3.422 \(\int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=102 \[ -\frac {\cos ^5(c+d x)}{5 a^2 d}+\frac {\cos ^3(c+d x)}{a^2 d}-\frac {2 \cos (c+d x)}{a^2 d}+\frac {\sin ^3(c+d x) \cos (c+d x)}{2 a^2 d}+\frac {3 \sin (c+d x) \cos (c+d x)}{4 a^2 d}-\frac {3 x}{4 a^2} \]

[Out]

-3/4*x/a^2-2*cos(d*x+c)/a^2/d+cos(d*x+c)^3/a^2/d-1/5*cos(d*x+c)^5/a^2/d+3/4*cos(d*x+c)*sin(d*x+c)/a^2/d+1/2*co
s(d*x+c)*sin(d*x+c)^3/a^2/d

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Rubi [A]  time = 0.20, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2869, 2757, 2633, 2635, 8} \[ -\frac {\cos ^5(c+d x)}{5 a^2 d}+\frac {\cos ^3(c+d x)}{a^2 d}-\frac {2 \cos (c+d x)}{a^2 d}+\frac {\sin ^3(c+d x) \cos (c+d x)}{2 a^2 d}+\frac {3 \sin (c+d x) \cos (c+d x)}{4 a^2 d}-\frac {3 x}{4 a^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]

[Out]

(-3*x)/(4*a^2) - (2*Cos[c + d*x])/(a^2*d) + Cos[c + d*x]^3/(a^2*d) - Cos[c + d*x]^5/(5*a^2*d) + (3*Cos[c + d*x
]*Sin[c + d*x])/(4*a^2*d) + (Cos[c + d*x]*Sin[c + d*x]^3)/(2*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 ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \sin ^3(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac {\int \left (a^2 \sin ^3(c+d x)-2 a^2 \sin ^4(c+d x)+a^2 \sin ^5(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \sin ^3(c+d x) \, dx}{a^2}+\frac {\int \sin ^5(c+d x) \, dx}{a^2}-\frac {2 \int \sin ^4(c+d x) \, dx}{a^2}\\ &=\frac {\cos (c+d x) \sin ^3(c+d x)}{2 a^2 d}-\frac {3 \int \sin ^2(c+d x) \, dx}{2 a^2}-\frac {\operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{a^2 d}-\frac {\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 {\cos ^3(c+d x)}{a^2 d}-\frac {\cos ^5(c+d x)}{5 a^2 d}+\frac {3 \cos (c+d x) \sin (c+d x)}{4 a^2 d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{2 a^2 d}-\frac {3 \int 1 \, dx}{4 a^2}\\ &=-\frac {3 x}{4 a^2}-\frac {2 \cos (c+d x)}{a^2 d}+\frac {\cos ^3(c+d x)}{a^2 d}-\frac {\cos ^5(c+d x)}{5 a^2 d}+\frac {3 \cos (c+d x) \sin (c+d x)}{4 a^2 d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{2 a^2 d}\\ \end {align*}

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Mathematica [B]  time = 1.32, size = 308, normalized size = 3.02 \[ -\frac {120 d x \sin \left (\frac {c}{2}\right )-110 \sin \left (\frac {c}{2}+d x\right )+110 \sin \left (\frac {3 c}{2}+d x\right )-40 \sin \left (\frac {3 c}{2}+2 d x\right )-40 \sin \left (\frac {5 c}{2}+2 d x\right )+15 \sin \left (\frac {5 c}{2}+3 d x\right )-15 \sin \left (\frac {7 c}{2}+3 d x\right )+5 \sin \left (\frac {7 c}{2}+4 d x\right )+5 \sin \left (\frac {9 c}{2}+4 d x\right )-\sin \left (\frac {9 c}{2}+5 d x\right )+\sin \left (\frac {11 c}{2}+5 d x\right )+5 \cos \left (\frac {c}{2}\right ) (24 d x+1)+110 \cos \left (\frac {c}{2}+d x\right )+110 \cos \left (\frac {3 c}{2}+d x\right )-40 \cos \left (\frac {3 c}{2}+2 d x\right )+40 \cos \left (\frac {5 c}{2}+2 d x\right )-15 \cos \left (\frac {5 c}{2}+3 d x\right )-15 \cos \left (\frac {7 c}{2}+3 d x\right )+5 \cos \left (\frac {7 c}{2}+4 d x\right )-5 \cos \left (\frac {9 c}{2}+4 d x\right )+\cos \left (\frac {9 c}{2}+5 d x\right )+\cos \left (\frac {11 c}{2}+5 d x\right )-5 \sin \left (\frac {c}{2}\right )}{160 a^2 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^2,x]

[Out]

