3.4.30 \(\int \frac {(e+f x)^2 \cos ^2(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx\) [330]

Optimal. Leaf size=566 \[ -\frac {i (e+f x)^3}{3 a f}-\frac {i \left (a^2-b^2\right ) (e+f x)^3}{3 a b^2 f}-\frac {2 f (e+f x) \cos (c+d x)}{b d^2}+\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {(e+f x)^2 \log \left (1-e^{2 i (c+d x)}\right )}{a d}-\frac {2 i \left (a^2-b^2\right ) f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d^2}-\frac {2 i \left (a^2-b^2\right ) f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^2 d^2}-\frac {i f (e+f x) \text {Li}_2\left (e^{2 i (c+d x)}\right )}{a d^2}+\frac {2 \left (a^2-b^2\right ) f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d^3}+\frac {2 \left (a^2-b^2\right ) f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^2 d^3}+\frac {f^2 \text {Li}_3\left (e^{2 i (c+d x)}\right )}{2 a d^3}+\frac {2 f^2 \sin (c+d x)}{b d^3}-\frac {(e+f x)^2 \sin (c+d x)}{b d} \]

[Out]

-1/3*I*(f*x+e)^3/a/f-1/3*I*(a^2-b^2)*(f*x+e)^3/a/b^2/f-2*f*(f*x+e)*cos(d*x+c)/b/d^2+(f*x+e)^2*ln(1-exp(2*I*(d*
x+c)))/a/d+(a^2-b^2)*(f*x+e)^2*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/a/b^2/d+(a^2-b^2)*(f*x+e)^2*ln(1-I
*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/a/b^2/d-I*f*(f*x+e)*polylog(2,exp(2*I*(d*x+c)))/a/d^2-2*I*(a^2-b^2)*f*(
f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/a/b^2/d^2-2*I*(a^2-b^2)*f*(f*x+e)*polylog(2,I*b*exp(I
*(d*x+c))/(a+(a^2-b^2)^(1/2)))/a/b^2/d^2+1/2*f^2*polylog(3,exp(2*I*(d*x+c)))/a/d^3+2*(a^2-b^2)*f^2*polylog(3,I
*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/a/b^2/d^3+2*(a^2-b^2)*f^2*polylog(3,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/
2)))/a/b^2/d^3+2*f^2*sin(d*x+c)/b/d^3-(f*x+e)^2*sin(d*x+c)/b/d

________________________________________________________________________________________

Rubi [A]
time = 0.72, antiderivative size = 566, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 13, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.382, Rules used = {4639, 4493, 4489, 3391, 3798, 2221, 2611, 2320, 6724, 4621, 3377, 2717, 4615} \begin {gather*} \frac {2 f^2 \left (a^2-b^2\right ) \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d^3}+\frac {2 f^2 \left (a^2-b^2\right ) \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a b^2 d^3}-\frac {2 i f \left (a^2-b^2\right ) (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d^2}-\frac {2 i f \left (a^2-b^2\right ) (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a b^2 d^2}+\frac {f^2 \text {PolyLog}\left (3,e^{2 i (c+d x)}\right )}{2 a d^3}-\frac {i f (e+f x) \text {PolyLog}\left (2,e^{2 i (c+d x)}\right )}{a d^2}+\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a b^2 d}-\frac {i \left (a^2-b^2\right ) (e+f x)^3}{3 a b^2 f}+\frac {(e+f x)^2 \log \left (1-e^{2 i (c+d x)}\right )}{a d}-\frac {i (e+f x)^3}{3 a f}+\frac {2 f^2 \sin (c+d x)}{b d^3}-\frac {2 f (e+f x) \cos (c+d x)}{b d^2}-\frac {(e+f x)^2 \sin (c+d x)}{b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^2*Cos[c + d*x]^2*Cot[c + d*x])/(a + b*Sin[c + d*x]),x]

[Out]

