Integrand size = 31, antiderivative size = 262 \[ \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {(5 A b+3 a B) \sqrt {x}}{64 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{3/2}}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(5 A b+3 a B) \sqrt {x}}{24 a b^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(5 A b+3 a B) \sqrt {x}}{96 a^2 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(5 A b+3 a B) (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{64 a^{7/2} b^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
1/4*(A*b-B*a)*x^(3/2)/a/b/(b*x+a)^3/((b*x+a)^2)^(1/2)+1/64*(5*A*b+3*B*a)*( b*x+a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/a^(7/2)/b^(5/2)/((b*x+a)^2)^(1/2)+1 /64*(5*A*b+3*B*a)*x^(1/2)/a^3/b^2/((b*x+a)^2)^(1/2)-1/24*(5*A*b+3*B*a)*x^( 1/2)/a/b^2/(b*x+a)^2/((b*x+a)^2)^(1/2)+1/96*(5*A*b+3*B*a)*x^(1/2)/a^2/b^2/ (b*x+a)/((b*x+a)^2)^(1/2)
Time = 0.22 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.56 \[ \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {a} \sqrt {b} \sqrt {x} \left (-9 a^4 B+15 A b^4 x^3+a b^3 x^2 (55 A+9 B x)-3 a^3 b (5 A+11 B x)+a^2 b^2 x (73 A+33 B x)\right )+3 (5 A b+3 a B) (a+b x)^4 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{192 a^{7/2} b^{5/2} (a+b x)^3 \sqrt {(a+b x)^2}} \]
(Sqrt[a]*Sqrt[b]*Sqrt[x]*(-9*a^4*B + 15*A*b^4*x^3 + a*b^3*x^2*(55*A + 9*B* x) - 3*a^3*b*(5*A + 11*B*x) + a^2*b^2*x*(73*A + 33*B*x)) + 3*(5*A*b + 3*a* B)*(a + b*x)^4*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(192*a^(7/2)*b^(5/2)*(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, 52, 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.
| \(\displaystyle \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b^5 (a+b x) \int \frac {\sqrt {x} (A+B x)}{b^5 (a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {\sqrt {x} (A+B x)}{(a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(3 a B+5 A b) \int \frac {\sqrt {x}}{(a+b x)^4}dx}{8 a b}+\frac {x^{3/2} (A b-a B)}{4 a b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(3 a B+5 A b) \left (\frac {\int \frac {1}{\sqrt {x} (a+b x)^3}dx}{6 b}-\frac {\sqrt {x}}{3 b (a+b x)^3}\right )}{8 a b}+\frac {x^{3/2} (A b-a B)}{4 a b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(3 a B+5 A b) \left (\frac {\frac {3 \int \frac {1}{\sqrt {x} (a+b x)^2}dx}{4 a}+\frac {\sqrt {x}}{2 a (a+b x)^2}}{6 b}-\frac {\sqrt {x}}{3 b (a+b x)^3}\right )}{8 a b}+\frac {x^{3/2} (A b-a B)}{4 a b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(3 a B+5 A b) \left (\frac {\frac {3 \left (\frac {\int \frac {1}{\sqrt {x} (a+b x)}dx}{2 a}+\frac {\sqrt {x}}{a (a+b x)}\right )}{4 a}+\frac {\sqrt {x}}{2 a (a+b x)^2}}{6 b}-\frac {\sqrt {x}}{3 b (a+b x)^3}\right )}{8 a b}+\frac {x^{3/2} (A b-a B)}{4 a b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(3 