3.409 \(\int \frac {\cos ^4(c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx\)
Optimal. Leaf size=135 \[ \frac {\cos ^7(c+d x)}{7 a d}-\frac {2 \cos ^5(c+d x)}{5 a d}+\frac {\cos ^3(c+d x)}{3 a d}-\frac {\sin ^3(c+d x) \cos ^3(c+d x)}{6 a d}-\frac {\sin (c+d x) \cos ^3(c+d x)}{8 a d}+\frac {\sin (c+d x) \cos (c+d x)}{16 a d}+\frac {x}{16 a} \]
[Out]
1/16*x/a+1/3*cos(d*x+c)^3/a/d-2/5*cos(d*x+c)^5/a/d+1/7*cos(d*x+c)^7/a/d+1/16*cos(d*x+c)*sin(d*x+c)/a/d-1/8*cos
(d*x+c)^3*sin(d*x+c)/a/d-1/6*cos(d*x+c)^3*sin(d*x+c)^3/a/d
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Rubi [A] time = 0.20, antiderivative size = 135, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used =
{2839, 2568, 2635, 8, 2565, 270} \[ \frac {\cos ^7(c+d x)}{7 a d}-\frac {2 \cos ^5(c+d x)}{5 a d}+\frac {\cos ^3(c+d x)}{3 a d}-\frac {\sin ^3(c+d x) \cos ^3(c+d x)}{6 a d}-\frac {\sin (c+d x) \cos ^3(c+d x)}{8 a d}+\frac {\sin (c+d x) \cos (c+d x)}{16 a d}+\frac {x}{16 a} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^4*Sin[c + d*x]^4)/(a + a*Sin[c + d*x]),x]
[Out]
x/(16*a) + Cos[c + d*x]^3/(3*a*d) - (2*Cos[c + d*x]^5)/(5*a*d) + Cos[c + d*x]^7/(7*a*d) + (Cos[c + d*x]*Sin[c
+ d*x])/(16*a*d) - (Cos[c + d*x]^3*Sin[c + d*x])/(8*a*d) - (Cos[c + d*x]^3*Sin[c + d*x]^3)/(6*a*d)
Rule 8
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
Rule 270
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]
Rule 2565
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&& !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
Rule 2568
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]
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 2839
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]
Rubi steps
\begin {align*} \int \frac {\cos ^4(c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\int \cos ^2(c+d x) \sin ^4(c+d x) \, dx}{a}-\frac {\int \cos ^2(c+d x) \sin ^5(c+d x) \, dx}{a}\\ &=-\frac {\cos ^3(c+d x) \sin ^3(c+d x)}{6 a d}+\frac {\int \cos ^2(c+d x) \sin ^2(c+d x) \, dx}{2 a}+\frac {\operatorname {Subst}\left (\int x^2 \left (1-x^2\right )^2 \, dx,x,\cos (c+d x)\right )}{a d}\\ &=-\frac {\cos ^3(c+d x) \sin (c+d x)}{8 a d}-\frac {\cos ^3(c+d x) \sin ^3(c+d x)}{6 a d}+\frac {\int \cos ^2(c+d x) \, dx}{8 a}+\frac {\operatorname {Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\cos (c+d x)\right )}{a d}\\ &=\frac {\cos ^3(c+d x)}{3 a d}-\frac {2 \cos ^5(c+d x)}{5 a d}+\frac {\cos ^7(c+d x)}{7 a d}+\frac {\cos (c+d x) \sin (c+d x)}{16 a d}-\frac {\cos ^3(c+d x) \sin (c+d x)}{8 a d}-\frac {\cos ^3(c+d x) \sin ^3(c+d x)}{6 a d}+\frac {\int 1 \, dx}{16 a}\\ &=\frac {x}{16 a}+\frac {\cos ^3(c+d x)}{3 a d}-\frac {2 \cos ^5(c+d x)}{5 a d}+\frac {\cos ^7(c+d x)}{7 a d}+\frac {\cos (c+d x) \sin (c+d x)}{16 a d}-\frac {\cos ^3(c+d x) \sin (c+d x)}{8 a d}-\frac {\cos ^3(c+d x) \sin ^3(c+d x)}{6 a d}\\ \end {align*}
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Mathematica [A] time = 0.26, size = 86, normalized size = 0.64 \[ \frac {-105 \sin (2 (c+d x))-105 \sin (4 (c+d x))+35 \sin (6 (c+d x))+525 \cos (c+d x)+35 \cos (3 (c+d x))-63 \cos (5 (c+d x))+15 \cos (7 (c+d x))+420 c+420 d x}{6720 a d} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]^4*Sin[c + d*x]^4)/(a + a*Sin[c + d*x]),x]
