\(\int \frac {(a+b x)^{5/2} (A+B x)}{x^{9/2}} \, dx\) [508]

Optimal result

Integrand size = 20, antiderivative size = 111 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{9/2}} \, dx=-\frac {2 b^2 B \sqrt {a+b x}}{\sqrt {x}}-\frac {2 b B (a+b x)^{3/2}}{3 x^{3/2}}-\frac {2 B (a+b x)^{5/2}}{5 x^{5/2}}-\frac {2 A (a+b x)^{7/2}}{7 a x^{7/2}}+2 b^{5/2} B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right ) \]

[Out]

-2/3*b*B*(b*x+a)^(3/2)/x^(3/2)-2/5*B*(b*x+a)^(5/2)/x^(5/2)-2/7*A*(b*x+a)^(7/2)/a/x^(7/2)+2*b^(5/2)*B*arctanh(b
^(1/2)*x^(1/2)/(b*x+a)^(1/2))-2*b^2*B*(b*x+a)^(1/2)/x^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {79, 49, 65, 223, 212} \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{9/2}} \, dx=-\frac {2 A (a+b x)^{7/2}}{7 a x^{7/2}}+2 b^{5/2} B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )-\frac {2 b^2 B \sqrt {a+b x}}{\sqrt {x}}-\frac {2 B (a+b x)^{5/2}}{5 x^{5/2}}-\frac {2 b B (a+b x)^{3/2}}{3 x^{3/2}} \]

[In]

Int[((a + b*x)^(5/2)*(A + B*x))/x^(9/2),x]

[Out]

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

Rule 49

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 65

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 79

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] || L
tQ[p, n]))))

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.95 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{9/2}} \, dx=-\frac {2 \sqrt {a+b x} \left (15 A b^3 x^3+3 a^3 (5 A+7 B x)+a^2 b x (45 A+77 B x)+a b^2 x^2 (45 A+161 B x)\right )}{105 a x^{7/2}}-2 b^{5/2} B \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right ) \]

[In]

Integrate[((a + b*x)^(5/2)*(A + B*x))/x^(9/2),x]

[Out]

(-2*Sqrt[a + b*x]*(15*A*b^3*x^3 + 3*a^3*(5*A + 7*B*x) + a^2*b*x*(45*A + 77*B*x) + a*b^2*x^2*(45*A + 161*B*x)))
/(105*a*x^(7/2)) - 2*b^(5/2)*B*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]]

Maple [A] (verified)

Time = 1.45 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.14

method	result	size
risch	\(-\frac {2 \sqrt {b x +a}\, \left (15 A \,b^{3} x^{3}+161 B a \,b^{2} x^{3}+45 a A \,b^{2} x^{2}+77 B \,a^{2} b \,x^{2}+45 a^{2} A b x +21 a^{3} B x +15 a^{3} A \right )}{105 x^{\frac {7}{2}} a}+\frac {b^{\frac {5}{2}} B \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{\sqrt {x}\, \sqrt {b x +a}}\)	\(126\)
default	\(-\frac {\sqrt {b x +a}\, \left (-105 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a \,b^{3} x^{4}+30 A \sqrt {x \left (b x +a \right )}\, b^{\frac {7}{2}} x^{3}+322 B \sqrt {x \left (b x +a \right )}\, b^{\frac {5}{2}} a \,x^{3}+90 A \sqrt {x \left (b x +a \right )}\, b^{\frac {5}{2}} a \,x^{2}+154 B \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a^{2} x^{2}+90 A \,a^{2} b^{\frac {3}{2}} x \sqrt {x \left (b x +a \right )}+42 B \,a^{3} x \sqrt {x \left (b x +a \right )}\, \sqrt {b}+30 A \,a^{3} \sqrt {x \left (b x +a \right )}\, \sqrt {b}\right )}{105 x^{\frac {7}{2}} a \sqrt {x \left (b x +a \right )}\, \sqrt {b}}\)	\(198\)

[In]

int((b*x+a)^(5/2)*(B*x+A)/x^(9/2),x,method=_RETURNVERBOSE)

[Out]

-2/105*(b*x+a)^(1/2)*(15*A*b^3*x^3+161*B*a*b^2*x^3+45*A*a*b^2*x^2+77*B*a^2*b*x^2+45*A*a^2*b*x+21*B*a^3*x+15*A*
a^3)/x^(7/2)/a+b^(5/2)*B*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))*(x*(b*x+a))^(1/2)/x^(1/2)/(b*x+a)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.08 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{9/2}} \, dx=\left [\frac {105 \, B a b^{\frac {5}{2}} x^{4} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (15 \, A a^{3} + {\left (161 \, B a b^{2} + 15 \, A b^{3}\right )} x^{3} + {\left (77 \, B a^{2} b + 45 \, A a b^{2}\right )} x^{2} + 3 \, {\left (7 \, B a^{3} + 15 \, A a^{2} b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{105 \, a x^{4}}, -\frac {2 \, {\left (105 \, B a \sqrt {-b} b^{2} x^{4} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (15 \, A a^{3} + {\left (161 \, B a b^{2} + 15 \, A b^{3}\right )} x^{3} + {\left (77 \, B a^{2} b + 45 \, A a b^{2}\right )} x^{2} + 3 \, {\left (7 \, B a^{3} + 15 \, A a^{2} b\right )} x\right )} \sqrt {b x + a} \sqrt {x}\right )}}{105 \, a x^{4}}\right ] \]

