3.285 \(\int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=132 \[ -\frac {a^3 \cos ^7(c+d x)}{7 d}+\frac {a^3 \cos ^5(c+d x)}{d}-\frac {4 a^3 \cos ^3(c+d x)}{3 d}-\frac {a^3 \sin ^3(c+d x) \cos ^3(c+d x)}{2 d}-\frac {5 a^3 \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac {5 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5 a^3 x}{16} \]

[Out]

5/16*a^3*x-4/3*a^3*cos(d*x+c)^3/d+a^3*cos(d*x+c)^5/d-1/7*a^3*cos(d*x+c)^7/d+5/16*a^3*cos(d*x+c)*sin(d*x+c)/d-5
/8*a^3*cos(d*x+c)^3*sin(d*x+c)/d-1/2*a^3*cos(d*x+c)^3*sin(d*x+c)^3/d

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Rubi [A]  time = 0.30, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2873, 2568, 2635, 8, 2565, 14, 270} \[ -\frac {a^3 \cos ^7(c+d x)}{7 d}+\frac {a^3 \cos ^5(c+d x)}{d}-\frac {4 a^3 \cos ^3(c+d x)}{3 d}-\frac {a^3 \sin ^3(c+d x) \cos ^3(c+d x)}{2 d}-\frac {5 a^3 \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac {5 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5 a^3 x}{16} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^3*x)/16 - (4*a^3*Cos[c + d*x]^3)/(3*d) + (a^3*Cos[c + d*x]^5)/d - (a^3*Cos[c + d*x]^7)/(7*d) + (5*a^3*Cos
[c + d*x]*Sin[c + d*x])/(16*d) - (5*a^3*Cos[c + d*x]^3*Sin[c + d*x])/(8*d) - (a^3*Cos[c + d*x]^3*Sin[c + d*x]^
3)/(2*d)

Rule 8

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]

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 2873

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^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] && IGtQ[m, 0]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^3 \, dx &=\int \left (a^3 \cos ^2(c+d x) \sin ^2(c+d x)+3 a^3 \cos ^2(c+d x) \sin ^3(c+d x)+3 a^3 \cos ^2(c+d x) \sin ^4(c+d x)+a^3 \cos ^2(c+d x) \sin ^5(c+d x)\right ) \, dx\\ &=a^3 \int \cos ^2(c+d x) \sin ^2(c+d x) \, dx+a^3 \int \cos ^2(c+d x) \sin ^5(c+d x) \, dx+\left (3 a^3\right ) \int \cos ^2(c+d x) \sin ^3(c+d x) \, dx+\left (3 a^3\right ) \int \cos ^2(c+d x) \sin ^4(c+d x) \, dx\\ &=-\frac {a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {a^3 \cos ^3(c+d x) \sin ^3(c+d x)}{2 d}+\frac {1}{4} a^3 \int \cos ^2(c+d x) \, dx+\frac {1}{2} \left (3 a^3\right ) \int \cos ^2(c+d x) \sin ^2(c+d x) \, dx-\frac {a^3 \operatorname {Subst}\left (\int x^2 \left (1-x^2\right )^2 \, dx,x,\cos (c+d x)\right )}{d}-\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {a^3 \cos (c+d x) \sin (c+d x)}{8 d}-\frac {5 a^3 \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac {a^3 \cos ^3(c+d x) \sin ^3(c+d x)}{2 d}+\frac {1}{8} a^3 \int 1 \, dx+\frac {1}{8} \left (3 a^3\right ) \int \cos ^2(c+d x) \, dx-\frac {a^3 \operatorname {Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {a^3 x}{8}-\frac {4 a^3 \cos ^3(c+d x)}{3 d}+\frac {a^3 \cos ^5(c+d x)}{d}-\frac {a^3 \cos ^7(c+d x)}{7 d}+\frac {5 a^3 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {5 a^3 \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac {a^3 \cos ^3(c+d x) \sin ^3(c+d x)}{2 d}+\frac {1}{16} \left (3 a^3\right ) \int 1 \, dx\\ &=\frac {5 a^3 x}{16}-\frac {4 a^3 \cos ^3(c+d x)}{3 d}+\frac {a^3 \cos ^5(c+d x)}{d}-\frac {a^3 \cos ^7(c+d x)}{7 d}+\frac {5 a^3 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {5 a^3 \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac {a^3 \cos ^3(c+d x) \sin ^3(c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 0.67, size = 86, normalized size = 0.65 \[ \frac {a^3 (-63 \sin (2 (c+d x))-105 \sin (4 (c+d x))+21 \sin (6 (c+d x))-609 \cos (c+d x)-91 \cos (3 (c+d x))+63 \cos (5 (c+d x))-3 \cos (7 (c+d x))+420 c+420 d x)}{1344 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(420*c + 420*d*x - 609*Cos[c + d*x] - 91*Cos[3*(c + d*x)] + 63*Cos[5*(c + d*x)] - 3*Cos[7*(c + d*x)] - 63
*Sin[2*(c + d*x)] - 105*Sin[4*(c + d*x)] + 21*Sin[6*(c + d*x)]))/(1344*d)

