3.774 \(\int \frac {x^{9/2} (A+B x)}{(a^2+2 a b x+b^2 x^2)^3} \, dx\)

Optimal. Leaf size=213 \[ \frac {63 (A b-11 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{128 \sqrt {a} b^{13/2}}-\frac {63 \sqrt {x} (A b-11 a B)}{128 a b^6}+\frac {21 x^{3/2} (A b-11 a B)}{128 a b^5 (a+b x)}+\frac {21 x^{5/2} (A b-11 a B)}{320 a b^4 (a+b x)^2}+\frac {3 x^{7/2} (A b-11 a B)}{80 a b^3 (a+b x)^3}+\frac {x^{9/2} (A b-11 a B)}{40 a b^2 (a+b x)^4}+\frac {x^{11/2} (A b-a B)}{5 a b (a+b x)^5} \]

[Out]

1/5*(A*b-B*a)*x^(11/2)/a/b/(b*x+a)^5+1/40*(A*b-11*B*a)*x^(9/2)/a/b^2/(b*x+a)^4+3/80*(A*b-11*B*a)*x^(7/2)/a/b^3
/(b*x+a)^3+21/320*(A*b-11*B*a)*x^(5/2)/a/b^4/(b*x+a)^2+21/128*(A*b-11*B*a)*x^(3/2)/a/b^5/(b*x+a)+63/128*(A*b-1
1*B*a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/b^(13/2)/a^(1/2)-63/128*(A*b-11*B*a)*x^(1/2)/a/b^6

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Rubi [A]  time = 0.10, antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 9, 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} \[ \frac {x^{9/2} (A b-11 a B)}{40 a b^2 (a+b x)^4}+\frac {3 x^{7/2} (A b-11 a B)}{80 a b^3 (a+b x)^3}+\frac {21 x^{5/2} (A b-11 a B)}{320 a b^4 (a+b x)^2}+\frac {21 x^{3/2} (A b-11 a B)}{128 a b^5 (a+b x)}-\frac {63 \sqrt {x} (A b-11 a B)}{128 a b^6}+\frac {63 (A b-11 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{128 \sqrt {a} b^{13/2}}+\frac {x^{11/2} (A b-a B)}{5 a b (a+b x)^5} \]

Antiderivative was successfully verified.

[In]

Int[(x^(9/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

(-63*(A*b - 11*a*B)*Sqrt[x])/(128*a*b^6) + ((A*b - a*B)*x^(11/2))/(5*a*b*(a + b*x)^5) + ((A*b - 11*a*B)*x^(9/2
))/(40*a*b^2*(a + b*x)^4) + (3*(A*b - 11*a*B)*x^(7/2))/(80*a*b^3*(a + b*x)^3) + (21*(A*b - 11*a*B)*x^(5/2))/(3
20*a*b^4*(a + b*x)^2) + (21*(A*b - 11*a*B)*x^(3/2))/(128*a*b^5*(a + b*x)) + (63*(A*b - 11*a*B)*ArcTan[(Sqrt[b]
*Sqrt[x])/Sqrt[a]])/(128*Sqrt[a]*b^(13/2))

