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

Optimal. Leaf size=122 \[ \frac {(3 B-4 C) \sin ^3(c+d x)}{3 a d}-\frac {(3 B-4 C) \sin (c+d x)}{a d}+\frac {(B-C) \sin (c+d x) \cos ^3(c+d x)}{d (a \cos (c+d x)+a)}+\frac {3 (B-C) \sin (c+d x) \cos (c+d x)}{2 a d}+\frac {3 x (B-C)}{2 a} \]

[Out]

3/2*(B-C)*x/a-(3*B-4*C)*sin(d*x+c)/a/d+3/2*(B-C)*cos(d*x+c)*sin(d*x+c)/a/d+(B-C)*cos(d*x+c)^3*sin(d*x+c)/d/(a+
a*cos(d*x+c))+1/3*(3*B-4*C)*sin(d*x+c)^3/a/d

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Rubi [A]  time = 0.26, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3029, 2977, 2748, 2635, 8, 2633} \[ \frac {(3 B-4 C) \sin ^3(c+d x)}{3 a d}-\frac {(3 B-4 C) \sin (c+d x)}{a d}+\frac {(B-C) \sin (c+d x) \cos ^3(c+d x)}{d (a \cos (c+d x)+a)}+\frac {3 (B-C) \sin (c+d x) \cos (c+d x)}{2 a d}+\frac {3 x (B-C)}{2 a} \]

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]),x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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

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Mathematica [B]  time = 0.57, size = 249, normalized size = 2.04 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (36 d x (B-C) \cos \left (c+\frac {d x}{2}\right )-12 B \sin \left (c+\frac {d x}{2}\right )-9 B \sin \left (c+\frac {3 d x}{2}\right )-9 B \sin \left (2 c+\frac {3 d x}{2}\right )+3 B \sin \left (2 c+\frac {5 d x}{2}\right )+3 B \sin \left (3 c+\frac {5 d x}{2}\right )+36 d x (B-C) \cos \left (\frac {d x}{2}\right )-60 B \sin \left (\frac {d x}{2}\right )+21 C \sin \left (c+\frac {d x}{2}\right )+18 C \sin \left (c+\frac {3 d x}{2}\right )+18 C \sin \left (2 c+\frac {3 d x}{2}\right )-2 C \sin \left (2 c+\frac {5 d x}{2}\right )-2 C \sin \left (3 c+\frac {5 d x}{2}\right )+C \sin \left (3 c+\frac {7 d x}{2}\right )+C \sin \left (4 c+\frac {7 d x}{2}\right )+69 C \sin \left (\frac {d x}{2}\right )\right )}{24 a d (\cos (c+d x)+1)} \]

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]),x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(36*(B - C)*d*x*Cos[(d*x)/2] + 36*(B - C)*d*x*Cos[c + (d*x)/2] - 60*B*Sin[(d*x)/2]
+ 69*C*Sin[(d*x)/2] - 12*B*Sin[c + (d*x)/2] + 21*C*Sin[c + (d*x)/2] - 9*B*Sin[c + (3*d*x)/2] + 18*C*Sin[c + (3
*d*x)/2] - 9*B*Sin[2*c + (3*d*x)/2] + 18*C*Sin[2*c + (3*d*x)/2] + 3*B*Sin[2*c + (5*d*x)/2] - 2*C*Sin[2*c + (5*
d*x)/2] + 3*B*Sin[3*c + (5*d*x)/2] - 2*C*Sin[3*c + (5*d*x)/2] + C*Sin[3*c + (7*d*x)/2] + C*Sin[4*c + (7*d*x)/2
]))/(24*a*d*(1 + Cos[c + d*x]))

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fricas [A]  time = 0.44, size = 98, normalized size = 0.80 \[ \frac {9 \, {\left (B - C\right )} d x \cos \left (d x + c\right ) + 9 \, {\left (B - C\right )} d x + {\left (2 \, C \cos \left (d x + c\right )^{3} + {\left (3 \, B - C\right )} \cos \left (d x + c\right )^{2} - {\left (3 \, B - 7 \, C\right )} \cos \left (d x + c\right ) - 12 \, B + 16 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

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)),x, algorithm="fricas")

[Out]

1/6*(9*(B - C)*d*x*cos(d*x + c) + 9*(B - C)*d*x + (2*C*cos(d*x + c)^3 + (3*B - C)*cos(d*x + c)^2 - (3*B - 7*C)
*cos(d*x + c) - 12*B + 16*C)*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)

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giac [A]  time = 0.59, size = 151, normalized size = 1.24 \[ \frac {\frac {9 \, {\left (d x + c\right )} {\left (B - C\right )}}{a} - \frac {6 \, {\left (B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} - \frac {2 \, {\left (9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a}}{6 \, d} \]

