3.307 \(\int \frac {\cos ^2(c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=111 \[ \frac {2 \cos ^3(c+d x)}{3 a^2 d}-\frac {4 \cos (c+d x)}{a^2 d}+\frac {\sin ^3(c+d x) \cos (c+d x)}{4 a^2 d}+\frac {11 \sin (c+d x) \cos (c+d x)}{8 a^2 d}-\frac {2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}-\frac {27 x}{8 a^2} \]

[Out]

-27/8*x/a^2-4*cos(d*x+c)/a^2/d+2/3*cos(d*x+c)^3/a^2/d+11/8*cos(d*x+c)*sin(d*x+c)/a^2/d+1/4*cos(d*x+c)*sin(d*x+
c)^3/a^2/d-2*cos(d*x+c)/a^2/d/(1+sin(d*x+c))

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Rubi [A]  time = 0.25, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2874, 2966, 2638, 2635, 8, 2633, 2648} \[ \frac {2 \cos ^3(c+d x)}{3 a^2 d}-\frac {4 \cos (c+d x)}{a^2 d}+\frac {\sin ^3(c+d x) \cos (c+d x)}{4 a^2 d}+\frac {11 \sin (c+d x) \cos (c+d x)}{8 a^2 d}-\frac {2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}-\frac {27 x}{8 a^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*Sin[c + d*x]^4)/(a + a*Sin[c + d*x])^2,x]

[Out]

(-27*x)/(8*a^2) - (4*Cos[c + d*x])/(a^2*d) + (2*Cos[c + d*x]^3)/(3*a^2*d) + (11*Cos[c + d*x]*Sin[c + d*x])/(8*
a^2*d) + (Cos[c + d*x]*Sin[c + d*x]^3)/(4*a^2*d) - (2*Cos[c + d*x])/(a^2*d*(1 + Sin[c + d*x]))

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 2638

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

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2874

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/b^2, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && (ILtQ[m, 0] ||  !IGtQ[n, 0])

Rule 2966

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \frac {\sin ^4(c+d x) (a-a \sin (c+d x))}{a+a \sin (c+d x)} \, dx}{a^2}\\ &=\frac {\int \left (-2+2 \sin (c+d x)-2 \sin ^2(c+d x)+2 \sin ^3(c+d x)-\sin ^4(c+d x)+\frac {2}{1+\sin (c+d x)}\right ) \, dx}{a^2}\\ &=-\frac {2 x}{a^2}-\frac {\int \sin ^4(c+d x) \, dx}{a^2}+\frac {2 \int \sin (c+d x) \, dx}{a^2}-\frac {2 \int \sin ^2(c+d x) \, dx}{a^2}+\frac {2 \int \sin ^3(c+d x) \, dx}{a^2}+\frac {2 \int \frac {1}{1+\sin (c+d x)} \, dx}{a^2}\\ &=-\frac {2 x}{a^2}-\frac {2 \cos (c+d x)}{a^2 d}+\frac {\cos (c+d x) \sin (c+d x)}{a^2 d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a^2 d}-\frac {2 \cos (c+d x)}{a^2 d (1+\sin (c+d x))}-\frac {3 \int \sin ^2(c+d x) \, dx}{4 a^2}-\frac {\int 1 \, dx}{a^2}-\frac {2 \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=-\frac {3 x}{a^2}-\frac {4 \cos (c+d x)}{a^2 d}+\frac {2 \cos ^3(c+d x)}{3 a^2 d}+\frac {11 \cos (c+d x) \sin (c+d x)}{8 a^2 d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a^2 d}-\frac {2 \cos (c+d x)}{a^2 d (1+\sin (c+d x))}-\frac {3 \int 1 \, dx}{8 a^2}\\ &=-\frac {27 x}{8 a^2}-\frac {4 \cos (c+d x)}{a^2 d}+\frac {2 \cos ^3(c+d x)}{3 a^2 d}+\frac {11 \cos (c+d x) \sin (c+d x)}{8 a^2 d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a^2 d}-\frac {2 \cos (c+d x)}{a^2 d (1+\sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 1.49, size = 209, normalized size = 1.88 \[ \frac {-648 d x \sin \left (c+\frac {d x}{2}\right )+4 \sin \left (c+\frac {d x}{2}\right )-264 \sin \left (2 c+\frac {3 d x}{2}\right )+56 \sin \left (2 c+\frac {5 d x}{2}\right )+13 \sin \left (4 c+\frac {7 d x}{2}\right )-3 \sin \left (4 c+\frac {9 d x}{2}\right )-340 \cos \left (c+\frac {d x}{2}\right )-264 \cos \left (c+\frac {3 d x}{2}\right )-56 \cos \left (3 c+\frac {5 d x}{2}\right )+13 \cos \left (3 c+\frac {7 d x}{2}\right )+3 \cos \left (5 c+\frac {9 d x}{2}\right )+1100 \sin \left (\frac {d x}{2}\right )+(4-648 d x) \cos \left (\frac {d x}{2}\right )}{192 a^2 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*Sin[c + d*x]^4)/(a + a*Sin[c + d*x])^2,x]