[Out]
(420*c + 420*d*x + 525*Cos[c + d*x] + 35*Cos[3*(c + d*x)] - 63*Cos[5*(c + d*x)] + 15*Cos[7*(c + d*x)] - 105*Si
n[2*(c + d*x)] - 105*Sin[4*(c + d*x)] + 35*Sin[6*(c + d*x)])/(6720*a*d)
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fricas [A] time = 0.44, size = 80, normalized size = 0.59 \[ \frac {240 \, \cos \left (d x + c\right )^{7} - 672 \, \cos \left (d x + c\right )^{5} + 560 \, \cos \left (d x + c\right )^{3} + 105 \, d x + 35 \, {\left (8 \, \cos \left (d x + c\right )^{5} - 14 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{1680 \, a d} \]
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)),x, algorithm="fricas")
[Out]
1/1680*(240*cos(d*x + c)^7 - 672*cos(d*x + c)^5 + 560*cos(d*x + c)^3 + 105*d*x + 35*(8*cos(d*x + c)^5 - 14*cos
(d*x + c)^3 + 3*cos(d*x + c))*sin(d*x + c))/(a*d)
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giac [A] time = 0.20, size = 166, normalized size = 1.23 \[ \frac {\frac {105 \, {\left (d x + c\right )}}{a} + \frac {2 \, {\left (105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 700 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 3395 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 8960 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 4480 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 3395 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2688 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 700 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 896 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 128\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{7} a}}{1680 \, d} \]
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)),x, algorithm="giac")
[Out]
1/1680*(105*(d*x + c)/a + 2*(105*tan(1/2*d*x + 1/2*c)^13 + 700*tan(1/2*d*x + 1/2*c)^11 - 3395*tan(1/2*d*x + 1/
2*c)^9 + 8960*tan(1/2*d*x + 1/2*c)^8 - 4480*tan(1/2*d*x + 1/2*c)^6 + 3395*tan(1/2*d*x + 1/2*c)^5 + 2688*tan(1/
2*d*x + 1/2*c)^4 - 700*tan(1/2*d*x + 1/2*c)^3 + 896*tan(1/2*d*x + 1/2*c)^2 - 105*tan(1/2*d*x + 1/2*c) + 128)/(
(tan(1/2*d*x + 1/2*c)^2 + 1)^7*a))/d
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maple [B] time = 0.27, size = 381, normalized size = 2.82 \[ \frac {\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {5 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}-\frac {97 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {32 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}-\frac {16 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {97 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {16 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}-\frac {5 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {16 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {16}{105 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d} \]
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)),x)
[Out]
1/8/a/d/(1+tan(1/2*d*x+1/2*c)^2)^7*tan(1/2*d*x+1/2*c)^13+5/6/a/d/(1+tan(1/2*d*x+1/2*c)^2)^7*tan(1/2*d*x+1/2*c)
^11-97/24/a/d/(1+tan(1/2*d*x+1/2*c)^2)^7*tan(1/2*d*x+1/2*c)^9+32/3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^7*tan(1/2*d*x+
