3.636 \(\int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=100 \[ -\frac {2 \cos ^5(c+d x)}{15 a^2 d}-\frac {\sin (c+d x) \cos ^3(c+d x)}{6 a^2 d}-\frac {\sin (c+d x) \cos (c+d x)}{4 a^2 d}-\frac {x}{4 a^2}-\frac {\cos ^7(c+d x)}{3 d (a \sin (c+d x)+a)^2} \]

[Out]

-1/4*x/a^2-2/15*cos(d*x+c)^5/a^2/d-1/4*cos(d*x+c)*sin(d*x+c)/a^2/d-1/6*cos(d*x+c)^3*sin(d*x+c)/a^2/d-1/3*cos(d
*x+c)^7/d/(a+a*sin(d*x+c))^2

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Rubi [A]  time = 0.13, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2859, 2682, 2635, 8} \[ -\frac {2 \cos ^5(c+d x)}{15 a^2 d}-\frac {\sin (c+d x) \cos ^3(c+d x)}{6 a^2 d}-\frac {\sin (c+d x) \cos (c+d x)}{4 a^2 d}-\frac {x}{4 a^2}-\frac {\cos ^7(c+d x)}{3 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^6*Sin[c + d*x])/(a + a*Sin[c + d*x])^2,x]

[Out]

-x/(4*a^2) - (2*Cos[c + d*x]^5)/(15*a^2*d) - (Cos[c + d*x]*Sin[c + d*x])/(4*a^2*d) - (Cos[c + d*x]^3*Sin[c + d
*x])/(6*a^2*d) - Cos[c + d*x]^7/(3*d*(a + a*Sin[c + d*x])^2)

Rule 8

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

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 2682

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e
+ f*x])^(p - 1))/(b*f*(p - 1)), x] + Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g
}, x] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]

Rule 2859

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(2*m +
p + 1)), x] + Dist[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^
(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[
m + p], 0]) && NeQ[2*m + p + 1, 0]

Rubi steps

\begin {align*} \int \frac {\cos ^6(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx &=-\frac {\cos ^7(c+d x)}{3 d (a+a \sin (c+d x))^2}-\frac {2 \int \frac {\cos ^6(c+d x)}{a+a \sin (c+d x)} \, dx}{3 a}\\ &=-\frac {2 \cos ^5(c+d x)}{15 a^2 d}-\frac {\cos ^7(c+d x)}{3 d (a+a \sin (c+d x))^2}-\frac {2 \int \cos ^4(c+d x) \, dx}{3 a^2}\\ &=-\frac {2 \cos ^5(c+d x)}{15 a^2 d}-\frac {\cos ^3(c+d x) \sin (c+d x)}{6 a^2 d}-\frac {\cos ^7(c+d x)}{3 d (a+a \sin (c+d x))^2}-\frac {\int \cos ^2(c+d x) \, dx}{2 a^2}\\ &=-\frac {2 \cos ^5(c+d x)}{15 a^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{4 a^2 d}-\frac {\cos ^3(c+d x) \sin (c+d x)}{6 a^2 d}-\frac {\cos ^7(c+d x)}{3 d (a+a \sin (c+d x))^2}-\frac {\int 1 \, dx}{4 a^2}\\ &=-\frac {x}{4 a^2}-\frac {2 \cos ^5(c+d x)}{15 a^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{4 a^2 d}-\frac {\cos ^3(c+d x) \sin (c+d x)}{6 a^2 d}-\frac {\cos ^7(c+d x)}{3 d (a+a \sin (c+d x))^2}\\ \end {align*}

