Integrand size = 33, antiderivative size = 175 \[ \int \frac {(a+b x) (d+e x)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {63 e^2 (b d-a e)^2 \sqrt {d+e x}}{4 b^5}+\frac {21 e^2 (b d-a e) (d+e x)^{3/2}}{4 b^4}+\frac {63 e^2 (d+e x)^{5/2}}{20 b^3}-\frac {9 e (d+e x)^{7/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{9/2}}{2 b (a+b x)^2}-\frac {63 e^2 (b d-a e)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{11/2}} \]
21/4*e^2*(-a*e+b*d)*(e*x+d)^(3/2)/b^4+63/20*e^2*(e*x+d)^(5/2)/b^3-9/4*e*(e *x+d)^(7/2)/b^2/(b*x+a)-1/2*(e*x+d)^(9/2)/b/(b*x+a)^2-63/4*e^2*(-a*e+b*d)^ (5/2)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(11/2)+63/4*e^2*(- a*e+b*d)^2*(e*x+d)^(1/2)/b^5
Time = 0.13 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.22 \[ \int \frac {(a+b x) (d+e x)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {\sqrt {d+e x} \left (315 a^4 e^4+105 a^3 b e^3 (-7 d+5 e x)+21 a^2 b^2 e^2 \left (23 d^2-59 d e x+8 e^2 x^2\right )-3 a b^3 e \left (15 d^3-277 d^2 e x+136 d e^2 x^2+8 e^3 x^3\right )+b^4 \left (-10 d^4-85 d^3 e x+288 d^2 e^2 x^2+56 d e^3 x^3+8 e^4 x^4\right )\right )}{20 b^5 (a+b x)^2}-\frac {63 e^2 (-b d+a e)^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{4 b^{11/2}} \]
(Sqrt[d + e*x]*(315*a^4*e^4 + 105*a^3*b*e^3*(-7*d + 5*e*x) + 21*a^2*b^2*e^ 2*(23*d^2 - 59*d*e*x + 8*e^2*x^2) - 3*a*b^3*e*(15*d^3 - 277*d^2*e*x + 136* d*e^2*x^2 + 8*e^3*x^3) + b^4*(-10*d^4 - 85*d^3*e*x + 288*d^2*e^2*x^2 + 56* d*e^3*x^3 + 8*e^4*x^4)))/(20*b^5*(a + b*x)^2) - (63*e^2*(-(b*d) + a*e)^(5/ 2)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(4*b^(11/2))
Time = 0.29 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1184, 27, 51, 51, 60, 60, 60, 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.
| \(\displaystyle \int \frac {(a+b x) (d+e x)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 1184 |
| \(\displaystyle b^4 \int \frac {(d+e x)^{9/2}}{b^4 (a+b x)^3}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {(d+e x)^{9/2}}{(a+b x)^3}dx\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {9 e \int \frac {(d+e x)^{7/2}}{(a+b x)^2}dx}{4 b}-\frac {(d+e x)^{9/2}}{2 b (a+b x)^2}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {9 e \left (\frac {7 e \int \frac {(d+e x)^{5/2}}{a+b x}dx}{2 b}-\frac {(d+e x)^{7/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{9/2}}{2 b (a+b x)^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {9 e \left (\frac {7 e \left (\frac {(b d-a e) \int \frac {(d+e x)^{3/2}}{a+b x}dx}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\right )}{2 b}-\frac {(d+e x)^{7/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{9/2}}{2 b (a+b x)^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {9 e \left (\frac {7 e \left (\frac {(b d-a e) \left (\frac {(b d-a e) \int \frac {\sqrt {d+e x}}{a+b x}dx}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\right )}{2 b}-\frac {(d+e x)^{7/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{9/2}}{2 b (a+b x)^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {9 e \left (\frac {7 e \left (\frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\right )}{2 b}-\frac {(d+e x)^{7/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{9/2}}{2 