3.843 \(\int \frac {\sec (c+d x) \tan ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=105 \[ -\frac {4 \tan ^9(c+d x)}{9 a^3 d}-\frac {\tan ^7(c+d x)}{a^3 d}-\frac {3 \tan ^5(c+d x)}{5 a^3 d}+\frac {4 \sec ^9(c+d x)}{9 a^3 d}-\frac {\sec ^7(c+d x)}{a^3 d}+\frac {3 \sec ^5(c+d x)}{5 a^3 d} \]

[Out]

3/5*sec(d*x+c)^5/a^3/d-sec(d*x+c)^7/a^3/d+4/9*sec(d*x+c)^9/a^3/d-3/5*tan(d*x+c)^5/a^3/d-tan(d*x+c)^7/a^3/d-4/9
*tan(d*x+c)^9/a^3/d

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Rubi [A]  time = 0.34, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2875, 2873, 2606, 14, 2607, 270} \[ -\frac {4 \tan ^9(c+d x)}{9 a^3 d}-\frac {\tan ^7(c+d x)}{a^3 d}-\frac {3 \tan ^5(c+d x)}{5 a^3 d}+\frac {4 \sec ^9(c+d x)}{9 a^3 d}-\frac {\sec ^7(c+d x)}{a^3 d}+\frac {3 \sec ^5(c+d x)}{5 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sec[c + d*x]^5)/(5*a^3*d) - Sec[c + d*x]^7/(a^3*d) + (4*Sec[c + d*x]^9)/(9*a^3*d) - (3*Tan[c + d*x]^5)/(5*a
^3*d) - Tan[c + d*x]^7/(a^3*d) - (4*Tan[c + d*x]^9)/(9*a^3*d)

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 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

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]

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 {\sec (c+d x) \tan ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\int \sec ^7(c+d x) (a-a \sin (c+d x))^3 \tan ^3(c+d x) \, dx}{a^6}\\ &=\frac {\int \left (a^3 \sec ^7(c+d x) \tan ^3(c+d x)-3 a^3 \sec ^6(c+d x) \tan ^4(c+d x)+3 a^3 \sec ^5(c+d x) \tan ^5(c+d x)-a^3 \sec ^4(c+d x) \tan ^6(c+d x)\right ) \, dx}{a^6}\\ &=\frac {\int \sec ^7(c+d x) \tan ^3(c+d x) \, dx}{a^3}-\frac {\int \sec ^4(c+d x) \tan ^6(c+d x) \, dx}{a^3}-\frac {3 \int \sec ^6(c+d x) \tan ^4(c+d x) \, dx}{a^3}+\frac {3 \int \sec ^5(c+d x) \tan ^5(c+d x) \, dx}{a^3}\\ &=\frac {\operatorname {Subst}\left (\int x^6 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{a^3 d}-\frac {\operatorname {Subst}\left (\int x^6 \left (1+x^2\right ) \, dx,x,\tan (c+d x)\right )}{a^3 d}+\frac {3 \operatorname {Subst}\left (\int x^4 \left (-1+x^2\right )^2 \, dx,x,\sec (c+d x)\right )}{a^3 d}-\frac {3 \operatorname {Subst}\left (\int x^4 \left (1+x^2\right )^2 \, dx,x,\tan (c+d x)\right )}{a^3 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-x^6+x^8\right ) \, dx,x,\sec (c+d x)\right )}{a^3 d}-\frac {\operatorname {Subst}\left (\int \left (x^6+x^8\right ) \, dx,x,\tan (c+d x)\right )}{a^3 d}+\frac {3 \operatorname {Subst}\left (\int \left (x^4-2 x^6+x^8\right ) \, dx,x,\sec (c+d x)\right )}{a^3 d}-\frac {3 \operatorname {Subst}\left (\int \left (x^4+2 x^6+x^8\right ) \, dx,x,\tan (c+d x)\right )}{a^3 d}\\ &=\frac {3 \sec ^5(c+d x)}{5 a^3 d}-\frac {\sec ^7(c+d x)}{a^3 d}+\frac {4 \sec ^9(c+d x)}{9 a^3 d}-\frac {3 \tan ^5(c+d x)}{5 a^3 d}-\frac {\tan ^7(c+d x)}{a^3 d}-\frac {4 \tan ^9(c+d x)}{9 a^3 d}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 185, normalized size = 1.76 \[ \frac {4608 \sin (c+d x)-1323 \sin (2 (c+d x))-128 \sin (3 (c+d x))-588 \sin (4 (c+d x))+384 \sin (5 (c+d x))+49 \sin (6 (c+d x))-1764 \cos (c+d x)-4032 \cos (2 (c+d x))-98 \cos (3 (c+d x))+768 \cos (4 (c+d x))+294 \cos (5 (c+d x))-64 \cos (6 (c+d x))+5376}{46080 d (a \sin (c+d x)+a)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5376 - 1764*Cos[c + d*x] - 4032*Cos[2*(c + d*x)] - 98*Cos[3*(c + d*x)] + 768*Cos[4*(c + d*x)] + 294*Cos[5*(c
+ d*x)] - 64*Cos[6*(c + d*x)] + 4608*Sin[c + d*x] - 1323*Sin[2*(c + d*x)] - 128*Sin[3*(c + d*x)] - 588*Sin[4*(
c + d*x)] + 384*Sin[5*(c + d*x)] + 49*Sin[6*(c + d*x)])/(46080*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(Cos[
(c + d*x)/2] + Sin[(c + d*x)/2])^3*(a + a*Sin[c + d*x])^3)