[Out]

((4 - 648*d*x)*Cos[(d*x)/2] - 340*Cos[c + (d*x)/2] - 264*Cos[c + (3*d*x)/2] - 56*Cos[3*c + (5*d*x)/2] + 13*Cos
[3*c + (7*d*x)/2] + 3*Cos[5*c + (9*d*x)/2] + 1100*Sin[(d*x)/2] + 4*Sin[c + (d*x)/2] - 648*d*x*Sin[c + (d*x)/2]
- 264*Sin[2*c + (3*d*x)/2] + 56*Sin[2*c + (5*d*x)/2] + 13*Sin[4*c + (7*d*x)/2] - 3*Sin[4*c + (9*d*x)/2])/(192
*a^2*d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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fricas [A]  time = 0.51, size = 144, normalized size = 1.30 \[ \frac {6 \, \cos \left (d x + c\right )^{5} + 16 \, \cos \left (d x + c\right )^{4} - 29 \, \cos \left (d x + c\right )^{3} - 81 \, d x - 3 \, {\left (27 \, d x + 35\right )} \cos \left (d x + c\right ) - 96 \, \cos \left (d x + c\right )^{2} - {\left (6 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{3} + 81 \, d x - 39 \, \cos \left (d x + c\right )^{2} + 57 \, \cos \left (d x + c\right ) - 48\right )} \sin \left (d x + c\right ) - 48}{24 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/24*(6*cos(d*x + c)^5 + 16*cos(d*x + c)^4 - 29*cos(d*x + c)^3 - 81*d*x - 3*(27*d*x + 35)*cos(d*x + c) - 96*co
s(d*x + c)^2 - (6*cos(d*x + c)^4 - 10*cos(d*x + c)^3 + 81*d*x - 39*cos(d*x + c)^2 + 57*cos(d*x + c) - 48)*sin(
d*x + c) - 48)/(a^2*d*cos(d*x + c) + a^2*d*sin(d*x + c) + a^2*d)

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giac [A]  time = 0.20, size = 145, normalized size = 1.31 \[ -\frac {\frac {81 \, {\left (d x + c\right )}}{a^{2}} + \frac {96}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}} + \frac {2 \, {\left (33 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 48 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 57 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 240 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 57 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 272 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 33 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 80\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a^{2}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/24*(81*(d*x + c)/a^2 + 96/(a^2*(tan(1/2*d*x + 1/2*c) + 1)) + 2*(33*tan(1/2*d*x + 1/2*c)^7 + 48*tan(1/2*d*x
+ 1/2*c)^6 + 57*tan(1/2*d*x + 1/2*c)^5 + 240*tan(1/2*d*x + 1/2*c)^4 - 57*tan(1/2*d*x + 1/2*c)^3 + 272*tan(1/2*
d*x + 1/2*c)^2 - 33*tan(1/2*d*x + 1/2*c) + 80)/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a^2))/d

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maple [B]  time = 0.46, size = 300, normalized size = 2.70 \[ -\frac {11 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {4 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {19 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {20 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {19 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {68 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {20}{3 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {27 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{2}}-\frac {4}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*sin(d*x+c)^4/(a+a*sin(d*x+c))^2,x)

[Out]