-1/160*(5*(1 + 24*d*x)*Cos[c/2] + 110*Cos[c/2 + d*x] + 110*Cos[(3*c)/2 + d*x] - 40*Cos[(3*c)/2 + 2*d*x] + 40*C
os[(5*c)/2 + 2*d*x] - 15*Cos[(5*c)/2 + 3*d*x] - 15*Cos[(7*c)/2 + 3*d*x] + 5*Cos[(7*c)/2 + 4*d*x] - 5*Cos[(9*c)
/2 + 4*d*x] + Cos[(9*c)/2 + 5*d*x] + Cos[(11*c)/2 + 5*d*x] - 5*Sin[c/2] + 120*d*x*Sin[c/2] - 110*Sin[c/2 + d*x
] + 110*Sin[(3*c)/2 + d*x] - 40*Sin[(3*c)/2 + 2*d*x] - 40*Sin[(5*c)/2 + 2*d*x] + 15*Sin[(5*c)/2 + 3*d*x] - 15*
Sin[(7*c)/2 + 3*d*x] + 5*Sin[(7*c)/2 + 4*d*x] + 5*Sin[(9*c)/2 + 4*d*x] - Sin[(9*c)/2 + 5*d*x] + Sin[(11*c)/2 +
5*d*x])/(a^2*d*(Cos[c/2] + Sin[c/2]))

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fricas [A]  time = 0.44, size = 68, normalized size = 0.67 \[ -\frac {4 \, \cos \left (d x + c\right )^{5} - 20 \, \cos \left (d x + c\right )^{3} + 15 \, d x + 5 \, {\left (2 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 40 \, \cos \left (d x + c\right )}{20 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/20*(4*cos(d*x + c)^5 - 20*cos(d*x + c)^3 + 15*d*x + 5*(2*cos(d*x + c)^3 - 5*cos(d*x + c))*sin(d*x + c) + 40
*cos(d*x + c))/(a^2*d)

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giac [A]  time = 0.22, size = 127, normalized size = 1.25 \[ -\frac {\frac {15 \, {\left (d x + c\right )}}{a^{2}} + \frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 70 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 200 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 70 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5} a^{2}}}{20 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/20*(15*(d*x + c)/a^2 + 2*(15*tan(1/2*d*x + 1/2*c)^9 + 70*tan(1/2*d*x + 1/2*c)^7 + 40*tan(1/2*d*x + 1/2*c)^6
+ 200*tan(1/2*d*x + 1/2*c)^4 - 70*tan(1/2*d*x + 1/2*c)^3 + 120*tan(1/2*d*x + 1/2*c)^2 - 15*tan(1/2*d*x + 1/2*
c) + 24)/((tan(1/2*d*x + 1/2*c)^2 + 1)^5*a^2))/d

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maple [B]  time = 0.40, size = 279, normalized size = 2.74 \[ -\frac {3 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {7 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {4 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {20 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {7 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {12 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {12}{5 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x)

[Out]

-3/2/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^9-7/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*
c)^7-4/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^6-20/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1
/2*c)^4+7/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^3-12/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*
x+1/2*c)^2+3/2/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)-12/5/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^5-3/2/d
/a^2*arctan(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.49, size = 290, normalized size = 2.84 \[ \frac {\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {120 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {70 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {200 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {40 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {70 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {15 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - 24}{a^{2} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {10 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} - \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{10 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/10*((15*sin(d*x + c)/(cos(d*x + c) + 1) - 120*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 70*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 - 200*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 40*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 70*sin(d*x
+ c)^7/(cos(d*x + c) + 1)^7 - 15*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 24)/(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*sin(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 = 8.72, size = 89, normalized size = 0.87 \[ \frac {3\,\cos \left (3\,c+3\,d\,x\right )}{16\,a^2\,d}-\frac {11\,\cos \left (c+d\,x\right )}{8\,a^2\,d}-\frac {3\,x}{4\,a^2}-\frac {\cos \left (5\,c+5\,d\,x\right )}{80\,a^2\,d}+\frac {\sin \left (2\,c+2\,d\,x\right )}{2\,a^2\,d}-\frac {\sin \left (4\,c+4\,d\,x\right )}{16\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^4*sin(c + d*x)^3)/(a + a*sin(c + d*x))^2,x)

[Out]

(3*cos(3*c + 3*d*x))/(16*a^2*d) - (11*cos(c + d*x))/(8*a^2*d) - (3*x)/(4*a^2) - cos(5*c + 5*d*x)/(80*a^2*d) +
sin(2*c + 2*d*x)/(2*a^2*d) - sin(4*c + 4*d*x)/(16*a^2*d)

