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

Optimal. Leaf size=185 \[ \frac {3 (a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{128 a^{7/2} b^{7/2}}+\frac {3 \sqrt {x} (a B+A b)}{128 a^3 b^3 (a+b x)}+\frac {\sqrt {x} (a B+A b)}{64 a^2 b^3 (a+b x)^2}-\frac {\sqrt {x} (a B+A b)}{16 a b^3 (a+b x)^3}-\frac {x^{3/2} (a B+A b)}{8 a b^2 (a+b x)^4}+\frac {x^{5/2} (A b-a B)}{5 a b (a+b x)^5} \]

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Rubi [A]  time = 0.09, antiderivative size = 185, 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, 51, 63, 205} \begin {gather*} \frac {3 \sqrt {x} (a B+A b)}{128 a^3 b^3 (a+b x)}+\frac {\sqrt {x} (a B+A b)}{64 a^2 b^3 (a+b x)^2}+\frac {3 (a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{128 a^{7/2} b^{7/2}}-\frac {x^{3/2} (a B+A b)}{8 a b^2 (a+b x)^4}-\frac {\sqrt {x} (a B+A b)}{16 a b^3 (a+b x)^3}+\frac {x^{5/2} (A b-a B)}{5 a b (a+b x)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*x^(5/2))/(5*a*b*(a + b*x)^5) - ((A*b + a*B)*x^(3/2))/(8*a*b^2*(a + b*x)^4) - ((A*b + a*B)*Sqrt[x]
)/(16*a*b^3*(a + b*x)^3) + ((A*b + a*B)*Sqrt[x])/(64*a^2*b^3*(a + b*x)^2) + (3*(A*b + a*B)*Sqrt[x])/(128*a^3*b
^3*(a + b*x)) + (3*(A*b + a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(128*a^(7/2)*b^(7/2))

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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^{3/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx &=\int \frac {x^{3/2} (A+B x)}{(a+b x)^6} \, dx\\ &=\frac {(A b-a B) x^{5/2}}{5 a b (a+b x)^5}+\frac {(A b+a B) \int \frac {x^{3/2}}{(a+b x)^5} \, dx}{2 a b}\\ &=\frac {(A b-a B) x^{5/2}}{5 a b (a+b x)^5}-\frac {(A b+a B) x^{3/2}}{8 a b^2 (a+b x)^4}+\frac {(3 (A b+a B)) \int \frac {\sqrt {x}}{(a+b x)^4} \, dx}{16 a b^2}\\ &=\frac {(A b-a B) x^{5/2}}{5 a b (a+b x)^5}-\frac {(A b+a B) x^{3/2}}{8 a b^2 (a+b x)^4}-\frac {(A b+a B) \sqrt {x}}{16 a b^3 (a+b x)^3}+\frac {(A b+a B) \int \frac {1}{\sqrt {x} (a+b x)^3} \, dx}{32 a b^3}\\ &=\frac {(A b-a B) x^{5/2}}{5 a b (a+b x)^5}-\frac {(A b+a B) x^{3/2}}{8 a b^2 (a+b x)^4}-\frac {(A b+a B) \sqrt {x}}{16 a b^3 (a+b x)^3}+\frac {(A b+a B) \sqrt {x}}{64 a^2 b^3 (a+b x)^2}+\frac {(3 (A b+a B)) \int \frac {1}{\sqrt {x} (a+b x)^2} \, dx}{128 a^2 b^3}\\ &=\frac {(A b-a B) x^{5/2}}{5 a b (a+b x)^5}-\frac {(A b+a B) x^{3/2}}{8 a b^2 (a+b x)^4}-\frac {(A b+a B) \sqrt {x}}{16 a b^3 (a+b x)^3}+\frac {(A b+a B) \sqrt {x}}{64 a^2 b^3 (a+b x)^2}+\frac {3 (A b+a B) \sqrt {x}}{128 a^3 b^3 (a+b x)}+\frac {(3 (A b+a B)) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{256 a^3 b^3}\\ &=\frac {(A b-a B) x^{5/2}}{5 a b (a+b x)^5}-\frac {(A b+a B) x^{3/2}}{8 a b^2 (a+b x)^4}-\frac {(A b+a B) \sqrt {x}}{16 a b^3 (a+b x)^3}+\frac {(A b+a B) \sqrt {x}}{64 a^2 b^3 (a+b x)^2}+\frac {3 (A b+a B) \sqrt {x}}{128 a^3 b^3 (a+b x)}+\frac {(3 (A b+a B)) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{128 a^3 b^3}\\ &=\frac {(A b-a B) x^{5/2}}{5 a b (a+b x)^5}-\frac {(A b+a B) x^{3/2}}{8 a b^2 (a+b x)^4}-\frac {(A b+a B) \sqrt {x}}{16 a b^3 (a+b x)^3}+\frac {(A b+a B) \sqrt {x}}{64 a^2 b^3 (a+b x)^2}+\frac {3 (A b+a B) \sqrt {x}}{128 a^3 b^3 (a+b x)}+\frac {3 (A b+a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{128 a^{7/2} b^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 60, normalized size = 0.32 \begin {gather*} \frac {x^{5/2} \left (\frac {5 a^5 (A b-a B)}{(a+b x)^5}+5 (a B+A b) \, _2F_1\left (\frac {5}{2},5;\frac {7}{2};-\frac {b x}{a}\right )\right )}{25 a^6 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(5/2)*((5*a^5*(A*b - a*B))/(a + b*x)^5 + 5*(A*b + a*B)*Hypergeometric2F1[5/2, 5, 7/2, -((b*x)/a)]))/(25*a^6
*b)