1/2*c)^8-16/3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^7*tan(1/2*d*x+1/2*c)^6+97/24/a/d/(1+tan(1/2*d*x+1/2*c)^2)^7*tan(1/2
*d*x+1/2*c)^5+16/5/a/d/(1+tan(1/2*d*x+1/2*c)^2)^7*tan(1/2*d*x+1/2*c)^4-5/6/a/d/(1+tan(1/2*d*x+1/2*c)^2)^7*tan(
1/2*d*x+1/2*c)^3+16/15/a/d/(1+tan(1/2*d*x+1/2*c)^2)^7*tan(1/2*d*x+1/2*c)^2-1/8/a/d/(1+tan(1/2*d*x+1/2*c)^2)^7*
tan(1/2*d*x+1/2*c)+16/105/a/d/(1+tan(1/2*d*x+1/2*c)^2)^7+1/8/a/d*arctan(tan(1/2*d*x+1/2*c))
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maxima [B] time = 0.44, size = 380, normalized size = 2.81 \[ -\frac {\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {896 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {700 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {2688 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {3395 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {4480 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {8960 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {3395 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {700 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac {105 \, \sin \left (d x + c\right )^{13}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{13}} - 128}{a + \frac {7 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {21 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {35 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {35 \, a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {21 \, a \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {7 \, a \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} + \frac {a \sin \left (d x + c\right )^{14}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{14}}} - \frac {105 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{840 \, d} \]
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)),x, algorithm="maxima")
[Out]
-1/840*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 896*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 700*sin(d*x + c)^3/(c
os(d*x + c) + 1)^3 - 2688*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 3395*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 448
0*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 8960*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 3395*sin(d*x + c)^9/(cos(d*
x + c) + 1)^9 - 700*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - 105*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - 128)/(
a + 7*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 21*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 35*a*sin(d*x + c)^6/(
cos(d*x + c) + 1)^6 + 35*a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 21*a*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 +
7*a*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 + a*sin(d*x + c)^14/(cos(d*x + c) + 1)^14) - 105*arctan(sin(d*x + c)
/(cos(d*x + c) + 1))/a)/d
