Integrand size = 31, antiderivative size = 302 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {(5 A b-9 a B) x^{7/2}}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{9/2}}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 a (5 A b-9 a B) \sqrt {x} (a+b x)}{4 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 (5 A b-9 a B) x^{3/2} (a+b x)}{12 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 (5 A b-9 a B) x^{5/2} (a+b x)}{20 a b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 a^{3/2} (5 A b-9 a B) (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
1/4*(5*A*b-9*B*a)*x^(7/2)/a/b^2/((b*x+a)^2)^(1/2)+1/2*(A*b-B*a)*x^(9/2)/a/ b/(b*x+a)/((b*x+a)^2)^(1/2)+7/12*(5*A*b-9*B*a)*x^(3/2)*(b*x+a)/b^4/((b*x+a )^2)^(1/2)-7/20*(5*A*b-9*B*a)*x^(5/2)*(b*x+a)/a/b^3/((b*x+a)^2)^(1/2)+7/4* a^(3/2)*(5*A*b-9*B*a)*(b*x+a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/b^(11/2)/((b *x+a)^2)^(1/2)-7/4*a*(5*A*b-9*B*a)*(b*x+a)*x^(1/2)/b^5/((b*x+a)^2)^(1/2)
Time = 0.05 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.49 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {b} \sqrt {x} \left (945 a^4 B-525 a^3 b (A-3 B x)+8 b^4 x^3 (5 A+3 B x)-8 a b^3 x^2 (35 A+9 B x)+7 a^2 b^2 x (-125 A+72 B x)\right )-105 a^{3/2} (-5 A b+9 a B) (a+b x)^2 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{60 b^{11/2} (a+b x) \sqrt {(a+b x)^2}} \]
(Sqrt[b]*Sqrt[x]*(945*a^4*B - 525*a^3*b*(A - 3*B*x) + 8*b^4*x^3*(5*A + 3*B *x) - 8*a*b^3*x^2*(35*A + 9*B*x) + 7*a^2*b^2*x*(-125*A + 72*B*x)) - 105*a^ (3/2)*(-5*A*b + 9*a*B)*(a + b*x)^2*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(60* b^(11/2)*(a + b*x)*Sqrt[(a + b*x)^2])
Time = 0.28 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.59, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {1187, 27, 87, 51, 60, 60, 60, 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.
| \(\displaystyle \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b^3 (a+b x) \int \frac {x^{7/2} (A+B x)}{b^3 (a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {x^{7/2} (A+B x)}{(a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(a+b x) \left (\frac {x^{9/2} (A b-a B)}{2 a b (a+b x)^2}-\frac {(5 A b-9 a B) \int \frac {x^{7/2}}{(a+b x)^2}dx}{4 a b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {(a+b x) \left (\frac {x^{9/2} (A b-a B)}{2 a b (a+b x)^2}-\frac {(5 A b-9 a B) \left (\frac {7 \int \frac {x^{5/2}}{a+b x}dx}{2 b}-\frac {x^{7/2}}{b (a+b x)}\right )}{4 a b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x) \left (\frac {x^{9/2} (A b-a B)}{2 a b (a+b x)^2}-\frac {(5 A b-9 a B) \left (\frac {7 \left (\frac {2 x^{5/2}}{5 b}-\frac {a \int \frac {x^{3/2}}{a+b x}dx}{b}\right )}{2 b}-\frac {x^{7/2}}{b (a+b x)}\right )}{4 a b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x) \left (\frac {x^{9/2} (A b-a B)}{2 a b (a+b x)^2}-\frac {(5 A b-9 a B) \left (\frac {7 \left (\frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \int \frac {\sqrt {x}}{a+b x}dx}{b}\right )}{b}\right )}{2 b}-\frac {x^{7/2}}{b (a+b x)}\right )}{4 a b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x) \left (\frac {x^{9/2} (A b-a B)}{2 a b (a+b x)^2}-\frac {(5 A b-9 a B) \left (\frac {7 \left (\frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {a \int \frac {1}{\sqrt {x} (a+b x)}dx}{b}\right )}{b}\right )}{b}\right )}{2 b}-\frac {x^{7/2}}{b (a+b x)}\right )}{4 a b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (\frac {x^{9/2} (A b-a B)}{2 a b (a+b x)^2}-\frac {(5 A b-9 a B) \left (\frac {7 \left (\frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \int \frac {1}{a+b x}d\sqrt {x}}{b}\right )}{b}\right )}{b}\right )}{2 b}-\frac {x^{7/2}}{b (a+b x)}\right )}{4 a b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {(a+b