3.6.14 \(\int \frac {x^{3/2} (A+B x)}{(a+b x)^{5/2}} \, dx\)

Optimal. Leaf size=133 \[ \frac {(2 A b-5 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{7/2}}-\frac {\sqrt {x} \sqrt {a+b x} (2 A b-5 a B)}{a b^3}+\frac {2 x^{3/2} (2 A b-5 a B)}{3 a b^2 \sqrt {a+b x}}+\frac {2 x^{5/2} (A b-a B)}{3 a b (a+b x)^{3/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {78, 47, 50, 63, 217, 206} \begin {gather*} \frac {2 x^{3/2} (2 A b-5 a B)}{3 a b^2 \sqrt {a+b x}}-\frac {\sqrt {x} \sqrt {a+b x} (2 A b-5 a B)}{a b^3}+\frac {(2 A b-5 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{7/2}}+\frac {2 x^{5/2} (A b-a B)}{3 a b (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(3/2)*(A + B*x))/(a + b*x)^(5/2),x]

[Out]

(2*(A*b - a*B)*x^(5/2))/(3*a*b*(a + b*x)^(3/2)) + (2*(2*A*b - 5*a*B)*x^(3/2))/(3*a*b^2*Sqrt[a + b*x]) - ((2*A*
b - 5*a*B)*Sqrt[x]*Sqrt[a + b*x])/(a*b^3) + ((2*A*b - 5*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/b^(7/2)

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {x^{3/2} (A+B x)}{(a+b x)^{5/2}} \, dx &=\frac {2 (A b-a B) x^{5/2}}{3 a b (a+b x)^{3/2}}-\frac {\left (2 \left (A b-\frac {5 a B}{2}\right )\right ) \int \frac {x^{3/2}}{(a+b x)^{3/2}} \, dx}{3 a b}\\ &=\frac {2 (A b-a B) x^{5/2}}{3 a b (a+b x)^{3/2}}+\frac {2 (2 A b-5 a B) x^{3/2}}{3 a b^2 \sqrt {a+b x}}-\frac {(2 A b-5 a B) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{a b^2}\\ &=\frac {2 (A b-a B) x^{5/2}}{3 a b (a+b x)^{3/2}}+\frac {2 (2 A b-5 a B) x^{3/2}}{3 a b^2 \sqrt {a+b x}}-\frac {(2 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{a b^3}+\frac {(2 A b-5 a B) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{2 b^3}\\ &=\frac {2 (A b-a B) x^{5/2}}{3 a b (a+b x)^{3/2}}+\frac {2 (2 A b-5 a B) x^{3/2}}{3 a b^2 \sqrt {a+b x}}-\frac {(2 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{a b^3}+\frac {(2 A b-5 a B) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{b^3}\\ &=\frac {2 (A b-a B) x^{5/2}}{3 a b (a+b x)^{3/2}}+\frac {2 (2 A b-5 a B) x^{3/2}}{3 a b^2 \sqrt {a+b x}}-\frac {(2 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{a b^3}+\frac {(2 A b-5 a B) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )}{b^3}\\ &=\frac {2 (A b-a B) x^{5/2}}{3 a b (a+b x)^{3/2}}+\frac {2 (2 A b-5 a B) x^{3/2}}{3 a b^2 \sqrt {a+b x}}-\frac {(2 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{a b^3}+\frac {(2 A b-5 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.08, size = 80, normalized size = 0.60 \begin {gather*} \frac {2 x^{5/2} \left ((a+b x) \sqrt {\frac {b x}{a}+1} (5 a B-2 A b) \, _2F_1\left (\frac {3}{2},\frac {5}{2};\frac {7}{2};-\frac {b x}{a}\right )+5 a (A b-a B)\right )}{15 a^2 b (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(3/2)*(A + B*x))/(a + b*x)^(5/2),x]

[Out]

(2*x^(5/2)*(5*a*(A*b - a*B) + (-2*A*b + 5*a*B)*(a + b*x)*Sqrt[1 + (b*x)/a]*Hypergeometric2F1[3/2, 5/2, 7/2, -(
(b*x)/a)]))/(15*a^2*b*(a + b*x)^(3/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.25, size = 109, normalized size = 0.82 \begin {gather*} \frac {15 a^2 B \sqrt {x}-6 a A b \sqrt {x}+20 a b B x^{3/2}-8 A b^2 x^{3/2}+3 b^2 B x^{5/2}}{3 b^3 (a+b x)^{3/2}}+\frac {(5 a B-2 A b) \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right )}{b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^(3/2)*(A + B*x))/(a + b*x)^(5/2),x]

