Integrand size = 33, antiderivative size = 291 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {35 e^2 (2 b B d-3 A b e+a B e)}{24 b (b d-a e)^4 (d+e x)^{3/2}}-\frac {A b-a B}{3 b (b d-a e) (a+b x)^3 (d+e x)^{3/2}}-\frac {2 b B d-3 A b e+a B e}{4 b (b d-a e)^2 (a+b x)^2 (d+e x)^{3/2}}+\frac {7 e (2 b B d-3 A b e+a B e)}{8 b (b d-a e)^3 (a+b x) (d+e x)^{3/2}}+\frac {35 e^2 (2 b B d-3 A b e+a B e)}{8 (b d-a e)^5 \sqrt {d+e x}}-\frac {35 \sqrt {b} e^2 (2 b B d-3 A b e+a B e) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 (b d-a e)^{11/2}} \]
35/24*e^2*(-3*A*b*e+B*a*e+2*B*b*d)/b/(-a*e+b*d)^4/(e*x+d)^(3/2)+1/3*(-A*b+ B*a)/b/(-a*e+b*d)/(b*x+a)^3/(e*x+d)^(3/2)+1/4*(3*A*b*e-B*a*e-2*B*b*d)/b/(- a*e+b*d)^2/(b*x+a)^2/(e*x+d)^(3/2)+7/8*e*(-3*A*b*e+B*a*e+2*B*b*d)/b/(-a*e+ b*d)^3/(b*x+a)/(e*x+d)^(3/2)-35/8*e^2*(-3*A*b*e+B*a*e+2*B*b*d)*arctanh(b^( 1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))*b^(1/2)/(-a*e+b*d)^(11/2)+35/8*e^2*(- 3*A*b*e+B*a*e+2*B*b*d)/(-a*e+b*d)^5/(e*x+d)^(1/2)
Time = 1.90 (sec) , antiderivative size = 408, normalized size of antiderivative = 1.40 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {-B \left (16 a^4 e^3 (2 d+3 e x)+a^3 b e^2 \left (247 d^2+414 d e x+231 e^2 x^2\right )+8 a^2 b^2 e \left (5 d^3+87 d^2 e x+105 d e^2 x^2+35 e^3 x^3\right )+2 b^4 d x \left (-6 d^3+21 d^2 e x+140 d e^2 x^2+105 e^3 x^3\right )+a b^3 \left (-4 d^4+114 d^3 e x+777 d^2 e^2 x^2+700 d e^3 x^3+105 e^4 x^4\right )\right )+A \left (-16 a^4 e^4+16 a^3 b e^3 (13 d+9 e x)+3 a^2 b^2 e^2 \left (55 d^2+318 d e x+231 e^2 x^2\right )+2 a b^3 e \left (-25 d^3+90 d^2 e x+567 d e^2 x^2+420 e^3 x^3\right )+b^4 \left (8 d^4-18 d^3 e x+63 d^2 e^2 x^2+420 d e^3 x^3+315 e^4 x^4\right )\right )}{24 (-b d+a e)^5 (a+b x)^3 (d+e x)^{3/2}}-\frac {35 \sqrt {b} e^2 (2 b B d-3 A b e+a B e) \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{8 (-b d+a e)^{11/2}} \]
(-(B*(16*a^4*e^3*(2*d + 3*e*x) + a^3*b*e^2*(247*d^2 + 414*d*e*x + 231*e^2* x^2) + 8*a^2*b^2*e*(5*d^3 + 87*d^2*e*x + 105*d*e^2*x^2 + 35*e^3*x^3) + 2*b ^4*d*x*(-6*d^3 + 21*d^2*e*x + 140*d*e^2*x^2 + 105*e^3*x^3) + a*b^3*(-4*d^4 + 114*d^3*e*x + 777*d^2*e^2*x^2 + 700*d*e^3*x^3 + 105*e^4*x^4))) + A*(-16 *a^4*e^4 + 16*a^3*b*e^3*(13*d + 9*e*x) + 3*a^2*b^2*e^2*(55*d^2 + 318*d*e*x + 231*e^2*x^2) + 2*a*b^3*e*(-25*d^3 + 90*d^2*e*x + 567*d*e^2*x^2 + 420*e^ 3*x^3) + b^4*(8*d^4 - 18*d^3*e*x + 63*d^2*e^2*x^2 + 420*d*e^3*x^3 + 315*e^ 4*x^4)))/(24*(-(b*d) + a*e)^5*(a + b*x)^3*(d + e*x)^(3/2)) - (35*Sqrt[b]*e ^2*(2*b*B*d - 3*A*b*e + a*B*e)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(8*(-(b*d) + a*e)^(11/2))
Time = 0.39 (sec) , antiderivative size = 269, normalized size of antiderivative = 0.92, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1184, 27, 87, 52, 52, 61, 61, 73, 221}
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.