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IntegrateAlgebraic [A]  time = 0.33, size = 158, normalized size = 0.85 \begin {gather*} \frac {3 (a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{128 a^{7/2} b^{7/2}}-\frac {\sqrt {x} \left (15 a^5 B+15 a^4 A b+70 a^4 b B x+70 a^3 A b^2 x+128 a^3 b^2 B x^2-128 a^2 A b^3 x^2-70 a^2 b^3 B x^3-70 a A b^4 x^3-15 a b^4 B x^4-15 A b^5 x^4\right )}{640 a^3 b^3 (a+b x)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/640*(Sqrt[x]*(15*a^4*A*b + 15*a^5*B + 70*a^3*A*b^2*x + 70*a^4*b*B*x - 128*a^2*A*b^3*x^2 + 128*a^3*b^2*B*x^2
- 70*a*A*b^4*x^3 - 70*a^2*b^3*B*x^3 - 15*A*b^5*x^4 - 15*a*b^4*B*x^4))/(a^3*b^3*(a + b*x)^5) + (3*(A*b + a*B)*
ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(128*a^(7/2)*b^(7/2))

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fricas [A]  time = 0.46, size = 619, normalized size = 3.35 \begin {gather*} \left [-\frac {15 \, {\left (B a^{6} + A a^{5} b + {\left (B a b^{5} + A b^{6}\right )} x^{5} + 5 \, {\left (B a^{2} b^{4} + A a b^{5}\right )} x^{4} + 10 \, {\left (B a^{3} b^{3} + A a^{2} b^{4}\right )} x^{3} + 10 \, {\left (B a^{4} b^{2} + A a^{3} b^{3}\right )} x^{2} + 5 \, {\left (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 (15 \, B a^{6} b + 15 \, A a^{5} b^{2} - 15 \, {\left (B a^{2} b^{5} + A a b^{6}\right )} x^{4} - 70 \, {\left (B a^{3} b^{4} + A a^{2} b^{5}\right )} x^{3} + 128 \, {\left (B a^{4} b^{3} - A a^{3} b^{4}\right )} x^{2} + 70 \, {\left (B a^{5} b^{2} + A a^{4} b^{3}\right )} x\right )} \sqrt {x}}{1280 \, {\left (a^{4} b^{9} x^{5} + 5 \, a^{5} b^{8} x^{4} + 10 \, a^{6} b^{7} x^{3} + 10 \, a^{7} b^{6} x^{2} + 5 \, a^{8} b^{5} x + a^{9} b^{4}\right )}}, -\frac {15 \, {\left (B a^{6} + A a^{5} b + {\left (B a b^{5} + A b^{6}\right )} x^{5} + 5 \, {\left (B a^{2} b^{4} + A a b^{5}\right )} x^{4} + 10 \, {\left (B a^{3} b^{3} + A a^{2} b^{4}\right )} x^{3} + 10 \, {\left (B a^{4} b^{2} + A a^{3} b^{3}\right )} x^{2} + 5 \, {\left (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 (15 \, B a^{6} b + 15 \, A a^{5} b^{2} - 15 \, {\left (B a^{2} b^{5} + A a b^{6}\right )} x^{4} - 70 \, {\left (B a^{3} b^{4} + A a^{2} b^{5}\right )} x^{3} + 128 \, {\left (B a^{4} b^{3} - A a^{3} b^{4}\right )} x^{2} + 70 \, {\left (B a^{5} b^{2} + A a^{4} b^{3}\right )} x\right )} \sqrt {x}}{640 \, {\left (a^{4} b^{9} x^{5} + 5 \, a^{5} b^{8} x^{4} + 10 \, a^{6} b^{7} x^{3} + 10 \, a^{7} b^{6} x^{2} + 5 \, a^{8} b^{5} x + a^{9} 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^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")

