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

Optimal. Leaf size=129 \[ \frac {2 \cos ^5(c+d x)}{5 a^2 d}-\frac {4 \cos ^3(c+d x)}{3 a^2 d}+\frac {2 \cos (c+d x)}{a^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x)}{6 a^2 d}-\frac {11 \sin ^3(c+d x) \cos (c+d x)}{24 a^2 d}-\frac {11 \sin (c+d x) \cos (c+d x)}{16 a^2 d}+\frac {11 x}{16 a^2} \]

[Out]

11/16*x/a^2+2*cos(d*x+c)/a^2/d-4/3*cos(d*x+c)^3/a^2/d+2/5*cos(d*x+c)^5/a^2/d-11/16*cos(d*x+c)*sin(d*x+c)/a^2/d
-11/24*cos(d*x+c)*sin(d*x+c)^3/a^2/d-1/6*cos(d*x+c)*sin(d*x+c)^5/a^2/d

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Rubi [A]  time = 0.23, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2869, 2757, 2635, 8, 2633} \[ \frac {2 \cos ^5(c+d x)}{5 a^2 d}-\frac {4 \cos ^3(c+d x)}{3 a^2 d}+\frac {2 \cos (c+d x)}{a^2 d}-\frac {\sin ^5(c+d x) \cos (c+d x)}{6 a^2 d}-\frac {11 \sin ^3(c+d x) \cos (c+d x)}{24 a^2 d}-\frac {11 \sin (c+d x) \cos (c+d x)}{16 a^2 d}+\frac {11 x}{16 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*x)/(16*a^2) + (2*Cos[c + d*x])/(a^2*d) - (4*Cos[c + d*x]^3)/(3*a^2*d) + (2*Cos[c + d*x]^5)/(5*a^2*d) - (11
*Cos[c + d*x]*Sin[c + d*x])/(16*a^2*d) - (11*Cos[c + d*x]*Sin[c + d*x]^3)/(24*a^2*d) - (Cos[c + d*x]*Sin[c + d
*x]^5)/(6*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 2757

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

Rule 2869

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

Rubi steps

\begin {align*} \int \frac {\cos ^4(c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \sin ^4(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac {\int \left (a^2 \sin ^4(c+d x)-2 a^2 \sin ^5(c+d x)+a^2 \sin ^6(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \sin ^4(c+d x) \, dx}{a^2}+\frac {\int \sin ^6(c+d x) \, dx}{a^2}-\frac {2 \int \sin ^5(c+d x) \, dx}{a^2}\\ &=-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a^2 d}-\frac {\cos (c+d x) \sin ^5(c+d x)}{6 a^2 d}+\frac {3 \int \sin ^2(c+d x) \, dx}{4 a^2}+\frac {5 \int \sin ^4(c+d x) \, dx}{6 a^2}+\frac {2 \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=\frac {2 \cos (c+d x)}{a^2 d}-\frac {4 \cos ^3(c+d x)}{3 a^2 d}+\frac {2 \cos ^5(c+d x)}{5 a^2 d}-\frac {3 \cos (c+d x) \sin (c+d x)}{8 a^2 d}-\frac {11 \cos (c+d x) \sin ^3(c+d x)}{24 a^2 d}-\frac {\cos (c+d x) \sin ^5(c+d x)}{6 a^2 d}+\frac {3 \int 1 \, dx}{8 a^2}+\frac {5 \int \sin ^2(c+d x) \, dx}{8 a^2}\\ &=\frac {3 x}{8 a^2}+\frac {2 \cos (c+d x)}{a^2 d}-\frac {4 \cos ^3(c+d x)}{3 a^2 d}+\frac {2 \cos ^5(c+d x)}{5 a^2 d}-\frac {11 \cos (c+d x) \sin (c+d x)}{16 a^2 d}-\frac {11 \cos (c+d x) \sin ^3(c+d x)}{24 a^2 d}-\frac {\cos (c+d x) \sin ^5(c+d x)}{6 a^2 d}+\frac {5 \int 1 \, dx}{16 a^2}\\ &=\frac {11 x}{16 a^2}+\frac {2 \cos (c+d x)}{a^2 d}-\frac {4 \cos ^3(c+d x)}{3 a^2 d}+\frac {2 \cos ^5(c+d x)}{5 a^2 d}-\frac {11 \cos (c+d x) \sin (c+d x)}{16 a^2 d}-\frac {11 \cos (c+d x) \sin ^3(c+d x)}{24 a^2 d}-\frac {\cos (c+d x) \sin ^5(c+d x)}{6 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 76, normalized size = 0.59 \[ \frac {-465 \sin (2 (c+d x))+75 \sin (4 (c+d x))-5 \sin (6 (c+d x))+1200 \cos (c+d x)-200 \cos (3 (c+d x))+24 \cos (5 (c+d x))+660 c+660 d x}{960 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(660*c + 660*d*x + 1200*Cos[c + d*x] - 200*Cos[3*(c + d*x)] + 24*Cos[5*(c + d*x)] - 465*Sin[2*(c + d*x)] + 75*
Sin[4*(c + d*x)] - 5*Sin[6*(c + d*x)])/(960*a^2*d)