| \(\displaystyle \int \frac {A+B x}{\left (a^2+2 a b x+b^2 x^2\right )^2 (d+e x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1184 |
| \(\displaystyle b^4 \int \frac {A+B x}{b^4 (a+b x)^4 (d+e x)^{5/2}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {A+B x}{(a+b x)^4 (d+e x)^{5/2}}dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(a B e-3 A b e+2 b B d) \int \frac {1}{(a+b x)^3 (d+e x)^{5/2}}dx}{2 b (b d-a e)}-\frac {A b-a B}{3 b (a+b x)^3 (d+e x)^{3/2} (b d-a e)}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {(a B e-3 A b e+2 b B d) \left (-\frac {7 e \int \frac {1}{(a+b x)^2 (d+e x)^{5/2}}dx}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 (d+e x)^{3/2} (b d-a e)}\right )}{2 b (b d-a e)}-\frac {A b-a B}{3 b (a+b x)^3 (d+e x)^{3/2} (b d-a e)}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {(a B e-3 A b e+2 b B d) \left (-\frac {7 e \left (-\frac {5 e \int \frac {1}{(a+b x) (d+e x)^{5/2}}dx}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{3/2} (b d-a e)}\right )}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 (d+e x)^{3/2} (b d-a e)}\right )}{2 b (b d-a e)}-\frac {A b-a B}{3 b (a+b x)^3 (d+e x)^{3/2} (b d-a e)}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(a B e-3 A b e+2 b B d) \left (-\frac {7 e \left (-\frac {5 e \left (\frac {b \int \frac {1}{(a+b x) (d+e x)^{3/2}}dx}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{3/2} (b d-a e)}\right )}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 (d+e x)^{3/2} (b d-a e)}\right )}{2 b (b d-a e)}-\frac {A b-a B}{3 b (a+b x)^3 (d+e x)^{3/2} (b d-a e)}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(a B e-3 A b e+2 b B d) \left (-\frac {7 e \left (-\frac {5 e \left (\frac {b \left (\frac {b \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b d-a e}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{3/2} (b d-a e)}\right )}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 (d+e x)^{3/2} (b d-a e)}\right )}{2 b (b d-a e)}-\frac {A b-a B}{3 b (a+b x)^3 (d+e x)^{3/2} (b d-a e)}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a B e-3 A b e+2 b B d) \left (-\frac {7 e \left (-\frac {5 e \left (\frac {b \left (\frac {2 b \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{e (b d-a e)}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{3/2} (b d-a e)}\right )}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 (d+e x)^{3/2} (b d-a e)}\right )}{2 b (b d-a e)}-\frac {A b-a B}{3 b (a+b x)^3 (d+e x)^{3/2} (b d-a e)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a B e-3 A b e+2 b B d) \left (-\frac {7 e \left (-\frac {5 e \left (\frac {b \left (\frac {2}{\sqrt {d+e x} (b d-a e)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{3/2} (b d-a e)}\right )}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 (d+e x)^{3/2} (b d-a e)}\right )}{2 b (b d-a e)}-\frac {A b-a B}{3 b (a+b x)^3 (d+e x)^{3/2} (b d-a e)}\) |