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Mathematica [B]  time = 1.13, size = 262, normalized size = 2.62 \[ \frac {-120 d x \sin \left (\frac {c}{2}\right )+90 \sin \left (\frac {c}{2}+d x\right )-90 \sin \left (\frac {3 c}{2}+d x\right )+25 \sin \left (\frac {5 c}{2}+3 d x\right )-25 \sin \left (\frac {7 c}{2}+3 d x\right )+15 \sin \left (\frac {7 c}{2}+4 d x\right )+15 \sin \left (\frac {9 c}{2}+4 d x\right )-3 \sin \left (\frac {9 c}{2}+5 d x\right )+3 \sin \left (\frac {11 c}{2}+5 d x\right )-5 \cos \left (\frac {c}{2}\right ) (24 d x+5)-90 \cos \left (\frac {c}{2}+d x\right )-90 \cos \left (\frac {3 c}{2}+d x\right )-25 \cos \left (\frac {5 c}{2}+3 d x\right )-25 \cos \left (\frac {7 c}{2}+3 d x\right )+15 \cos \left (\frac {7 c}{2}+4 d x\right )-15 \cos \left (\frac {9 c}{2}+4 d x\right )+3 \cos \left (\frac {9 c}{2}+5 d x\right )+3 \cos \left (\frac {11 c}{2}+5 d x\right )+25 \sin \left (\frac {c}{2}\right )}{480 a^2 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^6*Sin[c + d*x])/(a + a*Sin[c + d*x])^2,x]

[Out]

(-5*(5 + 24*d*x)*Cos[c/2] - 90*Cos[c/2 + d*x] - 90*Cos[(3*c)/2 + d*x] - 25*Cos[(5*c)/2 + 3*d*x] - 25*Cos[(7*c)
/2 + 3*d*x] + 15*Cos[(7*c)/2 + 4*d*x] - 15*Cos[(9*c)/2 + 4*d*x] + 3*Cos[(9*c)/2 + 5*d*x] + 3*Cos[(11*c)/2 + 5*
d*x] + 25*Sin[c/2] - 120*d*x*Sin[c/2] + 90*Sin[c/2 + d*x] - 90*Sin[(3*c)/2 + d*x] + 25*Sin[(5*c)/2 + 3*d*x] -
25*Sin[(7*c)/2 + 3*d*x] + 15*Sin[(7*c)/2 + 4*d*x] + 15*Sin[(9*c)/2 + 4*d*x] - 3*Sin[(9*c)/2 + 5*d*x] + 3*Sin[(
11*c)/2 + 5*d*x])/(480*a^2*d*(Cos[c/2] + Sin[c/2]))

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fricas [A]  time = 0.53, size = 60, normalized size = 0.60 \[ \frac {12 \, \cos \left (d x + c\right )^{5} - 40 \, \cos \left (d x + c\right )^{3} - 15 \, d x + 15 \, {\left (2 \, \cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{60 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/60*(12*cos(d*x + c)^5 - 40*cos(d*x + c)^3 - 15*d*x + 15*(2*cos(d*x + c)^3 - cos(d*x + c))*sin(d*x + c))/(a^2
*d)

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giac [A]  time = 0.19, size = 140, normalized size = 1.40 \[ -\frac {\frac {15 \, {\left (d x + c\right )}}{a^{2}} + \frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 90 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 240 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 90 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 80 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 28\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5} a^{2}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/60*(15*(d*x + c)/a^2 + 2*(15*tan(1/2*d*x + 1/2*c)^9 + 60*tan(1/2*d*x + 1/2*c)^8 - 90*tan(1/2*d*x + 1/2*c)^7
+ 240*tan(1/2*d*x + 1/2*c)^6 + 40*tan(1/2*d*x + 1/2*c)^4 + 90*tan(1/2*d*x + 1/2*c)^3 + 80*tan(1/2*d*x + 1/2*c
)^2 - 15*tan(1/2*d*x + 1/2*c) + 28)/((tan(1/2*d*x + 1/2*c)^2 + 1)^5*a^2))/d

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maple [B]  time = 0.30, size = 313, normalized size = 3.13 \[ -\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {2 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {3 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {8 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {14}{15 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*sin(d*x+c)/(a+a*sin(d*x+c))^2,x)

[Out]