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^{9/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx &=\int \frac {x^{9/2} (A+B x)}{(a+b x)^6} \, dx\\ &=\frac {(A b-a B) x^{11/2}}{5 a b (a+b x)^5}-\frac {(A b-11 a B) \int \frac {x^{9/2}}{(a+b x)^5} \, dx}{10 a b}\\ &=\frac {(A b-a B) x^{11/2}}{5 a b (a+b x)^5}+\frac {(A b-11 a B) x^{9/2}}{40 a b^2 (a+b x)^4}-\frac {(9 (A b-11 a B)) \int \frac {x^{7/2}}{(a+b x)^4} \, dx}{80 a b^2}\\ &=\frac {(A b-a B) x^{11/2}}{5 a b (a+b x)^5}+\frac {(A b-11 a B) x^{9/2}}{40 a b^2 (a+b x)^4}+\frac {3 (A b-11 a B) x^{7/2}}{80 a b^3 (a+b x)^3}-\frac {(21 (A b-11 a B)) \int \frac {x^{5/2}}{(a+b x)^3} \, dx}{160 a b^3}\\ &=\frac {(A b-a B) x^{11/2}}{5 a b (a+b x)^5}+\frac {(A b-11 a B) x^{9/2}}{40 a b^2 (a+b x)^4}+\frac {3 (A b-11 a B) x^{7/2}}{80 a b^3 (a+b x)^3}+\frac {21 (A b-11 a B) x^{5/2}}{320 a b^4 (a+b x)^2}-\frac {(21 (A b-11 a B)) \int \frac {x^{3/2}}{(a+b x)^2} \, dx}{128 a b^4}\\ &=\frac {(A b-a B) x^{11/2}}{5 a b (a+b x)^5}+\frac {(A b-11 a B) x^{9/2}}{40 a b^2 (a+b x)^4}+\frac {3 (A b-11 a B) x^{7/2}}{80 a b^3 (a+b x)^3}+\frac {21 (A b-11 a B) x^{5/2}}{320 a b^4 (a+b x)^2}+\frac {21 (A b-11 a B) x^{3/2}}{128 a b^5 (a+b x)}-\frac {(63 (A b-11 a B)) \int \frac {\sqrt {x}}{a+b x} \, dx}{256 a b^5}\\ &=-\frac {63 (A b-11 a B) \sqrt {x}}{128 a b^6}+\frac {(A b-a B) x^{11/2}}{5 a b (a+b x)^5}+\frac {(A b-11 a B) x^{9/2}}{40 a b^2 (a+b x)^4}+\frac {3 (A b-11 a B) x^{7/2}}{80 a b^3 (a+b x)^3}+\frac {21 (A b-11 a B) x^{5/2}}{320 a b^4 (a+b x)^2}+\frac {21 (A b-11 a B) x^{3/2}}{128 a b^5 (a+b x)}+\frac {(63 (A b-11 a B)) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{256 b^6}\\ &=-\frac {63 (A b-11 a B) \sqrt {x}}{128 a b^6}+\frac {(A b-a B) x^{11/2}}{5 a b (a+b x)^5}+\frac {(A b-11 a B) x^{9/2}}{40 a b^2 (a+b x)^4}+\frac {3 (A b-11 a B) x^{7/2}}{80 a b^3 (a+b x)^3}+\frac {21 (A b-11 a B) x^{5/2}}{320 a b^4 (a+b x)^2}+\frac {21 (A b-11 a B) x^{3/2}}{128 a b^5 (a+b x)}+\frac {(63 (A b-11 a B)) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{128 b^6}\\ &=-\frac {63 (A b-11 a B) \sqrt {x}}{128 a b^6}+\frac {(A b-a B) x^{11/2}}{5 a b (a+b x)^5}+\frac {(A b-11 a B) x^{9/2}}{40 a b^2 (a+b x)^4}+\frac {3 (A b-11 a B) x^{7/2}}{80 a b^3 (a+b x)^3}+\frac {21 (A b-11 a B) x^{5/2}}{320 a b^4 (a+b x)^2}+\frac {21 (A b-11 a B) x^{3/2}}{128 a b^5 (a+b x)}+\frac {63 (A b-11 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{128 \sqrt {a} b^{13/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 61, normalized size = 0.29 \[ \frac {x^{11/2} \left (\frac {11 a^5 (A b-a B)}{(a+b x)^5}+(11 a B-A b) \, _2F_1\left (5,\frac {11}{2};\frac {13}{2};-\frac {b x}{a}\right )\right )}{55 a^6 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(9/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

(x^(11/2)*((11*a^5*(A*b - a*B))/(a + b*x)^5 + (-(A*b) + 11*a*B)*Hypergeometric2F1[5, 11/2, 13/2, -((b*x)/a)]))
/(55*a^6*b)