[Out]

[-1/1280*(15*(B*a^6 + A*a^5*b + (B*a*b^5 + A*b^6)*x^5 + 5*(B*a^2*b^4 + A*a*b^5)*x^4 + 10*(B*a^3*b^3 + A*a^2*b^
4)*x^3 + 10*(B*a^4*b^2 + A*a^3*b^3)*x^2 + 5*(B*a^5*b + A*a^4*b^2)*x)*sqrt(-a*b)*log((b*x - a - 2*sqrt(-a*b)*sq
rt(x))/(b*x + a)) + 2*(15*B*a^6*b + 15*A*a^5*b^2 - 15*(B*a^2*b^5 + A*a*b^6)*x^4 - 70*(B*a^3*b^4 + A*a^2*b^5)*x
^3 + 128*(B*a^4*b^3 - A*a^3*b^4)*x^2 + 70*(B*a^5*b^2 + A*a^4*b^3)*x)*sqrt(x))/(a^4*b^9*x^5 + 5*a^5*b^8*x^4 + 1
0*a^6*b^7*x^3 + 10*a^7*b^6*x^2 + 5*a^8*b^5*x + a^9*b^4), -1/640*(15*(B*a^6 + A*a^5*b + (B*a*b^5 + A*b^6)*x^5 +
5*(B*a^2*b^4 + A*a*b^5)*x^4 + 10*(B*a^3*b^3 + A*a^2*b^4)*x^3 + 10*(B*a^4*b^2 + A*a^3*b^3)*x^2 + 5*(B*a^5*b +
A*a^4*b^2)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (15*B*a^6*b + 15*A*a^5*b^2 - 15*(B*a^2*b^5 + A*a*b^6)*
x^4 - 70*(B*a^3*b^4 + A*a^2*b^5)*x^3 + 128*(B*a^4*b^3 - A*a^3*b^4)*x^2 + 70*(B*a^5*b^2 + A*a^4*b^3)*x)*sqrt(x)
)/(a^4*b^9*x^5 + 5*a^5*b^8*x^4 + 10*a^6*b^7*x^3 + 10*a^7*b^6*x^2 + 5*a^8*b^5*x + a^9*b^4)]

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giac [A]  time = 0.16, size = 154, normalized size = 0.83 \begin {gather*} \frac {3 \, {\left (B a + A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{128 \, \sqrt {a b} a^{3} b^{3}} + \frac {15 \, B a b^{4} x^{\frac {9}{2}} + 15 \, A b^{5} x^{\frac {9}{2}} + 70 \, B a^{2} b^{3} x^{\frac {7}{2}} + 70 \, A a b^{4} x^{\frac {7}{2}} - 128 \, B a^{3} b^{2} x^{\frac {5}{2}} + 128 \, A a^{2} b^{3} x^{\frac {5}{2}} - 70 \, B a^{4} b x^{\frac {3}{2}} - 70 \, A a^{3} b^{2} x^{\frac {3}{2}} - 15 \, B a^{5} \sqrt {x} - 15 \, A a^{4} b \sqrt {x}}{640 \, {\left (b x + a\right )}^{5} a^{3} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/128*(B*a + A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3*b^3) + 1/640*(15*B*a*b^4*x^(9/2) + 15*A*b^5*x^(9/
2) + 70*B*a^2*b^3*x^(7/2) + 70*A*a*b^4*x^(7/2) - 128*B*a^3*b^2*x^(5/2) + 128*A*a^2*b^3*x^(5/2) - 70*B*a^4*b*x^
(3/2) - 70*A*a^3*b^2*x^(3/2) - 15*B*a^5*sqrt(x) - 15*A*a^4*b*sqrt(x))/((b*x + a)^5*a^3*b^3)

