Integrand size = 31, antiderivative size = 262 \[ \int \frac {x^{3/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {(3 A b+5 a B) \sqrt {x}}{64 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{5/2}}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(3 A b+5 a B) x^{3/2}}{24 a b^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(3 A b+5 a B) \sqrt {x}}{32 a b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(3 A b+5 a B) (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{64 a^{5/2} b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
1/4*(A*b-B*a)*x^(5/2)/a/b/(b*x+a)^3/((b*x+a)^2)^(1/2)-1/24*(3*A*b+5*B*a)*x ^(3/2)/a/b^2/(b*x+a)^2/((b*x+a)^2)^(1/2)+1/64*(3*A*b+5*B*a)*(b*x+a)*arctan (b^(1/2)*x^(1/2)/a^(1/2))/a^(5/2)/b^(7/2)/((b*x+a)^2)^(1/2)+1/64*(3*A*b+5* B*a)*x^(1/2)/a^2/b^3/((b*x+a)^2)^(1/2)-1/32*(3*A*b+5*B*a)*x^(1/2)/a/b^3/(b *x+a)/((b*x+a)^2)^(1/2)
Time = 0.25 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.56 \[ \int \frac {x^{3/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {a} \sqrt {b} \sqrt {x} \left (-15 a^4 B+9 A b^4 x^3+3 a b^3 x^2 (11 A+5 B x)-a^3 b (9 A+55 B x)-a^2 b^2 x (33 A+73 B x)\right )+3 (3 A b+5 a B) (a+b x)^4 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{192 a^{5/2} b^{7/2} (a+b x)^3 \sqrt {(a+b x)^2}} \]
(Sqrt[a]*Sqrt[b]*Sqrt[x]*(-15*a^4*B + 9*A*b^4*x^3 + 3*a*b^3*x^2*(11*A + 5* B*x) - a^3*b*(9*A + 55*B*x) - a^2*b^2*x*(33*A + 73*B*x)) + 3*(3*A*b + 5*a* B)*(a + b*x)^4*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(192*a^(5/2)*b^(7/2)*(a + b*x)^3*Sqrt[(a + b*x)^2])
Time = 0.28 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.67, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {1187, 27, 87, 51, 51, 52, 73, 218}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {x^{3/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b^5 (a+b x) \int \frac {x^{3/2} (A+B x)}{b^5 (a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {x^{3/2} (A+B x)}{(a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(5 a B+3 A b) \int \frac {x^{3/2}}{(a+b x)^4}dx}{8 a b}+\frac {x^{5/2} (A b-a B)}{4 a b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(5 a B+3 A b) \left (\frac {\int \frac {\sqrt {x}}{(a+b x)^3}dx}{2 b}-\frac {x^{3/2}}{3 b (a+b x)^3}\right )}{8 a b}+\frac {x^{5/2} (A b-a B)}{4 a b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(5 a B+3 A b) \left (\frac {\frac {\int \frac {1}{\sqrt {x} (a+b x)^2}dx}{4 b}-\frac {\sqrt {x}}{2 b (a+b x)^2}}{2 b}-\frac {x^{3/2}}{3 b (a+b x)^3}\right )}{8 a b}+\frac {x^{5/2} (A b-a B)}{4 a b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(5 a B+3 A b) \left (\frac {\frac {\frac {\int \frac {1}{\sqrt {x} (a+b x)}dx}{2 a}+\frac {\sqrt {x}}{a (a+b x)}}{4 b}-\frac {\sqrt {x}}{2 b (a+b x)^2}}{2 b}-\frac {x^{3/2}}{3 b (a+b x)^3}\right )}{8 a b}+\frac {x^{5/2} (A b-a B)}{4 a b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(5 a B+3 A b) \left (\frac {\frac {\frac {\int \frac {1}{a+b x}d\sqrt {x}}{a}+\frac {\sqrt {x}}{a (a+b x)}}{4 b}-\frac {\sqrt {x}}{2 b (a+b x)^2}}{2 b}-\frac {x^{3/2}}{3 b (a+b x)^3}\right )}{8 a b}+\frac {x^{5/2} (A b-a B)}{4 a b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(5 a B+3 A b) \left (\frac {\frac {\frac {\arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b}}+\frac {\sqrt {x}}{a (a+b x)}}{4 b}-\frac {\sqrt {x}}{2 b (a+b x)^2}}{2 b}-\frac {x^{3/2}}{3 b (a+b x)^3}\right )}{8 a b}+\frac {x^{5/2} (A b-a B)}{4 a b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(((A*b - a*B)*x^(5/2))/(4*a*b*(a + b*x)^4) + ((3*A*b + 5*a*B)*( -1/3*x^(3/2)/(b*(a + b*x)^3) + (-1/2*Sqrt[x]/(b*(a + b*x)^2) + (Sqrt[x]/(a *(a + b*x)) + ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]]/(a^(3/2)*Sqrt[b]))/(4*b))/ (2*b)))/(8*a*b)))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.9.30.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(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] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(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)^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p]
Time = 0.14 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.36
| method | result | size |
| default | \(\frac {\left (9 A \,x^{\frac {7}{2}} \sqrt {b a}\, b^{4}+15 B \,x^{\frac {7}{2}} \sqrt {b a}\, a \,b^{3}+33 A \,x^{\frac {5}{2}} \sqrt {b a}\, a \,b^{3}+9 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) b^{5} x^{4}-73 B \,x^{\frac {5}{2}} \sqrt {b a}\, a^{2} b^{2}+15 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a \,b^{4} x^{4}+36 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a \,b^{4} x^{3}+60 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{2} b^{3} x^{3}-33 A \,x^{\frac {3}{2}} \sqrt {b a}\, a^{2} b^{2}+54 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{2} b^{3} x^{2}-55 B \,x^{\frac {3}{2}} \sqrt {b a}\, a^{3} b +90 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{3} b^{2} x^{2}+36 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{3} b^{2} x +60 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{4} b x -9 A \sqrt {x}\, \sqrt {b a}\, a^{3} b +9 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{4} b -15 B \sqrt {x}\, \sqrt {b a}\, a^{4}+15 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{5}\right ) \left (b x +a \right )}{192 \sqrt {b a}\, b^{3} a^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(357\) |
1/192*(9*A*x^(7/2)*(b*a)^(1/2)*b^4+15*B*x^(7/2)*(b*a)^(1/2)*a*b^3+33*A*x^( 5/2)*(b*a)^(1/2)*a*b^3+9*A*arctan(b*x^(1/2)/(b*a)^(1/2))*b^5*x^4-73*B*x^(5 /2)*(b*a)^(1/2)*a^2*b^2+15*B*arctan(b*x^(1/2)/(b*a)^(1/2))*a*b^4*x^4+36*A* arctan(b*x^(1/2)/(b*a)^(1/2))*a*b^4*x^3+60*B*arctan(b*x^(1/2)/(b*a)^(1/2)) *a^2*b^3*x^3-33*A*x^(3/2)*(b*a)^(1/2)*a^2*b^2+54*A*arctan(b*x^(1/2)/(b*a)^ (1/2))*a^2*b^3*x^2-55*B*x^(3/2)*(b*a)^(1/2)*a^3*b+90*B*arctan(b*x^(1/2)/(b *a)^(1/2))*a^3*b^2*x^2+36*A*arctan(b*x^(1/2)/(b*a)^(1/2))*a^3*b^2*x+60*B*a rctan(b*x^(1/2)/(b*a)^(1/2))*a^4*b*x-9*A*x^(1/2)*(b*a)^(1/2)*a^3*b+9*A*arc tan(b*x^(1/2)/(b*a)^(1/2))*a^4*b-15*B*x^(1/2)*(b*a)^(1/2)*a^4+15*B*arctan( b*x^(1/2)/(b*a)^(1/2))*a^5)*(b*x+a)/(b*a)^(1/2)/b^3/a^2/((b*x+a)^2)^(5/2)