-1/3*(A*b - a*B)/(b*(b*d - a*e)*(a + b*x)^3*(d + e*x)^(3/2)) + ((2*b*B*d - 3*A*b*e + a*B*e)*(-1/2*1/((b*d - a*e)*(a + b*x)^2*(d + e*x)^(3/2)) - (7*e *(-(1/((b*d - a*e)*(a + b*x)*(d + e*x)^(3/2))) - (5*e*(2/(3*(b*d - a*e)*(d + e*x)^(3/2)) + (b*(2/((b*d - a*e)*Sqrt[d + e*x]) - (2*Sqrt[b]*ArcTanh[(S qrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(b*d - a*e)^(3/2)))/(b*d - a*e)))/ (2*(b*d - a*e))))/(4*(b*d - a*e))))/(2*b*(b*d - a*e))
3.19.23.3.1 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 + 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] :> 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] && 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]
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)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^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}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.39 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.08
| method | result | size |
| derivativedivides | \(2 e^{2} \left (-\frac {A e -B d}{3 \left (a e -b d \right )^{4} \left (e x +d \right )^{\frac {3}{2}}}-\frac {-4 A b e +B a e +3 B b d}{\left (a e -b d \right )^{5} \sqrt {e x +d}}+\frac {b \left (\frac {\left (\frac {41}{16} A \,b^{3} e -\frac {19}{16} B e \,b^{2} a -\frac {11}{8} B \,b^{3} d \right ) \left (e x +d \right )^{\frac {5}{2}}+\frac {b \left (35 A a b \,e^{2}-35 A \,b^{2} d e -17 a^{2} B \,e^{2}-B a b d e +18 B \,b^{2} d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{6}+\left (\frac {55}{16} A \,a^{2} b \,e^{3}-\frac {55}{8} A a \,b^{2} d \,e^{2}+\frac {55}{16} A \,b^{3} d^{2} e -\frac {29}{16} B \,e^{3} a^{3}+\frac {23}{16} B a \,b^{2} d^{2} e -\frac {13}{8} B \,b^{3} d^{3}+2 B \,a^{2} b d \,e^{2}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{3}}+\frac {35 \left (3 A b e -B a e -2 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{16 \sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{5}}\right )\) | \(315\) |