((-1/3*I)*(e + f*x)^3)/(a*f) - ((I/3)*(a^2 - b^2)*(e + f*x)^3)/(a*b^2*f) - (2*f*(e + f*x)*Cos[c + d*x])/(b*d^2
) + ((a^2 - b^2)*(e + f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(a*b^2*d) + ((a^2 - b^2)*(e
+ f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(a*b^2*d) + ((e + f*x)^2*Log[1 - E^((2*I)*(c +
d*x))])/(a*d) - ((2*I)*(a^2 - b^2)*f*(e + f*x)*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(a*b^
2*d^2) - ((2*I)*(a^2 - b^2)*f*(e + f*x)*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(a*b^2*d^2) -
(I*f*(e + f*x)*PolyLog[2, E^((2*I)*(c + d*x))])/(a*d^2) + (2*(a^2 - b^2)*f^2*PolyLog[3, (I*b*E^(I*(c + d*x)))
/(a - Sqrt[a^2 - b^2])])/(a*b^2*d^3) + (2*(a^2 - b^2)*f^2*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2
])])/(a*b^2*d^3) + (f^2*PolyLog[3, E^((2*I)*(c + d*x))])/(2*a*d^3) + (2*f^2*Sin[c + d*x])/(b*d^3) - ((e + f*x)
^2*Sin[c + d*x])/(b*d)

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2717

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

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3391

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

Rule 3798

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[I*((c + d*x)^(m + 1)/(d*(
m + 1))), x] - Dist[2*I, Int[(c + d*x)^m*E^(2*I*k*Pi)*(E^(2*I*(e + f*x))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x))))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4489

Int[Cos[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Simp[(c + d
*x)^m*(Sin[a + b*x]^(n + 1)/(b*(n + 1))), x] - Dist[d*(m/(b*(n + 1))), Int[(c + d*x)^(m - 1)*Sin[a + b*x]^(n +
1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]

Rule 4493

Int[Cos[(a_.) + (b_.)*(x_)]^(n_.)*Cot[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Int[
(c + d*x)^m*Cos[a + b*x]^n*Cot[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Cos[a + b*x]^(n - 2)*Cot[a + b*x]^p, x]
/; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]

Rule 4615

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
Simp[(-I)*((e + f*x)^(m + 1)/(b*f*(m + 1))), x] + (Int[(e + f*x)^m*(E^(I*(c + d*x))/(a - Rt[a^2 - b^2, 2] - I*
b*E^(I*(c + d*x)))), x] + Int[(e + f*x)^m*(E^(I*(c + d*x))/(a + Rt[a^2 - b^2, 2] - I*b*E^(I*(c + d*x)))), x])
/; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