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fricas [A]  time = 0.44, size = 78, normalized size = 0.60 \[ \frac {96 \, \cos \left (d x + c\right )^{5} - 320 \, \cos \left (d x + c\right )^{3} + 165 \, d x - 5 \, {\left (8 \, \cos \left (d x + c\right )^{5} - 38 \, \cos \left (d x + c\right )^{3} + 63 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 480 \, \cos \left (d x + c\right )}{240 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(96*cos(d*x + c)^5 - 320*cos(d*x + c)^3 + 165*d*x - 5*(8*cos(d*x + c)^5 - 38*cos(d*x + c)^3 + 63*cos(d*x
+ c))*sin(d*x + c) + 480*cos(d*x + c))/(a^2*d)

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giac [A]  time = 0.20, size = 153, normalized size = 1.19 \[ \frac {\frac {165 \, {\left (d x + c\right )}}{a^{2}} + \frac {2 \, {\left (165 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 935 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 1410 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 2560 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 1410 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3840 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 935 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1536 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 165 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 256\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6} a^{2}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(165*(d*x + c)/a^2 + 2*(165*tan(1/2*d*x + 1/2*c)^11 + 935*tan(1/2*d*x + 1/2*c)^9 + 1410*tan(1/2*d*x + 1/
2*c)^7 + 2560*tan(1/2*d*x + 1/2*c)^6 - 1410*tan(1/2*d*x + 1/2*c)^5 + 3840*tan(1/2*d*x + 1/2*c)^4 - 935*tan(1/2
*d*x + 1/2*c)^3 + 1536*tan(1/2*d*x + 1/2*c)^2 - 165*tan(1/2*d*x + 1/2*c) + 256)/((tan(1/2*d*x + 1/2*c)^2 + 1)^
6*a^2))/d

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maple [B]  time = 0.41, size = 347, normalized size = 2.69 \[ \frac {11 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {187 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {47 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {64 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {47 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {32 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {187 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {64 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {32}{15 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {11 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/8/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^11+187/24/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*
x+1/2*c)^9+47/4/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^7+64/3/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^6*ta
n(1/2*d*x+1/2*c)^6-47/4/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^5+32/a^2/d/(1+tan(1/2*d*x+1/2*c)^2
)^6*tan(1/2*d*x+1/2*c)^4-187/24/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^3+64/5/a^2/d/(1+tan(1/2*d*
x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^2-11/8/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)+32/15/a^2/d/(1+tan
(1/2*d*x+1/2*c)^2)^6+11/8/d/a^2*arctan(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.54, size = 353, normalized size = 2.74 \[ -\frac {\frac {\frac {165 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {1536 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {935 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {3840 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {1410 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {2560 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {1410 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {935 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {165 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - 256}{a^{2} + \frac {6 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {15 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {20 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {15 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {6 \, a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {a^{2} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}} - \frac {165 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*((165*sin(d*x + c)/(cos(d*x + c) + 1) - 1536*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 935*sin(d*x + c)^3/(
cos(d*x + c) + 1)^3 - 3840*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1410*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 25
60*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 1410*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 935*sin(d*x + c)^9/(cos(d*
x + c) + 1)^9 - 165*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - 256)/(a^2 + 6*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1
)^2 + 15*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 20*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 15*a^2*sin(d*x
+ c)^8/(cos(d*x + c) + 1)^8 + 6*a^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + a^2*sin(d*x + c)^12/(cos(d*x + c)
+ 1)^12) - 165*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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mupad [B]  time = 11.24, size = 146, normalized size = 1.13 \[ \frac {11\,x}{16\,a^2}+\frac {\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+\frac {187\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+\frac {47\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}-\frac {47\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {187\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+\frac {64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}-\frac {11\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {32}{15}}{a^2\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(11*x)/(16*a^2) + ((64*tan(c/2 + (d*x)/2)^2)/5 - (11*tan(c/2 + (d*x)/2))/8 - (187*tan(c/2 + (d*x)/2)^3)/24 + 3
2*tan(c/2 + (d*x)/2)^4 - (47*tan(c/2 + (d*x)/2)^5)/4 + (64*tan(c/2 + (d*x)/2)^6)/3 + (47*tan(c/2 + (d*x)/2)^7)
/4 + (187*tan(c/2 + (d*x)/2)^9)/24 + (11*tan(c/2 + (d*x)/2)^11)/8 + 32/15)/(a^2*d*(tan(c/2 + (d*x)/2)^2 + 1)^6
)

