3.591 \(\int \cos ^6(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=153 \[ -\frac {a^2 \cos ^7(c+d x)}{28 d}-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{36 d}+\frac {a^2 \sin (c+d x) \cos ^5(c+d x)}{24 d}+\frac {5 a^2 \sin (c+d x) \cos ^3(c+d x)}{96 d}+\frac {5 a^2 \sin (c+d x) \cos (c+d x)}{64 d}+\frac {5 a^2 x}{64}-\frac {\cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d} \]

[Out]

5/64*a^2*x-1/28*a^2*cos(d*x+c)^7/d+5/64*a^2*cos(d*x+c)*sin(d*x+c)/d+5/96*a^2*cos(d*x+c)^3*sin(d*x+c)/d+1/24*a^
2*cos(d*x+c)^5*sin(d*x+c)/d-1/9*cos(d*x+c)^7*(a+a*sin(d*x+c))^2/d-1/36*cos(d*x+c)^7*(a^2+a^2*sin(d*x+c))/d

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Rubi [A]  time = 0.15, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2860, 2678, 2669, 2635, 8} \[ -\frac {a^2 \cos ^7(c+d x)}{28 d}-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{36 d}+\frac {a^2 \sin (c+d x) \cos ^5(c+d x)}{24 d}+\frac {5 a^2 \sin (c+d x) \cos ^3(c+d x)}{96 d}+\frac {5 a^2 \sin (c+d x) \cos (c+d x)}{64 d}+\frac {5 a^2 x}{64}-\frac {\cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*Sin[c + d*x]*(a + a*Sin[c + d*x])^2,x]

[Out]

(5*a^2*x)/64 - (a^2*Cos[c + d*x]^7)/(28*d) + (5*a^2*Cos[c + d*x]*Sin[c + d*x])/(64*d) + (5*a^2*Cos[c + d*x]^3*
Sin[c + d*x])/(96*d) + (a^2*Cos[c + d*x]^5*Sin[c + d*x])/(24*d) - (Cos[c + d*x]^7*(a + a*Sin[c + d*x])^2)/(9*d
) - (Cos[c + d*x]^7*(a^2 + a^2*Sin[c + d*x]))/(36*d)

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 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
e + f*x])^(p + 1))/(f*g*(p + 1)), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x]
&& (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2678

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

Rule 2860

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 1, 0]

Rubi steps

\begin {align*} \int \cos ^6(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx &=-\frac {\cos ^7(c+d x) (a+a \sin (c+d x))^2}{9 d}+\frac {2}{9} \int \cos ^6(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=-\frac {\cos ^7(c+d x) (a+a \sin (c+d x))^2}{9 d}-\frac {\cos ^7(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{36 d}+\frac {1}{4} a \int \cos ^6(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac {a^2 \cos ^7(c+d x)}{28 d}-\frac {\cos ^7(c+d x) (a+a \sin (c+d x))^2}{9 d}-\frac {\cos ^7(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{36 d}+\frac {1}{4} a^2 \int \cos ^6(c+d x) \, dx\\ &=-\frac {a^2 \cos ^7(c+d x)}{28 d}+\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{24 d}-\frac {\cos ^7(c+d x) (a+a \sin (c+d x))^2}{9 d}-\frac {\cos ^7(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{36 d}+\frac {1}{24} \left (5 a^2\right ) \int \cos ^4(c+d x) \, dx\\ &=-\frac {a^2 \cos ^7(c+d x)}{28 d}+\frac {5 a^2 \cos ^3(c+d x) \sin (c+d x)}{96 d}+\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{24 d}-\frac {\cos ^7(c+d x) (a+a \sin (c+d x))^2}{9 d}-\frac {\cos ^7(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{36 d}+\frac {1}{32} \left (5 a^2\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac {a^2 \cos ^7(c+d x)}{28 d}+\frac {5 a^2 \cos (c+d x) \sin (c+d x)}{64 d}+\frac {5 a^2 \cos ^3(c+d x) \sin (c+d x)}{96 d}+\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{24 d}-\frac {\cos ^7(c+d x) (a+a \sin (c+d x))^2}{9 d}-\frac {\cos ^7(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{36 d}+\frac {1}{64} \left (5 a^2\right ) \int 1 \, dx\\ &=\frac {5 a^2 x}{64}-\frac {a^2 \cos ^7(c+d x)}{28 d}+\frac {5 a^2 \cos (c+d x) \sin (c+d x)}{64 d}+\frac {5 a^2 \cos ^3(c+d x) \sin (c+d x)}{96 d}+\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{24 d}-\frac {\cos ^7(c+d x) (a+a \sin (c+d x))^2}{9 d}-\frac {\cos ^7(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{36 d}\\ \end {align*}

