3.261 \(\int \frac {(c-c \sin (e+f x))^4}{a+a \sin (e+f x)} \, dx\)
Optimal. Leaf size=118 \[ -\frac {2 a^3 c^4 \cos ^7(e+f x)}{f (a \sin (e+f x)+a)^4}-\frac {35 c^4 \cos ^3(e+f x)}{3 a f}-\frac {14 a c^4 \cos ^5(e+f x)}{f (a \sin (e+f x)+a)^2}-\frac {35 c^4 \sin (e+f x) \cos (e+f x)}{2 a f}-\frac {35 c^4 x}{2 a} \]
[Out]
-35/2*c^4*x/a-35/3*c^4*cos(f*x+e)^3/a/f-35/2*c^4*cos(f*x+e)*sin(f*x+e)/a/f-2*a^3*c^4*cos(f*x+e)^7/f/(a+a*sin(f
*x+e))^4-14*a*c^4*cos(f*x+e)^5/f/(a+a*sin(f*x+e))^2
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Rubi [A] time = 0.20, antiderivative size = 118, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used =
{2736, 2680, 2682, 2635, 8} \[ -\frac {2 a^3 c^4 \cos ^7(e+f x)}{f (a \sin (e+f x)+a)^4}-\frac {35 c^4 \cos ^3(e+f x)}{3 a f}-\frac {14 a c^4 \cos ^5(e+f x)}{f (a \sin (e+f x)+a)^2}-\frac {35 c^4 \sin (e+f x) \cos (e+f x)}{2 a f}-\frac {35 c^4 x}{2 a} \]
Antiderivative was successfully verified.
[In]
Int[(c - c*Sin[e + f*x])^4/(a + a*Sin[e + f*x]),x]
[Out]
(-35*c^4*x)/(2*a) - (35*c^4*Cos[e + f*x]^3)/(3*a*f) - (35*c^4*Cos[e + f*x]*Sin[e + f*x])/(2*a*f) - (2*a^3*c^4*
Cos[e + f*x]^7)/(f*(a + a*Sin[e + f*x])^4) - (14*a*c^4*Cos[e + f*x]^5)/(f*(a + a*Sin[e + f*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 2680
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] && !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]
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 2736
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 0]))
Rubi steps
\begin {align*} \int \frac {(c-c \sin (e+f x))^4}{a+a \sin (e+f x)} \, dx &=\left (a^4 c^4\right ) \int \frac {\cos ^8(e+f x)}{(a+a \sin (e+f x))^5} \, dx\\ &=-\frac {2 a^3 c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^4}-\left (7 a^2 c^4\right ) \int \frac {\cos ^6(e+f x)}{(a+a \sin (e+f x))^3} \, dx\\ &=-\frac {2 a^3 c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^4}-\frac {14 a c^4 \cos ^5(e+f x)}{f (a+a \sin (e+f x))^2}-\left (35 c^4\right ) \int \frac {\cos ^4(e+f x)}{a+a \sin (e+f x)} \, dx\\ &=-\frac {35 c^4 \cos ^3(e+f x)}{3 a f}-\frac {2 a^3 c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^4}-\frac {14 a c^4 \cos ^5(e+f x)}{f (a+a \sin (e+f x))^2}-\frac {\left (35 c^4\right ) \int \cos ^2(e+f x) \, dx}{a}\\ &=-\frac {35 c^4 \cos ^3(e+f x)}{3 a f}-\frac {35 c^4 \cos (e+f x) \sin (e+f x)}{2 a f}-\frac {2 a^3 c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^4}-\frac {14 a c^4 \cos ^5(e+f x)}{f (a+a \sin (e+f x))^2}-\frac {\left (35 c^4\right ) \int 1 \, dx}{2 a}\\ &=-\frac {35 c^4 x}{2 a}-\frac {35 c^4 \cos ^3(e+f x)}{3 a f}-\frac {35 c^4 \cos (e+f x) \sin (e+f x)}{2 a f}-\frac {2 a^3 c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^4}-\frac {14 a c^4 \cos ^5(e+f x)}{f (a+a \sin (e+f x))^2}\\ \end {align*}
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Mathematica [A] time = 1.44, size = 175, normalized size = 1.48 \[ -\frac {c^4 (\sin (e+f x)-1)^4 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right ) (-15 \sin (2 (e+f x))+141 \cos (e+f x)-\cos (3 (e+f x))+210 e+210 f x-384)+\cos \left (\frac {1}{2} (e+f x)\right ) (-15 \sin (2 (e+f x))+141 \cos (e+f x)-\cos (3 (e+f x))+210 e+210 f x)\right )}{12 a f (\sin (e+f x)+1) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^8} \]
Antiderivative was successfully verified.