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mupad [B] time = 11.38, size = 159, normalized size = 1.18 \[ \frac {x}{16\,a}+\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{8}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{6}-\frac {97\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+\frac {32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{3}-\frac {16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {97\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{24}+\frac {16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}+\frac {16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {16}{105}}{a\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7} \]
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)),x)
[Out]
x/(16*a) + ((16*tan(c/2 + (d*x)/2)^2)/15 - tan(c/2 + (d*x)/2)/8 - (5*tan(c/2 + (d*x)/2)^3)/6 + (16*tan(c/2 + (
d*x)/2)^4)/5 + (97*tan(c/2 + (d*x)/2)^5)/24 - (16*tan(c/2 + (d*x)/2)^6)/3 + (32*tan(c/2 + (d*x)/2)^8)/3 - (97*
tan(c/2 + (d*x)/2)^9)/24 + (5*tan(c/2 + (d*x)/2)^11)/6 + tan(c/2 + (d*x)/2)^13/8 + 16/105)/(a*d*(tan(c/2 + (d*
x)/2)^2 + 1)^7)
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sympy [A] time = 88.92, size = 2635, normalized size = 19.52 \[ \text {result too large to display} \]
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)),x)
[Out]
Piecewise((105*d*x*tan(c/2 + d*x/2)**14/(1680*a*d*tan(c/2 + d*x/2)**14 + 11760*a*d*tan(c/2 + d*x/2)**12 + 3528
0*a*d*tan(c/2 + d*x/2)**10 + 58800*a*d*tan(c/2 + d*x/2)**8 + 58800*a*d*tan(c/2 + d*x/2)**6 + 35280*a*d*tan(c/2
+ d*x/2)**4 + 11760*a*d*tan(c/2 + d*x/2)**2 + 1680*a*d) + 735*d*x*tan(c/2 + d*x/2)**12/(1680*a*d*tan(c/2 + d*
x/2)**14 + 11760*a*d*tan(c/2 + d*x/2)**12 + 35280*a*d*tan(c/2 + d*x/2)**10 + 58800*a*d*tan(c/2 + d*x/2)**8 + 5
8800*a*d*tan(c/2 + d*x/2)**6 + 35280*a*d*tan(c/2 + d*x/2)**4 + 11760*a*d*tan(c/2 + d*x/2)**2 + 1680*a*d) + 220
5*d*x*tan(c/2 + d*x/2)**10/(1680*a*d*tan(c/2 + d*x/2)**14 + 11760*a*d*tan(c/2 + d*x/2)**12 + 35280*a*d*tan(c/2
+ d*x/2)**10 + 58800*a*d*tan(c/2 + d*x/2)**8 + 58800*a*d*tan(c/2 + d*x/2)**6 + 35280*a*d*tan(c/2 + d*x/2)**4
+ 11760*a*d*tan(c/2 + d*x/2)**2 + 1680*a*d) + 3675*d*x*tan(c/2 + d*x/2)**8/(1680*a*d*tan(c/2 + d*x/2)**14 + 11
760*a*d*tan(c/2 + d*x/2)**12 + 35280*a*d*tan(c/2 + d*x/2)**10 + 58800*a*d*tan(c/2 + d*x/2)**8 + 58800*a*d*tan(
c/2 + d*x/2)**6 + 35280*a*d*tan(c/2 + d*x/2)**4 + 11760*a*d*tan(c/2 + d*x/2)**2 + 1680*a*d) + 3675*d*x*tan(c/2
+ d*x/2)**6/(1680*a*d*tan(c/2 + d*x/2)**14 + 11760*a*d*tan(c/2 + d*x/2)**12 + 35280*a*d*tan(c/2 + d*x/2)**10
+ 58800*a*d*tan(c/2 + d*x/2)**8 + 58800*a*d*tan(c/2 + d*x/2)**6 + 35280*a*d*tan(c/2 + d*x/2)**4 + 11760*a*d*ta
n(c/2 + d*x/2)**2 + 1680*a*d) + 2205*d*x*tan(c/2 + d*x/2)**4/(1680*a*d*tan(c/2 + d*x/2)**14 + 11760*a*d*tan(c/
2 + d*x/2)**12 + 35280*a*d*tan(c/2 + d*x/2)**10 + 58800*a*d*tan(c/2 + d*x/2)**8 + 58800*a*d*tan(c/2 + d*x/2)**
6 + 35280*a*d*tan(c/2 + d*x/2)**4 + 11760*a*d*tan(c/2 + d*x/2)**2 + 1680*a*d) + 735*d*x*tan(c/2 + d*x/2)**2/(1
680*a*d*tan(c/2 + d*x/2)**14 + 11760*a*d*tan(c/2 + d*x/2)**12 + 35280*a*d*tan(c/2 + d*x/2)**10 + 58800*a*d*tan
(c/2 + d*x/2)**8 + 58800*a*d*tan(c/2 + d*x/2)**6 + 35280*a*d*tan(c/2 + d*x/2)**4 + 11760*a*d*tan(c/2 + d*x/2)*
*2 + 1680*a*d) + 105*d*x/(1680*a*d*tan(c/2 + d*x/2)**14 + 11760*a*d*tan(c/2 + d*x/2)**12 + 35280*a*d*tan(c/2 +