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maple [A]  time = 0.07, size = 143, normalized size = 0.77 \begin {gather*} \frac {3 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{128 \sqrt {a b}\, a^{3} b^{2}}+\frac {3 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{128 \sqrt {a b}\, a^{2} b^{3}}+\frac {\frac {3 \left (A b +B a \right ) b \,x^{\frac {9}{2}}}{128 a^{3}}+\frac {7 \left (A b +B a \right ) x^{\frac {7}{2}}}{64 a^{2}}+\frac {\left (A b -B a \right ) x^{\frac {5}{2}}}{5 a b}-\frac {7 \left (A b +B a \right ) x^{\frac {3}{2}}}{64 b^{2}}-\frac {3 \left (A b +B a \right ) a \sqrt {x}}{128 b^{3}}}{\left (b x +a \right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(3/256*(A*b+B*a)/a^3*b*x^(9/2)+7/128*(A*b+B*a)/a^2*x^(7/2)+1/10*(A*b-B*a)/a/b*x^(5/2)-7/128*(A*b+B*a)/b^2*x^
(3/2)-3/256*(A*b+B*a)*a/b^3*x^(1/2))/(b*x+a)^5+3/128/a^3/b^2/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*A+3/1
28/a^2/b^3/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*B

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maxima [A]  time = 1.19, size = 194, normalized size = 1.05 \begin {gather*} \frac {15 \, {\left (B a b^{4} + A b^{5}\right )} x^{\frac {9}{2}} + 70 \, {\left (B a^{2} b^{3} + A a b^{4}\right )} x^{\frac {7}{2}} - 128 \, {\left (B a^{3} b^{2} - A a^{2} b^{3}\right )} x^{\frac {5}{2}} - 70 \, {\left (B a^{4} b + A a^{3} b^{2}\right )} x^{\frac {3}{2}} - 15 \, {\left (B a^{5} + A a^{4} b\right )} \sqrt {x}}{640 \, {\left (a^{3} b^{8} x^{5} + 5 \, a^{4} b^{7} x^{4} + 10 \, a^{5} b^{6} x^{3} + 10 \, a^{6} b^{5} x^{2} + 5 \, a^{7} b^{4} x + a^{8} b^{3}\right )}} + \frac {3 \, {\left (B a + A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{128 \, \sqrt {a b} a^{3} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/640*(15*(B*a*b^4 + A*b^5)*x^(9/2) + 70*(B*a^2*b^3 + A*a*b^4)*x^(7/2) - 128*(B*a^3*b^2 - A*a^2*b^3)*x^(5/2) -
70*(B*a^4*b + A*a^3*b^2)*x^(3/2) - 15*(B*a^5 + A*a^4*b)*sqrt(x))/(a^3*b^8*x^5 + 5*a^4*b^7*x^4 + 10*a^5*b^6*x^
3 + 10*a^6*b^5*x^2 + 5*a^7*b^4*x + a^8*b^3) + 3/128*(B*a + A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3*b^3
)

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mupad [B]  time = 1.23, size = 161, normalized size = 0.87 \begin {gather*} \frac {\frac {7\,x^{7/2}\,\left (A\,b+B\,a\right )}{64\,a^2}-\frac {7\,x^{3/2}\,\left (A\,b+B\,a\right )}{64\,b^2}+\frac {x^{5/2}\,\left (A\,b-B\,a\right )}{5\,a\,b}-\frac {3\,a\,\sqrt {x}\,\left (A\,b+B\,a\right )}{128\,b^3}+\frac {3\,b\,x^{9/2}\,\left (A\,b+B\,a\right )}{128\,a^3}}{a^5+5\,a^4\,b\,x+10\,a^3\,b^2\,x^2+10\,a^2\,b^3\,x^3+5\,a\,b^4\,x^4+b^5\,x^5}+\frac {3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b+B\,a\right )}{128\,a^{7/2}\,b^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((7*x^(7/2)*(A*b + B*a))/(64*a^2) - (7*x^(3/2)*(A*b + B*a))/(64*b^2) + (x^(5/2)*(A*b - B*a))/(5*a*b) - (3*a*x^
(1/2)*(A*b + B*a))/(128*b^3) + (3*b*x^(9/2)*(A*b + B*a))/(128*a^3))/(a^5 + b^5*x^5 + 5*a*b^4*x^4 + 10*a^3*b^2*
x^2 + 10*a^2*b^3*x^3 + 5*a^4*b*x) + (3*atan((b^(1/2)*x^(1/2))/a^(1/2))*(A*b + B*a))/(128*a^(7/2)*b^(7/2))

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

[Out]

Timed out

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