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fricas [A]  time = 1.02, size = 673, normalized size = 3.16 \[ \left [\frac {315 \, {\left (11 \, B a^{6} - A a^{5} b + {\left (11 \, B a b^{5} - A b^{6}\right )} x^{5} + 5 \, {\left (11 \, B a^{2} b^{4} - A a b^{5}\right )} x^{4} + 10 \, {\left (11 \, B a^{3} b^{3} - A a^{2} b^{4}\right )} x^{3} + 10 \, {\left (11 \, B a^{4} b^{2} - A a^{3} b^{3}\right )} x^{2} + 5 \, {\left (11 \, B a^{5} b - A a^{4} b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (1280 \, B a b^{6} x^{5} + 3465 \, B a^{6} b - 315 \, A a^{5} b^{2} + 965 \, {\left (11 \, B a^{2} b^{5} - A a b^{6}\right )} x^{4} + 2370 \, {\left (11 \, B a^{3} b^{4} - A a^{2} b^{5}\right )} x^{3} + 2688 \, {\left (11 \, B a^{4} b^{3} - A a^{3} b^{4}\right )} x^{2} + 1470 \, {\left (11 \, B a^{5} b^{2} - A a^{4} b^{3}\right )} x\right )} \sqrt {x}}{1280 \, {\left (a b^{12} x^{5} + 5 \, a^{2} b^{11} x^{4} + 10 \, a^{3} b^{10} x^{3} + 10 \, a^{4} b^{9} x^{2} + 5 \, a^{5} b^{8} x + a^{6} b^{7}\right )}}, \frac {315 \, {\left (11 \, B a^{6} - A a^{5} b + {\left (11 \, B a b^{5} - A b^{6}\right )} x^{5} + 5 \, {\left (11 \, B a^{2} b^{4} - A a b^{5}\right )} x^{4} + 10 \, {\left (11 \, B a^{3} b^{3} - A a^{2} b^{4}\right )} x^{3} + 10 \, {\left (11 \, B a^{4} b^{2} - A a^{3} b^{3}\right )} x^{2} + 5 \, {\left (11 \, B a^{5} b - A a^{4} b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (1280 \, B a b^{6} x^{5} + 3465 \, B a^{6} b - 315 \, A a^{5} b^{2} + 965 \, {\left (11 \, B a^{2} b^{5} - A a b^{6}\right )} x^{4} + 2370 \, {\left (11 \, B a^{3} b^{4} - A a^{2} b^{5}\right )} x^{3} + 2688 \, {\left (11 \, B a^{4} b^{3} - A a^{3} b^{4}\right )} x^{2} + 1470 \, {\left (11 \, B a^{5} b^{2} - A a^{4} b^{3}\right )} x\right )} \sqrt {x}}{640 \, {\left (a b^{12} x^{5} + 5 \, a^{2} b^{11} x^{4} + 10 \, a^{3} b^{10} x^{3} + 10 \, a^{4} b^{9} x^{2} + 5 \, a^{5} b^{8} x + a^{6} b^{7}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")

[Out]

[1/1280*(315*(11*B*a^6 - A*a^5*b + (11*B*a*b^5 - A*b^6)*x^5 + 5*(11*B*a^2*b^4 - A*a*b^5)*x^4 + 10*(11*B*a^3*b^
3 - A*a^2*b^4)*x^3 + 10*(11*B*a^4*b^2 - A*a^3*b^3)*x^2 + 5*(11*B*a^5*b - A*a^4*b^2)*x)*sqrt(-a*b)*log((b*x - a
- 2*sqrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(1280*B*a*b^6*x^5 + 3465*B*a^6*b - 315*A*a^5*b^2 + 965*(11*B*a^2*b^5 -
A*a*b^6)*x^4 + 2370*(11*B*a^3*b^4 - A*a^2*b^5)*x^3 + 2688*(11*B*a^4*b^3 - A*a^3*b^4)*x^2 + 1470*(11*B*a^5*b^2
- A*a^4*b^3)*x)*sqrt(x))/(a*b^12*x^5 + 5*a^2*b^11*x^4 + 10*a^3*b^10*x^3 + 10*a^4*b^9*x^2 + 5*a^5*b^8*x + a^6*
b^7), 1/640*(315*(11*B*a^6 - A*a^5*b + (11*B*a*b^5 - A*b^6)*x^5 + 5*(11*B*a^2*b^4 - A*a*b^5)*x^4 + 10*(11*B*a^
3*b^3 - A*a^2*b^4)*x^3 + 10*(11*B*a^4*b^2 - A*a^3*b^3)*x^2 + 5*(11*B*a^5*b - A*a^4*b^2)*x)*sqrt(a*b)*arctan(sq
rt(a*b)/(b*sqrt(x))) + (1280*B*a*b^6*x^5 + 3465*B*a^6*b - 315*A*a^5*b^2 + 965*(11*B*a^2*b^5 - A*a*b^6)*x^4 + 2
370*(11*B*a^3*b^4 - A*a^2*b^5)*x^3 + 2688*(11*B*a^4*b^3 - A*a^3*b^4)*x^2 + 1470*(11*B*a^5*b^2 - A*a^4*b^3)*x)*
sqrt(x))/(a*b^12*x^5 + 5*a^2*b^11*x^4 + 10*a^3*b^10*x^3 + 10*a^4*b^9*x^2 + 5*a^5*b^8*x + a^6*b^7)]

