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

Optimal. Leaf size=255 \[ \frac {x^{7/2} (A b-a B)}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^{5/2} (3 A b-7 a B)}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 \sqrt {a} (a+b x) (3 A b-7 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 \sqrt {x} (a+b x) (3 A b-7 a B)}{4 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 x^{3/2} (a+b x) (3 A b-7 a B)}{12 a b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {770, 78, 47, 50, 63, 205} \[ \frac {x^{7/2} (A b-a B)}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^{5/2} (3 A b-7 a B)}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 x^{3/2} (a+b x) (3 A b-7 a B)}{12 a b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 \sqrt {x} (a+b x) (3 A b-7 a B)}{4 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 \sqrt {a} (a+b x) (3 A b-7 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*A*b - 7*a*B)*x^(5/2))/(4*a*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((A*b - a*B)*x^(7/2))/(2*a*b*(a + b*x)*Sqr
t[a^2 + 2*a*b*x + b^2*x^2]) + (5*(3*A*b - 7*a*B)*Sqrt[x]*(a + b*x))/(4*b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5
*(3*A*b - 7*a*B)*x^(3/2)*(a + b*x))/(12*a*b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*Sqrt[a]*(3*A*b - 7*a*B)*(a +
b*x)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*b^(9/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^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 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]

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {x^{5/2} (A+B x)}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(A b-a B) x^{7/2}}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left ((3 A b-7 a B) \left (a b+b^2 x\right )\right ) \int \frac {x^{5/2}}{\left (a b+b^2 x\right )^2} \, dx}{4 a \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(3 A b-7 a B) x^{5/2}}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{7/2}}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (5 (3 A b-7 a B) \left (a b+b^2 x\right )\right ) \int \frac {x^{3/2}}{a b+b^2 x} \, dx}{8 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(3 A b-7 a B) x^{5/2}}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{7/2}}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 (3 A b-7 a B) x^{3/2} (a+b x)}{12 a b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (5 (3 A b-7 a B) \left (a b+b^2 x\right )\right ) \int \frac {\sqrt {x}}{a b+b^2 x} \, dx}{8 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(3 A b-7 a B) x^{5/2}}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{7/2}}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 (3 A b-7 a B) \sqrt {x} (a+b x)}{4 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 (3 A b-7 a B) x^{3/2} (a+b x)}{12 a b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (5 a (3 A b-7 a B) \left (a b+b^2 x\right )\right ) \int \frac {1}{\sqrt {x} \left (a b+b^2 x\right )} \, dx}{8 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(3 A b-7 a B) x^{5/2}}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{7/2}}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 (3 A b-7 a B) \sqrt {x} (a+b x)}{4 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 (3 A b-7 a B) x^{3/2} (a+b x)}{12 a b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (5 a (3 A b-7 a B) \left (a b+b^2 x\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a b+b^2 x^2} \, dx,x,\sqrt {x}\right )}{4 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(3 A b-7 a B) x^{5/2}}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{7/2}}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 (3 A b-7 a B) \sqrt {x} (a+b x)}{4 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 (3 A b-7 a B) x^{3/2} (a+b x)}{12 a b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 \sqrt {a} (3 A b-7 a B) (a+b x) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 79, normalized size = 0.31 \[ \frac {x^{7/2} \left (7 a^2 (A b-a B)+(a+b x)^2 (7 a B-3 A b) \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};-\frac {b x}{a}\right )\right )}{14 a^3 b (a+b x) \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(7/2)*(7*a^2*(A*b - a*B) + (-3*A*b + 7*a*B)*(a + b*x)^2*Hypergeometric2F1[2, 7/2, 9/2, -((b*x)/a)]))/(14*a^
3*b*(a + b*x)*Sqrt[(a + b*x)^2])

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fricas [A]  time = 1.04, size = 349, normalized size = 1.37 \[ \left [-\frac {15 \, {\left (7 \, B a^{3} - 3 \, A a^{2} b + {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} + 2 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (8 \, B b^{3} x^{3} - 105 \, B a^{3} + 45 \, A a^{2} b - 8 \, {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} - 25 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt {x}}{24 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, \frac {15 \, {\left (7 \, B a^{3} - 3 \, A a^{2} b + {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} + 2 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (8 \, B b^{3} x^{3} - 105 \, B a^{3} + 45 \, A a^{2} b - 8 \, {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} - 25 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt {x}}{12 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/24*(15*(7*B*a^3 - 3*A*a^2*b + (7*B*a*b^2 - 3*A*b^3)*x^2 + 2*(7*B*a^2*b - 3*A*a*b^2)*x)*sqrt(-a/b)*log((b*x
- 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) - 2*(8*B*b^3*x^3 - 105*B*a^3 + 45*A*a^2*b - 8*(7*B*a*b^2 - 3*A*b^3)*
x^2 - 25*(7*B*a^2*b - 3*A*a*b^2)*x)*sqrt(x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4), 1/12*(15*(7*B*a^3 - 3*A*a^2*b +
(7*B*a*b^2 - 3*A*b^3)*x^2 + 2*(7*B*a^2*b - 3*A*a*b^2)*x)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (8*B*b^3*x^
3 - 105*B*a^3 + 45*A*a^2*b - 8*(7*B*a*b^2 - 3*A*b^3)*x^2 - 25*(7*B*a^2*b - 3*A*a*b^2)*x)*sqrt(x))/(b^6*x^2 + 2
*a*b^5*x + a^2*b^4)]