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(9/2),x, algorithm="fricas")

[Out]

[1/105*(105*B*a*b^(5/2)*x^4*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(15*A*a^3 + (161*B*a*b^2 + 15
*A*b^3)*x^3 + (77*B*a^2*b + 45*A*a*b^2)*x^2 + 3*(7*B*a^3 + 15*A*a^2*b)*x)*sqrt(b*x + a)*sqrt(x))/(a*x^4), -2/1
05*(105*B*a*sqrt(-b)*b^2*x^4*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (15*A*a^3 + (161*B*a*b^2 + 15*A*b^3)
*x^3 + (77*B*a^2*b + 45*A*a*b^2)*x^2 + 3*(7*B*a^3 + 15*A*a^2*b)*x)*sqrt(b*x + a)*sqrt(x))/(a*x^4)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 581 vs. \(2 (107) = 214\).

Time = 14.07 (sec) , antiderivative size = 581, normalized size of antiderivative = 5.23 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{9/2}} \, dx=- \frac {30 A a^{7} b^{\frac {9}{2}} \sqrt {\frac {a}{b x} + 1}}{105 a^{5} b^{4} x^{3} + 210 a^{4} b^{5} x^{4} + 105 a^{3} b^{6} x^{5}} - \frac {66 A a^{6} b^{\frac {11}{2}} x \sqrt {\frac {a}{b x} + 1}}{105 a^{5} b^{4} x^{3} + 210 a^{4} b^{5} x^{4} + 105 a^{3} b^{6} x^{5}} - \frac {34 A a^{5} b^{\frac {13}{2}} x^{2} \sqrt {\frac {a}{b x} + 1}}{105 a^{5} b^{4} x^{3} + 210 a^{4} b^{5} x^{4} + 105 a^{3} b^{6} x^{5}} - \frac {6 A a^{4} b^{\frac {15}{2}} x^{3} \sqrt {\frac {a}{b x} + 1}}{105 a^{5} b^{4} x^{3} + 210 a^{4} b^{5} x^{4} + 105 a^{3} b^{6} x^{5}} - \frac {24 A a^{3} b^{\frac {17}{2}} x^{4} \sqrt {\frac {a}{b x} + 1}}{105 a^{5} b^{4} x^{3} + 210 a^{4} b^{5} x^{4} + 105 a^{3} b^{6} x^{5}} - \frac {16 A a^{2} b^{\frac {19}{2}} x^{5} \sqrt {\frac {a}{b x} + 1}}{105 a^{5} b^{4} x^{3} + 210 a^{4} b^{5} x^{4} + 105 a^{3} b^{6} x^{5}} - \frac {4 A a b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{5 x^{2}} - \frac {14 A b^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}}{15 x} - \frac {2 A b^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}}{15 a} - \frac {2 B \sqrt {a} b^{2}}{\sqrt {x} \sqrt {1 + \frac {b x}{a}}} - \frac {2 B a^{2} \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{5 x^{2}} - \frac {22 B a b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{15 x} - \frac {16 B b^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}}{15} + 2 B b^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} - \frac {2 B b^{3} \sqrt {x}}{\sqrt {a} \sqrt {1 + \frac {b x}{a}}} \]

[In]

integrate((b*x+a)**(5/2)*(B*x+A)/x**(9/2),x)

[Out]