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)),x, algorithm="giac")

[Out]

1/6*(9*(d*x + c)*(B - C)/a - 6*(B*tan(1/2*d*x + 1/2*c) - C*tan(1/2*d*x + 1/2*c))/a - 2*(9*B*tan(1/2*d*x + 1/2*
c)^5 - 15*C*tan(1/2*d*x + 1/2*c)^5 + 12*B*tan(1/2*d*x + 1/2*c)^3 - 16*C*tan(1/2*d*x + 1/2*c)^3 + 3*B*tan(1/2*d
*x + 1/2*c) - 9*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a))/d

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maple [B]  time = 0.13, size = 281, normalized size = 2.30 \[ -\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {5 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {16 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{3 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {3 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{a d}-\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{a d} \]

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)),x)

[Out]

-1/a/d*B*tan(1/2*d*x+1/2*c)+1/a/d*C*tan(1/2*d*x+1/2*c)-3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*B
+5/a/d/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*C-4/a/d/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3
*B+16/3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*C-1/a/d/(1+tan(1/2*d*x+1/2*c)^2)^3*B*tan(1/2*d*x+1
/2*c)+3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^3*C*tan(1/2*d*x+1/2*c)+3/a/d*arctan(tan(1/2*d*x+1/2*c))*B-3/a/d*arctan(ta
n(1/2*d*x+1/2*c))*C

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maxima [B]  time = 1.08, size = 310, normalized size = 2.54 \[ \frac {C {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a + \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {9 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {3 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 3 \, B {\left (\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a + \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{3 \, d} \]

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)),x, algorithm="maxima")

[Out]

1/3*(C*((9*sin(d*x + c)/(cos(d*x + c) + 1) + 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d
*x + c) + 1)^5)/(a + 3*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a*sin
(d*x + c)^6/(cos(d*x + c) + 1)^6) - 9*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + 3*sin(d*x + c)/(a*(cos(d*x +
c) + 1))) - 3*B*((sin(d*x + c)/(cos(d*x + c) + 1) + 3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a + 2*a*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 + a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) - 3*arctan(sin(d*x + c)/(cos(d*x + c) + 1)
)/a + sin(d*x + c)/(a*(cos(d*x + c) + 1))))/d

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mupad [B]  time = 2.22, size = 138, normalized size = 1.13 \[ \frac {3\,x\,\left (B-C\right )}{2\,a}-\frac {\left (3\,B-5\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (4\,B-\frac {16\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (B-3\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B-C\right )}{a\,d} \]

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)),x)

[Out]

(3*x*(B - C))/(2*a) - (tan(c/2 + (d*x)/2)^5*(3*B - 5*C) + tan(c/2 + (d*x)/2)^3*(4*B - (16*C)/3) + tan(c/2 + (d
*x)/2)*(B - 3*C))/(d*(a + 3*a*tan(c/2 + (d*x)/2)^2 + 3*a*tan(c/2 + (d*x)/2)^4 + a*tan(c/2 + (d*x)/2)^6)) - (ta
n(c/2 + (d*x)/2)*(B - C))/(a*d)

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sympy [A]  time = 6.48, size = 1166, normalized size = 9.56 \[ \text {result too large to display} \]

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)),x)

[Out]

Piecewise((9*B*d*x*tan(c/2 + d*x/2)**6/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/
2 + d*x/2)**2 + 6*a*d) + 27*B*d*x*tan(c/2 + d*x/2)**4/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4
+ 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 27*B*d*x*tan(c/2 + d*x/2)**2/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c
/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 9*B*d*x/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d
*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 6*B*tan(c/2 + d*x/2)**7/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*t
an(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 36*B*tan(c/2 + d*x/2)**5/(6*a*d*tan(c/2 + d*x/2)**6
+ 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 42*B*tan(c/2 + d*x/2)**3/(6*a*d*tan(c/2
+ d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 12*B*tan(c/2 + d*x/2)/(6*a*d*
tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 9*C*d*x*tan(c/2 + d*x
/2)**6/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 27*C*d*
x*tan(c/2 + d*x/2)**4/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6
*a*d) - 27*C*d*x*tan(c/2 + d*x/2)**2/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2
+ d*x/2)**2 + 6*a*d) - 9*C*d*x/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/
2)**2 + 6*a*d) + 6*C*tan(c/2 + d*x/2)**7/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(
c/2 + d*x/2)**2 + 6*a*d) + 48*C*tan(c/2 + d*x/2)**5/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 +
18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 50*C*tan(c/2 + d*x/2)**3/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d
*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 24*C*tan(c/2 + d*x/2)/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan
(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d), Ne(d, 0)), (x*(B*cos(c) + C*cos(c)**2)*cos(c)**2/(a*co
s(c) + a), True))

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