Rule 4621

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

Rule 4639

Int[(Cos[(c_.) + (d_.)*(x_)]^(p_.)*Cot[(c_.) + (d_.)*(x_)]^(n_.)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin
[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[(e + f*x)^m*Cos[c + d*x]^p*Cot[c + d*x]^n, x], x] - Dist[b/a
, Int[(e + f*x)^m*Cos[c + d*x]^(p + 1)*(Cot[c + d*x]^(n - 1)/(a + b*Sin[c + d*x])), x], x] /; FreeQ[{a, b, c,
d, e, f}, x] && IGtQ[m, 0] && IGtQ[n, 0] && IGtQ[p, 0]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {(e+f x)^2 \cos ^2(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {\int (e+f x)^2 \cos ^2(c+d x) \cot (c+d x) \, dx}{a}-\frac {b \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx}{a}\\ &=\frac {\int (e+f x)^2 \cot (c+d x) \, dx}{a}-\frac {\int (e+f x)^2 \cos (c+d x) \, dx}{b}+\left (\frac {a}{b}-\frac {b}{a}\right ) \int \frac {(e+f x)^2 \cos (c+d x)}{a+b \sin (c+d x)} \, dx\\ &=-\frac {i (e+f x)^3}{3 a f}-\frac {i \left (a^2-b^2\right ) (e+f x)^3}{3 a b^2 f}-\frac {(e+f x)^2 \sin (c+d x)}{b d}-\frac {(2 i) \int \frac {e^{2 i (c+d x)} (e+f x)^2}{1-e^{2 i (c+d x)}} \, dx}{a}+\left (\frac {a}{b}-\frac {b}{a}\right ) \int \frac {e^{i (c+d x)} (e+f x)^2}{a-\sqrt {a^2-b^2}-i b e^{i (c+d x)}} \, dx+\left (\frac {a}{b}-\frac {b}{a}\right ) \int \frac {e^{i (c+d x)} (e+f x)^2}{a+\sqrt {a^2-b^2}-i b e^{i (c+d x)}} \, dx+\frac {(2 f) \int (e+f x) \sin (c+d x) \, dx}{b d}\\ &=-\frac {i (e+f x)^3}{3 a f}-\frac {i \left (a^2-b^2\right ) (e+f x)^3}{3 a b^2 f}-\frac {2 f (e+f x) \cos (c+d x)}{b d^2}+\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {(e+f x)^2 \log \left (1-e^{2 i (c+d x)}\right )}{a d}-\frac {(e+f x)^2 \sin (c+d x)}{b d}-\frac {(2 f) \int (e+f x) \log \left (1-e^{2 i (c+d x)}\right ) \, dx}{a d}+\frac {\left (2 \left (-\frac {a}{b}+\frac {b}{a}\right ) f\right ) \int (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{b d}+\frac {\left (2 \left (-\frac {a}{b}+\frac {b}{a}\right ) f\right ) \int (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{b d}+\frac {\left (2 f^2\right ) \int \cos (c+d x) \, dx}{b d^2}\\ &=-\frac {i (e+f x)^3}{3 a f}-\frac {i \left (a^2-b^2\right ) (e+f x)^3}{3 a b^2 f}-\frac {2 f (e+f x) \cos (c+d x)}{b d^2}+\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {(e+f x)^2 \log \left (1-e^{2 i (c+d x)}\right )}{a d}-\frac {2 i \left (a^2-b^2\right ) f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d^2}-\frac {2 i \left (a^2-b^2\right ) f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^2 d^2}-\frac {i f (e+f x) \text {Li}_2\left (e^{2 i (c+d x)}\right )}{a d^2}+\frac {2 f^2 \sin (c+d x)}{b d^3}-\frac {(e+f x)^2 \sin (c+d x)}{b d}+\frac {\left (i f^2\right ) \int \text {Li}_2\left (e^{2 i (c+d x)}\right ) \, dx}{a d^2}+\frac {\left (2 i \left (a^2-b^2\right ) f^2\right ) \int \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{a b^2 d^2}+\frac {\left (2 i \left (a^2-b^2\right ) f^2\right ) \int \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{a