[Out]

(-6*a*A*b*Sqrt[x] + 15*a^2*B*Sqrt[x] - 8*A*b^2*x^(3/2) + 20*a*b*B*x^(3/2) + 3*b^2*B*x^(5/2))/(3*b^3*(a + b*x)^
(3/2)) + ((-2*A*b + 5*a*B)*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]])/b^(7/2)

________________________________________________________________________________________

fricas [A]  time = 1.31, size = 314, normalized size = 2.36 \begin {gather*} \left [-\frac {3 \, {\left (5 \, B a^{3} - 2 \, A a^{2} b + {\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} x^{2} + 2 \, {\left (5 \, B a^{2} b - 2 \, A a b^{2}\right )} x\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (3 \, B b^{3} x^{2} + 15 \, B a^{2} b - 6 \, A a b^{2} + 4 \, {\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{6 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, \frac {3 \, {\left (5 \, B a^{3} - 2 \, A a^{2} b + {\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} x^{2} + 2 \, {\left (5 \, B a^{2} b - 2 \, A a b^{2}\right )} x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (3 \, B b^{3} x^{2} + 15 \, B a^{2} b - 6 \, A a b^{2} + 4 \, {\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{3 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x+A)/(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

[-1/6*(3*(5*B*a^3 - 2*A*a^2*b + (5*B*a*b^2 - 2*A*b^3)*x^2 + 2*(5*B*a^2*b - 2*A*a*b^2)*x)*sqrt(b)*log(2*b*x + 2
*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(3*B*b^3*x^2 + 15*B*a^2*b - 6*A*a*b^2 + 4*(5*B*a*b^2 - 2*A*b^3)*x)*sqr
t(b*x + a)*sqrt(x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4), 1/3*(3*(5*B*a^3 - 2*A*a^2*b + (5*B*a*b^2 - 2*A*b^3)*x^2 +
2*(5*B*a^2*b - 2*A*a*b^2)*x)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (3*B*b^3*x^2 + 15*B*a^2*b
- 6*A*a*b^2 + 4*(5*B*a*b^2 - 2*A*b^3)*x)*sqrt(b*x + a)*sqrt(x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4)]

________________________________________________________________________________________

giac [B]  time = 108.51, size = 309, normalized size = 2.32 \begin {gather*} \frac {\sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} B {\left | b \right |}}{b^{5}} + \frac {{\left (5 \, B a \sqrt {b} {\left | b \right |} - 2 \, A b^{\frac {3}{2}} {\left | b \right |}\right )} \log \left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{2 \, b^{5}} + \frac {4 \, {\left (9 \, B a^{2} {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{4} \sqrt {b} {\left | b \right |} + 12 \, B a^{3} {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {3}{2}} {\left | b \right |} - 6 \, A a {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{4} b^{\frac {3}{2}} {\left | b \right |} + 7 \, B a^{4} b^{\frac {5}{2}} {\left | b \right |} - 6 \, A a^{2} {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {5}{2}} {\left | b \right |} - 4 \, A a^{3} b^{\frac {7}{2}} {\left | b \right |}\right )}}{3 \, {\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )}^{3} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x+A)/(b*x+a)^(5/2),x, algorithm="giac")

[Out]

sqrt((b*x + a)*b - a*b)*sqrt(b*x + a)*B*abs(b)/b^5 + 1/2*(5*B*a*sqrt(b)*abs(b) - 2*A*b^(3/2)*abs(b))*log((sqrt
(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2)/b^5 + 4/3*(9*B*a^2*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b -
a*b))^4*sqrt(b)*abs(b) + 12*B*a^3*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(3/2)*abs(b) - 6*A*a*
(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^4*b^(3/2)*abs(b) + 7*B*a^4*b^(5/2)*abs(b) - 6*A*a^2*(sqrt(b*
x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(5/2)*abs(b) - 4*A*a^3*b^(7/2)*abs(b))/(((sqrt(b*x + a)*sqrt(b)
- sqrt((b*x + a)*b - a*b))^2 + a*b)^3*b^4)