[In]
Integrate[(c - c*Sin[e + f*x])^4/(a + a*Sin[e + f*x]),x]
[Out]
-1/12*(c^4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^4*(Sin[(e + f*x)/2]*(-384 + 210*e + 210*f
*x + 141*Cos[e + f*x] - Cos[3*(e + f*x)] - 15*Sin[2*(e + f*x)]) + Cos[(e + f*x)/2]*(210*e + 210*f*x + 141*Cos[
e + f*x] - Cos[3*(e + f*x)] - 15*Sin[2*(e + f*x)])))/(a*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^8*(1 + Sin[e +
f*x]))
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fricas [A] time = 0.44, size = 156, normalized size = 1.32 \[ \frac {2 \, c^{4} \cos \left (f x + e\right )^{4} - 13 \, c^{4} \cos \left (f x + e\right )^{3} - 105 \, c^{4} f x - 72 \, c^{4} \cos \left (f x + e\right )^{2} - 96 \, c^{4} - 3 \, {\left (35 \, c^{4} f x + 51 \, c^{4}\right )} \cos \left (f x + e\right ) + {\left (2 \, c^{4} \cos \left (f x + e\right )^{3} - 105 \, c^{4} f x + 15 \, c^{4} \cos \left (f x + e\right )^{2} - 57 \, c^{4} \cos \left (f x + e\right ) + 96 \, c^{4}\right )} \sin \left (f x + e\right )}{6 \, {\left (a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c-c*sin(f*x+e))^4/(a+a*sin(f*x+e)),x, algorithm="fricas")
[Out]
1/6*(2*c^4*cos(f*x + e)^4 - 13*c^4*cos(f*x + e)^3 - 105*c^4*f*x - 72*c^4*cos(f*x + e)^2 - 96*c^4 - 3*(35*c^4*f
*x + 51*c^4)*cos(f*x + e) + (2*c^4*cos(f*x + e)^3 - 105*c^4*f*x + 15*c^4*cos(f*x + e)^2 - 57*c^4*cos(f*x + e)
+ 96*c^4)*sin(f*x + e))/(a*f*cos(f*x + e) + a*f*sin(f*x + e) + a*f)
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giac [A] time = 0.22, size = 135, normalized size = 1.14 \[ -\frac {\frac {105 \, {\left (f x + e\right )} c^{4}}{a} + \frac {192 \, c^{4}}{a {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}} + \frac {2 \, {\left (15 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 66 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 144 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 15 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 70 \, c^{4}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{3} a}}{6 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c-c*sin(f*x+e))^4/(a+a*sin(f*x+e)),x, algorithm="giac")
[Out]
-1/6*(105*(f*x + e)*c^4/a + 192*c^4/(a*(tan(1/2*f*x + 1/2*e) + 1)) + 2*(15*c^4*tan(1/2*f*x + 1/2*e)^5 + 66*c^4
*tan(1/2*f*x + 1/2*e)^4 + 144*c^4*tan(1/2*f*x + 1/2*e)^2 - 15*c^4*tan(1/2*f*x + 1/2*e) + 70*c^4)/((tan(1/2*f*x
+ 1/2*e)^2 + 1)^3*a))/f
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maple [A] time = 0.32, size = 219, normalized size = 1.86 \[ -\frac {5 c^{4} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3}}-\frac {22 c^{4} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3}}-\frac {48 c^{4} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3}}+\frac {5 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f a \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3}}-\frac {70 c^{4}}{3 f a \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3}}-\frac {35 c^{4} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}-\frac {32 c^{4}}{f a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c-c*sin(f*x+e))^4/(a+a*sin(f*x+e)),x)
[Out]
-5/f*c^4/a/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)^5-22/f*c^4/a/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1
/2*e)^4-48/f*c^4/a/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)^2+5/f*c^4/a/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/