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sympy [A]  time = 143.14, size = 2271, normalized size = 17.60 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((165*d*x*tan(c/2 + d*x/2)**12/(240*a**2*d*tan(c/2 + d*x/2)**12 + 1440*a**2*d*tan(c/2 + d*x/2)**10 +
3600*a**2*d*tan(c/2 + d*x/2)**8 + 4800*a**2*d*tan(c/2 + d*x/2)**6 + 3600*a**2*d*tan(c/2 + d*x/2)**4 + 1440*a**
2*d*tan(c/2 + d*x/2)**2 + 240*a**2*d) + 990*d*x*tan(c/2 + d*x/2)**10/(240*a**2*d*tan(c/2 + d*x/2)**12 + 1440*a
**2*d*tan(c/2 + d*x/2)**10 + 3600*a**2*d*tan(c/2 + d*x/2)**8 + 4800*a**2*d*tan(c/2 + d*x/2)**6 + 3600*a**2*d*t
an(c/2 + d*x/2)**4 + 1440*a**2*d*tan(c/2 + d*x/2)**2 + 240*a**2*d) + 2475*d*x*tan(c/2 + d*x/2)**8/(240*a**2*d*
tan(c/2 + d*x/2)**12 + 1440*a**2*d*tan(c/2 + d*x/2)**10 + 3600*a**2*d*tan(c/2 + d*x/2)**8 + 4800*a**2*d*tan(c/
2 + d*x/2)**6 + 3600*a**2*d*tan(c/2 + d*x/2)**4 + 1440*a**2*d*tan(c/2 + d*x/2)**2 + 240*a**2*d) + 3300*d*x*tan
(c/2 + d*x/2)**6/(240*a**2*d*tan(c/2 + d*x/2)**12 + 1440*a**2*d*tan(c/2 + d*x/2)**10 + 3600*a**2*d*tan(c/2 + d
*x/2)**8 + 4800*a**2*d*tan(c/2 + d*x/2)**6 + 3600*a**2*d*tan(c/2 + d*x/2)**4 + 1440*a**2*d*tan(c/2 + d*x/2)**2
+ 240*a**2*d) + 2475*d*x*tan(c/2 + d*x/2)**4/(240*a**2*d*tan(c/2 + d*x/2)**12 + 1440*a**2*d*tan(c/2 + d*x/2)*
*10 + 3600*a**2*d*tan(c/2 + d*x/2)**8 + 4800*a**2*d*tan(c/2 + d*x/2)**6 + 3600*a**2*d*tan(c/2 + d*x/2)**4 + 14
40*a**2*d*tan(c/2 + d*x/2)**2 + 240*a**2*d) + 990*d*x*tan(c/2 + d*x/2)**2/(240*a**2*d*tan(c/2 + d*x/2)**12 + 1
440*a**2*d*tan(c/2 + d*x/2)**10 + 3600*a**2*d*tan(c/2 + d*x/2)**8 + 4800*a**2*d*tan(c/2 + d*x/2)**6 + 3600*a**
2*d*tan(c/2 + d*x/2)**4 + 1440*a**2*d*tan(c/2 + d*x/2)**2 + 240*a**2*d) + 165*d*x/(240*a**2*d*tan(c/2 + d*x/2)
**12 + 1440*a**2*d*tan(c/2 + d*x/2)**10 + 3600*a**2*d*tan(c/2 + d*x/2)**8 + 4800*a**2*d*tan(c/2 + d*x/2)**6 +
3600*a**2*d*tan(c/2 + d*x/2)**4 + 1440*a**2*d*tan(c/2 + d*x/2)**2 + 240*a**2*d) + 330*tan(c/2 + d*x/2)**11/(24
0*a**2*d*tan(c/2 + d*x/2)**12 + 1440*a**2*d*tan(c/2 + d*x/2)**10 + 3600*a**2*d*tan(c/2 + d*x/2)**8 + 4800*a**2
*d*tan(c/2 + d*x/2)**6 + 3600*a**2*d*tan(c/2 + d*x/2)**4 + 1440*a**2*d*tan(c/2 + d*x/2)**2 + 240*a**2*d) + 187
0*tan(c/2 + d*x/2)**9/(240*a**2*d*tan(c/2 + d*x/2)**12 + 1440*a**2*d*tan(c/2 + d*x/2)**10 + 3600*a**2*d*tan(c/
