Integrand size = 25, antiderivative size = 237 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\left (3 a^2-30 a b+35 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{8 a^{9/2} f}-\frac {(5 a-7 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {(23 a-35 b) b \sec (e+f x)}{24 a^3 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {5 (11 a-21 b) b \sec (e+f x)}{24 a^4 f \sqrt {a-b+b \sec ^2(e+f x)}} \]
-1/8*(3*a^2-30*a*b+35*b^2)*arctanh(sec(f*x+e)*a^(1/2)/(a-b+b*sec(f*x+e)^2) ^(1/2))/a^(9/2)/f-1/8*(5*a-7*b)*cot(f*x+e)*csc(f*x+e)/a^2/f/(a-b+b*sec(f*x +e)^2)^(3/2)-1/4*cot(f*x+e)^3*csc(f*x+e)/a/f/(a-b+b*sec(f*x+e)^2)^(3/2)-1/ 24*(23*a-35*b)*b*sec(f*x+e)/a^3/f/(a-b+b*sec(f*x+e)^2)^(3/2)-5/24*(11*a-21 *b)*b*sec(f*x+e)/a^4/f/(a-b+b*sec(f*x+e)^2)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(1132\) vs. \(2(237)=474\).
Time = 8.49 (sec) , antiderivative size = 1132, normalized size of antiderivative = 4.78 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \]
(Sqrt[(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f* x)])]*((4*b^2*Cos[e + f*x])/(3*a^3*(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*( e + f*x)])^2) - (2*(2*a*b*Cos[e + f*x] - 3*b^2*Cos[e + f*x]))/(a^4*(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])) + ((-3*a*Cos[e + f*x] + 11*b*C os[e + f*x])*Csc[e + f*x]^2)/(8*a^4) - (Cot[e + f*x]*Csc[e + f*x]^3)/(4*a^ 3)))/f + ((3*a^2 - 30*a*b + 35*b^2)*(((1 + Cos[e + f*x])*Sqrt[(1 + Cos[2*( e + f*x)])/(1 + Cos[e + f*x])^2]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/( 1 + Cos[2*(e + f*x)])]*(4*Sqrt[a]*ArcTanh[(-(Sqrt[a]*(-1 + Tan[(e + f*x)/2 ]^2)) + Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2])/(2*S qrt[b])] - Sqrt[b]*(2*ArcTanh[Tan[(e + f*x)/2]^2 - Sqrt[4*b*Tan[(e + f*x)/ 2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]/Sqrt[a]] + Log[a - 2*b - a*Tan[(e + f*x)/2]^2 + Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2] ^2)^2]]))*(-1 + Tan[(e + f*x)/2]^2)*(1 + Tan[(e + f*x)/2]^2)*Sqrt[(4*b*Tan [(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2)/(1 + Tan[(e + f*x)/2]^2)^ 2])/(4*Sqrt[a]*Sqrt[b]*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]*Sqrt[(-1 + T an[(e + f*x)/2]^2)^2]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/ 2]^2)^2]) - ((1 + Cos[e + f*x])*Sqrt[(1 + Cos[2*(e + f*x)])/(1 + Cos[e + f *x])^2]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*(4 *Sqrt[a]*ArcTanh[(-(Sqrt[a]*(-1 + Tan[(e + f*x)/2]^2)) + Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2])/(2*Sqrt[b])] + Sqrt[b]*(2*...
Time = 0.50 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.10, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 4147, 25, 372, 402, 25, 402, 25, 27, 402, 27, 291, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x)^5 \left (a+b \tan (e+f x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int -\frac {\sec ^4(e+f x)}{\left (1-\sec ^2(e+f x)\right )^3 \left (b \sec ^2(e+f x)+a-b\right )^{5/2}}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {\sec ^4(e+f x)}{\left (1-\sec ^2(e+f x)\right )^3 \left (b \sec ^2(e+f x)+a-b\right )^{5/2}}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\int \frac {2 (2 a-3 b) \sec ^2(e+f x)+a-b}{\left (1-\sec ^2(e+f x)\right )^2 \left (b \sec ^2(e+f x)+a-b\right )^{5/2}}d\sec (e+f x)}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\int -\frac {(3 a-7 b) (a-b)-4 (5 a-7 b) b \sec ^2(e+f x)}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{5/2}}d\sec (e+f x)}{2 a}+\frac {(5 a-7 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {(5 a-7 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\int \frac {(3 a-7 b) (a-b)-4 (5 a-7 b) b \sec ^2(e+f x)}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{5/2}}d\sec (e+f x)}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {(5 a-7 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {b (23 a-35 b) \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\int -\frac {(a-b) \left ((9 a-35 b) (a-b)-2 (23 a-35 b) b \sec ^2(e+f x)\right )}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}}d\sec (e+f x)}{3 a (a-b)}}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {(5 a-7 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {\int \frac {(a-b) \left ((9 a-35 b) (a-b)-2 (23 a-35 b) b \sec ^2(e+f x)\right )}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}}d\sec (e+f x)}{3 a (a-b)}+\frac {b (23 a-35 b) \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {(5 a-7 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {\int \frac {(9 a-35 b) (a-b)-2 (23 a-35 b) b \sec ^2(e+f