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giac [A]  time = 0.17, size = 159, normalized size = 0.75 \[ \frac {2 \, B \sqrt {x}}{b^{6}} - \frac {63 \, {\left (11 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{128 \, \sqrt {a b} b^{6}} + \frac {4215 \, B a b^{4} x^{\frac {9}{2}} - 965 \, A b^{5} x^{\frac {9}{2}} + 13270 \, B a^{2} b^{3} x^{\frac {7}{2}} - 2370 \, A a b^{4} x^{\frac {7}{2}} + 16768 \, B a^{3} b^{2} x^{\frac {5}{2}} - 2688 \, A a^{2} b^{3} x^{\frac {5}{2}} + 9770 \, B a^{4} b x^{\frac {3}{2}} - 1470 \, A a^{3} b^{2} x^{\frac {3}{2}} + 2185 \, B a^{5} \sqrt {x} - 315 \, A a^{4} b \sqrt {x}}{640 \, {\left (b x + a\right )}^{5} b^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="giac")

[Out]

2*B*sqrt(x)/b^6 - 63/128*(11*B*a - A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^6) + 1/640*(4215*B*a*b^4*x^(9
/2) - 965*A*b^5*x^(9/2) + 13270*B*a^2*b^3*x^(7/2) - 2370*A*a*b^4*x^(7/2) + 16768*B*a^3*b^2*x^(5/2) - 2688*A*a^
2*b^3*x^(5/2) + 9770*B*a^4*b*x^(3/2) - 1470*A*a^3*b^2*x^(3/2) + 2185*B*a^5*sqrt(x) - 315*A*a^4*b*sqrt(x))/((b*
x + a)^5*b^6)

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maple [A]  time = 0.08, size = 239, normalized size = 1.12 \[ -\frac {193 A \,x^{\frac {9}{2}}}{128 \left (b x +a \right )^{5} b}+\frac {843 B a \,x^{\frac {9}{2}}}{128 \left (b x +a \right )^{5} b^{2}}-\frac {237 A a \,x^{\frac {7}{2}}}{64 \left (b x +a \right )^{5} b^{2}}+\frac {1327 B \,a^{2} x^{\frac {7}{2}}}{64 \left (b x +a \right )^{5} b^{3}}-\frac {21 A \,a^{2} x^{\frac {5}{2}}}{5 \left (b x +a \right )^{5} b^{3}}+\frac {131 B \,a^{3} x^{\frac {5}{2}}}{5 \left (b x +a \right )^{5} b^{4}}-\frac {147 A \,a^{3} x^{\frac {3}{2}}}{64 \left (b x +a \right )^{5} b^{4}}+\frac {977 B \,a^{4} x^{\frac {3}{2}}}{64 \left (b x +a \right )^{5} b^{5}}-\frac {63 A \,a^{4} \sqrt {x}}{128 \left (b x +a \right )^{5} b^{5}}+\frac {437 B \,a^{5} \sqrt {x}}{128 \left (b x +a \right )^{5} b^{6}}+\frac {63 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{128 \sqrt {a b}\, b^{5}}-\frac {693 B a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{128 \sqrt {a b}\, b^{6}}+\frac {2 B \sqrt {x}}{b^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(9/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x)

[Out]

2*B/b^6*x^(1/2)-193/128/b/(b*x+a)^5*x^(9/2)*A+843/128/b^2/(b*x+a)^5*x^(9/2)*B*a+1327/64/b^3/(b*x+a)^5*x^(7/2)*
B*a^2-237/64/b^2/(b*x+a)^5*x^(7/2)*A*a+131/5/b^4/(b*x+a)^5*x^(5/2)*B*a^3-21/5/b^3/(b*x+a)^5*x^(5/2)*A*a^2-147/
64/b^4/(b*x+a)^5*A*x^(3/2)*a^3+977/64/b^5/(b*x+a)^5*B*x^(3/2)*a^4+437/128/b^6/(b*x+a)^5*x^(1/2)*B*a^5-63/128/b
^5/(b*x+a)^5*x^(1/2)*A*a^4+63/128/b^5/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*A-693/128/b^6/(a*b)^(1/2)*ar
ctan(1/(a*b)^(1/2)*b*x^(1/2))*B*a