-11/4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7-4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2
*c)^6-19/4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5-20/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d
*x+1/2*c)^4+19/4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3-68/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*t
an(1/2*d*x+1/2*c)^2+11/4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)-20/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^
2)^4-27/4/d/a^2*arctan(tan(1/2*d*x+1/2*c))-4/d/a^2/(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.44, size = 398, normalized size = 3.59 \[ -\frac {\frac {\frac {47 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {431 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {215 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {471 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {297 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {297 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {81 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {81 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 128}{a^{2} + \frac {a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {6 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {6 \, a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {a^{2} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}} + \frac {81 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*sin(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/12*((47*sin(d*x + c)/(cos(d*x + c) + 1) + 431*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 215*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 + 471*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 297*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 297*sin
(d*x + c)^6/(cos(d*x + c) + 1)^6 + 81*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 81*sin(d*x + c)^8/(cos(d*x + c) +
1)^8 + 128)/(a^2 + a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 4*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 4*a^2*sin
(d*x + c)^3/(cos(d*x + c) + 1)^3 + 6*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 6*a^2*sin(d*x + c)^5/(cos(d*x +
c) + 1)^5 + 4*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 4*a^2*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + a^2*sin(d
*x + c)^8/(cos(d*x + c) + 1)^8 + a^2*sin(d*x + c)^9/(cos(d*x + c) + 1)^9) + 81*arctan(sin(d*x + c)/(cos(d*x +
c) + 1))/a^2)/d

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mupad [B]  time = 12.34, size = 147, normalized size = 1.32 \[ -\frac {27\,x}{8\,a^2}-\frac {\frac {27\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{4}+\frac {27\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {99\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{4}+\frac {99\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {157\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{4}+\frac {215\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12}+\frac {431\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{12}+\frac {47\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{12}+\frac {32}{3}}{a^2\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^2*sin(c + d*x)^4)/(a + a*sin(c + d*x))^2,x)

[Out]

- (27*x)/(8*a^2) - ((47*tan(c/2 + (d*x)/2))/12 + (431*tan(c/2 + (d*x)/2)^2)/12 + (215*tan(c/2 + (d*x)/2)^3)/12
+ (157*tan(c/2 + (d*x)/2)^4)/4 + (99*tan(c/2 + (d*x)/2)^5)/4 + (99*tan(c/2 + (d*x)/2)^6)/4 + (27*tan(c/2 + (d
*x)/2)^7)/4 + (27*tan(c/2 + (d*x)/2)^8)/4 + 32/3)/(a^2*d*(tan(c/2 + (d*x)/2) + 1)*(tan(c/2 + (d*x)/2)^2 + 1)^4
)

