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

Optimal. Leaf size=87 \[ \frac {\cos ^3(c+d x)}{3 a^3 d}-\frac {5 \cos (c+d x)}{a^3 d}+\frac {3 \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac {4 \cos (c+d x)}{a^3 d (\sin (c+d x)+1)}-\frac {11 x}{2 a^3} \]

[Out]

-11/2*x/a^3-5*cos(d*x+c)/a^3/d+1/3*cos(d*x+c)^3/a^3/d+3/2*cos(d*x+c)*sin(d*x+c)/a^3/d-4*cos(d*x+c)/a^3/d/(1+si
n(d*x+c))

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Rubi [A]  time = 0.23, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2875, 2709, 2638, 2635, 8, 2633, 2648} \[ \frac {\cos ^3(c+d x)}{3 a^3 d}-\frac {5 \cos (c+d x)}{a^3 d}+\frac {3 \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac {4 \cos (c+d x)}{a^3 d (\sin (c+d x)+1)}-\frac {11 x}{2 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*x)/(2*a^3) - (5*Cos[c + d*x])/(a^3*d) + Cos[c + d*x]^3/(3*a^3*d) + (3*Cos[c + d*x]*Sin[c + d*x])/(2*a^3*d
) - (4*Cos[c + d*x])/(a^3*d*(1 + Sin[c + d*x]))

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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2709

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[(Sin[e + f*x]^p*(a + b*Sin[e + f*x])^(m - p/2))/(a - b*Sin[e + f*x])^(p/2), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

Rule 2875

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

Rubi steps

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

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Mathematica [B]  time = 1.14, size = 181, normalized size = 2.08 \[ \frac {-660 d x \sin \left (c+\frac {d x}{2}\right )+\sin \left (c+\frac {d x}{2}\right )-240 \sin \left (2 c+\frac {3 d x}{2}\right )+40 \sin \left (2 c+\frac {5 d x}{2}\right )+5 \sin \left (4 c+\frac {7 d x}{2}\right )-286 \cos \left (c+\frac {d x}{2}\right )-240 \cos \left (c+\frac {3 d x}{2}\right )-40 \cos \left (3 c+\frac {5 d x}{2}\right )+5 \cos \left (3 c+\frac {7 d x}{2}\right )+1244 \sin \left (\frac {d x}{2}\right )+(1-660 d x) \cos \left (\frac {d x}{2}\right )}{120 a^3 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - 660*d*x)*Cos[(d*x)/2] - 286*Cos[c + (d*x)/2] - 240*Cos[c + (3*d*x)/2] - 40*Cos[3*c + (5*d*x)/2] + 5*Cos[
3*c + (7*d*x)/2] + 1244*Sin[(d*x)/2] + Sin[c + (d*x)/2] - 660*d*x*Sin[c + (d*x)/2] - 240*Sin[2*c + (3*d*x)/2]
+ 40*Sin[2*c + (5*d*x)/2] + 5*Sin[4*c + (7*d*x)/2])/(120*a^3*d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(
c + d*x)/2]))

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fricas [A]  time = 0.43, size = 123, normalized size = 1.41 \[ \frac {2 \, \cos \left (d x + c\right )^{4} - 7 \, \cos \left (d x + c\right )^{3} - 33 \, d x - 3 \, {\left (11 \, d x + 15\right )} \cos \left (d x + c\right ) - 30 \, \cos \left (d x + c\right )^{2} + {\left (2 \, \cos \left (d x + c\right )^{3} - 33 \, d x + 9 \, \cos \left (d x + c\right )^{2} - 21 \, \cos \left (d x + c\right ) + 24\right )} \sin \left (d x + c\right ) - 24}{6 \, {\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*cos(d*x + c)^4 - 7*cos(d*x + c)^3 - 33*d*x - 3*(11*d*x + 15)*cos(d*x + c) - 30*cos(d*x + c)^2 + (2*cos(
d*x + c)^3 - 33*d*x + 9*cos(d*x + c)^2 - 21*cos(d*x + c) + 24)*sin(d*x + c) - 24)/(a^3*d*cos(d*x + c) + a^3*d*
sin(d*x + c) + a^3*d)

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giac [A]  time = 0.23, size = 106, normalized size = 1.22 \[ -\frac {\frac {33 \, {\left (d x + c\right )}}{a^{3}} + \frac {48}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}} + \frac {2 \, {\left (9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 60 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 28\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(33*(d*x + c)/a^3 + 48/(a^3*(tan(1/2*d*x + 1/2*c) + 1)) + 2*(9*tan(1/2*d*x + 1/2*c)^5 + 24*tan(1/2*d*x +
1/2*c)^4 + 60*tan(1/2*d*x + 1/2*c)^2 - 9*tan(1/2*d*x + 1/2*c) + 28)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^3))/d