x) \left (\frac {x^{9/2} (A b-a B)}{2 a b (a+b x)^2}-\frac {(5 A b-9 a B) \left (\frac {7 \left (\frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}}\right )}{b}\right )}{b}\right )}{2 b}-\frac {x^{7/2}}{b (a+b x)}\right )}{4 a b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(((A*b - a*B)*x^(9/2))/(2*a*b*(a + b*x)^2) - ((5*A*b - 9*a*B)*( -(x^(7/2)/(b*(a + b*x))) + (7*((2*x^(5/2))/(5*b) - (a*((2*x^(3/2))/(3*b) - (a*((2*Sqrt[x])/b - (2*Sqrt[a]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(3/2) ))/b))/b))/(2*b)))/(4*a*b)))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.9.18.3.1 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/(b*(m + n + 1))), x] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
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 = 0.18 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.50
| method | result | size |
| risch | \(-\frac {2 \left (-3 b^{2} B \,x^{2}-5 A \,b^{2} x +15 B a b x +45 A b a -90 B \,a^{2}\right ) \sqrt {x}\, \sqrt {\left (b x +a \right )^{2}}}{15 b^{5} \left (b x +a \right )}+\frac {a^{2} \left (\frac {2 \left (-\frac {13}{8} A \,b^{2}+\frac {17}{8} a b B \right ) x^{\frac {3}{2}}-\frac {a \left (11 A b -15 B a \right ) \sqrt {x}}{4}}{\left (b x +a \right )^{2}}+\frac {7 \left (5 A b -9 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right )}{4 \sqrt {b a}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b^{5} \left (b x +a \right )}\) | \(152\) |
| default | \(\frac {\left (24 B \,x^{\frac {9}{2}} \sqrt {b a}\, b^{4}+40 A \,x^{\frac {7}{2}} \sqrt {b a}\, b^{4}-72 B \,x^{\frac {7}{2}} \sqrt {b a}\, a \,b^{3}-280 A \,x^{\frac {5}{2}} \sqrt {b a}\, a \,b^{3}+504 B \,x^{\frac {5}{2}} \sqrt {b a}\, a^{2} b^{2}-875 A \,x^{\frac {3}{2}} \sqrt {b a}\, a^{2} b^{2}+525 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{2} b^{3} x^{2}+1575 B \,x^{\frac {3}{2}} \sqrt {b a}\, a^{3} b -945 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{3} b^{2} x^{2}+1050 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{3} b^{2} x -1890 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{4} b x -525 A \sqrt {x}\, \sqrt {b a}\, a^{3} b +525 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{4} b +945 B \sqrt {x}\, \sqrt {b a}\, a^{4}-945 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{5}\right ) \left (b x +a \right )}{60 \sqrt {b a}\, b^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\) | \(283\) |
-2/15*(-3*B*b^2*x^2-5*A*b^2*x+15*B*a*b*x+45*A*a*b-90*B*a^2)*x^(1/2)/b^5*(( b*x+a)^2)^(1/2)/(b*x+a)+a^2/b^5*(2*((-13/8*A*b^2+17/8*a*b*B)*x^(3/2)-1/8*a *(11*A*b-15*B*a)*x^(1/2))/(b*x+a)^2+7/4*(5*A*b-9*B*a)/(b*a)^(1/2)*arctan(b *x^(1/2)/(b*a)^(1/2)))*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.33 (sec) , antiderivative size = 408, normalized size of antiderivative = 1.35 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\left [-\frac {105 \, {\left (9 \, B a^{4} - 5 \, A a^{3} b + {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} + 2 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x + 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (24 \, B b^{4} x^{4} + 945 \, B a^{4} - 525 \, A a^{3} b - 8 \, {\left (9 \, B a b^{3} - 5 \, A b^{4}\right )} x^{3} + 56 \, {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} + 175 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{120 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}, -\frac {105 \, {\left (9 \, B a^{4} - 5 \, A a^{3} b + {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} + 2 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (24 \, B b^{4} x^{4} + 945 \, B a^{4} - 525 \, A a^{3} b - 8 \, {\left (9 \, B a b^{3} - 5 \, A b^{4}\right )} x^{3} + 56 \, {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} + 175 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{60 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}\right ] \]