| default | \(2 e^{2} \left (-\frac {A e -B d}{3 \left (a e -b d \right )^{4} \left (e x +d \right )^{\frac {3}{2}}}-\frac {-4 A b e +B a e +3 B b d}{\left (a e -b d \right )^{5} \sqrt {e x +d}}+\frac {b \left (\frac {\left (\frac {41}{16} A \,b^{3} e -\frac {19}{16} B e \,b^{2} a -\frac {11}{8} B \,b^{3} d \right ) \left (e x +d \right )^{\frac {5}{2}}+\frac {b \left (35 A a b \,e^{2}-35 A \,b^{2} d e -17 a^{2} B \,e^{2}-B a b d e +18 B \,b^{2} d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{6}+\left (\frac {55}{16} A \,a^{2} b \,e^{3}-\frac {55}{8} A a \,b^{2} d \,e^{2}+\frac {55}{16} A \,b^{3} d^{2} e -\frac {29}{16} B \,e^{3} a^{3}+\frac {23}{16} B a \,b^{2} d^{2} e -\frac {13}{8} B \,b^{3} d^{3}+2 B \,a^{2} b d \,e^{2}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{3}}+\frac {35 \left (3 A b e -B a e -2 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{16 \sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{5}}\right )\) | \(315\) |
| pseudoelliptic | \(-\frac {2 \left (-\frac {315 b \left (e x +d \right )^{\frac {3}{2}} \left (b x +a \right )^{3} e^{2} \left (\left (A e -\frac {2 B d}{3}\right ) b -\frac {B a e}{3}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{16}+\left (\left (-\frac {315 A \,e^{4} x^{4}}{16}-\frac {105 \left (-\frac {B x}{2}+A \right ) x^{3} d \,e^{3}}{4}-\frac {63 x^{2} d^{2} \left (-\frac {40 B x}{9}+A \right ) e^{2}}{16}+\frac {9 \left (\frac {7 B x}{3}+A \right ) x \,d^{3} e}{8}-\frac {d^{4} \left (\frac {3 B x}{2}+A \right )}{2}\right ) b^{4}+\frac {25 \left (-\frac {84 x^{3} \left (-\frac {B x}{8}+A \right ) e^{4}}{5}-\frac {567 x^{2} \left (-\frac {50 B x}{81}+A \right ) d \,e^{3}}{25}-\frac {18 x \left (-\frac {259 B x}{60}+A \right ) d^{2} e^{2}}{5}+d^{3} \left (\frac {57 B x}{25}+A \right ) e -\frac {2 B \,d^{4}}{25}\right ) a \,b^{3}}{8}-\frac {165 \left (\frac {21 x^{2} \left (-\frac {40 B x}{99}+A \right ) e^{3}}{5}+\frac {318 x d \left (-\frac {140 B x}{159}+A \right ) e^{2}}{55}+d^{2} \left (-\frac {232 B x}{55}+A \right ) e -\frac {8 B \,d^{3}}{33}\right ) e \,a^{2} b^{2}}{16}-13 \left (\frac {9 \left (-\frac {77 B x}{48}+A \right ) x \,e^{2}}{13}+d \left (-\frac {207 B x}{104}+A \right ) e -\frac {19 B \,d^{2}}{16}\right ) e^{2} a^{3} b +\left (\left (3 B x +A \right ) e +2 B d \right ) e^{3} a^{4}\right ) \sqrt {\left (a e -b d \right ) b}\right )}{3 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} \left (b x +a \right )^{3} \left (a e -b d \right )^{5}}\) | \(361\) |
2*e^2*(-1/3*(A*e-B*d)/(a*e-b*d)^4/(e*x+d)^(3/2)-1/(a*e-b*d)^5*(-4*A*b*e+B* a*e+3*B*b*d)/(e*x+d)^(1/2)+1/(a*e-b*d)^5*b*(((41/16*A*b^3*e-19/16*B*e*b^2* a-11/8*B*b^3*d)*(e*x+d)^(5/2)+1/6*b*(35*A*a*b*e^2-35*A*b^2*d*e-17*B*a^2*e^ 2-B*a*b*d*e+18*B*b^2*d^2)*(e*x+d)^(3/2)+(55/16*A*a^2*b*e^3-55/8*A*a*b^2*d* e^2+55/16*A*b^3*d^2*e-29/16*B*e^3*a^3+23/16*B*a*b^2*d^2*e-13/8*B*b^3*d^3+2 *B*a^2*b*d*e^2)*(e*x+d)^(1/2))/(b*(e*x+d)+a*e-b*d)^3+35/16*(3*A*b*e-B*a*e- 2*B*b*d)/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 1319 vs. \(2 (259) = 518\).