2 + d*x/2)**8 + 4800*a**2*d*tan(c/2 + d*x/2)**6 + 3600*a**2*d*tan(c/2 + d*x/2)**4 + 1440*a**2*d*tan(c/2 + d*x/
2)**2 + 240*a**2*d) + 2820*tan(c/2 + d*x/2)**7/(240*a**2*d*tan(c/2 + d*x/2)**12 + 1440*a**2*d*tan(c/2 + d*x/2)
**10 + 3600*a**2*d*tan(c/2 + d*x/2)**8 + 4800*a**2*d*tan(c/2 + d*x/2)**6 + 3600*a**2*d*tan(c/2 + d*x/2)**4 + 1
440*a**2*d*tan(c/2 + d*x/2)**2 + 240*a**2*d) + 5120*tan(c/2 + d*x/2)**6/(240*a**2*d*tan(c/2 + d*x/2)**12 + 144
0*a**2*d*tan(c/2 + d*x/2)**10 + 3600*a**2*d*tan(c/2 + d*x/2)**8 + 4800*a**2*d*tan(c/2 + d*x/2)**6 + 3600*a**2*
d*tan(c/2 + d*x/2)**4 + 1440*a**2*d*tan(c/2 + d*x/2)**2 + 240*a**2*d) - 2820*tan(c/2 + d*x/2)**5/(240*a**2*d*t
an(c/2 + d*x/2)**12 + 1440*a**2*d*tan(c/2 + d*x/2)**10 + 3600*a**2*d*tan(c/2 + d*x/2)**8 + 4800*a**2*d*tan(c/2
+ d*x/2)**6 + 3600*a**2*d*tan(c/2 + d*x/2)**4 + 1440*a**2*d*tan(c/2 + d*x/2)**2 + 240*a**2*d) + 7680*tan(c/2
+ d*x/2)**4/(240*a**2*d*tan(c/2 + d*x/2)**12 + 1440*a**2*d*tan(c/2 + d*x/2)**10 + 3600*a**2*d*tan(c/2 + d*x/2)
**8 + 4800*a**2*d*tan(c/2 + d*x/2)**6 + 3600*a**2*d*tan(c/2 + d*x/2)**4 + 1440*a**2*d*tan(c/2 + d*x/2)**2 + 24
0*a**2*d) - 1870*tan(c/2 + d*x/2)**3/(240*a**2*d*tan(c/2 + d*x/2)**12 + 1440*a**2*d*tan(c/2 + d*x/2)**10 + 360
0*a**2*d*tan(c/2 + d*x/2)**8 + 4800*a**2*d*tan(c/2 + d*x/2)**6 + 3600*a**2*d*tan(c/2 + d*x/2)**4 + 1440*a**2*d
*tan(c/2 + d*x/2)**2 + 240*a**2*d) + 3072*tan(c/2 + d*x/2)**2/(240*a**2*d*tan(c/2 + d*x/2)**12 + 1440*a**2*d*t
an(c/2 + d*x/2)**10 + 3600*a**2*d*tan(c/2 + d*x/2)**8 + 4800*a**2*d*tan(c/2 + d*x/2)**6 + 3600*a**2*d*tan(c/2
+ d*x/2)**4 + 1440*a**2*d*tan(c/2 + d*x/2)**2 + 240*a**2*d) - 330*tan(c/2 + d*x/2)/(240*a**2*d*tan(c/2 + d*x/2
)**12 + 1440*a**2*d*tan(c/2 + d*x/2)**10 + 3600*a**2*d*tan(c/2 + d*x/2)**8 + 4800*a**2*d*tan(c/2 + d*x/2)**6 +
3600*a**2*d*tan(c/2 + d*x/2)**4 + 1440*a**2*d*tan(c/2 + d*x/2)**2 + 240*a**2*d) + 512/(240*a**2*d*tan(c/2 + d
*x/2)**12 + 1440*a**2*d*tan(c/2 + d*x/2)**10 + 3600*a**2*d*tan(c/2 + d*x/2)**8 + 4800*a**2*d*tan(c/2 + d*x/2)*
*6 + 3600*a**2*d*tan(c/2 + d*x/2)**4 + 1440*a**2*d*tan(c/2 + d*x/2)**2 + 240*a**2*d), Ne(d, 0)), (x*sin(c)**4*
cos(c)**4/(a*sin(c) + a)**2, True))

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