b^2 d^2}\\ &=-\frac {i (e+f x)^3}{3 a f}-\frac {i \left (a^2-b^2\right ) (e+f x)^3}{3 a b^2 f}-\frac {2 f (e+f x) \cos (c+d x)}{b d^2}+\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {(e+f x)^2 \log \left (1-e^{2 i (c+d x)}\right )}{a d}-\frac {2 i \left (a^2-b^2\right ) f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d^2}-\frac {2 i \left (a^2-b^2\right ) f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^2 d^2}-\frac {i f (e+f x) \text {Li}_2\left (e^{2 i (c+d x)}\right )}{a d^2}+\frac {2 f^2 \sin (c+d x)}{b d^3}-\frac {(e+f x)^2 \sin (c+d x)}{b d}+\frac {f^2 \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{2 i (c+d x)}\right )}{2 a d^3}+\frac {\left (2 \left (a^2-b^2\right ) f^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i b x}{a-\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{a b^2 d^3}+\frac {\left (2 \left (a^2-b^2\right ) f^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i b x}{a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{a b^2 d^3}\\ &=-\frac {i (e+f x)^3}{3 a f}-\frac {i \left (a^2-b^2\right ) (e+f x)^3}{3 a b^2 f}-\frac {2 f (e+f x) \cos (c+d x)}{b d^2}+\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {(e+f x)^2 \log \left (1-e^{2 i (c+d x)}\right )}{a d}-\frac {2 i \left (a^2-b^2\right ) f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d^2}-\frac {2 i \left (a^2-b^2\right ) f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^2 d^2}-\frac {i f (e+f x) \text {Li}_2\left (e^{2 i (c+d x)}\right )}{a d^2}+\frac {2 \left (a^2-b^2\right ) f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d^3}+\frac {2 \left (a^2-b^2\right ) f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^2 d^3}+\frac {f^2 \text {Li}_3\left (e^{2 i (c+d x)}\right )}{2 a d^3}+\frac {2 f^2 \sin (c+d x)}{b d^3}-\frac {(e+f x)^2 \sin (c+d x)}{b d}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1740\) vs. \(2(566)=1132\).
time = 7.99, size = 1740, normalized size = 3.07 \begin {gather*} -\frac {e^{-i c} f^2 \csc (c) \left (2 d^2 x^2 \left (2 d e^{2 i c} x+3 i \left (-1+e^{2 i c}\right ) \log \left (1-e^{2 i (c+d x)}\right )\right )+6 d \left (-1+e^{2 i c}\right ) x \text {Li}_2\left (e^{2 i (c+d x)}\right )+3 i \left (-1+e^{2 i c}\right ) \text {Li}_3\left (e^{2 i (c+d x)}\right )\right )}{12 a d^3}+\frac {\left (a^2-b^2\right ) \left (-12 i d^3 e^2 e^{2 i c} x-12 i d^3 e e^{2 i c} f x^2-4 i d^3 e^{2 i c} f^2 x^3-6 i d^2 e^2 \tan ^{-1}\left (\frac {2 a e^{i (c+d x)}}{b \left (-1+e^{2 i (c+d x)}\right )}\right )+6 i d^2 e^2 e^{2 i c} \tan ^{-1}\left (\frac {2 a e^{i (c+d x)}}{b \left (-1+e^{2 i (c+d x)}\right )}\right )-3 d^2 e^2 \log \left (4 a^2 e^{2 i (c+d x)}+b^2 \left (-1+e^{2 i (c+d x)}\right )^2\right )+3 d^2 e^2 e^{2 i c} \log \left (4 a^2 e^{2 i (c+d x)}+b^2 \left (-1+e^{2 i (c+d x)}\right )^2\right )-12 d^2 e f x \log \left (1+\frac {b e^{i (2 c+d x)}}{i a e^{i c}-\sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )+12 d^2 e e^{2 i c} f x \log \left (1+\frac {b e^{i (2 c+d x)}}{i a e^{i c}-\sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )-6 d^2 f^2 x^2 \log \left (1+\frac {b e^{i (2 c+d x)}}{i a e^{i c}-\sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )+6 d^2 e^{2 i c} f^2 x^2 \log \left (1+\frac {b e^{i (2 c+d x)}}{i a e^{i c}-\sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )-12 d^2 e f x \log \left (1+\frac {b e^{i (2 c+d x)}}{i a e^{i c}+\sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )+12 d^2 e e^{2 i c} f x \log \left (1+\frac {b e^{i (2 c+d x)}}{i a e^{i c}+\sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )-6 d^2 f^2 x^2 \log \left (1+\frac {b e^{i (2 c+d x)}}{i a e^{i c}+\sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )+6 d^2 e^{2 i c} f^2 x^2 \log \left (1+\frac {b e^{i (2 c+d x)}}{i a e^{i c}+\sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )-12 i d \left (-1+e^{2 i c}\right ) f (e+f x) \text {Li}_2\left (\frac {i b e^{i (2 c+d x)}}{a e^{i c}+i \sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )-12 i d \left (-1+e^{2 i c}\right ) f (e+f x) \text {Li}_2\left (-\frac {b e^{i (2 c+d x)}}{i a e^{i c}+\sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )-12 f^2 \text {Li}_3\left (\frac {i b e^{i (2 c+d x)}}{a e^{i c}+i \sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )+12 e^{2 i c} f^2 \text {Li}_3\left (\frac {i b e^{i (2 c+d x)}}{a e^{i c}+i \sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )-12 f^2 \text {Li}_3\left (-\frac {b e^{i (2 c+d x)}}{i a e^{i c}+\sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )+12 e^{2 i c} f^2 \text {Li}_3\left (-\frac {b e^{i (2 c+d x)}}{i a e^{i c}+\sqrt {\left (-a^2+b^2\right ) e^{2 i c}}}\right )\right )}{6 a b^2 d^3 \left (-1+e^{2 i c}\right )}+\frac {a x \left (3 e^2+3 e f x+f^2 x^2\right ) \cos (c) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )}{6 b^2}-\frac {\cos (d x) \left (2 d e f \cos (c)+2 d f^2 x \cos (c)+d^2 e^2 \sin (c)-2 f^2 \sin (c)+2 d^2 e f x \sin (c)+d^2 f^2 x^2 \sin (c)\right )}{b d^3}+\frac {e^2 \csc (c) (-d x \cos (c)+\log (\cos (d x) \sin (c)+\cos (c) \sin (d x)) \sin (c))}{a d \left (\cos ^2(c)+\sin ^2(c)\right )}-\frac {\left (d^2 e^2 \cos (c)-2 f^2 \cos (c)+2 d^2 e f x \cos (c)+d^2 f^2 x^2 \cos (c)-2 d e f \sin (c)-2 d f^2 x \sin (c)\right ) \sin (d x)}{b d^3}-\frac {e f \csc (c) \sec (c) \left (d^2 e^{i \tan ^{-1}(\tan (c))} x^2+\frac {\left (i d x \left (-\pi +2 \tan ^{-1}(\tan (c))\right )-\pi \log \left (1+e^{-2 i d x}\right )-2 \left (d x+\tan ^{-1}(\tan (c))\right ) \log \left (1-e^{2 i \left (d x+\tan ^{-1}(\tan (c))\right )}\right )+\pi \log (\cos (d x))+2 \tan ^{-1}(\tan (c)) \log \left (\sin \left (d x+\tan ^{-1}(\tan (c))\right )\right )+i \text {Li}_2\left (e^{2 i \left (d x+\tan ^{-1}(\tan (c))\right )}\right )\right ) \tan (c)}{\sqrt {1+\tan ^2(c)}}\right )}{a d^2 \sqrt {\sec ^2(c) \left (\cos ^2(c)+\sin ^2(c)\right )}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((e + f*x)^2*Cos[c + d*x]^2*Cot[c + d*x])/(a + b*Sin[c + d*x]),x]