x)}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}}d\sec (e+f x)}{3 a}+\frac {b (23 a-35 b) \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {(5 a-7 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {\frac {5 b (11 a-21 b) \sec (e+f x)}{a \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\int -\frac {3 (a-b) \left (3 a^2-30 b a+35 b^2\right )}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{a (a-b)}}{3 a}+\frac {b (23 a-35 b) \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {(5 a-7 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {\frac {3 \left (3 a^2-30 a b+35 b^2\right ) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{a}+\frac {5 b (11 a-21 b) \sec (e+f x)}{a \sqrt {a+b \sec ^2(e+f x)-b}}}{3 a}+\frac {b (23 a-35 b) \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\frac {(5 a-7 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {\frac {3 \left (3 a^2-30 a b+35 b^2\right ) \int \frac {1}{1-\frac {a \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}}{a}+\frac {5 b (11 a-21 b) \sec (e+f x)}{a \sqrt {a+b \sec ^2(e+f x)-b}}}{3 a}+\frac {b (23 a-35 b) \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {(5 a-7 b) \sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {\frac {3 \left (3 a^2-30 a b+35 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{a^{3/2}}+\frac {5 b (11 a-21 b) \sec (e+f x)}{a \sqrt {a+b \sec ^2(e+f x)-b}}}{3 a}+\frac {b (23 a-35 b) \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{2 a}}{4 a}-\frac {\sec (e+f x)}{4 a \left (1-\sec ^2(e+f x)\right )^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{f}\) |
(-1/4*Sec[e + f*x]/(a*(1 - Sec[e + f*x]^2)^2*(a - b + b*Sec[e + f*x]^2)^(3 /2)) + (((5*a - 7*b)*Sec[e + f*x])/(2*a*(1 - Sec[e + f*x]^2)*(a - b + b*Se c[e + f*x]^2)^(3/2)) - (((23*a - 35*b)*b*Sec[e + f*x])/(3*a*(a - b + b*Sec [e + f*x]^2)^(3/2)) + ((3*(3*a^2 - 30*a*b + 35*b^2)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/a^(3/2) + (5*(11*a - 21*b)*b*Sec [e + f*x])/(a*Sqrt[a - b + b*Sec[e + f*x]^2]))/(3*a))/(2*a))/(4*a))/f
3.2.45.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 )^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 )) Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a , b, c, d, e, m, 2, p, q, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[1/(f*ff^ m) Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m + 1 )), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[( m - 1)/2]
Leaf count of result is larger than twice the leaf count of optimal. \(18431\) vs. \(2(213)=426\).
Time = 4.46 (sec) , antiderivative size = 18432, normalized size of antiderivative = 77.77
Leaf count of result is larger than twice the leaf count of optimal. 505 vs. \(2 (213) = 426\).
Time = 0.51 (sec) , antiderivative size = 1037, normalized size of antiderivative = 4.38 \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \]
[1/48*(3*((3*a^4 - 36*a^3*b + 98*a^2*b^2 - 100*a*b^3 + 35*b^4)*cos(f*x + e )^8 - 2*(3*a^4 - 39*a^3*b + 131*a^2*b^2 - 165*a*b^3 + 70*b^4)*cos(f*x + e) ^6 + (3*a^4 - 48*a^3*b + 233*a^2*b^2 - 390*a*b^3 + 210*b^4)*cos(f*x + e)^4 + 3*a^2*b^2 - 30*a*b^3 + 35*b^4 + 2*(3*a^3*b - 36*a^2*b^2 + 95*a*b^3 - 70 *b^4)*cos(f*x + e)^2)*sqrt(a)*log(-2*((a - b)*cos(f*x + e)^2 - 2*sqrt(a)*s qrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + b)/(co s(f*x + e)^2 - 1)) + 2*(3*(3*a^4 - 33*a^3*b + 65*a^2*b^2 - 35*a*b^3)*cos(f *x + e)^7 - (15*a^4 - 177*a^3*b + 445*a^2*b^2 - 315*a*b^3)*cos(f*x + e)^5 - (78*a^3*b - 305*a^2*b^2 + 315*a*b^3)*cos(f*x + e)^3 - 5*(11*a^2*b^2 - 21 *a*b^3)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/( (a^7 - 2*a^6*b + a^5*b^2)*f*cos(f*x + e)^8 + a^5*b^2*f - 2*(a^7 - 3*a^6*b + 2*a^5*b^2)*f*cos(f*x + e)^6 + (a^7 - 6*a^6*b + 6*a^5*b^2)*f*cos(f*x + e) ^4 + 2*(a^6*b - 2*a^5*b^2)*f*cos(f*x + e)^2), 1/24*(3*((3*a^4 - 36*a^3*b + 98*a^2*b^2 - 100*a*b^3 + 35*b^4)*cos(f*x + e)^8 - 2*(3*a^4 - 39*a^3*b + 1 31*a^2*b^2 - 165*a*b^3 + 70*b^4)*cos(f*x + e)^6 + (3*a^4 - 48*a^3*b + 233* a^2*b^2 - 390*a*b^3 + 210*b^4)*cos(f*x + e)^4 + 3*a^2*b^2 - 30*a*b^3 + 35* b^4 + 2*(3*a^3*b - 36*a^2*b^2 + 95*a*b^3 - 70*b^4)*cos(f*x + e)^2)*sqrt(-a )*arctan(sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f* x + e)/a) + (3*(3*a^4 - 33*a^3*b + 65*a^2*b^2 - 35*a*b^3)*cos(f*x + e)^7 - (15*a^4 - 177*a^3*b + 445*a^2*b^2 - 315*a*b^3)*cos(f*x + e)^5 - (78*a^...
\[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\csc ^{5}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\csc \left (f x + e\right )^{5}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Hanged} \]