a B+5 A b) \left (\frac {\frac {3 \left (\frac {\int \frac {1}{a+b x}d\sqrt {x}}{a}+\frac {\sqrt {x}}{a (a+b x)}\right )}{4 a}+\frac {\sqrt {x}}{2 a (a+b x)^2}}{6 b}-\frac {\sqrt {x}}{3 b (a+b x)^3}\right )}{8 a b}+\frac {x^{3/2} (A b-a B)}{4 a b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(3 a B+5 A b) \left (\frac {\frac {3 \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b}}+\frac {\sqrt {x}}{a (a+b x)}\right )}{4 a}+\frac {\sqrt {x}}{2 a (a+b x)^2}}{6 b}-\frac {\sqrt {x}}{3 b (a+b x)^3}\right )}{8 a b}+\frac {x^{3/2} (A b-a B)}{4 a b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(((A*b - a*B)*x^(3/2))/(4*a*b*(a + b*x)^4) + ((5*A*b + 3*a*B)*( -1/3*Sqrt[x]/(b*(a + b*x)^3) + (Sqrt[x]/(2*a*(a + b*x)^2) + (3*(Sqrt[x]/(a *(a + b*x)) + ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]]/(a^(3/2)*Sqrt[b])))/(4*a)) /(6*b)))/(8*a*b)))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.9.31.3.1 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 = 0.15 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.36
| method | result | size |
| default | \(\frac {\left (15 A \,x^{\frac {7}{2}} \sqrt {b a}\, b^{4}+9 B \,x^{\frac {7}{2}} \sqrt {b a}\, a \,b^{3}+55 A \,x^{\frac {5}{2}} \sqrt {b a}\, a \,b^{3}+15 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) b^{5} x^{4}+33 B \,x^{\frac {5}{2}} \sqrt {b a}\, a^{2} b^{2}+9 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a \,b^{4} x^{4}+60 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a \,b^{4} x^{3}+36 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{2} b^{3} x^{3}+73 A \,x^{\frac {3}{2}} \sqrt {b a}\, a^{2} b^{2}+90 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{2} b^{3} x^{2}-33 B \,x^{\frac {3}{2}} \sqrt {b a}\, a^{3} b +54 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{3} b^{2} x^{2}+60 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{3} b^{2} x +36 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{4} b x -15 A \sqrt {x}\, \sqrt {b a}\, a^{3} b +15 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{4} b -9 B \sqrt {x}\, \sqrt {b a}\, a^{4}+9 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{5}\right ) \left (b x +a \right )}{192 \sqrt {b a}\, b^{2} a^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(357\) |
1/192*(15*A*x^(7/2)*(b*a)^(1/2)*b^4+9*B*x^(7/2)*(b*a)^(1/2)*a*b^3+55*A*x^( 5/2)*(b*a)^(1/2)*a*b^3+15*A*arctan(b*x^(1/2)/(b*a)^(1/2))*b^5*x^4+33*B*x^( 5/2)*(b*a)^(1/2)*a^2*b^2+9*B*arctan(b*x^(1/2)/(b*a)^(1/2))*a*b^4*x^4+60*A* arctan(b*x^(1/2)/(b*a)^(1/2))*a*b^4*x^3+36*B*arctan(b*x^(1/2)/(b*a)^(1/2)) *a^2*b^3*x^3+73*A*x^(3/2)*(b*a)^(1/2)*a^2*b^2+90*A*arctan(b*x^(1/2)/(b*a)^ (1/2))*a^2*b^3*x^2-33*B*x^(3/2)*(b*a)^(1/2)*a^3*b+54*B*arctan(b*x^(1/2)/(b *a)^(1/2))*a^3*b^2*x^2+60*A*arctan(b*x^(1/2)/(b*a)^(1/2))*a^3*b^2*x+36*B*a rctan(b*x^(1/2)/(b*a)^(1/2))*a^4*b*x-15*A*x^(1/2)*(b*a)^(1/2)*a^3*b+15*A*a rctan(b*x^(1/2)/(b*a)^(1/2))*a^4*b-9*B*x^(1/2)*(b*a)^(1/2)*a^4+9*B*arctan( b*x^(1/2)/(b*a)^(1/2))*a^5)*(b*x+a)/(b*a)^(1/2)/b^2/a^3/((b*x+a)^2)^(5/2)