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fricas [A]  time = 0.46, size = 130, normalized size = 1.24 \[ \frac {2 \, \cos \left (d x + c\right )^{6} - 9 \, \cos \left (d x + c\right )^{4} + 15 \, \cos \left (d x + c\right )^{2} - {\left (6 \, \cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{2} + 5\right )} \sin \left (d x + c\right ) - 10}{45 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3} + {\left (a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/45*(2*cos(d*x + c)^6 - 9*cos(d*x + c)^4 + 15*cos(d*x + c)^2 - (6*cos(d*x + c)^4 - 5*cos(d*x + c)^2 + 5)*sin(
d*x + c) - 10)/(3*a^3*d*cos(d*x + c)^5 - 4*a^3*d*cos(d*x + c)^3 + (a^3*d*cos(d*x + c)^5 - 4*a^3*d*cos(d*x + c)
^3)*sin(d*x + c))

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giac [A]  time = 0.35, size = 161, normalized size = 1.53 \[ -\frac {\frac {15 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} - \frac {45 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 540 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 5940 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 8298 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 6372 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3528 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 972 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 113}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{9}}}{1440 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1440*(15*(3*tan(1/2*d*x + 1/2*c)^2 + 1)/(a^3*(tan(1/2*d*x + 1/2*c) - 1)^3) - (45*tan(1/2*d*x + 1/2*c)^8 + 5
40*tan(1/2*d*x + 1/2*c)^7 + 3120*tan(1/2*d*x + 1/2*c)^6 + 5940*tan(1/2*d*x + 1/2*c)^5 + 8298*tan(1/2*d*x + 1/2
*c)^4 + 6372*tan(1/2*d*x + 1/2*c)^3 + 3528*tan(1/2*d*x + 1/2*c)^2 + 972*tan(1/2*d*x + 1/2*c) + 113)/(a^3*(tan(
1/2*d*x + 1/2*c) + 1)^9))/d

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maple [A]  time = 0.56, size = 190, normalized size = 1.81 \[ \frac {-\frac {1}{24 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {8}{9 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}+\frac {8}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {28}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {67}{10 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {11}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {5}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {16}{512 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+512}}{a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/d/a^3*(-1/384/(tan(1/2*d*x+1/2*c)-1)^3-1/256/(tan(1/2*d*x+1/2*c)-1)^2-1/512/(tan(1/2*d*x+1/2*c)-1)+1/18/(ta
n(1/2*d*x+1/2*c)+1)^9-1/4/(tan(1/2*d*x+1/2*c)+1)^8+1/2/(tan(1/2*d*x+1/2*c)+1)^7-7/12/(tan(1/2*d*x+1/2*c)+1)^6+
67/160/(tan(1/2*d*x+1/2*c)+1)^5-11/64/(tan(1/2*d*x+1/2*c)+1)^4+5/192/(tan(1/2*d*x+1/2*c)+1)^3+1/128/(tan(1/2*d
*x+1/2*c)+1)^2+1/512/(tan(1/2*d*x+1/2*c)+1))

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maxima [B]  time = 0.34, size = 422, normalized size = 4.02 \[ \frac {4 \, {\left (\frac {6 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {12 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {18 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {18 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {84 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {54 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {45 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 1\right )}}{45 \, {\left (a^{3} + \frac {6 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {12 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {27 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {36 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {36 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {27 \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {12 \, a^{3} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac {6 \, a^{3} \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac {a^{3} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/45*(6*sin(d*x + c)/(cos(d*x + c) + 1) + 12*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 2*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 + 18*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 18*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 84*sin(d*x + c)
^6/(cos(d*x + c) + 1)^6 + 54*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 45*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 1)
/((a^3 + 6*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 12*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 2*a^3*sin(d*x +
c)^3/(cos(d*x + c) + 1)^3 - 27*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 36*a^3*sin(d*x + c)^5/(cos(d*x + c) +
1)^5 + 36*a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 27*a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 2*a^3*sin(d*
x + c)^9/(cos(d*x + c) + 1)^9 - 12*a^3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 6*a^3*sin(d*x + c)^11/(cos(d*x
+ c) + 1)^11 - a^3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12)*d)

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mupad [B]  time = 16.12, size = 255, normalized size = 2.43 \[ \frac {\frac {4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{45}+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}+\frac {16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{45}+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}+\frac {112\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{15}+\frac {24\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{5}+4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{a^3\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^3\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((4*cos(c/2 + (d*x)/2)^12)/45 + (8*cos(c/2 + (d*x)/2)^11*sin(c/2 + (d*x)/2))/15 + 4*cos(c/2 + (d*x)/2)^4*sin(c
/2 + (d*x)/2)^8 + (24*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^7)/5 + (112*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)
/2)^6)/15 + (8*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)^5)/5 + (8*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^4)/5
+ (8*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2)^3)/45 + (16*cos(c/2 + (d*x)/2)^10*sin(c/2 + (d*x)/2)^2)/15)/(a^3*
d*(cos(c/2 + (d*x)/2) - sin(c/2 + (d*x)/2))^3*(cos(c/2 + (d*x)/2) + sin(c/2 + (d*x)/2))^9)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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