3.53 \(\int \frac {\cos ^5(c+d x)}{(a+a \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=124 \[ -\frac {4 \sin ^3(c+d x)}{a^2 d}+\frac {12 \sin (c+d x)}{a^2 d}-\frac {10 \sin (c+d x) \cos ^3(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac {5 \sin (c+d x) \cos (c+d x)}{a^2 d}-\frac {5 x}{a^2}-\frac {\sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

-5*x/a^2+12*sin(d*x+c)/a^2/d-5*cos(d*x+c)*sin(d*x+c)/a^2/d-10/3*cos(d*x+c)^3*sin(d*x+c)/a^2/d/(1+cos(d*x+c))-1
/3*cos(d*x+c)^4*sin(d*x+c)/d/(a+a*cos(d*x+c))^2-4*sin(d*x+c)^3/a^2/d

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Rubi [A]  time = 0.18, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2765, 2977, 2748, 2635, 8, 2633} \[ -\frac {4 \sin ^3(c+d x)}{a^2 d}+\frac {12 \sin (c+d x)}{a^2 d}-\frac {10 \sin (c+d x) \cos ^3(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac {5 \sin (c+d x) \cos (c+d x)}{a^2 d}-\frac {5 x}{a^2}-\frac {\sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5/(a + a*Cos[c + d*x])^2,x]

[Out]

(-5*x)/a^2 + (12*Sin[c + d*x])/(a^2*d) - (5*Cos[c + d*x]*Sin[c + d*x])/(a^2*d) - (10*Cos[c + d*x]^3*Sin[c + d*
x])/(3*a^2*d*(1 + Cos[c + d*x])) - (Cos[c + d*x]^4*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) - (4*Sin[c + d*x
]^3)/(a^2*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 2765

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

Rubi steps

\begin {align*} \int \frac {\cos ^5(c+d x)}{(a+a \cos (c+d x))^2} \, dx &=-\frac {\cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {\int \frac {\cos ^3(c+d x) (4 a-6 a \cos (c+d x))}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac {10 \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {\cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {\int \cos ^2(c+d x) \left (30 a^2-36 a^2 \cos (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {10 \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {\cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {10 \int \cos ^2(c+d x) \, dx}{a^2}+\frac {12 \int \cos ^3(c+d x) \, dx}{a^2}\\ &=-\frac {5 \cos (c+d x) \sin (c+d x)}{a^2 d}-\frac {10 \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {\cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {5 \int 1 \, dx}{a^2}-\frac {12 \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a^2 d}\\ &=-\frac {5 x}{a^2}+\frac {12 \sin (c+d x)}{a^2 d}-\frac {5 \cos (c+d x) \sin (c+d x)}{a^2 d}-\frac {10 \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {\cos ^4(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {4 \sin ^3(c+d x)}{a^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 199, normalized size = 1.60 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (-156 \sin \left (c+\frac {d x}{2}\right )+342 \sin \left (c+\frac {3 d x}{2}\right )+118 \sin \left (2 c+\frac {3 d x}{2}\right )+30 \sin \left (2 c+\frac {5 d x}{2}\right )+30 \sin \left (3 c+\frac {5 d x}{2}\right )-3 \sin \left (3 c+\frac {7 d x}{2}\right )-3 \sin \left (4 c+\frac {7 d x}{2}\right )+\sin \left (4 c+\frac {9 d x}{2}\right )+\sin \left (5 c+\frac {9 d x}{2}\right )-360 d x \cos \left (c+\frac {d x}{2}\right )-120 d x \cos \left (c+\frac {3 d x}{2}\right )-120 d x \cos \left (2 c+\frac {3 d x}{2}\right )+516 \sin \left (\frac {d x}{2}\right )-360 d x \cos \left (\frac {d x}{2}\right )\right )}{192 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5/(a + a*Cos[c + d*x])^2,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^3*(-360*d*x*Cos[(d*x)/2] - 360*d*x*Cos[c + (d*x)/2] - 120*d*x*Cos[c + (3*d*x)/2] -
120*d*x*Cos[2*c + (3*d*x)/2] + 516*Sin[(d*x)/2] - 156*Sin[c + (d*x)/2] + 342*Sin[c + (3*d*x)/2] + 118*Sin[2*c
+ (3*d*x)/2] + 30*Sin[2*c + (5*d*x)/2] + 30*Sin[3*c + (5*d*x)/2] - 3*Sin[3*c + (7*d*x)/2] - 3*Sin[4*c + (7*d*x
)/2] + Sin[4*c + (9*d*x)/2] + Sin[5*c + (9*d*x)/2]))/(192*a^2*d)