-30*A*a**7*b**(9/2)*sqrt(a/(b*x) + 1)/(105*a**5*b**4*x**3 + 210*a**4*b**5*x**4 + 105*a**3*b**6*x**5) - 66*A*a*
*6*b**(11/2)*x*sqrt(a/(b*x) + 1)/(105*a**5*b**4*x**3 + 210*a**4*b**5*x**4 + 105*a**3*b**6*x**5) - 34*A*a**5*b*
*(13/2)*x**2*sqrt(a/(b*x) + 1)/(105*a**5*b**4*x**3 + 210*a**4*b**5*x**4 + 105*a**3*b**6*x**5) - 6*A*a**4*b**(1
5/2)*x**3*sqrt(a/(b*x) + 1)/(105*a**5*b**4*x**3 + 210*a**4*b**5*x**4 + 105*a**3*b**6*x**5) - 24*A*a**3*b**(17/
2)*x**4*sqrt(a/(b*x) + 1)/(105*a**5*b**4*x**3 + 210*a**4*b**5*x**4 + 105*a**3*b**6*x**5) - 16*A*a**2*b**(19/2)
*x**5*sqrt(a/(b*x) + 1)/(105*a**5*b**4*x**3 + 210*a**4*b**5*x**4 + 105*a**3*b**6*x**5) - 4*A*a*b**(3/2)*sqrt(a
/(b*x) + 1)/(5*x**2) - 14*A*b**(5/2)*sqrt(a/(b*x) + 1)/(15*x) - 2*A*b**(7/2)*sqrt(a/(b*x) + 1)/(15*a) - 2*B*sq
rt(a)*b**2/(sqrt(x)*sqrt(1 + b*x/a)) - 2*B*a**2*sqrt(b)*sqrt(a/(b*x) + 1)/(5*x**2) - 22*B*a*b**(3/2)*sqrt(a/(b
*x) + 1)/(15*x) - 16*B*b**(5/2)*sqrt(a/(b*x) + 1)/15 + 2*B*b**(5/2)*asinh(sqrt(b)*sqrt(x)/sqrt(a)) - 2*B*b**3*
sqrt(x)/(sqrt(a)*sqrt(1 + b*x/a))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 258 vs. \(2 (81) = 162\).

Time = 0.21 (sec) , antiderivative size = 258, normalized size of antiderivative = 2.32 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{9/2}} \, dx=B b^{\frac {5}{2}} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - \frac {38 \, \sqrt {b x^{2} + a x} B b^{2}}{15 \, x} - \frac {2 \, \sqrt {b x^{2} + a x} A b^{3}}{7 \, a x} - \frac {7 \, \sqrt {b x^{2} + a x} B a b}{30 \, x^{2}} + \frac {\sqrt {b x^{2} + a x} A b^{2}}{7 \, x^{2}} + \frac {3 \, \sqrt {b x^{2} + a x} B a^{2}}{10 \, x^{3}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B b}{3 \, x^{3}} - \frac {3 \, \sqrt {b x^{2} + a x} A a b}{28 \, x^{3}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a}{2 \, x^{4}} - \frac {15 \, \sqrt {b x^{2} + a x} A a^{2}}{28 \, x^{4}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} B}{5 \, x^{5}} + \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} A a}{4 \, x^{5}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} A}{x^{6}} \]

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(9/2),x, algorithm="maxima")

[Out]

B*b^(5/2)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) - 38/15*sqrt(b*x^2 + a*x)*B*b^2/x - 2/7*sqrt(b*x^2 + a*
x)*A*b^3/(a*x) - 7/30*sqrt(b*x^2 + a*x)*B*a*b/x^2 + 1/7*sqrt(b*x^2 + a*x)*A*b^2/x^2 + 3/10*sqrt(b*x^2 + a*x)*B
*a^2/x^3 - 1/3*(b*x^2 + a*x)^(3/2)*B*b/x^3 - 3/28*sqrt(b*x^2 + a*x)*A*a*b/x^3 - 1/2*(b*x^2 + a*x)^(3/2)*B*a/x^
4 - 15/28*sqrt(b*x^2 + a*x)*A*a^2/x^4 - 1/5*(b*x^2 + a*x)^(5/2)*B/x^5 + 5/4*(b*x^2 + a*x)^(3/2)*A*a/x^5 - (b*x
^2 + a*x)^(5/2)*A/x^6

Giac [A] (verification not implemented)

none

Time = 75.55 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.23 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{9/2}} \, dx=-\frac {2 \, {\left (105 \, B b^{\frac {5}{2}} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right ) - \frac {{\left (105 \, B a^{3} b^{6} - {\left (350 \, B a^{2} b^{6} - {\left (406 \, B a b^{6} - \frac {{\left (161 \, B a^{3} b^{6} + 15 \, A a^{2} b^{7}\right )} {\left (b x + a\right )}}{a^{3}}\right )} {\left (b x + a\right )}\right )} {\left (b x + a\right )}\right )} \sqrt {b x + a}}{{\left ({\left (b x + a\right )} b - a b\right )}^{\frac {7}{2}}}\right )} b}{105 \, {\left | b \right |}} \]

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(9/2),x, algorithm="giac")

[Out]

-2/105*(105*B*b^(5/2)*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a*b))) - (105*B*a^3*b^6 - (350*B*a^2
*b^6 - (406*B*a*b^6 - (161*B*a^3*b^6 + 15*A*a^2*b^7)*(b*x + a)/a^3)*(b*x + a))*(b*x + a))*sqrt(b*x + a)/((b*x
+ a)*b - a*b)^(7/2))*b/abs(b)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^{9/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a+b\,x\right )}^{5/2}}{x^{9/2}} \,d x \]

[In]

int(((A + B*x)*(a + b*x)^(5/2))/x^(9/2),x)

[Out]

int(((A + B*x)*(a + b*x)^(5/2))/x^(9/2), x)