3.7.93 \(\int \frac {x^{7/2} (A+B x)}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=173 \[ -\frac {35 \sqrt {a} (A b-3 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 b^{11/2}}+\frac {35 \sqrt {x} (A b-3 a B)}{8 b^5}-\frac {35 x^{3/2} (A b-3 a B)}{24 a b^4}+\frac {7 x^{5/2} (A b-3 a B)}{8 a b^3 (a+b x)}+\frac {x^{7/2} (A b-3 a B)}{4 a b^2 (a+b x)^2}+\frac {x^{9/2} (A b-a B)}{3 a b (a+b x)^3} \]

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Rubi [A]  time = 0.08, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {27, 78, 47, 50, 63, 205} \begin {gather*} \frac {x^{7/2} (A b-3 a B)}{4 a b^2 (a+b x)^2}+\frac {7 x^{5/2} (A b-3 a B)}{8 a b^3 (a+b x)}-\frac {35 x^{3/2} (A b-3 a B)}{24 a b^4}+\frac {35 \sqrt {x} (A b-3 a B)}{8 b^5}-\frac {35 \sqrt {a} (A b-3 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 b^{11/2}}+\frac {x^{9/2} (A b-a B)}{3 a b (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(7/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(35*(A*b - 3*a*B)*Sqrt[x])/(8*b^5) - (35*(A*b - 3*a*B)*x^(3/2))/(24*a*b^4) + ((A*b - a*B)*x^(9/2))/(3*a*b*(a +
b*x)^3) + ((A*b - 3*a*B)*x^(7/2))/(4*a*b^2*(a + b*x)^2) + (7*(A*b - 3*a*B)*x^(5/2))/(8*a*b^3*(a + b*x)) - (35
*Sqrt[a]*(A*b - 3*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(8*b^(11/2))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin {align*} \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {x^{7/2} (A+B x)}{(a+b x)^4} \, dx\\ &=\frac {(A b-a B) x^{9/2}}{3 a b (a+b x)^3}-\frac {\left (\frac {3 A b}{2}-\frac {9 a B}{2}\right ) \int \frac {x^{7/2}}{(a+b x)^3} \, dx}{3 a b}\\ &=\frac {(A b-a B) x^{9/2}}{3 a b (a+b x)^3}+\frac {(A b-3 a B) x^{7/2}}{4 a b^2 (a+b x)^2}-\frac {(7 (A b-3 a B)) \int \frac {x^{5/2}}{(a+b x)^2} \, dx}{8 a b^2}\\ &=\frac {(A b-a B) x^{9/2}}{3 a b (a+b x)^3}+\frac {(A b-3 a B) x^{7/2}}{4 a b^2 (a+b x)^2}+\frac {7 (A b-3 a B) x^{5/2}}{8 a b^3 (a+b x)}-\frac {(35 (A b-3 a B)) \int \frac {x^{3/2}}{a+b x} \, dx}{16 a b^3}\\ &=-\frac {35 (A b-3 a B) x^{3/2}}{24 a b^4}+\frac {(A b-a B) x^{9/2}}{3 a b (a+b x)^3}+\frac {(A b-3 a B) x^{7/2}}{4 a b^2 (a+b x)^2}+\frac {7 (A b-3 a B) x^{5/2}}{8 a b^3 (a+b x)}+\frac {(35 (A b-3 a B)) \int \frac {\sqrt {x}}{a+b x} \, dx}{16 b^4}\\ &=\frac {35 (A b-3 a B) \sqrt {x}}{8 b^5}-\frac {35 (A b-3 a B) x^{3/2}}{24 a b^4}+\frac {(A b-a B) x^{9/2}}{3 a b (a+b x)^3}+\frac {(A b-3 a B) x^{7/2}}{4 a b^2 (a+b x)^2}+\frac {7 (A b-3 a B) x^{5/2}}{8 a b^3 (a+b x)}-\frac {(35 a (A b-3 a B)) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{16 b^5}\\ &=\frac {35 (A b-3 a B) \sqrt {x}}{8 b^5}-\frac {35 (A b-3 a B) x^{3/2}}{24 a b^4}+\frac {(A b-a B) x^{9/2}}{3 a b (a+b x)^3}+\frac {(A b-3 a B) x^{7/2}}{4 a b^2 (a+b x)^2}+\frac {7 (A b-3 a B) x^{5/2}}{8 a b^3 (a+b x)}-\frac {(35 a (A b-3 a B)) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{8 b^5}\\ &=\frac {35 (A b-3 a B) \sqrt {x}}{8 b^5}-\frac {35 (A b-3 a B) x^{3/2}}{24 a b^4}+\frac {(A b-a B) x^{9/2}}{3 a b (a+b x)^3}+\frac {(A b-3 a B) x^{7/2}}{4 a b^2 (a+b x)^2}+\frac {7 (A b-3 a B) x^{5/2}}{8 a b^3 (a+b x)}-\frac {35 \sqrt {a} (A b-3 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 b^{11/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 61, normalized size = 0.35 \begin {gather*} \frac {x^{9/2} \left (\frac {9 a^3 (A b-a B)}{(a+b x)^3}+(9 a B-3 A b) \, _2F_1\left (3,\frac {9}{2};\frac {11}{2};-\frac {b x}{a}\right )\right )}{27 a^4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(7/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(x^(9/2)*((9*a^3*(A*b - a*B))/(a + b*x)^3 + (-3*A*b + 9*a*B)*Hypergeometric2F1[3, 9/2, 11/2, -((b*x)/a)]))/(27
*a^4*b)