b (a+b x)^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {9 e \left (\frac {7 e \left (\frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {2 (b d-a e) \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b e}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\right )}{2 b}-\frac {(d+e x)^{7/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{9/2}}{2 b (a+b x)^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {9 e \left (\frac {7 e \left (\frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {2 \sqrt {d+e x}}{b}-\frac {2 \sqrt {b d-a e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2}}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\right )}{2 b}-\frac {(d+e x)^{7/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{9/2}}{2 b (a+b x)^2}\) |
-1/2*(d + e*x)^(9/2)/(b*(a + b*x)^2) + (9*e*(-((d + e*x)^(7/2)/(b*(a + b*x ))) + (7*e*((2*(d + e*x)^(5/2))/(5*b) + ((b*d - a*e)*((2*(d + e*x)^(3/2))/ (3*b) + ((b*d - a*e)*((2*Sqrt[d + e*x])/b - (2*Sqrt[b*d - a*e]*ArcTanh[(Sq rt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/b^(3/2)))/b))/b))/(2*b)))/(4*b)
3.21.74.3.1 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_)^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 = 0.44 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.11
| method | result | size |
| risch | \(\frac {2 e^{2} \left (b^{2} e^{2} x^{2}-5 a b \,e^{2} x +7 b^{2} d e x +30 e^{2} a^{2}-65 a b d e +36 b^{2} d^{2}\right ) \sqrt {e x +d}}{5 b^{5}}-\frac {\left (2 a^{3} e^{3}-6 a^{2} b d \,e^{2}+6 a \,b^{2} d^{2} e -2 b^{3} d^{3}\right ) e^{2} \left (\frac {-\frac {17 b \left (e x +d \right )^{\frac {3}{2}}}{8}+\left (-\frac {15 a e}{8}+\frac {15 b d}{8}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{2}}+\frac {63 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \sqrt {\left (a e -b d \right ) b}}\right )}{b^{5}}\) | \(194\) |
| pseudoelliptic | \(-\frac {63 \left (e^{2} \left (b x +a \right )^{2} \left (a e -b d \right )^{3} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )-\sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, \left (\left (\frac {8}{315} e^{4} x^{4}+\frac {8}{45} d \,e^{3} x^{3}+\frac {32}{35} d^{2} e^{2} x^{2}-\frac {17}{63} d^{3} e x -\frac {2}{63} d^{4}\right ) b^{4}-\frac {e \left (\frac {8}{15} e^{3} x^{3}+\frac {136}{15} d \,e^{2} x^{2}-\frac {277}{15} d^{2} e x +d^{3}\right ) a \,b^{3}}{7}+\frac {23 e^{2} \left (\frac {8}{23} e^{2} x^{2}-\frac {59}{23} d e x +d^{2}\right ) a^{2} b^{2}}{15}-\frac {7 e^{3} \left (-\frac {5 e x}{7}+d \right ) a^{3} b}{3}+e^{4} a^{4}\right )\right )}{4 \sqrt {\left (a e -b d \right ) b}\, b^{5} \left (b x +a \right )^{2}}\) | \(221\) |
| derivativedivides | \(2 e^{2} \left (\frac {\frac {\left (e x +d \right )^{\frac {5}{2}} b^{2}}{5}-a b e \left (e x +d \right )^{\frac {3}{2}}+b^{2} d \left (e x +d \right )^{\frac {3}{2}}+6 a^{2} e^{2} \sqrt {e x +d}-12 a b d e \sqrt {e x +d}+6 b^{2} d^{2} \sqrt {e x +d}}{b^{5}}-\frac {\frac {\left (-\frac {17}{8} e^{3} b \,a^{3}+\frac {51}{8} a^{2} b^{2} d \,e^{2}-\frac {51}{8} d^{2} e \,b^{3} a +\frac {17}{8} b^{4} d^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {15}{8} e^{4} a^{4}+\frac {15}{2} b d \,e^{3} a^{3}-\frac {45}{4} b^{2} d^{2} e^{2} a^{2}+\frac {15}{2} b^{3} d^{3} e a -\frac {15}{8} b^{4} d^{4}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{2}}+\frac {63 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \sqrt {\left (a e -b d \right ) b}}}{b^{5}}\right )\) | \(295\) |