[Out]

-1/12*(f^2*Csc[c]*(2*d^2*x^2*(2*d*E^((2*I)*c)*x + (3*I)*(-1 + E^((2*I)*c))*Log[1 - E^((2*I)*(c + d*x))]) + 6*d
*(-1 + E^((2*I)*c))*x*PolyLog[2, E^((2*I)*(c + d*x))] + (3*I)*(-1 + E^((2*I)*c))*PolyLog[3, E^((2*I)*(c + d*x)
)]))/(a*d^3*E^(I*c)) + ((a^2 - b^2)*((-12*I)*d^3*e^2*E^((2*I)*c)*x - (12*I)*d^3*e*E^((2*I)*c)*f*x^2 - (4*I)*d^
3*E^((2*I)*c)*f^2*x^3 - (6*I)*d^2*e^2*ArcTan[(2*a*E^(I*(c + d*x)))/(b*(-1 + E^((2*I)*(c + d*x))))] + (6*I)*d^2
*e^2*E^((2*I)*c)*ArcTan[(2*a*E^(I*(c + d*x)))/(b*(-1 + E^((2*I)*(c + d*x))))] - 3*d^2*e^2*Log[4*a^2*E^((2*I)*(
c + d*x)) + b^2*(-1 + E^((2*I)*(c + d*x)))^2] + 3*d^2*e^2*E^((2*I)*c)*Log[4*a^2*E^((2*I)*(c + d*x)) + b^2*(-1
+ E^((2*I)*(c + d*x)))^2] - 12*d^2*e*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*
I)*c)])] + 12*d^2*e*E^((2*I)*c)*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)
])] - 6*d^2*f^2*x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 6*d^2*E^((
2*I)*c)*f^2*x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 12*d^2*e*f*x*L
og[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 12*d^2*e*E^((2*I)*c)*f*x*Log[1
+ (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 6*d^2*f^2*x^2*Log[1 + (b*E^(I*(2*c +
d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 6*d^2*E^((2*I)*c)*f^2*x^2*Log[1 + (b*E^(I*(2*c + d*x
)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - (12*I)*d*(-1 + E^((2*I)*c))*f*(e + f*x)*PolyLog[2, (I*b*
E^(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - (12*I)*d*(-1 + E^((2*I)*c))*f*(e + f*x)*P
olyLog[2, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] - 12*f^2*PolyLog[3, (I*b*E^
(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 12*E^((2*I)*c)*f^2*PolyLog[3, (I*b*E^(I*(2*
c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 12*f^2*PolyLog[3, -((b*E^(I*(2*c + d*x)))/(I*a*E^
(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] + 12*E^((2*I)*c)*f^2*PolyLog[3, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c)
+ Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))]))/(6*a*b^2*d^3*(-1 + E^((2*I)*c))) + (a*x*(3*e^2 + 3*e*f*x + f^2*x^2)*Cos
[c]*Csc[c/2]*Sec[c/2])/(6*b^2) - (Cos[d*x]*(2*d*e*f*Cos[c] + 2*d*f^2*x*Cos[c] + d^2*e^2*Sin[c] - 2*f^2*Sin[c]
+ 2*d^2*e*f*x*Sin[c] + d^2*f^2*x^2*Sin[c]))/(b*d^3) + (e^2*Csc[c]*(-(d*x*Cos[c]) + Log[Cos[d*x]*Sin[c] + Cos[c
]*Sin[d*x]]*Sin[c]))/(a*d*(Cos[c]^2 + Sin[c]^2)) - ((d^2*e^2*Cos[c] - 2*f^2*Cos[c] + 2*d^2*e*f*x*Cos[c] + d^2*
f^2*x^2*Cos[c] - 2*d*e*f*Sin[c] - 2*d*f^2*x*Sin[c])*Sin[d*x])/(b*d^3) - (e*f*Csc[c]*Sec[c]*(d^2*E^(I*ArcTan[Ta
n[c]])*x^2 + ((I*d*x*(-Pi + 2*ArcTan[Tan[c]]) - Pi*Log[1 + E^((-2*I)*d*x)] - 2*(d*x + ArcTan[Tan[c]])*Log[1 -
E^((2*I)*(d*x + ArcTan[Tan[c]]))] + Pi*Log[Cos[d*x]] + 2*ArcTan[Tan[c]]*Log[Sin[d*x + ArcTan[Tan[c]]]] + I*Pol
yLog[2, E^((2*I)*(d*x + ArcTan[Tan[c]]))])*Tan[c])/Sqrt[1 + Tan[c]^2]))/(a*d^2*Sqrt[Sec[c]^2*(Cos[c]^2 + Sin[c
]^2)])