Time = 0.33 (sec) , antiderivative size = 537, normalized size of antiderivative = 2.05 \[ \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left (3 \, B a^{5} + 5 \, A a^{4} b + {\left (3 \, B a b^{4} + 5 \, A b^{5}\right )} x^{4} + 4 \, {\left (3 \, B a^{2} b^{3} + 5 \, A a b^{4}\right )} x^{3} + 6 \, {\left (3 \, B a^{3} b^{2} + 5 \, A a^{2} b^{3}\right )} x^{2} + 4 \, {\left (3 \, B a^{4} b + 5 \, A a^{3} b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (9 \, B a^{5} b + 15 \, A a^{4} b^{2} - 3 \, {\left (3 \, B a^{2} b^{4} + 5 \, A a b^{5}\right )} x^{3} - 11 \, {\left (3 \, B a^{3} b^{3} + 5 \, A a^{2} b^{4}\right )} x^{2} + {\left (33 \, B a^{4} b^{2} - 73 \, A a^{3} b^{3}\right )} x\right )} \sqrt {x}}{384 \, {\left (a^{4} b^{7} x^{4} + 4 \, a^{5} b^{6} x^{3} + 6 \, a^{6} b^{5} x^{2} + 4 \, a^{7} b^{4} x + a^{8} b^{3}\right )}}, -\frac {3 \, {\left (3 \, B a^{5} + 5 \, A a^{4} b + {\left (3 \, B a b^{4} + 5 \, A b^{5}\right )} x^{4} + 4 \, {\left (3 \, B a^{2} b^{3} + 5 \, A a b^{4}\right )} x^{3} + 6 \, {\left (3 \, B a^{3} b^{2} + 5 \, A a^{2} b^{3}\right )} x^{2} + 4 \, {\left (3 \, B a^{4} b + 5 \, A a^{3} b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (9 \, B a^{5} b + 15 \, A a^{4} b^{2} - 3 \, {\left (3 \, B a^{2} b^{4} + 5 \, A a b^{5}\right )} x^{3} - 11 \, {\left (3 \, B a^{3} b^{3} + 5 \, A a^{2} b^{4}\right )} x^{2} + {\left (33 \, B a^{4} b^{2} - 73 \, A a^{3} b^{3}\right )} x\right )} \sqrt {x}}{192 \, {\left (a^{4} b^{7} x^{4} + 4 \, a^{5} b^{6} x^{3} + 6 \, a^{6} b^{5} x^{2} + 4 \, a^{7} b^{4} x + a^{8} b^{3}\right )}}\right ] \]
[-1/384*(3*(3*B*a^5 + 5*A*a^4*b + (3*B*a*b^4 + 5*A*b^5)*x^4 + 4*(3*B*a^2*b ^3 + 5*A*a*b^4)*x^3 + 6*(3*B*a^3*b^2 + 5*A*a^2*b^3)*x^2 + 4*(3*B*a^4*b + 5 *A*a^3*b^2)*x)*sqrt(-a*b)*log((b*x - a - 2*sqrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(9*B*a^5*b + 15*A*a^4*b^2 - 3*(3*B*a^2*b^4 + 5*A*a*b^5)*x^3 - 11*(3*B* a^3*b^3 + 5*A*a^2*b^4)*x^2 + (33*B*a^4*b^2 - 73*A*a^3*b^3)*x)*sqrt(x))/(a^ 4*b^7*x^4 + 4*a^5*b^6*x^3 + 6*a^6*b^5*x^2 + 4*a^7*b^4*x + a^8*b^3), -1/192 *(3*(3*B*a^5 + 5*A*a^4*b + (3*B*a*b^4 + 5*A*b^5)*x^4 + 4*(3*B*a^2*b^3 + 5* A*a*b^4)*x^3 + 6*(3*B*a^3*b^2 + 5*A*a^2*b^3)*x^2 + 4*(3*B*a^4*b + 5*A*a^3* b^2)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (9*B*a^5*b + 15*A*a^4*b^ 2 - 3*(3*B*a^2*b^4 + 5*A*a*b^5)*x^3 - 11*(3*B*a^3*b^3 + 5*A*a^2*b^4)*x^2 + (33*B*a^4*b^2 - 73*A*a^3*b^3)*x)*sqrt(x))/(a^4*b^7*x^4 + 4*a^5*b^6*x^3 + 6*a^6*b^5*x^2 + 4*a^7*b^4*x + a^8*b^3)]
Timed out. \[ \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 368 vs. \(2 (179) = 358\).