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giac [A]  time = 0.26, size = 143, normalized size = 0.56 \[ \frac {5 \, {\left (7 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{4} \mathrm {sgn}\left (b x + a\right )} - \frac {13 \, B a^{2} b x^{\frac {3}{2}} - 9 \, A a b^{2} x^{\frac {3}{2}} + 11 \, B a^{3} \sqrt {x} - 7 \, A a^{2} b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} b^{4} \mathrm {sgn}\left (b x + a\right )} + \frac {2 \, {\left (B b^{6} x^{\frac {3}{2}} - 9 \, B a b^{5} \sqrt {x} + 3 \, A b^{6} \sqrt {x}\right )}}{3 \, b^{9} \mathrm {sgn}\left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/4*(7*B*a^2 - 3*A*a*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4*sgn(b*x + a)) - 1/4*(13*B*a^2*b*x^(3/2) - 9
*A*a*b^2*x^(3/2) + 11*B*a^3*sqrt(x) - 7*A*a^2*b*sqrt(x))/((b*x + a)^2*b^4*sgn(b*x + a)) + 2/3*(B*b^6*x^(3/2) -
9*B*a*b^5*sqrt(x) + 3*A*b^6*sqrt(x))/(b^9*sgn(b*x + a))

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maple [A]  time = 0.07, size = 247, normalized size = 0.97 \[ \frac {\left (-45 A a \,b^{3} x^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+105 B \,a^{2} b^{2} x^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+8 \sqrt {a b}\, B \,b^{3} x^{\frac {7}{2}}-90 A \,a^{2} b^{2} x \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+210 B \,a^{3} b x \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+24 \sqrt {a b}\, A \,b^{3} x^{\frac {5}{2}}-56 \sqrt {a b}\, B a \,b^{2} x^{\frac {5}{2}}-45 A \,a^{3} b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+105 B \,a^{4} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+75 \sqrt {a b}\, A a \,b^{2} x^{\frac {3}{2}}-175 \sqrt {a b}\, B \,a^{2} b \,x^{\frac {3}{2}}+45 \sqrt {a b}\, A \,a^{2} b \sqrt {x}-105 \sqrt {a b}\, B \,a^{3} \sqrt {x}\right ) \left (b x +a \right )}{12 \sqrt {a b}\, \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(8*(a*b)^(1/2)*B*b^3*x^(7/2)-56*(a*b)^(1/2)*B*a*b^2*x^(5/2)+75*(a*b)^(1/2)*A*a*b^2*x^(3/2)+24*(a*b)^(1/2)
*A*b^3*x^(5/2)-45*A*arctan(1/(a*b)^(1/2)*b*x^(1/2))*x^2*a*b^3-175*(a*b)^(1/2)*B*a^2*b*x^(3/2)+105*B*arctan(1/(
a*b)^(1/2)*b*x^(1/2))*x^2*a^2*b^2-90*A*arctan(1/(a*b)^(1/2)*b*x^(1/2))*x*a^2*b^2+210*B*arctan(1/(a*b)^(1/2)*b*
x^(1/2))*x*a^3*b+45*(a*b)^(1/2)*A*a^2*b*x^(1/2)-45*A*a^3*b*arctan(1/(a*b)^(1/2)*b*x^(1/2))-105*(a*b)^(1/2)*B*a
^3*x^(1/2)+105*B*a^4*arctan(1/(a*b)^(1/2)*b*x^(1/2)))*(b*x+a)/(a*b)^(1/2)/b^4/((b*x+a)^2)^(3/2)

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maxima [A]  time = 1.70, size = 252, normalized size = 0.99 \[ -\frac {{\left ({\left (89 \, B a b^{3} - 35 \, A b^{4}\right )} x^{2} + 3 \, {\left (19 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x\right )} x^{\frac {5}{2}} + 12 \, {\left (4 \, {\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} + {\left (7 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} x^{\frac {3}{2}} + {\left (21 \, {\left (3 \, B a^{3} b - A a^{2} b^{2}\right )} x^{2} + 5 \, {\left (7 \, B a^{4} - A a^{3} b\right )} x\right )} \sqrt {x}}{24 \, {\left (a b^{6} x^{3} + 3 \, a^{2} b^{5} x^{2} + 3 \, a^{3} b^{4} x + a^{4} b^{3}\right )}} + \frac {5 \, {\left (7 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{4}} + \frac {5 \, {\left (7 \, {\left (3 \, B a b - A b^{2}\right )} x^{\frac {3}{2}} - 6 \, {\left (7 \, B a^{2} - 3 \, A a b\right )} \sqrt {x}\right )}}{24 \, a b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(((89*B*a*b^3 - 35*A*b^4)*x^2 + 3*(19*B*a^2*b^2 - 5*A*a*b^3)*x)*x^(5/2) + 12*(4*(3*B*a^2*b^2 - A*a*b^3)*
x^2 + (7*B*a^3*b - A*a^2*b^2)*x)*x^(3/2) + (21*(3*B*a^3*b - A*a^2*b^2)*x^2 + 5*(7*B*a^4 - A*a^3*b)*x)*sqrt(x))
/(a*b^6*x^3 + 3*a^2*b^5*x^2 + 3*a^3*b^4*x + a^4*b^3) + 5/4*(7*B*a^2 - 3*A*a*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sq
rt(a*b)*b^4) + 5/24*(7*(3*B*a*b - A*b^2)*x^(3/2) - 6*(7*B*a^2 - 3*A*a*b)*sqrt(x))/(a*b^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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