Time = 0.64 (sec) , antiderivative size = 2648, normalized size of antiderivative = 9.10 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Too large to display} \]
[1/48*(105*(2*B*a^3*b*d^3*e^2 + (B*a^4 - 3*A*a^3*b)*d^2*e^3 + (2*B*b^4*d*e ^4 + (B*a*b^3 - 3*A*b^4)*e^5)*x^5 + (4*B*b^4*d^2*e^3 + 2*(4*B*a*b^3 - 3*A* b^4)*d*e^4 + 3*(B*a^2*b^2 - 3*A*a*b^3)*e^5)*x^4 + (2*B*b^4*d^3*e^2 + (13*B *a*b^3 - 3*A*b^4)*d^2*e^3 + 6*(2*B*a^2*b^2 - 3*A*a*b^3)*d*e^4 + 3*(B*a^3*b - 3*A*a^2*b^2)*e^5)*x^3 + (6*B*a*b^3*d^3*e^2 + 3*(5*B*a^2*b^2 - 3*A*a*b^3 )*d^2*e^3 + 2*(4*B*a^3*b - 9*A*a^2*b^2)*d*e^4 + (B*a^4 - 3*A*a^3*b)*e^5)*x ^2 + (6*B*a^2*b^2*d^3*e^2 + (7*B*a^3*b - 9*A*a^2*b^2)*d^2*e^3 + 2*(B*a^4 - 3*A*a^3*b)*d*e^4)*x)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e - 2*(b* d - a*e)*sqrt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) + 2*(16*A*a^4*e^4 - 4*(B*a*b^3 + 2*A*b^4)*d^4 + 10*(4*B*a^2*b^2 + 5*A*a*b^3)*d^3*e + (247*B*a ^3*b - 165*A*a^2*b^2)*d^2*e^2 + 16*(2*B*a^4 - 13*A*a^3*b)*d*e^3 + 105*(2*B *b^4*d*e^3 + (B*a*b^3 - 3*A*b^4)*e^4)*x^4 + 140*(2*B*b^4*d^2*e^2 + (5*B*a* b^3 - 3*A*b^4)*d*e^3 + 2*(B*a^2*b^2 - 3*A*a*b^3)*e^4)*x^3 + 21*(2*B*b^4*d^ 3*e + (37*B*a*b^3 - 3*A*b^4)*d^2*e^2 + 2*(20*B*a^2*b^2 - 27*A*a*b^3)*d*e^3 + 11*(B*a^3*b - 3*A*a^2*b^2)*e^4)*x^2 - 6*(2*B*b^4*d^4 - (19*B*a*b^3 + 3* A*b^4)*d^3*e - 2*(58*B*a^2*b^2 - 15*A*a*b^3)*d^2*e^2 - 3*(23*B*a^3*b - 53* A*a^2*b^2)*d*e^3 - 8*(B*a^4 - 3*A*a^3*b)*e^4)*x)*sqrt(e*x + d))/(a^3*b^5*d ^7 - 5*a^4*b^4*d^6*e + 10*a^5*b^3*d^5*e^2 - 10*a^6*b^2*d^4*e^3 + 5*a^7*b*d ^3*e^4 - a^8*d^2*e^5 + (b^8*d^5*e^2 - 5*a*b^7*d^4*e^3 + 10*a^2*b^6*d^3*e^4 - 10*a^3*b^5*d^2*e^5 + 5*a^4*b^4*d*e^6 - a^5*b^3*e^7)*x^5 + (2*b^8*d^6...
Timed out. \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 762 vs. \(2 (259) = 518\).
Time = 0.28 (sec) , antiderivative size = 762, normalized size of antiderivative = 2.62 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {35 \, {\left (2 \, B b^{2} d e^{2} + B a b e^{3} - 3 \, A b^{2} e^{3}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{8 \, {\left (b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}\right )} \sqrt {-b^{2} d + a b e}} + \frac {210 \, {\left (e x + d\right )}^{4} B b^{4} d e^{2} - 560 \, {\left (e x + d\right )}^{3} B b^{4} d^{2} e^{2} + 462 \, {\left (e x + d\right )}^{2} B b^{4} d^{3} e^{2} - 96 \, {\left (e x + d\right )} B