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sympy [A]  time = 61.08, size = 3580, normalized size = 32.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*sin(d*x+c)**4/(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((-81*d*x*tan(c/2 + d*x/2)**9/(24*a**2*d*tan(c/2 + d*x/2)**9 + 24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**
2*d*tan(c/2 + d*x/2)**7 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**5 + 144*a**2*d*tan(c/2
+ d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**3 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d*tan(c/2 + d*x/2) + 24*
a**2*d) - 81*d*x*tan(c/2 + d*x/2)**8/(24*a**2*d*tan(c/2 + d*x/2)**9 + 24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*
d*tan(c/2 + d*x/2)**7 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**5 + 144*a**2*d*tan(c/2 +
d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**3 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d*tan(c/2 + d*x/2) + 24*a*
*2*d) - 324*d*x*tan(c/2 + d*x/2)**7/(24*a**2*d*tan(c/2 + d*x/2)**9 + 24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d
*tan(c/2 + d*x/2)**7 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**5 + 144*a**2*d*tan(c/2 + d
*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**3 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d*tan(c/2 + d*x/2) + 24*a**
2*d) - 324*d*x*tan(c/2 + d*x/2)**6/(24*a**2*d*tan(c/2 + d*x/2)**9 + 24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*
tan(c/2 + d*x/2)**7 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**5 + 144*a**2*d*tan(c/2 + d*
x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**3 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d*tan(c/2 + d*x/2) + 24*a**2
*d) - 486*d*x*tan(c/2 + d*x/2)**5/(24*a**2*d*tan(c/2 + d*x/2)**9 + 24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*t
an(c/2 + d*x/2)**7 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**5 + 144*a**2*d*tan(c/2 + d*x
/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**3 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d*tan(c/2 + d*x/2) + 24*a**2*
d) - 486*d*x*tan(c/2 + d*x/2)**4/(24*a**2*d*tan(c/2 + d*x/2)**9 + 24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*ta
n(c/2 + d*x/2)**7 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**5 + 144*a**2*d*tan(c/2 + d*x/
2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**3 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d*tan(c/2 + d*x/2) + 24*a**2*d
) - 324*d*x*tan(c/2 + d*x/2)**3/(24*a**2*d*tan(c/2 + d*x/2)**9 + 24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan
(c/2 + d*x/2)**7 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**5 + 144*a**2*d*tan(c/2 + d*x/2
)**4 + 96*a**2*d*tan(c/2 + d*x/2)**3 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d*tan(c/2 + d*x/2) + 24*a**2*d)
- 324*d*x*tan(c/2 + d*x/2)**2/(24*a**2*d*tan(c/2 + d*x/2)**9 + 24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(
c/2 + d*x/2)**7 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**5 + 144*a**2*d*tan(c/2 + d*x/2)
**4 + 96*a**2*d*tan(c/2 + d*x/2)**3 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d*tan(c/2 + d*x/2) + 24*a**2*d)
- 81*d*x*tan(c/2 + d*x/2)/(24*a**2*d*tan(c/2 + d*x/2)**9 + 24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 +
d*x/2)**7 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**5 + 144*a**2*d*tan(c/2 + d*x/2)**4 +
96*a**2*d*tan(c/2 + d*x/2)**3 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d*tan(c/2 + d*x/2) + 24*a**2*d) - 81*
d*x/(24*a**2*d*tan(c/2 + d*x/2)**9 + 24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**7 + 96*a**2*d
*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**5 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d
*x/2)**3 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d*tan(c/2 + d*x/2) + 24*a**2*d) - 162*tan(c/2 + d*x/2)**8/(
24*a**2*d*tan(c/2 + d*x/2)**9 + 24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**7 + 96*a**2*d*tan(
c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**5 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)
**3 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d*tan(c/2 + d*x/2) + 24*a**2*d) - 162*tan(c/2 + d*x/2)**7/(24*a*
*2*d*tan(c/2 + d*x/2)**9 + 24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**7 + 96*a**2*d*tan(c/2 +
d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**5 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**3 +
96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d*tan(c/2 + d*x/2) + 24*a**2*d) - 594*tan(c/2 + d*x/2)**6/(24*a**2*d*
tan(c/2 + d*x/2)**9 + 24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**7 + 96*a**2*d*tan(c/2 + d*x/
2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**5 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**3 + 96*a
**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d*tan(c/2 + d*x/2) + 24*a**2*d) - 594*tan(c/2 + d*x/2)**5/(24*a**2*d*tan(c
/2 + d*x/2)**9 + 24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**7 + 96*a**2*d*tan(c/2 + d*x/2)**6
+ 144*a**2*d*tan(c/2 + d*x/2)**5 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**3 + 96*a**2*d
*tan(c/2 + d*x/2)**2 + 24*a**2*d*tan(c/2 + d*x/2) + 24*a**2*d) - 942*tan(c/2 + d*x/2)**4/(24*a**2*d*tan(c/2 +
d*x/2)**9 + 24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**7 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 14
4*a**2*d*tan(c/2 + d*x/2)**5 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**3 + 96*a**2*d*tan(
c/2 + d*x/2)**2 + 24*a**2*d*tan(c/2 + d*x/2) + 24*a**2*d) - 430*tan(c/2 + d*x/2)**3/(24*a**2*d*tan(c/2 + d*x/2
)**9 + 24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**7 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**
2*d*tan(c/2 + d*x/2)**5 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**3 + 96*a**2*d*tan(c/2 +
d*x/2)**2 + 24*a**2*d*tan(c/2 + d*x/2) + 24*a**2*d) - 862*tan(c/2 + d*x/2)**2/(24*a**2*d*tan(c/2 + d*x/2)**9
+ 24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**7 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*t
an(c/2 + d*x/2)**5 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**3 + 96*a**2*d*tan(c/2 + d*x/
2)**2 + 24*a**2*d*tan(c/2 + d*x/2) + 24*a**2*d) - 94*tan(c/2 + d*x/2)/(24*a**2*d*tan(c/2 + d*x/2)**9 + 24*a**2
*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**7 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 +
d*x/2)**5 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**3 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 2
4*a**2*d*tan(c/2 + d*x/2) + 24*a**2*d) - 256/(24*a**2*d*tan(c/2 + d*x/2)**9 + 24*a**2*d*tan(c/2 + d*x/2)**8 +
96*a**2*d*tan(c/2 + d*x/2)**7 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**5 + 144*a**2*d*ta
n(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**3 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d*tan(c/2 + d*x/2)
+ 24*a**2*d), Ne(d, 0)), (x*sin(c)**4*cos(c)**2/(a*sin(c) + a)**2, True))

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