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sympy [A]  time = 94.09, size = 1608, normalized size = 15.76 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)**3/(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((-15*d*x*tan(c/2 + d*x/2)**10/(20*a**2*d*tan(c/2 + d*x/2)**10 + 100*a**2*d*tan(c/2 + d*x/2)**8 + 200
*a**2*d*tan(c/2 + d*x/2)**6 + 200*a**2*d*tan(c/2 + d*x/2)**4 + 100*a**2*d*tan(c/2 + d*x/2)**2 + 20*a**2*d) - 7
5*d*x*tan(c/2 + d*x/2)**8/(20*a**2*d*tan(c/2 + d*x/2)**10 + 100*a**2*d*tan(c/2 + d*x/2)**8 + 200*a**2*d*tan(c/
2 + d*x/2)**6 + 200*a**2*d*tan(c/2 + d*x/2)**4 + 100*a**2*d*tan(c/2 + d*x/2)**2 + 20*a**2*d) - 150*d*x*tan(c/2
+ d*x/2)**6/(20*a**2*d*tan(c/2 + d*x/2)**10 + 100*a**2*d*tan(c/2 + d*x/2)**8 + 200*a**2*d*tan(c/2 + d*x/2)**6
+ 200*a**2*d*tan(c/2 + d*x/2)**4 + 100*a**2*d*tan(c/2 + d*x/2)**2 + 20*a**2*d) - 150*d*x*tan(c/2 + d*x/2)**4/
(20*a**2*d*tan(c/2 + d*x/2)**10 + 100*a**2*d*tan(c/2 + d*x/2)**8 + 200*a**2*d*tan(c/2 + d*x/2)**6 + 200*a**2*d
*tan(c/2 + d*x/2)**4 + 100*a**2*d*tan(c/2 + d*x/2)**2 + 20*a**2*d) - 75*d*x*tan(c/2 + d*x/2)**2/(20*a**2*d*tan
(c/2 + d*x/2)**10 + 100*a**2*d*tan(c/2 + d*x/2)**8 + 200*a**2*d*tan(c/2 + d*x/2)**6 + 200*a**2*d*tan(c/2 + d*x
/2)**4 + 100*a**2*d*tan(c/2 + d*x/2)**2 + 20*a**2*d) - 15*d*x/(20*a**2*d*tan(c/2 + d*x/2)**10 + 100*a**2*d*tan
(c/2 + d*x/2)**8 + 200*a**2*d*tan(c/2 + d*x/2)**6 + 200*a**2*d*tan(c/2 + d*x/2)**4 + 100*a**2*d*tan(c/2 + d*x/
2)**2 + 20*a**2*d) - 30*tan(c/2 + d*x/2)**9/(20*a**2*d*tan(c/2 + d*x/2)**10 + 100*a**2*d*tan(c/2 + d*x/2)**8 +
200*a**2*d*tan(c/2 + d*x/2)**6 + 200*a**2*d*tan(c/2 + d*x/2)**4 + 100*a**2*d*tan(c/2 + d*x/2)**2 + 20*a**2*d)
- 140*tan(c/2 + d*x/2)**7/(20*a**2*d*tan(c/2 + d*x/2)**10 + 100*a**2*d*tan(c/2 + d*x/2)**8 + 200*a**2*d*tan(c
/2 + d*x/2)**6 + 200*a**2*d*tan(c/2 + d*x/2)**4 + 100*a**2*d*tan(c/2 + d*x/2)**2 + 20*a**2*d) - 80*tan(c/2 + d
*x/2)**6/(20*a**2*d*tan(c/2 + d*x/2)**10 + 100*a**2*d*tan(c/2 + d*x/2)**8 + 200*a**2*d*tan(c/2 + d*x/2)**6 + 2
00*a**2*d*tan(c/2 + d*x/2)**4 + 100*a**2*d*tan(c/2 + d*x/2)**2 + 20*a**2*d) - 400*tan(c/2 + d*x/2)**4/(20*a**2
*d*tan(c/2 + d*x/2)**10 + 100*a**2*d*tan(c/2 + d*x/2)**8 + 200*a**2*d*tan(c/2 + d*x/2)**6 + 200*a**2*d*tan(c/2
+ d*x/2)**4 + 100*a**2*d*tan(c/2 + d*x/2)**2 + 20*a**2*d) + 140*tan(c/2 + d*x/2)**3/(20*a**2*d*tan(c/2 + d*x/
2)**10 + 100*a**2*d*tan(c/2 + d*x/2)**8 + 200*a**2*d*tan(c/2 + d*x/2)**6 + 200*a**2*d*tan(c/2 + d*x/2)**4 + 10
0*a**2*d*tan(c/2 + d*x/2)**2 + 20*a**2*d) - 240*tan(c/2 + d*x/2)**2/(20*a**2*d*tan(c/2 + d*x/2)**10 + 100*a**2
*d*tan(c/2 + d*x/2)**8 + 200*a**2*d*tan(c/2 + d*x/2)**6 + 200*a**2*d*tan(c/2 + d*x/2)**4 + 100*a**2*d*tan(c/2
+ d*x/2)**2 + 20*a**2*d) + 30*tan(c/2 + d*x/2)/(20*a**2*d*tan(c/2 + d*x/2)**10 + 100*a**2*d*tan(c/2 + d*x/2)**
8 + 200*a**2*d*tan(c/2 + d*x/2)**6 + 200*a**2*d*tan(c/2 + d*x/2)**4 + 100*a**2*d*tan(c/2 + d*x/2)**2 + 20*a**2
*d) - 48/(20*a**2*d*tan(c/2 + d*x/2)**10 + 100*a**2*d*tan(c/2 + d*x/2)**8 + 200*a**2*d*tan(c/2 + d*x/2)**6 + 2
00*a**2*d*tan(c/2 + d*x/2)**4 + 100*a**2*d*tan(c/2 + d*x/2)**2 + 20*a**2*d), Ne(d, 0)), (x*sin(c)**3*cos(c)**4
/(a*sin(c) + a)**2, True))

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