b^{4} d^{4} e^{2} - 16 \, B b^{4} d^{5} e^{2} + 105 \, {\left (e x + d\right )}^{4} B a b^{3} e^{3} - 315 \, {\left (e x + d\right )}^{4} A b^{4} e^{3} + 280 \, {\left (e x + d\right )}^{3} B a b^{3} d e^{3} + 840 \, {\left (e x + d\right )}^{3} A b^{4} d e^{3} - 693 \, {\left (e x + d\right )}^{2} B a b^{3} d^{2} e^{3} - 693 \, {\left (e x + d\right )}^{2} A b^{4} d^{2} e^{3} + 240 \, {\left (e x + d\right )} B a b^{3} d^{3} e^{3} + 144 \, {\left (e x + d\right )} A b^{4} d^{3} e^{3} + 64 \, B a b^{3} d^{4} e^{3} + 16 \, A b^{4} d^{4} e^{3} + 280 \, {\left (e x + d\right )}^{3} B a^{2} b^{2} e^{4} - 840 \, {\left (e x + d\right )}^{3} A a b^{3} e^{4} + 1386 \, {\left (e x + d\right )}^{2} A a b^{3} d e^{4} - 144 \, {\left (e x + d\right )} B a^{2} b^{2} d^{2} e^{4} - 432 \, {\left (e x + d\right )} A a b^{3} d^{2} e^{4} - 96 \, B a^{2} b^{2} d^{3} e^{4} - 64 \, A a b^{3} d^{3} e^{4} + 231 \, {\left (e x + d\right )}^{2} B a^{3} b e^{5} - 693 \, {\left (e x + d\right )}^{2} A a^{2} b^{2} e^{5} - 48 \, {\left (e x + d\right )} B a^{3} b d e^{5} + 432 \, {\left (e x + d\right )} A a^{2} b^{2} d e^{5} + 64 \, B a^{3} b d^{2} e^{5} + 96 \, A a^{2} b^{2} d^{2} e^{5} + 48 \, {\left (e x + d\right )} B a^{4} e^{6} - 144 \, {\left (e x + d\right )} A a^{3} b e^{6} - 16 \, B a^{4} d e^{6} - 64 \, A a^{3} b d e^{6} + 16 \, A a^{4} e^{7}}{24 \, {\left (b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}\right )} {\left ({\left (e x + d\right )}^{\frac {3}{2}} b - \sqrt {e x + d} b d + \sqrt {e x + d} a e\right )}^{3}} \]
35/8*(2*B*b^2*d*e^2 + B*a*b*e^3 - 3*A*b^2*e^3)*arctan(sqrt(e*x + d)*b/sqrt (-b^2*d + a*b*e))/((b^5*d^5 - 5*a*b^4*d^4*e + 10*a^2*b^3*d^3*e^2 - 10*a^3* b^2*d^2*e^3 + 5*a^4*b*d*e^4 - a^5*e^5)*sqrt(-b^2*d + a*b*e)) + 1/24*(210*( e*x + d)^4*B*b^4*d*e^2 - 560*(e*x + d)^3*B*b^4*d^2*e^2 + 462*(e*x + d)^2*B *b^4*d^3*e^2 - 96*(e*x + d)*B*b^4*d^4*e^2 - 16*B*b^4*d^5*e^2 + 105*(e*x + d)^4*B*a*b^3*e^3 - 315*(e*x + d)^4*A*b^4*e^3 + 280*(e*x + d)^3*B*a*b^3*d*e ^3 + 840*(e*x + d)^3*A*b^4*d*e^3 - 693*(e*x + d)^2*B*a*b^3*d^2*e^3 - 693*( e*x + d)^2*A*b^4*d^2*e^3 + 240*(e*x + d)*B*a*b^3*d^3*e^3 + 144*(e*x + d)*A *b^4*d^3*e^3 + 64*B*a*b^3*d^4*e^3 + 16*A*b^4*d^4*e^3 + 280*(e*x + d)^3*B*a ^2*b^2*e^4 - 840*(e*x + d)^3*A*a*b^3*e^4 + 1386*(e*x + d)^2*A*a*b^3*d*e^4 - 144*(e*x + d)*B*a^2*b^2*d^2*e^4 - 432*(e*x + d)*A*a*b^3*d^2*e^4 - 96*B*a ^2*b^2*d^3*e^4 - 64*A*a*b^3*d^3*e^4 + 231*(e*x + d)^2*B*a^3*b*e^5 - 693*(e *x + d)^2*A*a^2*b^2*e^5 - 48*(e*x + d)*B*a^3*b*d*e^5 + 432*(e*x + d)*A*a^2 *b^2*d*e^5 + 64*B*a^3*b*d^2*e^5 + 96*A*a^2*b^2*d^2*e^5 + 48*(e*x + d)*B*a^ 4*e^6 - 144*(e*x + d)*A*a^3*b*e^6 - 16*B*a^4*d*e^6 - 64*A*a^3*b*d*e^6 + 16 *A*a^4*e^7)/((b^5*d^5 - 5*a*b^4*d^4*e + 10*a^2*b^3*d^3*e^2 - 10*a^3*b^2*d^ 2*e^3 + 5*a^4*b*d*e^4 - a^5*e^5)*((e*x + d)^(3/2)*b - sqrt(e*x + d)*b*d + sqrt(e*x + d)*a*e)^3)