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fricas [A]  time = 0.51, size = 98, normalized size = 0.74 \[ -\frac {48 \, a^{3} \cos \left (d x + c\right )^{7} - 336 \, a^{3} \cos \left (d x + c\right )^{5} + 448 \, a^{3} \cos \left (d x + c\right )^{3} - 105 \, a^{3} d x - 21 \, {\left (8 \, a^{3} \cos \left (d x + c\right )^{5} - 18 \, a^{3} \cos \left (d x + c\right )^{3} + 5 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{336 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/336*(48*a^3*cos(d*x + c)^7 - 336*a^3*cos(d*x + c)^5 + 448*a^3*cos(d*x + c)^3 - 105*a^3*d*x - 21*(8*a^3*cos(
d*x + c)^5 - 18*a^3*cos(d*x + c)^3 + 5*a^3*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.23, size = 123, normalized size = 0.93 \[ \frac {5}{16} \, a^{3} x - \frac {a^{3} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {3 \, a^{3} \cos \left (5 \, d x + 5 \, c\right )}{64 \, d} - \frac {13 \, a^{3} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac {29 \, a^{3} \cos \left (d x + c\right )}{64 \, d} + \frac {a^{3} \sin \left (6 \, d x + 6 \, c\right )}{64 \, d} - \frac {5 \, a^{3} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} - \frac {3 \, a^{3} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/16*a^3*x - 1/448*a^3*cos(7*d*x + 7*c)/d + 3/64*a^3*cos(5*d*x + 5*c)/d - 13/192*a^3*cos(3*d*x + 3*c)/d - 29/6
4*a^3*cos(d*x + c)/d + 1/64*a^3*sin(6*d*x + 6*c)/d - 5/64*a^3*sin(4*d*x + 4*c)/d - 3/64*a^3*sin(2*d*x + 2*c)/d

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maple [A]  time = 0.16, size = 194, normalized size = 1.47 \[ \frac {a^{3} \left (-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{7}-\frac {4 \left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{35}-\frac {8 \left (\cos ^{3}\left (d x +c \right )\right )}{105}\right )+3 a^{3} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{8}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )+3 a^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+a^{3} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*(-1/7*sin(d*x+c)^4*cos(d*x+c)^3-4/35*sin(d*x+c)^2*cos(d*x+c)^3-8/105*cos(d*x+c)^3)+3*a^3*(-1/6*sin(d*
x+c)^3*cos(d*x+c)^3-1/8*cos(d*x+c)^3*sin(d*x+c)+1/16*cos(d*x+c)*sin(d*x+c)+1/16*d*x+1/16*c)+3*a^3*(-1/5*sin(d*
x+c)^2*cos(d*x+c)^3-2/15*cos(d*x+c)^3)+a^3*(-1/4*cos(d*x+c)^3*sin(d*x+c)+1/8*cos(d*x+c)*sin(d*x+c)+1/8*d*x+1/8
*c))

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maxima [A]  time = 0.33, size = 129, normalized size = 0.98 \[ -\frac {64 \, {\left (15 \, \cos \left (d x + c\right )^{7} - 42 \, \cos \left (d x + c\right )^{5} + 35 \, \cos \left (d x + c\right )^{3}\right )} a^{3} - 1344 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{3} + 105 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3} - 210 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3}}{6720 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6720*(64*(15*cos(d*x + c)^7 - 42*cos(d*x + c)^5 + 35*cos(d*x + c)^3)*a^3 - 1344*(3*cos(d*x + c)^5 - 5*cos(d
*x + c)^3)*a^3 + 105*(4*sin(2*d*x + 2*c)^3 - 12*d*x - 12*c + 3*sin(4*d*x + 4*c))*a^3 - 210*(4*d*x + 4*c - sin(
4*d*x + 4*c))*a^3)/d