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Mathematica [A]  time = 0.63, size = 106, normalized size = 0.69 \[ \frac {a^2 (1008 \sin (2 (c+d x))-504 \sin (4 (c+d x))-336 \sin (6 (c+d x))-63 \sin (8 (c+d x))-3276 \cos (c+d x)-1848 \cos (3 (c+d x))-504 \cos (5 (c+d x))-18 \cos (7 (c+d x))+14 \cos (9 (c+d x))+2520 c+2520 d x)}{32256 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*Sin[c + d*x]*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*(2520*c + 2520*d*x - 3276*Cos[c + d*x] - 1848*Cos[3*(c + d*x)] - 504*Cos[5*(c + d*x)] - 18*Cos[7*(c + d*x
)] + 14*Cos[9*(c + d*x)] + 1008*Sin[2*(c + d*x)] - 504*Sin[4*(c + d*x)] - 336*Sin[6*(c + d*x)] - 63*Sin[8*(c +
d*x)]))/(32256*d)

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fricas [A]  time = 0.76, size = 98, normalized size = 0.64 \[ \frac {448 \, a^{2} \cos \left (d x + c\right )^{9} - 1152 \, a^{2} \cos \left (d x + c\right )^{7} + 315 \, a^{2} d x - 21 \, {\left (48 \, a^{2} \cos \left (d x + c\right )^{7} - 8 \, a^{2} \cos \left (d x + c\right )^{5} - 10 \, a^{2} \cos \left (d x + c\right )^{3} - 15 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4032 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/4032*(448*a^2*cos(d*x + c)^9 - 1152*a^2*cos(d*x + c)^7 + 315*a^2*d*x - 21*(48*a^2*cos(d*x + c)^7 - 8*a^2*cos
(d*x + c)^5 - 10*a^2*cos(d*x + c)^3 - 15*a^2*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.26, size = 157, normalized size = 1.03 \[ \frac {5}{64} \, a^{2} x + \frac {a^{2} \cos \left (9 \, d x + 9 \, c\right )}{2304 \, d} - \frac {a^{2} \cos \left (7 \, d x + 7 \, c\right )}{1792 \, d} - \frac {a^{2} \cos \left (5 \, d x + 5 \, c\right )}{64 \, d} - \frac {11 \, a^{2} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac {13 \, a^{2} \cos \left (d x + c\right )}{128 \, d} - \frac {a^{2} \sin \left (8 \, d x + 8 \, c\right )}{512 \, d} - \frac {a^{2} \sin \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac {a^{2} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

5/64*a^2*x + 1/2304*a^2*cos(9*d*x + 9*c)/d - 1/1792*a^2*cos(7*d*x + 7*c)/d - 1/64*a^2*cos(5*d*x + 5*c)/d - 11/
192*a^2*cos(3*d*x + 3*c)/d - 13/128*a^2*cos(d*x + c)/d - 1/512*a^2*sin(8*d*x + 8*c)/d - 1/96*a^2*sin(6*d*x + 6
*c)/d - 1/64*a^2*sin(4*d*x + 4*c)/d + 1/32*a^2*sin(2*d*x + 2*c)/d