Time = 0.35 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.40 \[ \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {15 \, {\left ({\left (B a b^{5} + A b^{6}\right )} x^{2} - {\left (3 \, B a^{2} b^{4} + 7 \, A a b^{5}\right )} x\right )} x^{\frac {9}{2}} + 30 \, {\left ({\left (B a^{2} b^{4} + A a b^{5}\right )} x^{2} - 3 \, {\left (3 \, B a^{3} b^{3} + 7 \, A a^{2} b^{4}\right )} x\right )} x^{\frac {7}{2}} - 20 \, {\left (6 \, {\left (B a^{3} b^{3} + A a^{2} b^{4}\right )} x^{2} + 11 \, {\left (3 \, B a^{4} b^{2} + 7 \, A a^{3} b^{3}\right )} x\right )} x^{\frac {5}{2}} - 2 \, {\left (255 \, {\left (B a^{4} b^{2} + A a^{3} b^{3}\right )} x^{2} + 139 \, {\left (3 \, B a^{5} b + 7 \, A a^{4} b^{2}\right )} x\right )} x^{\frac {3}{2}} + {\left (3 \, {\left (3 \, B a^{5} b - 253 \, A a^{4} b^{2}\right )} x^{2} - 5 \, {\left (3 \, B a^{6} + 263 \, A a^{5} b\right )} x\right )} \sqrt {x}}{1920 \, {\left (a^{5} b^{6} x^{5} + 5 \, a^{6} b^{5} x^{4} + 10 \, a^{7} b^{4} x^{3} + 10 \, a^{8} b^{3} x^{2} + 5 \, a^{9} b^{2} x + a^{10} b\right )}} + \frac {{\left (3 \, B a + 5 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{64 \, \sqrt {a b} a^{3} b^{2}} + \frac {{\left (B a b + A b^{2}\right )} x^{\frac {3}{2}} - 2 \, {\left (3 \, B a^{2} + 5 \, A a b\right )} \sqrt {x}}{128 \, a^{5} b^{2}} \]
-1/1920*(15*((B*a*b^5 + A*b^6)*x^2 - (3*B*a^2*b^4 + 7*A*a*b^5)*x)*x^(9/2) + 30*((B*a^2*b^4 + A*a*b^5)*x^2 - 3*(3*B*a^3*b^3 + 7*A*a^2*b^4)*x)*x^(7/2) - 20*(6*(B*a^3*b^3 + A*a^2*b^4)*x^2 + 11*(3*B*a^4*b^2 + 7*A*a^3*b^3)*x)*x ^(5/2) - 2*(255*(B*a^4*b^2 + A*a^3*b^3)*x^2 + 139*(3*B*a^5*b + 7*A*a^4*b^2 )*x)*x^(3/2) + (3*(3*B*a^5*b - 253*A*a^4*b^2)*x^2 - 5*(3*B*a^6 + 263*A*a^5 *b)*x)*sqrt(x))/(a^5*b^6*x^5 + 5*a^6*b^5*x^4 + 10*a^7*b^4*x^3 + 10*a^8*b^3 *x^2 + 5*a^9*b^2*x + a^10*b) + 1/64*(3*B*a + 5*A*b)*arctan(b*sqrt(x)/sqrt( a*b))/(sqrt(a*b)*a^3*b^2) + 1/128*((B*a*b + A*b^2)*x^(3/2) - 2*(3*B*a^2 + 5*A*a*b)*sqrt(x))/(a^5*b^2)
Time = 0.27 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.56 \[ \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {{\left (3 \, B a + 5 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{64 \, \sqrt {a b} a^{3} b^{2} \mathrm {sgn}\left (b x + a\right )} + \frac {9 \, B a b^{3} x^{\frac {7}{2}} + 15 \, A b^{4} x^{\frac {7}{2}} + 33 \, B a^{2} b^{2} x^{\frac {5}{2}} + 55 \, A a b^{3} x^{\frac {5}{2}} - 33 \, B a^{3} b x^{\frac {3}{2}} + 73 \, A a^{2} b^{2} x^{\frac {3}{2}} - 9 \, B a^{4} \sqrt {x} - 15 \, A a^{3} b \sqrt {x}}{192 \, {\left (b x + a\right )}^{4} a^{3} b^{2} \mathrm {sgn}\left (b x + a\right )} \]
1/64*(3*B*a + 5*A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3*b^2*sgn(b* x + a)) + 1/192*(9*B*a*b^3*x^(7/2) + 15*A*b^4*x^(7/2) + 33*B*a^2*b^2*x^(5/ 2) + 55*A*a*b^3*x^(5/2) - 33*B*a^3*b*x^(3/2) + 73*A*a^2*b^2*x^(3/2) - 9*B* a^4*sqrt(x) - 15*A*a^3*b*sqrt(x))/((b*x + a)^4*a^3*b^2*sgn(b*x + a))
Timed out. \[ \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {\sqrt {x}\,\left (A+B\,x\right )}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]