Time = 0.36 (sec) , antiderivative size = 537, normalized size of antiderivative = 2.05 \[ \int \frac {x^{3/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left (5 \, B a^{5} + 3 \, A a^{4} b + {\left (5 \, B a b^{4} + 3 \, A b^{5}\right )} x^{4} + 4 \, {\left (5 \, B a^{2} b^{3} + 3 \, A a b^{4}\right )} x^{3} + 6 \, {\left (5 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\right )} x^{2} + 4 \, {\left (5 \, B a^{4} b + 3 \, A a^{3} b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (15 \, B a^{5} b + 9 \, A a^{4} b^{2} - 3 \, {\left (5 \, B a^{2} b^{4} + 3 \, A a b^{5}\right )} x^{3} + {\left (73 \, B a^{3} b^{3} - 33 \, A a^{2} b^{4}\right )} x^{2} + 11 \, {\left (5 \, B a^{4} b^{2} + 3 \, A a^{3} b^{3}\right )} x\right )} \sqrt {x}}{384 \, {\left (a^{3} b^{8} x^{4} + 4 \, a^{4} b^{7} x^{3} + 6 \, a^{5} b^{6} x^{2} + 4 \, a^{6} b^{5} x + a^{7} b^{4}\right )}}, -\frac {3 \, {\left (5 \, B a^{5} + 3 \, A a^{4} b + {\left (5 \, B a b^{4} + 3 \, A b^{5}\right )} x^{4} + 4 \, {\left (5 \, B a^{2} b^{3} + 3 \, A a b^{4}\right )} x^{3} + 6 \, {\left (5 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\right )} x^{2} + 4 \, {\left (5 \, B a^{4} b + 3 \, A a^{3} b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (15 \, B a^{5} b + 9 \, A a^{4} b^{2} - 3 \, {\left (5 \, B a^{2} b^{4} + 3 \, A a b^{5}\right )} x^{3} + {\left (73 \, B a^{3} b^{3} - 33 \, A a^{2} b^{4}\right )} x^{2} + 11 \, {\left (5 \, B a^{4} b^{2} + 3 \, A a^{3} b^{3}\right )} x\right )} \sqrt {x}}{192 \, {\left (a^{3} b^{8} x^{4} + 4 \, a^{4} b^{7} x^{3} + 6 \, a^{5} b^{6} x^{2} + 4 \, a^{6} b^{5} x + a^{7} b^{4}\right )}}\right ] \]
[-1/384*(3*(5*B*a^5 + 3*A*a^4*b + (5*B*a*b^4 + 3*A*b^5)*x^4 + 4*(5*B*a^2*b ^3 + 3*A*a*b^4)*x^3 + 6*(5*B*a^3*b^2 + 3*A*a^2*b^3)*x^2 + 4*(5*B*a^4*b + 3 *A*a^3*b^2)*x)*sqrt(-a*b)*log((b*x - a - 2*sqrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(15*B*a^5*b + 9*A*a^4*b^2 - 3*(5*B*a^2*b^4 + 3*A*a*b^5)*x^3 + (73*B*a^ 3*b^3 - 33*A*a^2*b^4)*x^2 + 11*(5*B*a^4*b^2 + 3*A*a^3*b^3)*x)*sqrt(x))/(a^ 3*b^8*x^4 + 4*a^4*b^7*x^3 + 6*a^5*b^6*x^2 + 4*a^6*b^5*x + a^7*b^4), -1/192 *(3*(5*B*a^5 + 3*A*a^4*b + (5*B*a*b^4 + 3*A*b^5)*x^4 + 4*(5*B*a^2*b^3 + 3* A*a*b^4)*x^3 + 6*(5*B*a^3*b^2 + 3*A*a^2*b^3)*x^2 + 4*(5*B*a^4*b + 3*A*a^3* b^2)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (15*B*a^5*b + 9*A*a^4*b^ 2 - 3*(5*B*a^2*b^4 + 3*A*a*b^5)*x^3 + (73*B*a^3*b^3 - 33*A*a^2*b^4)*x^2 + 11*(5*B*a^4*b^2 + 3*A*a^3*b^3)*x)*sqrt(x))/(a^3*b^8*x^4 + 4*a^4*b^7*x^3 + 6*a^5*b^6*x^2 + 4*a^6*b^5*x + a^7*b^4)]
Timed out. \[ \int \frac {x^{3/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 372 vs. \(2 (179) = 358\).