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maple [A]  time = 0.25, size = 116, normalized size = 0.76 \[ \frac {a^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{7}\left (d x +c \right )\right )}{9}-\frac {2 \left (\cos ^{7}\left (d x +c \right )\right )}{63}\right )+2 a^{2} \left (-\frac {\left (\cos ^{7}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{8}+\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{48}+\frac {5 d x}{128}+\frac {5 c}{128}\right )-\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{7}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*sin(d*x+c)*(a+a*sin(d*x+c))^2,x)

[Out]

1/d*(a^2*(-1/9*sin(d*x+c)^2*cos(d*x+c)^7-2/63*cos(d*x+c)^7)+2*a^2*(-1/8*cos(d*x+c)^7*sin(d*x+c)+1/48*(cos(d*x+
c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/128*d*x+5/128*c)-1/7*a^2*cos(d*x+c)^7)

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maxima [A]  time = 0.98, size = 93, normalized size = 0.61 \[ -\frac {4608 \, a^{2} \cos \left (d x + c\right )^{7} - 512 \, {\left (7 \, \cos \left (d x + c\right )^{9} - 9 \, \cos \left (d x + c\right )^{7}\right )} a^{2} - 21 \, {\left (64 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 120 \, d x + 120 \, c - 3 \, \sin \left (8 \, d x + 8 \, c\right ) - 24 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2}}{32256 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/32256*(4608*a^2*cos(d*x + c)^7 - 512*(7*cos(d*x + c)^9 - 9*cos(d*x + c)^7)*a^2 - 21*(64*sin(2*d*x + 2*c)^3
+ 120*d*x + 120*c - 3*sin(8*d*x + 8*c) - 24*sin(4*d*x + 4*c))*a^2)/d

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mupad [B]  time = 10.85, size = 501, normalized size = 3.27 \[ \frac {5\,a^2\,x}{64}-\frac {\frac {83\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{16}-\frac {191\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{48}-\frac {145\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{16}+\frac {145\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{16}-\frac {83\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{16}+\frac {191\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{48}-\frac {5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}}{32}+\frac {a^2\,\left (315\,c+315\,d\,x\right )}{4032}-\frac {a^2\,\left (315\,c+315\,d\,x-1408\right )}{4032}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2\,\left (315\,c+315\,d\,x\right )}{448}-\frac {a^2\,\left (2835\,c+2835\,d\,x-4608\right )}{4032}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}\,\left (\frac {a^2\,\left (315\,c+315\,d\,x\right )}{448}-\frac {a^2\,\left (2835\,c+2835\,d\,x-8064\right )}{4032}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^2\,\left (315\,c+315\,d\,x\right )}{112}-\frac {a^2\,\left (11340\,c+11340\,d\,x-18432\right )}{4032}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}\,\left (\frac {a^2\,\left (315\,c+315\,d\,x\right )}{112}-\frac {a^2\,\left (11340\,c+11340\,d\,x-32256\right )}{4032}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (\frac {a^2\,\left (315\,c+315\,d\,x\right )}{48}-\frac {a^2\,\left (26460\,c+26460\,d\,x-21504\right )}{4032}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {a^2\,\left (315\,c+315\,d\,x\right )}{32}-\frac {a^2\,\left (39690\,c+39690\,d\,x-16128\right )}{4032}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {a^2\,\left (315\,c+315\,d\,x\right )}{48}-\frac {a^2\,\left (26460\,c+26460\,d\,x-96768\right )}{4032}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {a^2\,\left (315\,c+315\,d\,x\right )}{32}-\frac {a^2\,\left (39690\,c+39690\,d\,x-161280\right )}{4032}\right )+\frac {5\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{32}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6*sin(c + d*x)*(a + a*sin(c + d*x))^2,x)

[Out]