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IntegrateAlgebraic [A]  time = 0.26, size = 146, normalized size = 0.84 \begin {gather*} \frac {35 \left (3 a^{3/2} B-\sqrt {a} A b\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 b^{11/2}}+\frac {\sqrt {x} \left (-315 a^4 B+105 a^3 A b-840 a^3 b B x+280 a^2 A b^2 x-693 a^2 b^2 B x^2+231 a A b^3 x^2-144 a b^3 B x^3+48 A b^4 x^3+16 b^4 B x^4\right )}{24 b^5 (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^(7/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(Sqrt[x]*(105*a^3*A*b - 315*a^4*B + 280*a^2*A*b^2*x - 840*a^3*b*B*x + 231*a*A*b^3*x^2 - 693*a^2*b^2*B*x^2 + 48
*A*b^4*x^3 - 144*a*b^3*B*x^3 + 16*b^4*B*x^4))/(24*b^5*(a + b*x)^3) + (35*(-(Sqrt[a]*A*b) + 3*a^(3/2)*B)*ArcTan
[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(8*b^(11/2))

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fricas [A]  time = 0.44, size = 467, normalized size = 2.70 \begin {gather*} \left [-\frac {105 \, {\left (3 \, B a^{4} - A a^{3} b + {\left (3 \, B a b^{3} - A b^{4}\right )} x^{3} + 3 \, {\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} + 3 \, {\left (3 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (16 \, B b^{4} x^{4} - 315 \, B a^{4} + 105 \, A a^{3} b - 48 \, {\left (3 \, B a b^{3} - A b^{4}\right )} x^{3} - 231 \, {\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} - 280 \, {\left (3 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{48 \, {\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}}, \frac {105 \, {\left (3 \, B a^{4} - A a^{3} b + {\left (3 \, B a b^{3} - A b^{4}\right )} x^{3} + 3 \, {\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} + 3 \, {\left (3 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (16 \, B b^{4} x^{4} - 315 \, B a^{4} + 105 \, A a^{3} b - 48 \, {\left (3 \, B a b^{3} - A b^{4}\right )} x^{3} - 231 \, {\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} - 280 \, {\left (3 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{24 \, {\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")

[Out]