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mupad [B]  time = 12.16, size = 331, normalized size = 2.51 \[ \frac {5\,a^3\,x}{16}-\frac {\frac {5\,a^3\,\left (c+d\,x\right )}{16}+\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-\frac {119\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{8}+\frac {119\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{8}-\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{2}-\frac {5\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{8}-\frac {a^3\,\left (105\,c+105\,d\,x-320\right )}{336}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {35\,a^3\,\left (c+d\,x\right )}{16}-\frac {a^3\,\left (735\,c+735\,d\,x-2240\right )}{336}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {105\,a^3\,\left (c+d\,x\right )}{16}-\frac {a^3\,\left (2205\,c+2205\,d\,x-2688\right )}{336}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {175\,a^3\,\left (c+d\,x\right )}{16}-\frac {a^3\,\left (3675\,c+3675\,d\,x-896\right )}{336}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {105\,a^3\,\left (c+d\,x\right )}{16}-\frac {a^3\,\left (2205\,c+2205\,d\,x-4032\right )}{336}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {175\,a^3\,\left (c+d\,x\right )}{16}-\frac {a^3\,\left (3675\,c+3675\,d\,x-10304\right )}{336}\right )+\frac {5\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*a^3*x)/16 - ((5*a^3*(c + d*x))/16 + (3*a^3*tan(c/2 + (d*x)/2)^3)/2 - (119*a^3*tan(c/2 + (d*x)/2)^5)/8 + (11
9*a^3*tan(c/2 + (d*x)/2)^9)/8 - (3*a^3*tan(c/2 + (d*x)/2)^11)/2 - (5*a^3*tan(c/2 + (d*x)/2)^13)/8 - (a^3*(105*
c + 105*d*x - 320))/336 + tan(c/2 + (d*x)/2)^2*((35*a^3*(c + d*x))/16 - (a^3*(735*c + 735*d*x - 2240))/336) +
tan(c/2 + (d*x)/2)^4*((105*a^3*(c + d*x))/16 - (a^3*(2205*c + 2205*d*x - 2688))/336) + tan(c/2 + (d*x)/2)^6*((
175*a^3*(c + d*x))/16 - (a^3*(3675*c + 3675*d*x - 896))/336) + tan(c/2 + (d*x)/2)^10*((105*a^3*(c + d*x))/16 -
(a^3*(2205*c + 2205*d*x - 4032))/336) + tan(c/2 + (d*x)/2)^8*((175*a^3*(c + d*x))/16 - (a^3*(3675*c + 3675*d*
x - 10304))/336) + (5*a^3*tan(c/2 + (d*x)/2))/8)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^7)

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sympy [A]  time = 7.84, size = 379, normalized size = 2.87 \[ \begin {cases} \frac {3 a^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {9 a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {9 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 a^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{3} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} - \frac {a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{2 d} + \frac {a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {4 a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac {a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{d} - \frac {3 a^{3} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {8 a^{3} \cos ^{7}{\left (c + d x \right )}}{105 d} - \frac {2 a^{3} \cos ^{5}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{3} \sin ^{2}{\relax (c )} \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*a**3*x*sin(c + d*x)**6/16 + 9*a**3*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + a**3*x*sin(c + d*x)**4/
8 + 9*a**3*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + a**3*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*a**3*x*cos(c +
d*x)**6/16 + a**3*x*cos(c + d*x)**4/8 + 3*a**3*sin(c + d*x)**5*cos(c + d*x)/(16*d) - a**3*sin(c + d*x)**4*cos(
c + d*x)**3/(3*d) - a**3*sin(c + d*x)**3*cos(c + d*x)**3/(2*d) + a**3*sin(c + d*x)**3*cos(c + d*x)/(8*d) - 4*a
**3*sin(c + d*x)**2*cos(c + d*x)**5/(15*d) - a**3*sin(c + d*x)**2*cos(c + d*x)**3/d - 3*a**3*sin(c + d*x)*cos(
c + d*x)**5/(16*d) - a**3*sin(c + d*x)*cos(c + d*x)**3/(8*d) - 8*a**3*cos(c + d*x)**7/(105*d) - 2*a**3*cos(c +
d*x)**5/(5*d), Ne(d, 0)), (x*(a*sin(c) + a)**3*sin(c)**2*cos(c)**2, True))

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