2*f*x+1/2*e)-70/3/f*c^4/a/(1+tan(1/2*f*x+1/2*e)^2)^3-35/f*c^4/a*arctan(tan(1/2*f*x+1/2*e))-32/f*c^4/a/(tan(1/2
*f*x+1/2*e)+1)
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maxima [B] time = 0.91, size = 720, normalized size = 6.10 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c-c*sin(f*x+e))^4/(a+a*sin(f*x+e)),x, algorithm="maxima")
[Out]
-1/3*(c^4*((7*sin(f*x + e)/(cos(f*x + e) + 1) + 39*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 24*sin(f*x + e)^3/(co
s(f*x + e) + 1)^3 + 24*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 9*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 9*sin(f*x
+ e)^6/(cos(f*x + e) + 1)^6 + 16)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a*sin(f*x + e)^2/(cos(f*x + e) +
1)^2 + 3*a*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*a*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 3*a*sin(f*x + e)^5
/(cos(f*x + e) + 1)^5 + a*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + a*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) + 9*arc
tan(sin(f*x + e)/(cos(f*x + e) + 1))/a) + 12*c^4*((sin(f*x + e)/(cos(f*x + e) + 1) + 5*sin(f*x + e)^2/(cos(f*x
+ e) + 1)^2 + 3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 4)/(a + a*sin(f
*x + e)/(cos(f*x + e) + 1) + 2*a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2*a*sin(f*x + e)^3/(cos(f*x + e) + 1)^3
+ a*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos
(f*x + e) + 1))/a) + 36*c^4*((sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2)/(a +
a*sin(f*x + e)/(cos(f*x + e) + 1) + a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a*sin(f*x + e)^3/(cos(f*x + e) + 1
)^3) + arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a) + 24*c^4*(arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a + 1/(a +
a*sin(f*x + e)/(cos(f*x + e) + 1))) + 6*c^4/(a + a*sin(f*x + e)/(cos(f*x + e) + 1)))/f
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mupad [B] time = 10.53, size = 290, normalized size = 2.46 \[ \frac {\frac {35\,c^4\,\left (e+f\,x\right )}{2}+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {35\,c^4\,\left (e+f\,x\right )}{2}-\frac {c^4\,\left (105\,e+105\,f\,x+110\right )}{6}\right )-\frac {c^4\,\left (105\,e+105\,f\,x+332\right )}{6}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (\frac {35\,c^4\,\left (e+f\,x\right )}{2}-\frac {c^4\,\left (105\,e+105\,f\,x+222\right )}{6}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (\frac {105\,c^4\,\left (e+f\,x\right )}{2}-\frac {c^4\,\left (315\,e+315\,f\,x+162\right )}{6}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (\frac {105\,c^4\,\left (e+f\,x\right )}{2}-\frac {c^4\,\left (315\,e+315\,f\,x+288\right )}{6}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {105\,c^4\,\left (e+f\,x\right )}{2}-\frac {c^4\,\left (315\,e+315\,f\,x+708\right )}{6}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {105\,c^4\,\left (e+f\,x\right )}{2}-\frac {c^4\,\left (315\,e+315\,f\,x+834\right )}{6}\right )}{a\,f\,\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^3}-\frac {35\,c^4\,x}{2\,a} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c - c*sin(e + f*x))^4/(a + a*sin(e + f*x)),x)
[Out]
((35*c^4*(e + f*x))/2 + tan(e/2 + (f*x)/2)*((35*c^4*(e + f*x))/2 - (c^4*(105*e + 105*f*x + 110))/6) - (c^4*(10
5*e + 105*f*x + 332))/6 + tan(e/2 + (f*x)/2)^6*((35*c^4*(e + f*x))/2 - (c^4*(105*e + 105*f*x + 222))/6) + tan(