[-1/120*(105*(9*B*a^4 - 5*A*a^3*b + (9*B*a^2*b^2 - 5*A*a*b^3)*x^2 + 2*(9*B *a^3*b - 5*A*a^2*b^2)*x)*sqrt(-a/b)*log((b*x + 2*b*sqrt(x)*sqrt(-a/b) - a) /(b*x + a)) - 2*(24*B*b^4*x^4 + 945*B*a^4 - 525*A*a^3*b - 8*(9*B*a*b^3 - 5 *A*b^4)*x^3 + 56*(9*B*a^2*b^2 - 5*A*a*b^3)*x^2 + 175*(9*B*a^3*b - 5*A*a^2* b^2)*x)*sqrt(x))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5), -1/60*(105*(9*B*a^4 - 5* A*a^3*b + (9*B*a^2*b^2 - 5*A*a*b^3)*x^2 + 2*(9*B*a^3*b - 5*A*a^2*b^2)*x)*s qrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) - (24*B*b^4*x^4 + 945*B*a^4 - 525*A *a^3*b - 8*(9*B*a*b^3 - 5*A*b^4)*x^3 + 56*(9*B*a^2*b^2 - 5*A*a*b^3)*x^2 + 175*(9*B*a^3*b - 5*A*a^2*b^2)*x)*sqrt(x))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)]
Timed out. \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\text {Timed out} \]
Time = 0.35 (sec) , antiderivative size = 273, normalized size of antiderivative = 0.90 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {16 \, {\left (3 \, B b^{4} x^{2} + 5 \, B a b^{3} x\right )} x^{\frac {7}{2}} + {\left (89 \, {\left (11 \, B a b^{3} - 5 \, A b^{4}\right )} x^{2} + 285 \, {\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} x^{\frac {5}{2}} + 12 \, {\left (12 \, {\left (11 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} + 35 \, {\left (3 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} x^{\frac {3}{2}} + 7 \, {\left (9 \, {\left (11 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} + 25 \, {\left (3 \, B a^{4} - A a^{3} b\right )} x\right )} \sqrt {x}}{120 \, {\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\right )}} - \frac {7 \, {\left (9 \, B a^{3} - 5 \, A a^{2} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{5}} - \frac {7 \, {\left ({\left (11 \, B a b - 5 \, A b^{2}\right )} x^{\frac {3}{2}} - 2 \, {\left (9 \, B a^{2} - 5 \, A a b\right )} \sqrt {x}\right )}}{8 \, b^{5}} \]
1/120*(16*(3*B*b^4*x^2 + 5*B*a*b^3*x)*x^(7/2) + (89*(11*B*a*b^3 - 5*A*b^4) *x^2 + 285*(3*B*a^2*b^2 - A*a*b^3)*x)*x^(5/2) + 12*(12*(11*B*a^2*b^2 - 5*A *a*b^3)*x^2 + 35*(3*B*a^3*b - A*a^2*b^2)*x)*x^(3/2) + 7*(9*(11*B*a^3*b - 5 *A*a^2*b^2)*x^2 + 25*(3*B*a^4 - A*a^3*b)*x)*sqrt(x))/(b^7*x^3 + 3*a*b^6*x^ 2 + 3*a^2*b^5*x + a^3*b^4) - 7/4*(9*B*a^3 - 5*A*a^2*b)*arctan(b*sqrt(x)/sq rt(a*b))/(sqrt(a*b)*b^5) - 7/8*((11*B*a*b - 5*A*b^2)*x^(3/2) - 2*(9*B*a^2 - 5*A*a*b)*sqrt(x))/b^5
Time = 0.27 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.56 \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {7 \, {\left (9 \, B a^{3} - 5 \, A a^{2} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {17 \, B a^{3} b x^{\frac {3}{2}} - 13 \, A a^{2} b^{2} x^{\frac {3}{2}} + 15 \, B a^{4} \sqrt {x} - 11 \, A a^{3} b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} b^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {2 \, {\left (3 \, B b^{12} x^{\frac {5}{2}} - 15 \, B a b^{11} x^{\frac {3}{2}} + 5 \, A b^{12} x^{\frac {3}{2}} + 90 \, B a^{2} b^{10} \sqrt {x} - 45 \, A a b^{11} \sqrt {x}\right )}}{15 \, b^{15} \mathrm {sgn}\left (b x + a\right )} \]
-7/4*(9*B*a^3 - 5*A*a^2*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^5*sgn( b*x + a)) + 1/4*(17*B*a^3*b*x^(3/2) - 13*A*a^2*b^2*x^(3/2) + 15*B*a^4*sqrt (x) - 11*A*a^3*b*sqrt(x))/((b*x + a)^2*b^5*sgn(b*x + a)) + 2/15*(3*B*b^12* x^(5/2) - 15*B*a*b^11*x^(3/2) + 5*A*b^12*x^(3/2) + 90*B*a^2*b^10*sqrt(x) - 45*A*a*b^11*sqrt(x))/(b^15*sgn(b*x + a))
Timed out. \[ \int \frac {x^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {x^{7/2}\,\left (A+B\,x\right )}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]