d*x/2)**10 + 58800*a*d*tan(c/2 + d*x/2)**8 + 58800*a*d*tan(c/2 + d*x/2)**6 + 35280*a*d*tan(c/2 + d*x/2)**4 +
11760*a*d*tan(c/2 + d*x/2)**2 + 1680*a*d) + 210*tan(c/2 + d*x/2)**13/(1680*a*d*tan(c/2 + d*x/2)**14 + 11760*a*
d*tan(c/2 + d*x/2)**12 + 35280*a*d*tan(c/2 + d*x/2)**10 + 58800*a*d*tan(c/2 + d*x/2)**8 + 58800*a*d*tan(c/2 +
d*x/2)**6 + 35280*a*d*tan(c/2 + d*x/2)**4 + 11760*a*d*tan(c/2 + d*x/2)**2 + 1680*a*d) + 1400*tan(c/2 + d*x/2)*
*11/(1680*a*d*tan(c/2 + d*x/2)**14 + 11760*a*d*tan(c/2 + d*x/2)**12 + 35280*a*d*tan(c/2 + d*x/2)**10 + 58800*a
*d*tan(c/2 + d*x/2)**8 + 58800*a*d*tan(c/2 + d*x/2)**6 + 35280*a*d*tan(c/2 + d*x/2)**4 + 11760*a*d*tan(c/2 + d
*x/2)**2 + 1680*a*d) - 6790*tan(c/2 + d*x/2)**9/(1680*a*d*tan(c/2 + d*x/2)**14 + 11760*a*d*tan(c/2 + d*x/2)**1
2 + 35280*a*d*tan(c/2 + d*x/2)**10 + 58800*a*d*tan(c/2 + d*x/2)**8 + 58800*a*d*tan(c/2 + d*x/2)**6 + 35280*a*d
*tan(c/2 + d*x/2)**4 + 11760*a*d*tan(c/2 + d*x/2)**2 + 1680*a*d) + 17920*tan(c/2 + d*x/2)**8/(1680*a*d*tan(c/2
+ d*x/2)**14 + 11760*a*d*tan(c/2 + d*x/2)**12 + 35280*a*d*tan(c/2 + d*x/2)**10 + 58800*a*d*tan(c/2 + d*x/2)**
8 + 58800*a*d*tan(c/2 + d*x/2)**6 + 35280*a*d*tan(c/2 + d*x/2)**4 + 11760*a*d*tan(c/2 + d*x/2)**2 + 1680*a*d)
- 8960*tan(c/2 + d*x/2)**6/(1680*a*d*tan(c/2 + d*x/2)**14 + 11760*a*d*tan(c/2 + d*x/2)**12 + 35280*a*d*tan(c/2
+ d*x/2)**10 + 58800*a*d*tan(c/2 + d*x/2)**8 + 58800*a*d*tan(c/2 + d*x/2)**6 + 35280*a*d*tan(c/2 + d*x/2)**4
+ 11760*a*d*tan(c/2 + d*x/2)**2 + 1680*a*d) + 6790*tan(c/2 + d*x/2)**5/(1680*a*d*tan(c/2 + d*x/2)**14 + 11760*
a*d*tan(c/2 + d*x/2)**12 + 35280*a*d*tan(c/2 + d*x/2)**10 + 58800*a*d*tan(c/2 + d*x/2)**8 + 58800*a*d*tan(c/2
+ d*x/2)**6 + 35280*a*d*tan(c/2 + d*x/2)**4 + 11760*a*d*tan(c/2 + d*x/2)**2 + 1680*a*d) + 5376*tan(c/2 + d*x/2
)**4/(1680*a*d*tan(c/2 + d*x/2)**14 + 11760*a*d*tan(c/2 + d*x/2)**12 + 35280*a*d*tan(c/2 + d*x/2)**10 + 58800*
a*d*tan(c/2 + d*x/2)**8 + 58800*a*d*tan(c/2 + d*x/2)**6 + 35280*a*d*tan(c/2 + d*x/2)**4 + 11760*a*d*tan(c/2 +
d*x/2)**2 + 1680*a*d) - 1400*tan(c/2 + d*x/2)**3/(1680*a*d*tan(c/2 + d*x/2)**14 + 11760*a*d*tan(c/2 + d*x/2)**
12 + 35280*a*d*tan(c/2 + d*x/2)**10 + 58800*a*d*tan(c/2 + d*x/2)**8 + 58800*a*d*tan(c/2 + d*x/2)**6 + 35280*a*
d*tan(c/2 + d*x/2)**4 + 11760*a*d*tan(c/2 + d*x/2)**2 + 1680*a*d) + 1792*tan(c/2 + d*x/2)**2/(1680*a*d*tan(c/2
+ d*x/2)**14 + 11760*a*d*tan(c/2 + d*x/2)**12 + 35280*a*d*tan(c/2 + d*x/2)**10 + 58800*a*d*tan(c/2 + d*x/2)**
8 + 58800*a*d*tan(c/2 + d*x/2)**6 + 35280*a*d*tan(c/2 + d*x/2)**4 + 11760*a*d*tan(c/2 + d*x/2)**2 + 1680*a*d)
- 210*tan(c/2 + d*x/2)/(1680*a*d*tan(c/2 + d*x/2)**14 + 11760*a*d*tan(c/2 + d*x/2)**12 + 35280*a*d*tan(c/2 + d
*x/2)**10 + 58800*a*d*tan(c/2 + d*x/2)**8 + 58800*a*d*tan(c/2 + d*x/2)**6 + 35280*a*d*tan(c/2 + d*x/2)**4 + 11
760*a*d*tan(c/2 + d*x/2)**2 + 1680*a*d) + 256/(1680*a*d*tan(c/2 + d*x/2)**14 + 11760*a*d*tan(c/2 + d*x/2)**12
+ 35280*a*d*tan(c/2 + d*x/2)**10 + 58800*a*d*tan(c/2 + d*x/2)**8 + 58800*a*d*tan(c/2 + d*x/2)**6 + 35280*a*d*t
an(c/2 + d*x/2)**4 + 11760*a*d*tan(c/2 + d*x/2)**2 + 1680*a*d), Ne(d, 0)), (x*sin(c)**4*cos(c)**4/(a*sin(c) +
a), True))
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