-1/2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^9-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*
c)^8+3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^7-8/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/
2*c)^6-4/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^4-3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*
x+1/2*c)^3-8/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^2+1/2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(
1/2*d*x+1/2*c)-14/15/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5-1/2/d/a^2*arctan(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.42, size = 310, normalized size = 3.10 \[ \frac {\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {80 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {90 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {40 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {240 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {90 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {60 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {15 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - 28}{a^{2} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {10 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} - \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/30*((15*sin(d*x + c)/(cos(d*x + c) + 1) - 80*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 90*sin(d*x + c)^3/(cos(d*
x + c) + 1)^3 - 40*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 240*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 90*sin(d*x
+ c)^7/(cos(d*x + c) + 1)^7 - 60*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 15*sin(d*x + c)^9/(cos(d*x + c) + 1)^9
- 28)/(a^2 + 5*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*a^2*s
in(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + a^2*sin(d*x + c)^10/(cos(d*x
+ c) + 1)^10) - 15*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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mupad [B]  time = 9.00, size = 81, normalized size = 0.81 \[ \frac {{\cos \left (c+d\,x\right )}^5}{5\,a^2\,d}-\frac {2\,{\cos \left (c+d\,x\right )}^3}{3\,a^2\,d}-\frac {x}{4\,a^2}+\frac {{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{2\,a^2\,d}-\frac {\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{4\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^6*sin(c + d*x))/(a + a*sin(c + d*x))^2,x)

[Out]