(5*a^2*x)/64 - ((83*a^2*tan(c/2 + (d*x)/2)^5)/16 - (191*a^2*tan(c/2 + (d*x)/2)^3)/48 - (145*a^2*tan(c/2 + (d*x
)/2)^7)/16 + (145*a^2*tan(c/2 + (d*x)/2)^11)/16 - (83*a^2*tan(c/2 + (d*x)/2)^13)/16 + (191*a^2*tan(c/2 + (d*x)
/2)^15)/48 - (5*a^2*tan(c/2 + (d*x)/2)^17)/32 + (a^2*(315*c + 315*d*x))/4032 - (a^2*(315*c + 315*d*x - 1408))/
4032 + tan(c/2 + (d*x)/2)^2*((a^2*(315*c + 315*d*x))/448 - (a^2*(2835*c + 2835*d*x - 4608))/4032) + tan(c/2 +
(d*x)/2)^16*((a^2*(315*c + 315*d*x))/448 - (a^2*(2835*c + 2835*d*x - 8064))/4032) + tan(c/2 + (d*x)/2)^4*((a^2
*(315*c + 315*d*x))/112 - (a^2*(11340*c + 11340*d*x - 18432))/4032) + tan(c/2 + (d*x)/2)^14*((a^2*(315*c + 315
*d*x))/112 - (a^2*(11340*c + 11340*d*x - 32256))/4032) + tan(c/2 + (d*x)/2)^12*((a^2*(315*c + 315*d*x))/48 - (
a^2*(26460*c + 26460*d*x - 21504))/4032) + tan(c/2 + (d*x)/2)^8*((a^2*(315*c + 315*d*x))/32 - (a^2*(39690*c +
39690*d*x - 16128))/4032) + tan(c/2 + (d*x)/2)^6*((a^2*(315*c + 315*d*x))/48 - (a^2*(26460*c + 26460*d*x - 967
68))/4032) + tan(c/2 + (d*x)/2)^10*((a^2*(315*c + 315*d*x))/32 - (a^2*(39690*c + 39690*d*x - 161280))/4032) +
(5*a^2*tan(c/2 + (d*x)/2))/32)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^9)

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sympy [A]  time = 16.87, size = 282, normalized size = 1.84 \[ \begin {cases} \frac {5 a^{2} x \sin ^{8}{\left (c + d x \right )}}{64} + \frac {5 a^{2} x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{32} + \frac {5 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 a^{2} x \cos ^{8}{\left (c + d x \right )}}{64} + \frac {5 a^{2} \sin ^{7}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{64 d} + \frac {55 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{192 d} + \frac {73 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{192 d} - \frac {a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{7 d} - \frac {5 a^{2} \sin {\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{64 d} - \frac {2 a^{2} \cos ^{9}{\left (c + d x \right )}}{63 d} - \frac {a^{2} \cos ^{7}{\left (c + d x \right )}}{7 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{2} \sin {\relax (c )} \cos ^{6}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*sin(d*x+c)*(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((5*a**2*x*sin(c + d*x)**8/64 + 5*a**2*x*sin(c + d*x)**6*cos(c + d*x)**2/16 + 15*a**2*x*sin(c + d*x)*
*4*cos(c + d*x)**4/32 + 5*a**2*x*sin(c + d*x)**2*cos(c + d*x)**6/16 + 5*a**2*x*cos(c + d*x)**8/64 + 5*a**2*sin
(c + d*x)**7*cos(c + d*x)/(64*d) + 55*a**2*sin(c + d*x)**5*cos(c + d*x)**3/(192*d) + 73*a**2*sin(c + d*x)**3*c
os(c + d*x)**5/(192*d) - a**2*sin(c + d*x)**2*cos(c + d*x)**7/(7*d) - 5*a**2*sin(c + d*x)*cos(c + d*x)**7/(64*
d) - 2*a**2*cos(c + d*x)**9/(63*d) - a**2*cos(c + d*x)**7/(7*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*sin(c)*cos(c)
**6, True))

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