e/2 + (f*x)/2)^5*((105*c^4*(e + f*x))/2 - (c^4*(315*e + 315*f*x + 162))/6) + tan(e/2 + (f*x)/2)^3*((105*c^4*(e
+ f*x))/2 - (c^4*(315*e + 315*f*x + 288))/6) + tan(e/2 + (f*x)/2)^4*((105*c^4*(e + f*x))/2 - (c^4*(315*e + 31
5*f*x + 708))/6) + tan(e/2 + (f*x)/2)^2*((105*c^4*(e + f*x))/2 - (c^4*(315*e + 315*f*x + 834))/6))/(a*f*(tan(e
/2 + (f*x)/2) + 1)*(tan(e/2 + (f*x)/2)^2 + 1)^3) - (35*c^4*x)/(2*a)
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sympy [A] time = 14.00, size = 2108, normalized size = 17.86 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c-c*sin(f*x+e))**4/(a+a*sin(f*x+e)),x)
[Out]
Piecewise((-105*c**4*f*x*tan(e/2 + f*x/2)**7/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*t
an(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*
a*f*tan(e/2 + f*x/2) + 6*a*f) - 105*c**4*f*x*tan(e/2 + f*x/2)**6/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 +
f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(
e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 315*c**4*f*x*tan(e/2 + f*x/2)**5/(6*a*f*tan(e/2 + f*x/2)**
7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x
/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 315*c**4*f*x*tan(e/2 + f*x/2)**4/(6*a*
f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 +
18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 315*c**4*f*x*tan(e
/2 + f*x/2)**3/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*ta
n(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f)
- 315*c**4*f*x*tan(e/2 + f*x/2)**2/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f
*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/
2 + f*x/2) + 6*a*f) - 105*c**4*f*x*tan(e/2 + f*x/2)/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 1
8*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)*
*2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 105*c**4*f*x/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 1
8*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)*
*2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 222*c**4*tan(e/2 + f*x/2)**6/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2
+ f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*t
an(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 162*c**4*tan(e/2 + f*x/2)**5/(6*a*f*tan(e/2 + f*x/2)**7
+ 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/
2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 708*c**4*tan(e/2 + f*x/2)**4/(6*a*f*tan
(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*
f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 288*c**4*tan(e/2 + f*x/
2)**3/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18*a*f*tan(e/2 + f
*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) + 6*a*f) - 834*c**
4*tan(e/2 + f*x/2)**2/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2)**5 + 18
*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 + f*x/2) +
6*a*f) - 110*c**4*tan(e/2 + f*x/2)/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f
*x/2)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/
2 + f*x/2) + 6*a*f) - 332*c**4/(6*a*f*tan(e/2 + f*x/2)**7 + 6*a*f*tan(e/2 + f*x/2)**6 + 18*a*f*tan(e/2 + f*x/2
)**5 + 18*a*f*tan(e/2 + f*x/2)**4 + 18*a*f*tan(e/2 + f*x/2)**3 + 18*a*f*tan(e/2 + f*x/2)**2 + 6*a*f*tan(e/2 +
f*x/2) + 6*a*f), Ne(f, 0)), (x*(-c*sin(e) + c)**4/(a*sin(e) + a), True))
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