3.1304 \(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=277 \[ \frac {(49 A-9 B-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{10 a^3 d}-\frac {(13 A-3 B-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{6 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(13 A-3 B-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac {(49 A-9 B-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {(8 A-3 B-2 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 a d (a \cos (c+d x)+a)^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

1/10*(49*A-9*B-C)*sin(d*x+c)*sec(d*x+c)^(1/2)/a^3/d-1/5*(A-B+C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))
^3-1/15*(8*A-3*B-2*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/a/d/(a+a*cos(d*x+c))^2-1/6*(13*A-3*B-C)*sin(d*x+c)*sec(d*x+c
)^(1/2)/d/(a^3+a^3*cos(d*x+c))-1/10*(49*A-9*B-C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin
(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^3/d-1/6*(13*A-3*B-C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)
/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^3/d

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Rubi [A]  time = 0.69, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4221, 3041, 2978, 2748, 2636, 2639, 2641} \[ \frac {(49 A-9 B-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{10 a^3 d}-\frac {(13 A-3 B-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{6 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(13 A-3 B-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac {(49 A-9 B-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {(8 A-3 B-2 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 a d (a \cos (c+d x)+a)^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2))/(a + a*Cos[c + d*x])^3,x]

[Out]

-((49*A - 9*B - C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) - ((13*A - 3*B
- C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) + ((49*A - 9*B - C)*Sqrt[Sec[c
+ d*x]]*Sin[c + d*x])/(10*a^3*d) - ((A - B + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3)
- ((8*A - 3*B - 2*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((13*A - 3*B - C)*Sqr
t[Sec[c + d*x]]*Sin[c + d*x])/(6*d*(a^3 + a^3*Cos[c + d*x]))

Rule 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 3041

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((a*A - b*B + a*C)*Cos[e + f*x]*(
a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(f*(b*c - a*d)*(2*m + 1)), x] + Dist[1/(b*(b*c - a*d)*(2*m
+ 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(
b*c*m + a*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c*(2*m + 1) - a*d*(m - n -
1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^
2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]

Rule 4221

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rubi steps

\begin {align*} \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx\\ &=-\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} a (11 A-B+C)-\frac {5}{2} a (A-B-C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(8 A-3 B-2 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} a^2 (41 A-6 B+C)-\frac {3}{2} a^2 (8 A-3 B-2 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))} \, dx}{15 a^4}\\ &=-\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(8 A-3 B-2 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(13 A-3 B-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3}{4} a^3 (49 A-9 B-C)-\frac {5}{4} a^3 (13 A-3 B-C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{15 a^6}\\ &=-\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(8 A-3 B-2 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(13 A-3 B-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\left ((49 A-9 B-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{20 a^3}-\frac {\left ((13 A-3 B-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3}\\ &=-\frac {(13 A-3 B-C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}+\frac {(49 A-9 B-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a^3 d}-\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(8 A-3 B-2 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(13 A-3 B-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )}-\frac {\left ((49 A-9 B-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3}\\ &=-\frac {(49 A-9 B-C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}-\frac {(13 A-3 B-C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}+\frac {(49 A-9 B-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a^3 d}-\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(8 A-3 B-2 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(13 A-3 B-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 2.07, size = 215, normalized size = 0.78 \[ -\frac {2 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (10 (13 A-3 B-C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+6 (49 A-9 B-C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )-\frac {1}{8} \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) ((1621 A-261 B+11 C) \cos (c+d x)+4 (188 A-33 B-2 C) \cos (2 (c+d x))+147 A \cos (3 (c+d x))+992 A-27 B \cos (3 (c+d x))-132 B-3 C \cos (3 (c+d x))-8 C)\right )}{15 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2))/(a + a*Cos[c + d*x])^3,x]

[Out]

(-2*Cos[(c + d*x)/2]^6*Sqrt[Sec[c + d*x]]*(6*(49*A - 9*B - C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 1
0*(13*A - 3*B - C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] - ((992*A - 132*B - 8*C + (1621*A - 261*B + 11
*C)*Cos[c + d*x] + 4*(188*A - 33*B - 2*C)*Cos[2*(c + d*x)] + 147*A*Cos[3*(c + d*x)] - 27*B*Cos[3*(c + d*x)] -
3*C*Cos[3*(c + d*x)])*Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2])/8))/(15*a^3*d*(1 + Cos[c + d*x])^3)

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{a^{3} \cos \left (d x + c\right )^{3} + 3 \, a^{3} \cos \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(3/2)/(a^3*cos(d*x + c)^3 + 3*a^3*cos(d*x + c)^2
+ 3*a^3*cos(d*x + c) + a^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(3/2)/(a*cos(d*x + c) + a)^3, x)

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maple [B]  time = 3.43, size = 793, normalized size = 2.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^3,x)

[Out]

-1/60*(-2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2
*c)^2-1)^(1/2)*(65*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-15*B*El
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))+27*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-5*C*EllipticF(cos(1/2*d*x+1/2*c
),2^(1/2))+3*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+4*(sin(1/2*d*x+1
/2*c)^2)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(65*A*Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-15*B*EllipticF(cos(1/2*d*x+1/2*
c),2^(1/2))+27*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-5*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*EllipticE
(cos(1/2*d*x+1/2*c),2^(1/2)))*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*sin(1
/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(65*A*EllipticF(cos(1/2*d*x+1/2*c
),2^(1/2))-147*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-15*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+27*B*Ellipti
cE(cos(1/2*d*x+1/2*c),2^(1/2))-5*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1
/2)))*cos(1/2*d*x+1/2*c)+12*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(49*A-9*B-C)*sin(1/2*d*x+1/2*
c)^8-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(817*A-147*B-13*C)*sin(1/2*d*x+1/2*c)^6+6*(-2*sin(
1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(248*A-43*B-2*C)*sin(1/2*d*x+1/2*c)^4-(-2*sin(1/2*d*x+1/2*c)^4+si
n(1/2*d*x+1/2*c)^2)^(1/2)*(439*A-69*B-C)*sin(1/2*d*x+1/2*c)^2)/a^3/cos(1/2*d*x+1/2*c)^5/(-2*sin(1/2*d*x+1/2*c)
^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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maxima [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((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1/cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^3,x)

[Out]

int(((1/cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^3, x)

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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((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(3/2)/(a+a*cos(d*x+c))**3,x)

[Out]

Timed out

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