*tan(1/2*d*x+1/2*c)-4/d/a^2*arctan(tan(1/2*d*x+1/2*c))*B+7/d/a^2*arctan(tan(1/2*d*x+1/2*c))*C

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maxima [B]  time = 0.96, size = 283, normalized size = 1.93 \[ -\frac {C {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {42 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - B {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(C*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) + 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 + 2*a^2*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) - s
in(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 42*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2) - B*((15*sin(d*x + c
)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a
^2 + 12*sin(d*x + c)/((a^2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))))/d

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mupad [B]  time = 1.15, size = 152, normalized size = 1.03 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (B-C\right )}{2\,a^2}+\frac {2\,B-4\,C}{2\,a^2}\right )}{d}-\frac {x\,\left (4\,B-7\,C\right )}{2\,a^2}+\frac {\left (2\,B-5\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,B-3\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (B-C\right )}{6\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((3*(B - C))/(2*a^2) + (2*B - 4*C)/(2*a^2)))/d - (x*(4*B - 7*C))/(2*a^2) + (tan(c/2 + (d*x
)/2)^3*(2*B - 5*C) + tan(c/2 + (d*x)/2)*(2*B - 3*C))/(d*(2*a^2*tan(c/2 + (d*x)/2)^2 + a^2*tan(c/2 + (d*x)/2)^4
+ a^2)) - (tan(c/2 + (d*x)/2)^3*(B - C))/(6*a^2*d)

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sympy [A]  time = 9.54, size = 848, normalized size = 5.77 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-12*B*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**
2*d) - 24*B*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d)
- 12*B*d*x/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - B*tan(c/2 + d*x/2)**7/(
6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 13*B*tan(c/2 + d*x/2)**5/(6*a**2*d*
tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 41*B*tan(c/2 + d*x/2)**3/(6*a**2*d*tan(c/2 +
d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 27*B*tan(c/2 + d*x/2)/(6*a**2*d*tan(c/2 + d*x/2)**4 +
12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 21*C*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a
**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 42*C*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d
*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 21*C*d*x/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*
a**2*d) + C*tan(c/2 + d*x/2)**7/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 19
*C*tan(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 71*C*tan(c/
2 + d*x/2)**3/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 39*C*tan(c/2 + d*x/2
)/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d), Ne(d, 0)), (x*(B*cos(c) + C*cos(c
)**2)*cos(c)**2/(a*cos(c) + a)**2, True))

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