Time = 11.36 (sec) , antiderivative size = 483, normalized size of antiderivative = 1.66 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {\frac {2\,\left (A\,e^3-B\,d\,e^2\right )}{3\,\left (a\,e-b\,d\right )}+\frac {77\,{\left (d+e\,x\right )}^2\,\left (-3\,A\,b^2\,e^3+2\,B\,d\,b^2\,e^2+B\,a\,b\,e^3\right )}{8\,{\left (a\,e-b\,d\right )}^3}+\frac {35\,{\left (d+e\,x\right )}^3\,\left (-3\,A\,b^3\,e^3+2\,B\,d\,b^3\,e^2+B\,a\,b^2\,e^3\right )}{3\,{\left (a\,e-b\,d\right )}^4}+\frac {2\,\left (d+e\,x\right )\,\left (B\,a\,e^3-3\,A\,b\,e^3+2\,B\,b\,d\,e^2\right )}{{\left (a\,e-b\,d\right )}^2}+\frac {35\,b^3\,{\left (d+e\,x\right )}^4\,\left (B\,a\,e^3-3\,A\,b\,e^3+2\,B\,b\,d\,e^2\right )}{8\,{\left (a\,e-b\,d\right )}^5}}{{\left (d+e\,x\right )}^{3/2}\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )+b^3\,{\left (d+e\,x\right )}^{9/2}-\left (3\,b^3\,d-3\,a\,b^2\,e\right )\,{\left (d+e\,x\right )}^{7/2}+{\left (d+e\,x\right )}^{5/2}\,\left (3\,a^2\,b\,e^2-6\,a\,b^2\,d\,e+3\,b^3\,d^2\right )}-\frac {35\,\sqrt {b}\,e^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,e^2\,\sqrt {d+e\,x}\,\left (B\,a\,e-3\,A\,b\,e+2\,B\,b\,d\right )\,\left (a^5\,e^5-5\,a^4\,b\,d\,e^4+10\,a^3\,b^2\,d^2\,e^3-10\,a^2\,b^3\,d^3\,e^2+5\,a\,b^4\,d^4\,e-b^5\,d^5\right )}{{\left (a\,e-b\,d\right )}^{11/2}\,\left (B\,a\,e^3-3\,A\,b\,e^3+2\,B\,b\,d\,e^2\right )}\right )\,\left (B\,a\,e-3\,A\,b\,e+2\,B\,b\,d\right )}{8\,{\left (a\,e-b\,d\right )}^{11/2}} \]
- ((2*(A*e^3 - B*d*e^2))/(3*(a*e - b*d)) + (77*(d + e*x)^2*(B*a*b*e^3 - 3* A*b^2*e^3 + 2*B*b^2*d*e^2))/(8*(a*e - b*d)^3) + (35*(d + e*x)^3*(B*a*b^2*e ^3 - 3*A*b^3*e^3 + 2*B*b^3*d*e^2))/(3*(a*e - b*d)^4) + (2*(d + e*x)*(B*a*e ^3 - 3*A*b*e^3 + 2*B*b*d*e^2))/(a*e - b*d)^2 + (35*b^3*(d + e*x)^4*(B*a*e^ 3 - 3*A*b*e^3 + 2*B*b*d*e^2))/(8*(a*e - b*d)^5))/((d + e*x)^(3/2)*(a^3*e^3 - b^3*d^3 + 3*a*b^2*d^2*e - 3*a^2*b*d*e^2) + b^3*(d + e*x)^(9/2) - (3*b^3 *d - 3*a*b^2*e)*(d + e*x)^(7/2) + (d + e*x)^(5/2)*(3*b^3*d^2 + 3*a^2*b*e^2 - 6*a*b^2*d*e)) - (35*b^(1/2)*e^2*atan((b^(1/2)*e^2*(d + e*x)^(1/2)*(B*a* e - 3*A*b*e + 2*B*b*d)*(a^5*e^5 - b^5*d^5 - 10*a^2*b^3*d^3*e^2 + 10*a^3*b^ 2*d^2*e^3 + 5*a*b^4*d^4*e - 5*a^4*b*d*e^4))/((a*e - b*d)^(11/2)*(B*a*e^3 - 3*A*b*e^3 + 2*B*b*d*e^2)))*(B*a*e - 3*A*b*e + 2*B*b*d))/(8*(a*e - b*d)^(1 1/2))