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fricas [A]  time = 1.08, size = 108, normalized size = 0.87 \[ -\frac {15 \, d x \cos \left (d x + c\right )^{2} + 30 \, d x \cos \left (d x + c\right ) + 15 \, d x - {\left (\cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} + 33 \, \cos \left (d x + c\right ) + 24\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+a*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/3*(15*d*x*cos(d*x + c)^2 + 30*d*x*cos(d*x + c) + 15*d*x - (cos(d*x + c)^4 - cos(d*x + c)^3 + 6*cos(d*x + c)
^2 + 33*cos(d*x + c) + 24)*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 = 0.88, size = 108, normalized size = 0.87 \[ -\frac {\frac {30 \, {\left (d x + c\right )}}{a^{2}} - \frac {4 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{2}} + \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 27 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a+a*cos(d*x+c))^2,x, algorithm="giac")

[Out]

-1/6*(30*(d*x + c)/a^2 - 4*(15*tan(1/2*d*x + 1/2*c)^5 + 20*tan(1/2*d*x + 1/2*c)^3 + 9*tan(1/2*d*x + 1/2*c))/((
tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^2) + (a^4*tan(1/2*d*x + 1/2*c)^3 - 27*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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maple [A]  time = 0.07, size = 156, normalized size = 1.26 \[ -\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 d \,a^{2}}+\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {10 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {40 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {10 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5/(a+a*cos(d*x+c))^2,x)

[Out]

-1/6/d/a^2*tan(1/2*d*x+1/2*c)^3+9/2/d/a^2*tan(1/2*d*x+1/2*c)+10/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1
/2*c)^5+40/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3+6/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*
d*x+1/2*c)-10/d/a^2*arctan(tan(1/2*d*x+1/2*c))

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maxima [A]  time = 1.36, size = 207, normalized size = 1.67 \[ \frac {\frac {4 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {60 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(4*(9*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d*
x + c) + 1)^5)/(a^2 + 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 +
a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (27*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3)/a^2 - 60*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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mupad [B]  time = 0.50, size = 135, normalized size = 1.09 \[ -\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-28\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-60\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+40\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+30\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (c+d\,x\right )}{6\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5/(a + a*cos(c + d*x))^2,x)

[Out]

-(sin(c/2 + (d*x)/2) - 28*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2) - 60*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)
+ 40*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2) - 16*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2) + 30*cos(c/2 + (d*x
)/2)^3*(c + d*x))/(6*a^2*d*cos(c/2 + (d*x)/2)^3)

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sympy [A]  time = 9.74, size = 700, normalized size = 5.65 \[ \begin {cases} - \frac {30 d x \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {90 d x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {90 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {30 d x}{6 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {\tan ^{9}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {24 \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {138 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {160 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {63 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{5}{\relax (c )}}{\left (a \cos {\relax (c )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5/(a+a*cos(d*x+c))**2,x)

[Out]

Piecewise((-30*d*x*tan(c/2 + d*x/2)**6/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2
*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 90*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(
c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 90*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d
*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 30*d*x/(6*a**2*d*tan(c/
2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - tan(c/2 + d*x/2)**
9/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) +
24*tan(c/2 + d*x/2)**7/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x
/2)**2 + 6*a**2*d) + 138*tan(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 1
8*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 160*tan(c/2 + d*x/2)**3/(6*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*t
an(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 63*tan(c/2 + d*x/2)/(6*a**2*d*tan(c/2 + d*x/2
)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 18*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d), Ne(d, 0)), (x*cos(c)**5/(a*c
os(c) + a)**2, True))

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