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

Optimal. Leaf size=83 \[ -\frac {\cos ^3(c+d x)}{3 a^2 d}+\frac {3 \cos (c+d x)}{a^2 d}-\frac {\sin (c+d x) \cos (c+d x)}{a^2 d}+\frac {2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}+\frac {3 x}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 83, 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 = {2874, 2966, 2638, 2635, 8, 2633, 2648} \[ -\frac {\cos ^3(c+d x)}{3 a^2 d}+\frac {3 \cos (c+d x)}{a^2 d}-\frac {\sin (c+d x) \cos (c+d x)}{a^2 d}+\frac {2 \cos (c+d x)}{a^2 d (\sin (c+d x)+1)}+\frac {3 x}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*x)/a^2 + (3*Cos[c + d*x])/(a^2*d) - Cos[c + d*x]^3/(3*a^2*d) - (Cos[c + d*x]*Sin[c + d*x])/(a^2*d) + (2*Cos
[c + d*x])/(a^2*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 2874

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/b^2, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && (ILtQ[m, 0] ||  !IGtQ[n, 0])

Rule 2966

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && IntegerQ[n]

Rubi steps

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

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Mathematica [A]  time = 0.97, size = 165, normalized size = 1.99 \[ -\frac {-72 d x \sin \left (c+\frac {d x}{2}\right )-27 \sin \left (2 c+\frac {3 d x}{2}\right )+5 \sin \left (2 c+\frac {5 d x}{2}\right )+\sin \left (4 c+\frac {7 d x}{2}\right )-31 \cos \left (c+\frac {d x}{2}\right )-27 \cos \left (c+\frac {3 d x}{2}\right )-5 \cos \left (3 c+\frac {5 d x}{2}\right )+\cos \left (3 c+\frac {7 d x}{2}\right )+131 \sin \left (\frac {d x}{2}\right )-72 d x \cos \left (\frac {d x}{2}\right )}{24 a^2 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(-72*d*x*Cos[(d*x)/2] - 31*Cos[c + (d*x)/2] - 27*Cos[c + (3*d*x)/2] - 5*Cos[3*c + (5*d*x)/2] + Cos[3*c +
(7*d*x)/2] + 131*Sin[(d*x)/2] - 72*d*x*Sin[c + (d*x)/2] - 27*Sin[2*c + (3*d*x)/2] + 5*Sin[2*c + (5*d*x)/2] +
Sin[4*c + (7*d*x)/2])/(a^2*d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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fricas [A]  time = 0.50, size = 119, normalized size = 1.43 \[ -\frac {\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} - 9 \, d x - 3 \, {\left (3 \, d x + 4\right )} \cos \left (d x + c\right ) - 9 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{3} - 9 \, d x + 3 \, \cos \left (d x + c\right )^{2} - 6 \, \cos \left (d x + c\right ) + 6\right )} \sin \left (d x + c\right ) - 6}{3 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(cos(d*x + c)^4 - 2*cos(d*x + c)^3 - 9*d*x - 3*(3*d*x + 4)*cos(d*x + c) - 9*cos(d*x + c)^2 + (cos(d*x + c
)^3 - 9*d*x + 3*cos(d*x + c)^2 - 6*cos(d*x + c) + 6)*sin(d*x + c) - 6)/(a^2*d*cos(d*x + c) + a^2*d*sin(d*x + c
) + a^2*d)

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giac [A]  time = 0.20, size = 106, normalized size = 1.28 \[ \frac {\frac {9 \, {\left (d x + c\right )}}{a^{2}} + \frac {12}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}} + \frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 18 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{2}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(9*(d*x + c)/a^2 + 12/(a^2*(tan(1/2*d*x + 1/2*c) + 1)) + 2*(3*tan(1/2*d*x + 1/2*c)^5 + 6*tan(1/2*d*x + 1/2
*c)^4 + 18*tan(1/2*d*x + 1/2*c)^2 - 3*tan(1/2*d*x + 1/2*c) + 8)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^2))/d