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maple [B]  time = 0.38, size = 198, normalized size = 2.28 \[ -\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {8 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {20 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {28}{3 d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {11 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}-\frac {8}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5-8/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)
^4-20/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^2+3/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2
*c)-28/3/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3-11/d/a^3*arctan(tan(1/2*d*x+1/2*c))-8/d/a^3/(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.42, size = 312, normalized size = 3.59 \[ -\frac {\frac {\frac {19 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {123 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {60 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {96 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {33 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {33 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 52}{a^{3} + \frac {a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}} + \frac {33 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*((19*sin(d*x + c)/(cos(d*x + c) + 1) + 123*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 60*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 + 96*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 33*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 33*sin(d*x
+ c)^6/(cos(d*x + c) + 1)^6 + 52)/(a^3 + a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 3*a^3*sin(d*x + c)^2/(cos(d*x +
c) + 1)^2 + 3*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 3*a^3*sin
(d*x + c)^5/(cos(d*x + c) + 1)^5 + a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^3*sin(d*x + c)^7/(cos(d*x + c)
+ 1)^7) + 33*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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mupad [B]  time = 12.35, size = 121, normalized size = 1.39 \[ -\frac {11\,x}{2\,a^3}-\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+41\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {19\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {52}{3}}{a^3\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (11*x)/(2*a^3) - ((19*tan(c/2 + (d*x)/2))/3 + 41*tan(c/2 + (d*x)/2)^2 + 20*tan(c/2 + (d*x)/2)^3 + 32*tan(c/2
+ (d*x)/2)^4 + 11*tan(c/2 + (d*x)/2)^5 + 11*tan(c/2 + (d*x)/2)^6 + 52/3)/(a^3*d*(tan(c/2 + (d*x)/2) + 1)*(tan
(c/2 + (d*x)/2)^2 + 1)^3)

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sympy [A]  time = 95.06, size = 2264, normalized size = 26.02 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-33*d*x*tan(c/2 + d*x/2)**7/(6*a**3*d*tan(c/2 + d*x/2)**7 + 6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*
d*tan(c/2 + d*x/2)**5 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**3 + 18*a**3*d*tan(c/2 + d*
x/2)**2 + 6*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 33*d*x*tan(c/2 + d*x/2)**6/(6*a**3*d*tan(c/2 + d*x/2)**7 + 6
*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**5 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/
2 + d*x/2)**3 + 18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 99*d*x*tan(c/2 + d*x/2
)**5/(6*a**3*d*tan(c/2 + d*x/2)**7 + 6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**5 + 18*a**3*d*
tan(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**3 + 18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d*tan(c/2 + d*x/2
) + 6*a**3*d) - 99*d*x*tan(c/2 + d*x/2)**4/(6*a**3*d*tan(c/2 + d*x/2)**7 + 6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a
**3*d*tan(c/2 + d*x/2)**5 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**3 + 18*a**3*d*tan(c/2
+ d*x/2)**2 + 6*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 99*d*x*tan(c/2 + d*x/2)**3/(6*a**3*d*tan(c/2 + d*x/2)**7
+ 6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**5 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*d*ta
n(c/2 + d*x/2)**3 + 18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 99*d*x*tan(c/2 + d
*x/2)**2/(6*a**3*d*tan(c/2 + d*x/2)**7 + 6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**5 + 18*a**
3*d*tan(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**3 + 18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d*tan(c/2 + d
*x/2) + 6*a**3*d) - 33*d*x*tan(c/2 + d*x/2)/(6*a**3*d*tan(c/2 + d*x/2)**7 + 6*a**3*d*tan(c/2 + d*x/2)**6 + 18*
a**3*d*tan(c/2 + d*x/2)**5 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**3 + 18*a**3*d*tan(c/2
+ d*x/2)**2 + 6*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 33*d*x/(6*a**3*d*tan(c/2 + d*x/2)**7 + 6*a**3*d*tan(c/2
+ d*x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**5 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**3 +
18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 66*tan(c/2 + d*x/2)**6/(6*a**3*d*tan(
c/2 + d*x/2)**7 + 6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**5 + 18*a**3*d*tan(c/2 + d*x/2)**4
+ 18*a**3*d*tan(c/2 + d*x/2)**3 + 18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 66*
tan(c/2 + d*x/2)**5/(6*a**3*d*tan(c/2 + d*x/2)**7 + 6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)*
*5 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**3 + 18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d*
tan(c/2 + d*x/2) + 6*a**3*d) - 192*tan(c/2 + d*x/2)**4/(6*a**3*d*tan(c/2 + d*x/2)**7 + 6*a**3*d*tan(c/2 + d*x/
2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**5 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**3 + 18*a**
3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 120*tan(c/2 + d*x/2)**3/(6*a**3*d*tan(c/2 +
d*x/2)**7 + 6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**5 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*
a**3*d*tan(c/2 + d*x/2)**3 + 18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 246*tan(c
/2 + d*x/2)**2/(6*a**3*d*tan(c/2 + d*x/2)**7 + 6*a**3*d*tan(c/2 + d*x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**5 +
18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**3 + 18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d*tan(c
/2 + d*x/2) + 6*a**3*d) - 38*tan(c/2 + d*x/2)/(6*a**3*d*tan(c/2 + d*x/2)**7 + 6*a**3*d*tan(c/2 + d*x/2)**6 + 1
8*a**3*d*tan(c/2 + d*x/2)**5 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**3 + 18*a**3*d*tan(c
/2 + d*x/2)**2 + 6*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d) - 104/(6*a**3*d*tan(c/2 + d*x/2)**7 + 6*a**3*d*tan(c/2
+ d*x/2)**6 + 18*a**3*d*tan(c/2 + d*x/2)**5 + 18*a**3*d*tan(c/2 + d*x/2)**4 + 18*a**3*d*tan(c/2 + d*x/2)**3 +
18*a**3*d*tan(c/2 + d*x/2)**2 + 6*a**3*d*tan(c/2 + d*x/2) + 6*a**3*d), Ne(d, 0)), (x*sin(c)**2*cos(c)**4/(a*si
n(c) + a)**3, True))

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