[-1/48*(105*(3*B*a^4 - A*a^3*b + (3*B*a*b^3 - A*b^4)*x^3 + 3*(3*B*a^2*b^2 - A*a*b^3)*x^2 + 3*(3*B*a^3*b - A*a^
2*b^2)*x)*sqrt(-a/b)*log((b*x - 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) - 2*(16*B*b^4*x^4 - 315*B*a^4 + 105*A*a
^3*b - 48*(3*B*a*b^3 - A*b^4)*x^3 - 231*(3*B*a^2*b^2 - A*a*b^3)*x^2 - 280*(3*B*a^3*b - A*a^2*b^2)*x)*sqrt(x))/
(b^8*x^3 + 3*a*b^7*x^2 + 3*a^2*b^6*x + a^3*b^5), 1/24*(105*(3*B*a^4 - A*a^3*b + (3*B*a*b^3 - A*b^4)*x^3 + 3*(3
*B*a^2*b^2 - A*a*b^3)*x^2 + 3*(3*B*a^3*b - A*a^2*b^2)*x)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (16*B*b^4*x
^4 - 315*B*a^4 + 105*A*a^3*b - 48*(3*B*a*b^3 - A*b^4)*x^3 - 231*(3*B*a^2*b^2 - A*a*b^3)*x^2 - 280*(3*B*a^3*b -
A*a^2*b^2)*x)*sqrt(x))/(b^8*x^3 + 3*a*b^7*x^2 + 3*a^2*b^6*x + a^3*b^5)]

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giac [A]  time = 0.17, size = 143, normalized size = 0.83 \begin {gather*} \frac {35 \, {\left (3 \, B a^{2} - A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{5}} - \frac {165 \, B a^{2} b^{2} x^{\frac {5}{2}} - 87 \, A a b^{3} x^{\frac {5}{2}} + 280 \, B a^{3} b x^{\frac {3}{2}} - 136 \, A a^{2} b^{2} x^{\frac {3}{2}} + 123 \, B a^{4} \sqrt {x} - 57 \, A a^{3} b \sqrt {x}}{24 \, {\left (b x + a\right )}^{3} b^{5}} + \frac {2 \, {\left (B b^{8} x^{\frac {3}{2}} - 12 \, B a b^{7} \sqrt {x} + 3 \, A b^{8} \sqrt {x}\right )}}{3 \, b^{12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")

[Out]

35/8*(3*B*a^2 - A*a*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^5) - 1/24*(165*B*a^2*b^2*x^(5/2) - 87*A*a*b^3*
x^(5/2) + 280*B*a^3*b*x^(3/2) - 136*A*a^2*b^2*x^(3/2) + 123*B*a^4*sqrt(x) - 57*A*a^3*b*sqrt(x))/((b*x + a)^3*b
^5) + 2/3*(B*b^8*x^(3/2) - 12*B*a*b^7*sqrt(x) + 3*A*b^8*sqrt(x))/b^12

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maple [A]  time = 0.07, size = 190, normalized size = 1.10 \begin {gather*} \frac {29 A a \,x^{\frac {5}{2}}}{8 \left (b x +a \right )^{3} b^{2}}-\frac {55 B \,a^{2} x^{\frac {5}{2}}}{8 \left (b x +a \right )^{3} b^{3}}+\frac {17 A \,a^{2} x^{\frac {3}{2}}}{3 \left (b x +a \right )^{3} b^{3}}-\frac {35 B \,a^{3} x^{\frac {3}{2}}}{3 \left (b x +a \right )^{3} b^{4}}+\frac {19 A \,a^{3} \sqrt {x}}{8 \left (b x +a \right )^{3} b^{4}}-\frac {41 B \,a^{4} \sqrt {x}}{8 \left (b x +a \right )^{3} b^{5}}-\frac {35 A a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{4}}+\frac {105 B \,a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{5}}+\frac {2 B \,x^{\frac {3}{2}}}{3 b^{4}}+\frac {2 A \sqrt {x}}{b^{4}}-\frac {8 B a \sqrt {x}}{b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x)

[Out]