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maple [B]  time = 0.46, size = 198, normalized size = 2.39 \[ \frac {2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {12 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {16}{3 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}+\frac {4}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5+4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^
4+12/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^2-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*
c)+16/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3+6/d/a^2*arctan(tan(1/2*d*x+1/2*c))+4/d/a^2/(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.43, size = 312, normalized size = 3.76 \[ \frac {2 \, {\left (\frac {\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {33 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {18 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {24 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {9 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {9 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 14}{a^{2} + \frac {a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}} + \frac {9 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*((5*sin(d*x + c)/(cos(d*x + c) + 1) + 33*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 18*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 + 24*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 9*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 9*sin(d*x + c)^
6/(cos(d*x + c) + 1)^6 + 14)/(a^2 + a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 3*a^2*sin(d*x + c)^2/(cos(d*x + c) +
1)^2 + 3*a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 3*a^2*sin(d*x
+ c)^5/(cos(d*x + c) + 1)^5 + a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^2*sin(d*x + c)^7/(cos(d*x + c) + 1)^
7) + 9*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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mupad [B]  time = 12.38, size = 120, normalized size = 1.45 \[ \frac {3\,x}{a^2}+\frac {6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+22\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {10\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {28}{3}}{a^2\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*x)/a^2 + ((10*tan(c/2 + (d*x)/2))/3 + 22*tan(c/2 + (d*x)/2)^2 + 12*tan(c/2 + (d*x)/2)^3 + 16*tan(c/2 + (d*x
)/2)^4 + 6*tan(c/2 + (d*x)/2)^5 + 6*tan(c/2 + (d*x)/2)^6 + 28/3)/(a^2*d*(tan(c/2 + (d*x)/2) + 1)*(tan(c/2 + (d
*x)/2)^2 + 1)^3)

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sympy [A]  time = 37.33, size = 2263, normalized size = 27.27 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((9*d*x*tan(c/2 + d*x/2)**7/(3*a**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*t
an(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)**
2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 9*d*x*tan(c/2 + d*x/2)**6/(3*a**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d
*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)
**3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 27*d*x*tan(c/2 + d*x/2)**5/(3*a**
2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x
/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d) +
27*d*x*tan(c/2 + d*x/2)**4/(3*a**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 +
d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**
2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 27*d*x*tan(c/2 + d*x/2)**3/(3*a**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2
+ d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**3 + 9*
a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 27*d*x*tan(c/2 + d*x/2)**2/(3*a**2*d*tan(
c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 +
9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 9*d*x*t
an(c/2 + d*x/2)/(3*a**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 +
9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2
+ d*x/2) + 3*a**2*d) + 9*d*x/(3*a**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 +
d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a*
*2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 18*tan(c/2 + d*x/2)**6/(3*a**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 +
d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**
2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 18*tan(c/2 + d*x/2)**5/(3*a**2*d*tan(c/2 + d
*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2
*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 48*tan(c/2 + d
*x/2)**4/(3*a**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 + 9*a**2*
d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2 + d*x/2
) + 3*a**2*d) + 36*tan(c/2 + d*x/2)**3/(3*a**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d
*tan(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)
**2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 66*tan(c/2 + d*x/2)**2/(3*a**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*
tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)*
*3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 10*tan(c/2 + d*x/2)/(3*a**2*d*tan(
c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d*x/2)**4 +
9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d) + 28/(3*a
**2*d*tan(c/2 + d*x/2)**7 + 3*a**2*d*tan(c/2 + d*x/2)**6 + 9*a**2*d*tan(c/2 + d*x/2)**5 + 9*a**2*d*tan(c/2 + d
*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**3 + 9*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d*tan(c/2 + d*x/2) + 3*a**2*d)
, Ne(d, 0)), (x*sin(c)**3*cos(c)**2/(a*sin(c) + a)**2, True))

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