________________________________________________________________________________________

maple [B]  time = 0.02, size = 315, normalized size = 2.37 \begin {gather*} \frac {\left (6 A \,b^{3} x^{2} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-15 B a \,b^{2} x^{2} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )+12 A a \,b^{2} x \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-30 B \,a^{2} b x \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )+6 \sqrt {\left (b x +a \right ) x}\, B \,b^{\frac {5}{2}} x^{2}+6 A \,a^{2} b \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-15 B \,a^{3} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-16 \sqrt {\left (b x +a \right ) x}\, A \,b^{\frac {5}{2}} x +40 \sqrt {\left (b x +a \right ) x}\, B a \,b^{\frac {3}{2}} x -12 \sqrt {\left (b x +a \right ) x}\, A a \,b^{\frac {3}{2}}+30 \sqrt {\left (b x +a \right ) x}\, B \,a^{2} \sqrt {b}\right ) \sqrt {x}}{6 \sqrt {\left (b x +a \right ) x}\, \left (b x +a \right )^{\frac {3}{2}} b^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(B*x+A)/(b*x+a)^(5/2),x)

[Out]

1/6*(6*A*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))*x^2*b^3-15*B*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2
)*b^(1/2))/b^(1/2))*x^2*a*b^2+6*((b*x+a)*x)^(1/2)*B*b^(5/2)*x^2+12*A*a*b^2*x*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/
2)*b^(1/2))/b^(1/2))-16*((b*x+a)*x)^(1/2)*A*b^(5/2)*x-30*B*a^2*b*x*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2)
)/b^(1/2))+40*((b*x+a)*x)^(1/2)*B*a*b^(3/2)*x+6*A*a^2*b*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))-
12*((b*x+a)*x)^(1/2)*A*a*b^(3/2)-15*B*a^3*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))+30*((b*x+a)*x)
^(1/2)*B*a^2*b^(1/2))*x^(1/2)/((b*x+a)*x)^(1/2)/b^(7/2)/(b*x+a)^(3/2)

________________________________________________________________________________________

maxima [B]  time = 1.07, size = 334, normalized size = 2.51 \begin {gather*} \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a}{3 \, {\left (b^{5} x^{3} + 3 \, a b^{4} x^{2} + 3 \, a^{2} b^{3} x + a^{3} b^{2}\right )}} - \frac {\sqrt {b x^{2} + a x} B a^{2}}{3 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A}{3 \, {\left (b^{4} x^{3} + 3 \, a b^{3} x^{2} + 3 \, a^{2} b^{2} x + a^{3} b\right )}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B}{b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}} + \frac {\sqrt {b x^{2} + a x} A a}{3 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} + \frac {16 \, \sqrt {b x^{2} + a x} B a}{3 \, {\left (b^{4} x + a b^{3}\right )}} - \frac {7 \, \sqrt {b x^{2} + a x} A}{3 \, {\left (b^{3} x + a b^{2}\right )}} - \frac {5 \, B a \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{2 \, b^{\frac {7}{2}}} + \frac {A \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x+A)/(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

1/3*(b*x^2 + a*x)^(3/2)*B*a/(b^5*x^3 + 3*a*b^4*x^2 + 3*a^2*b^3*x + a^3*b^2) - 1/3*sqrt(b*x^2 + a*x)*B*a^2/(b^5
*x^2 + 2*a*b^4*x + a^2*b^3) - 1/3*(b*x^2 + a*x)^(3/2)*A/(b^4*x^3 + 3*a*b^3*x^2 + 3*a^2*b^2*x + a^3*b) + (b*x^2
+ a*x)^(3/2)*B/(b^4*x^2 + 2*a*b^3*x + a^2*b^2) + 1/3*sqrt(b*x^2 + a*x)*A*a/(b^4*x^2 + 2*a*b^3*x + a^2*b^2) +
16/3*sqrt(b*x^2 + a*x)*B*a/(b^4*x + a*b^3) - 7/3*sqrt(b*x^2 + a*x)*A/(b^3*x + a*b^2) - 5/2*B*a*log(2*x + a/b +
2*sqrt(b*x^2 + a*x)/sqrt(b))/b^(7/2) + A*log(2*x + a/b + 2*sqrt(b*x^2 + a*x)/sqrt(b))/b^(5/2)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{3/2}\,\left (A+B\,x\right )}{{\left (a+b\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(3/2)*(A + B*x))/(a + b*x)^(5/2),x)

[Out]

int((x^(3/2)*(A + B*x))/(a + b*x)^(5/2), x)

________________________________________________________________________________________

sympy [F(-1)]  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(x**(3/2)*(B*x+A)/(b*x+a)**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________