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Maple [F]
time = 0.55, size = 0, normalized size = 0.00 \[\int \frac {\left (f x +e \right )^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \cot \left (d x +c \right )}{a +b \sin \left (d x +c \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^2*cos(d*x+c)^2*cot(d*x+c)/(a+b*sin(d*x+c)),x)

[Out]

int((f*x+e)^2*cos(d*x+c)^2*cot(d*x+c)/(a+b*sin(d*x+c)),x)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*cos(d*x+c)^2*cot(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2275 vs. \(2 (521) = 1042\).
time = 0.68, size = 2275, normalized size = 4.02 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*cos(d*x+c)^2*cot(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(2*b^2*f^2*polylog(3, cos(d*x + c) + I*sin(d*x + c)) + 2*b^2*f^2*polylog(3, cos(d*x + c) - I*sin(d*x + c))
+ 2*b^2*f^2*polylog(3, -cos(d*x + c) + I*sin(d*x + c)) + 2*b^2*f^2*polylog(3, -cos(d*x + c) - I*sin(d*x + c))
+ 2*(a^2 - b^2)*f^2*polylog(3, -(I*a*cos(d*x + c) + a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt
(-(a^2 - b^2)/b^2))/b) + 2*(a^2 - b^2)*f^2*polylog(3, -(I*a*cos(d*x + c) + a*sin(d*x + c) - (b*cos(d*x + c) -
I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 2*(a^2 - b^2)*f^2*polylog(3, -(-I*a*cos(d*x + c) + a*sin(d*x +
c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 2*(a^2 - b^2)*f^2*polylog(3, -(-I*a*cos(
d*x + c) + a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) - 4*(a*b*d*f^2*x +
a*b*d*f*e)*cos(d*x + c) - 2*(I*(a^2 - b^2)*d*f^2*x + I*(a^2 - b^2)*d*f*e)*dilog((I*a*cos(d*x + c) - a*sin(d*x
+ c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 2*(I*(a^2 - b^2)*d*f^2*x + I*(
a^2 - b^2)*d*f*e)*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 -
b^2)/b^2) - b)/b + 1) - 2*(-I*(a^2 - b^2)*d*f^2*x - I*(a^2 - b^2)*d*f*e)*dilog((-I*a*cos(d*x + c) - a*sin(d*x
+ c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 2*(-I*(a^2 - b^2)*d*f^2*x - I*
(a^2 - b^2)*d*f*e)*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2
- b^2)/b^2) - b)/b + 1) - 2*(I*b^2*d*f^2*x + I*b^2*d*f*e)*dilog(cos(d*x + c) + I*sin(d*x + c)) - 2*(-I*b^2*d*f
^2*x - I*b^2*d*f*e)*dilog(cos(d*x + c) - I*sin(d*x + c)) - 2*(-I*b^2*d*f^2*x - I*b^2*d*f*e)*dilog(-cos(d*x + c
) + I*sin(d*x + c)) - 2*(I*b^2*d*f^2*x + I*b^2*d*f*e)*dilog(-cos(d*x + c) - I*sin(d*x + c)) + ((a^2 - b^2)*c^2
*f^2 - 2*(a^2 - b^2)*c*d*f*e + (a^2 - b^2)*d^2*e^2)*log(2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2
- b^2)/b^2) + 2*I*a) + ((a^2 - b^2)*c^2*f^2 - 2*(a^2 - b^2)*c*d*f*e + (a^2 - b^2)*d^2*e^2)*log(2*b*cos(d*x +
c) - 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) + ((a^2 - b^2)*c^2*f^2 - 2*(a^2 - b^2)*c*d*f*e +
(a^2 - b^2)*d^2*e^2)*log(-2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) + ((a^2
- b^2)*c^2*f^2 - 2*(a^2 - b^2)*c*d*f*e + (a^2 - b^2)*d^2*e^2)*log(-2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*
b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) + ((a^2 - b^2)*d^2*f^2*x^2 - (a^2 - b^2)*c^2*f^2 + 2*((a^2 - b^2)*d^2*f*x +
(a^2 - b^2)*c*d*f)*e)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2
- b^2)/b^2) - b)/b) + ((a^2 - b^2)*d^2*f^2*x^2 - (a^2 - b^2)*c^2*f^2 + 2*((a^2 - b^2)*d^2*f*x + (a^2 - b^2)*c
*d*f)*e)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2)
- b)/b) + ((a^2 - b^2)*d^2*f^2*x^2 - (a^2 - b^2)*c^2*f^2 + 2*((a^2 - b^2)*d^2*f*x + (a^2 - b^2)*c*d*f)*e)*log(
-(-I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) + ((
a^2 - b^2)*d^2*f^2*x^2 - (a^2 - b^2)*c^2*f^2 + 2*((a^2 - b^2)*d^2*f*x + (a^2 - b^2)*c*d*f)*e)*log(-(-I*a*cos(d
*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) + (b^2*d^2*f^2*x
^2 + 2*b^2*d^2*f*x*e + b^2*d^2*e^2)*log(cos(d*x + c) + I*sin(d*x + c) + 1) + (b^2*d^2*f^2*x^2 + 2*b^2*d^2*f*x*
e + b^2*d^2*e^2)*log(cos(d*x + c) - I*sin(d*x + c) + 1) + (b^2*c^2*f^2 - 2*b^2*c*d*f*e + b^2*d^2*e^2)*log(-1/2
*cos(d*x + c) + 1/2*I*sin(d*x + c) + 1/2) + (b^2*c^2*f^2 - 2*b^2*c*d*f*e + b^2*d^2*e^2)*log(-1/2*cos(d*x + c)
- 1/2*I*sin(d*x + c) + 1/2) + (b^2*d^2*f^2*x^2 - b^2*c^2*f^2 + 2*(b^2*d^2*f*x + b^2*c*d*f)*e)*log(-cos(d*x + c
) + I*sin(d*x + c) + 1) + (b^2*d^2*f^2*x^2 - b^2*c^2*f^2 + 2*(b^2*d^2*f*x + b^2*c*d*f)*e)*log(-cos(d*x + c) -
I*sin(d*x + c) + 1) - 2*(a*b*d^2*f^2*x^2 + 2*a*b*d^2*f*x*e + a*b*d^2*e^2 - 2*a*b*f^2)*sin(d*x + c))/(a*b^2*d^3
)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (e + f x\right )^{2} \cos ^{2}{\left (c + d x \right )} \cot {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**2*cos(d*x+c)**2*cot(d*x+c)/(a+b*sin(d*x+c)),x)

[Out]

Integral((e + f*x)**2*cos(c + d*x)**2*cot(c + d*x)/(a + b*sin(c + d*x)), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*cos(d*x+c)^2*cot(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^2*cot(c + d*x)*(e + f*x)^2)/(a + b*sin(c + d*x)),x)

[Out]

\text{Hanged}

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