2/3/b^4*B*x^(3/2)+2/b^4*A*x^(1/2)-8/b^5*B*a*x^(1/2)+29/8*a/b^2/(b*x+a)^3*x^(5/2)*A-55/8*a^2/b^3/(b*x+a)^3*x^(5
/2)*B+17/3*a^2/b^3/(b*x+a)^3*A*x^(3/2)-35/3*a^3/b^4/(b*x+a)^3*B*x^(3/2)+19/8*a^3/b^4/(b*x+a)^3*x^(1/2)*A-41/8*
a^4/b^5/(b*x+a)^3*x^(1/2)*B-35/8*a/b^4/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*A+105/8*a^2/b^5/(a*b)^(1/2)
*arctan(1/(a*b)^(1/2)*b*x^(1/2))*B

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maxima [A]  time = 1.17, size = 161, normalized size = 0.93 \begin {gather*} -\frac {3 \, {\left (55 \, B a^{2} b^{2} - 29 \, A a b^{3}\right )} x^{\frac {5}{2}} + 8 \, {\left (35 \, B a^{3} b - 17 \, A a^{2} b^{2}\right )} x^{\frac {3}{2}} + 3 \, {\left (41 \, B a^{4} - 19 \, A a^{3} b\right )} \sqrt {x}}{24 \, {\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}} + \frac {35 \, {\left (3 \, B a^{2} - A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{5}} + \frac {2 \, {\left (B b x^{\frac {3}{2}} - 3 \, {\left (4 \, B a - A b\right )} \sqrt {x}\right )}}{3 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")

[Out]

-1/24*(3*(55*B*a^2*b^2 - 29*A*a*b^3)*x^(5/2) + 8*(35*B*a^3*b - 17*A*a^2*b^2)*x^(3/2) + 3*(41*B*a^4 - 19*A*a^3*
b)*sqrt(x))/(b^8*x^3 + 3*a*b^7*x^2 + 3*a^2*b^6*x + a^3*b^5) + 35/8*(3*B*a^2 - A*a*b)*arctan(b*sqrt(x)/sqrt(a*b
))/(sqrt(a*b)*b^5) + 2/3*(B*b*x^(3/2) - 3*(4*B*a - A*b)*sqrt(x))/b^5

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mupad [B]  time = 1.22, size = 176, normalized size = 1.02 \begin {gather*} \sqrt {x}\,\left (\frac {2\,A}{b^4}-\frac {8\,B\,a}{b^5}\right )-\frac {x^{5/2}\,\left (\frac {55\,B\,a^2\,b^2}{8}-\frac {29\,A\,a\,b^3}{8}\right )-x^{3/2}\,\left (\frac {17\,A\,a^2\,b^2}{3}-\frac {35\,B\,a^3\,b}{3}\right )+\sqrt {x}\,\left (\frac {41\,B\,a^4}{8}-\frac {19\,A\,a^3\,b}{8}\right )}{a^3\,b^5+3\,a^2\,b^6\,x+3\,a\,b^7\,x^2+b^8\,x^3}+\frac {2\,B\,x^{3/2}}{3\,b^4}+\frac {35\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,\sqrt {x}\,\left (A\,b-3\,B\,a\right )}{3\,B\,a^2-A\,a\,b}\right )\,\left (A\,b-3\,B\,a\right )}{8\,b^{11/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(7/2)*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^2,x)

[Out]

x^(1/2)*((2*A)/b^4 - (8*B*a)/b^5) - (x^(5/2)*((55*B*a^2*b^2)/8 - (29*A*a*b^3)/8) - x^(3/2)*((17*A*a^2*b^2)/3 -
(35*B*a^3*b)/3) + x^(1/2)*((41*B*a^4)/8 - (19*A*a^3*b)/8))/(a^3*b^5 + b^8*x^3 + 3*a^2*b^6*x + 3*a*b^7*x^2) +
(2*B*x^(3/2))/(3*b^4) + (35*a^(1/2)*atan((a^(1/2)*b^(1/2)*x^(1/2)*(A*b - 3*B*a))/(3*B*a^2 - A*a*b))*(A*b - 3*B
*a))/(8*b^(11/2))

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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**(7/2)*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

Timed out

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