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maxima [A]  time = 1.34, size = 206, normalized size = 0.97 \[ \frac {5 \, {\left (843 \, B a b^{4} - 193 \, A b^{5}\right )} x^{\frac {9}{2}} + 10 \, {\left (1327 \, B a^{2} b^{3} - 237 \, A a b^{4}\right )} x^{\frac {7}{2}} + 128 \, {\left (131 \, B a^{3} b^{2} - 21 \, A a^{2} b^{3}\right )} x^{\frac {5}{2}} + 10 \, {\left (977 \, B a^{4} b - 147 \, A a^{3} b^{2}\right )} x^{\frac {3}{2}} + 5 \, {\left (437 \, B a^{5} - 63 \, A a^{4} b\right )} \sqrt {x}}{640 \, {\left (b^{11} x^{5} + 5 \, a b^{10} x^{4} + 10 \, a^{2} b^{9} x^{3} + 10 \, a^{3} b^{8} x^{2} + 5 \, a^{4} b^{7} x + a^{5} b^{6}\right )}} + \frac {2 \, B \sqrt {x}}{b^{6}} - \frac {63 \, {\left (11 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{128 \, \sqrt {a b} b^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="maxima")

[Out]

1/640*(5*(843*B*a*b^4 - 193*A*b^5)*x^(9/2) + 10*(1327*B*a^2*b^3 - 237*A*a*b^4)*x^(7/2) + 128*(131*B*a^3*b^2 -
21*A*a^2*b^3)*x^(5/2) + 10*(977*B*a^4*b - 147*A*a^3*b^2)*x^(3/2) + 5*(437*B*a^5 - 63*A*a^4*b)*sqrt(x))/(b^11*x
^5 + 5*a*b^10*x^4 + 10*a^2*b^9*x^3 + 10*a^3*b^8*x^2 + 5*a^4*b^7*x + a^5*b^6) + 2*B*sqrt(x)/b^6 - 63/128*(11*B*
a - A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^6)

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mupad [B]  time = 0.18, size = 200, normalized size = 0.94 \[ \frac {2\,B\,\sqrt {x}}{b^6}-\frac {x^{3/2}\,\left (\frac {147\,A\,a^3\,b^2}{64}-\frac {977\,B\,a^4\,b}{64}\right )-x^{7/2}\,\left (\frac {1327\,B\,a^2\,b^3}{64}-\frac {237\,A\,a\,b^4}{64}\right )-\sqrt {x}\,\left (\frac {437\,B\,a^5}{128}-\frac {63\,A\,a^4\,b}{128}\right )+x^{9/2}\,\left (\frac {193\,A\,b^5}{128}-\frac {843\,B\,a\,b^4}{128}\right )+x^{5/2}\,\left (\frac {21\,A\,a^2\,b^3}{5}-\frac {131\,B\,a^3\,b^2}{5}\right )}{a^5\,b^6+5\,a^4\,b^7\,x+10\,a^3\,b^8\,x^2+10\,a^2\,b^9\,x^3+5\,a\,b^{10}\,x^4+b^{11}\,x^5}+\frac {63\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b-11\,B\,a\right )}{128\,\sqrt {a}\,b^{13/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(9/2)*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^3,x)

[Out]

(2*B*x^(1/2))/b^6 - (x^(3/2)*((147*A*a^3*b^2)/64 - (977*B*a^4*b)/64) - x^(7/2)*((1327*B*a^2*b^3)/64 - (237*A*a
*b^4)/64) - x^(1/2)*((437*B*a^5)/128 - (63*A*a^4*b)/128) + x^(9/2)*((193*A*b^5)/128 - (843*B*a*b^4)/128) + x^(
5/2)*((21*A*a^2*b^3)/5 - (131*B*a^3*b^2)/5))/(a^5*b^6 + b^11*x^5 + 5*a^4*b^7*x + 5*a*b^10*x^4 + 10*a^3*b^8*x^2
+ 10*a^2*b^9*x^3) + (63*atan((b^(1/2)*x^(1/2))/a^(1/2))*(A*b - 11*B*a))/(128*a^(1/2)*b^(13/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(9/2)*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**3,x)

[Out]

Timed out

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