Time = 0.35 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.42 \[ \int \frac {x^{3/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {5 \, {\left ({\left (7 \, B a b^{5} + 3 \, A b^{6}\right )} x^{2} - 9 \, {\left (B a^{2} b^{4} + A a b^{5}\right )} x\right )} x^{\frac {9}{2}} + 10 \, {\left ({\left (7 \, B a^{2} b^{4} + 3 \, A a b^{5}\right )} x^{2} - 27 \, {\left (B a^{3} b^{3} + A a^{2} b^{4}\right )} x\right )} x^{\frac {7}{2}} - 20 \, {\left (2 \, {\left (7 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{2} + 33 \, {\left (B a^{4} b^{2} + A a^{3} b^{3}\right )} x\right )} x^{\frac {5}{2}} + 6 \, {\left (5 \, {\left (3 \, B a^{4} b^{2} - 17 \, A a^{3} b^{3}\right )} x^{2} - {\left (11 \, B a^{5} b + 139 \, A a^{4} b^{2}\right )} x\right )} x^{\frac {3}{2}} + 3 \, {\left ({\left (7 \, B a^{5} b + 3 \, A a^{4} b^{2}\right )} x^{2} - 5 \, {\left (B a^{6} + A a^{5} b\right )} x\right )} \sqrt {x}}{1920 \, {\left (a^{4} b^{7} x^{5} + 5 \, a^{5} b^{6} x^{4} + 10 \, a^{6} b^{5} x^{3} + 10 \, a^{7} b^{4} x^{2} + 5 \, a^{8} b^{3} x + a^{9} b^{2}\right )}} + \frac {{\left (5 \, B a + 3 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{64 \, \sqrt {a b} a^{2} b^{3}} + \frac {{\left (7 \, B a b + 3 \, A b^{2}\right )} x^{\frac {3}{2}} - 6 \, {\left (5 \, B a^{2} + 3 \, A a b\right )} \sqrt {x}}{384 \, a^{4} b^{3}} \]
-1/1920*(5*((7*B*a*b^5 + 3*A*b^6)*x^2 - 9*(B*a^2*b^4 + A*a*b^5)*x)*x^(9/2) + 10*((7*B*a^2*b^4 + 3*A*a*b^5)*x^2 - 27*(B*a^3*b^3 + A*a^2*b^4)*x)*x^(7/ 2) - 20*(2*(7*B*a^3*b^3 + 3*A*a^2*b^4)*x^2 + 33*(B*a^4*b^2 + A*a^3*b^3)*x) *x^(5/2) + 6*(5*(3*B*a^4*b^2 - 17*A*a^3*b^3)*x^2 - (11*B*a^5*b + 139*A*a^4 *b^2)*x)*x^(3/2) + 3*((7*B*a^5*b + 3*A*a^4*b^2)*x^2 - 5*(B*a^6 + A*a^5*b)* x)*sqrt(x))/(a^4*b^7*x^5 + 5*a^5*b^6*x^4 + 10*a^6*b^5*x^3 + 10*a^7*b^4*x^2 + 5*a^8*b^3*x + a^9*b^2) + 1/64*(5*B*a + 3*A*b)*arctan(b*sqrt(x)/sqrt(a*b ))/(sqrt(a*b)*a^2*b^3) + 1/384*((7*B*a*b + 3*A*b^2)*x^(3/2) - 6*(5*B*a^2 + 3*A*a*b)*sqrt(x))/(a^4*b^3)
Time = 0.29 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.56 \[ \int \frac {x^{3/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {{\left (5 \, B a + 3 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{64 \, \sqrt {a b} a^{2} b^{3} \mathrm {sgn}\left (b x + a\right )} + \frac {15 \, B a b^{3} x^{\frac {7}{2}} + 9 \, A b^{4} x^{\frac {7}{2}} - 73 \, B a^{2} b^{2} x^{\frac {5}{2}} + 33 \, A a b^{3} x^{\frac {5}{2}} - 55 \, B a^{3} b x^{\frac {3}{2}} - 33 \, A a^{2} b^{2} x^{\frac {3}{2}} - 15 \, B a^{4} \sqrt {x} - 9 \, A a^{3} b \sqrt {x}}{192 \, {\left (b x + a\right )}^{4} a^{2} b^{3} \mathrm {sgn}\left (b x + a\right )} \]
1/64*(5*B*a + 3*A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^2*b^3*sgn(b* x + a)) + 1/192*(15*B*a*b^3*x^(7/2) + 9*A*b^4*x^(7/2) - 73*B*a^2*b^2*x^(5/ 2) + 33*A*a*b^3*x^(5/2) - 55*B*a^3*b*x^(3/2) - 33*A*a^2*b^2*x^(3/2) - 15*B *a^4*sqrt(x) - 9*A*a^3*b*sqrt(x))/((b*x + a)^4*a^2*b^3*sgn(b*x + a))
Timed out. \[ \int \frac {x^{3/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {x^{3/2}\,\left (A+B\,x\right )}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]