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

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sympy [A]  time = 87.47, size = 1720, normalized size = 17.20 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*sin(d*x+c)/(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((-15*d*x*tan(c/2 + d*x/2)**10/(60*a**2*d*tan(c/2 + d*x/2)**10 + 300*a**2*d*tan(c/2 + d*x/2)**8 + 600
*a**2*d*tan(c/2 + d*x/2)**6 + 600*a**2*d*tan(c/2 + d*x/2)**4 + 300*a**2*d*tan(c/2 + d*x/2)**2 + 60*a**2*d) - 7
5*d*x*tan(c/2 + d*x/2)**8/(60*a**2*d*tan(c/2 + d*x/2)**10 + 300*a**2*d*tan(c/2 + d*x/2)**8 + 600*a**2*d*tan(c/
2 + d*x/2)**6 + 600*a**2*d*tan(c/2 + d*x/2)**4 + 300*a**2*d*tan(c/2 + d*x/2)**2 + 60*a**2*d) - 150*d*x*tan(c/2
+ d*x/2)**6/(60*a**2*d*tan(c/2 + d*x/2)**10 + 300*a**2*d*tan(c/2 + d*x/2)**8 + 600*a**2*d*tan(c/2 + d*x/2)**6
+ 600*a**2*d*tan(c/2 + d*x/2)**4 + 300*a**2*d*tan(c/2 + d*x/2)**2 + 60*a**2*d) - 150*d*x*tan(c/2 + d*x/2)**4/
(60*a**2*d*tan(c/2 + d*x/2)**10 + 300*a**2*d*tan(c/2 + d*x/2)**8 + 600*a**2*d*tan(c/2 + d*x/2)**6 + 600*a**2*d
*tan(c/2 + d*x/2)**4 + 300*a**2*d*tan(c/2 + d*x/2)**2 + 60*a**2*d) - 75*d*x*tan(c/2 + d*x/2)**2/(60*a**2*d*tan
(c/2 + d*x/2)**10 + 300*a**2*d*tan(c/2 + d*x/2)**8 + 600*a**2*d*tan(c/2 + d*x/2)**6 + 600*a**2*d*tan(c/2 + d*x
/2)**4 + 300*a**2*d*tan(c/2 + d*x/2)**2 + 60*a**2*d) - 15*d*x/(60*a**2*d*tan(c/2 + d*x/2)**10 + 300*a**2*d*tan
(c/2 + d*x/2)**8 + 600*a**2*d*tan(c/2 + d*x/2)**6 + 600*a**2*d*tan(c/2 + d*x/2)**4 + 300*a**2*d*tan(c/2 + d*x/
2)**2 + 60*a**2*d) - 30*tan(c/2 + d*x/2)**9/(60*a**2*d*tan(c/2 + d*x/2)**10 + 300*a**2*d*tan(c/2 + d*x/2)**8 +
600*a**2*d*tan(c/2 + d*x/2)**6 + 600*a**2*d*tan(c/2 + d*x/2)**4 + 300*a**2*d*tan(c/2 + d*x/2)**2 + 60*a**2*d)
- 120*tan(c/2 + d*x/2)**8/(60*a**2*d*tan(c/2 + d*x/2)**10 + 300*a**2*d*tan(c/2 + d*x/2)**8 + 600*a**2*d*tan(c
/2 + d*x/2)**6 + 600*a**2*d*tan(c/2 + d*x/2)**4 + 300*a**2*d*tan(c/2 + d*x/2)**2 + 60*a**2*d) + 180*tan(c/2 +
d*x/2)**7/(60*a**2*d*tan(c/2 + d*x/2)**10 + 300*a**2*d*tan(c/2 + d*x/2)**8 + 600*a**2*d*tan(c/2 + d*x/2)**6 +
600*a**2*d*tan(c/2 + d*x/2)**4 + 300*a**2*d*tan(c/2 + d*x/2)**2 + 60*a**2*d) - 480*tan(c/2 + d*x/2)**6/(60*a**
2*d*tan(c/2 + d*x/2)**10 + 300*a**2*d*tan(c/2 + d*x/2)**8 + 600*a**2*d*tan(c/2 + d*x/2)**6 + 600*a**2*d*tan(c/
2 + d*x/2)**4 + 300*a**2*d*tan(c/2 + d*x/2)**2 + 60*a**2*d) - 80*tan(c/2 + d*x/2)**4/(60*a**2*d*tan(c/2 + d*x/
2)**10 + 300*a**2*d*tan(c/2 + d*x/2)**8 + 600*a**2*d*tan(c/2 + d*x/2)**6 + 600*a**2*d*tan(c/2 + d*x/2)**4 + 30
0*a**2*d*tan(c/2 + d*x/2)**2 + 60*a**2*d) - 180*tan(c/2 + d*x/2)**3/(60*a**2*d*tan(c/2 + d*x/2)**10 + 300*a**2
*d*tan(c/2 + d*x/2)**8 + 600*a**2*d*tan(c/2 + d*x/2)**6 + 600*a**2*d*tan(c/2 + d*x/2)**4 + 300*a**2*d*tan(c/2
+ d*x/2)**2 + 60*a**2*d) - 160*tan(c/2 + d*x/2)**2/(60*a**2*d*tan(c/2 + d*x/2)**10 + 300*a**2*d*tan(c/2 + d*x/
2)**8 + 600*a**2*d*tan(c/2 + d*x/2)**6 + 600*a**2*d*tan(c/2 + d*x/2)**4 + 300*a**2*d*tan(c/2 + d*x/2)**2 + 60*
a**2*d) + 30*tan(c/2 + d*x/2)/(60*a**2*d*tan(c/2 + d*x/2)**10 + 300*a**2*d*tan(c/2 + d*x/2)**8 + 600*a**2*d*ta
n(c/2 + d*x/2)**6 + 600*a**2*d*tan(c/2 + d*x/2)**4 + 300*a**2*d*tan(c/2 + d*x/2)**2 + 60*a**2*d) - 56/(60*a**2
*d*tan(c/2 + d*x/2)**10 + 300*a**2*d*tan(c/2 + d*x/2)**8 + 600*a**2*d*tan(c/2 + d*x/2)**6 + 600*a**2*d*tan(c/2
+ d*x/2)**4 + 300*a**2*d*tan(c/2 + d*x/2)**2 + 60*a**2*d), Ne(d, 0)), (x*sin(c)*cos(c)**6/(a*sin(c) + a)**2,
True))

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