| default | \(2 e^{2} \left (\frac {\frac {\left (e x +d \right )^{\frac {5}{2}} b^{2}}{5}-a b e \left (e x +d \right )^{\frac {3}{2}}+b^{2} d \left (e x +d \right )^{\frac {3}{2}}+6 a^{2} e^{2} \sqrt {e x +d}-12 a b d e \sqrt {e x +d}+6 b^{2} d^{2} \sqrt {e x +d}}{b^{5}}-\frac {\frac {\left (-\frac {17}{8} e^{3} b \,a^{3}+\frac {51}{8} a^{2} b^{2} d \,e^{2}-\frac {51}{8} d^{2} e \,b^{3} a +\frac {17}{8} b^{4} d^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {15}{8} e^{4} a^{4}+\frac {15}{2} b d \,e^{3} a^{3}-\frac {45}{4} b^{2} d^{2} e^{2} a^{2}+\frac {15}{2} b^{3} d^{3} e a -\frac {15}{8} b^{4} d^{4}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{2}}+\frac {63 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \sqrt {\left (a e -b d \right ) b}}}{b^{5}}\right )\) | \(295\) |
2/5*e^2*(b^2*e^2*x^2-5*a*b*e^2*x+7*b^2*d*e*x+30*a^2*e^2-65*a*b*d*e+36*b^2* d^2)*(e*x+d)^(1/2)/b^5-1/b^5*(2*a^3*e^3-6*a^2*b*d*e^2+6*a*b^2*d^2*e-2*b^3* d^3)*e^2*((-17/8*b*(e*x+d)^(3/2)+(-15/8*a*e+15/8*b*d)*(e*x+d)^(1/2))/(b*(e *x+d)+a*e-b*d)^2+63/8/((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. 360 vs. \(2 (143) = 286\).
Time = 0.34 (sec) , antiderivative size = 730, normalized size of antiderivative = 4.17 \[ \int \frac {(a+b x) (d+e x)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\left [\frac {315 \, {\left (a^{2} b^{2} d^{2} e^{2} - 2 \, a^{3} b d e^{3} + a^{4} e^{4} + {\left (b^{4} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 2 \, {\left (a b^{3} d^{2} e^{2} - 2 \, a^{2} b^{2} d e^{3} + a^{3} b e^{4}\right )} x\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) + 2 \, {\left (8 \, b^{4} e^{4} x^{4} - 10 \, b^{4} d^{4} - 45 \, a b^{3} d^{3} e + 483 \, a^{2} b^{2} d^{2} e^{2} - 735 \, a^{3} b d e^{3} + 315 \, a^{4} e^{4} + 8 \, {\left (7 \, b^{4} d e^{3} - 3 \, a b^{3} e^{4}\right )} x^{3} + 24 \, {\left (12 \, b^{4} d^{2} e^{2} - 17 \, a b^{3} d e^{3} + 7 \, a^{2} b^{2} e^{4}\right )} x^{2} - {\left (85 \, b^{4} d^{3} e - 831 \, a b^{3} d^{2} e^{2} + 1239 \, a^{2} b^{2} d e^{3} - 525 \, a^{3} b e^{4}\right )} x\right )} \sqrt {e x + d}}{40 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}, -\frac {315 \, {\left (a^{2} b^{2} d^{2} e^{2} - 2 \, a^{3} b d e^{3} + a^{4} e^{4} + {\left (b^{4} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 2 \, {\left (a b^{3} d^{2} e^{2} - 2 \, a^{2} b^{2} d e^{3} + a^{3} b e^{4}\right )} x\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (8 \, b^{4} e^{4} x^{4} - 10 \, b^{4} d^{4} - 45 \, a b^{3} d^{3} e + 483 \, a^{2} b^{2} d^{2} e^{2} - 735 \, a^{3} b d e^{3} + 315 \, a^{4} e^{4} + 8 \, {\left (7 \, b^{4} d e^{3} - 3 \, a b^{3} e^{4}\right )} x^{3} + 24 \, {\left (12 \, b^{4} d^{2} e^{2} - 17 \, a b^{3} d e^{3} + 7 \, a^{2} b^{2} e^{4}\right )} x^{2} - {\left (85 \, b^{4} d^{3} e - 831 \, a b^{3} d^{2} e^{2} + 1239 \, a^{2} b^{2} d e^{3} - 525 \, a^{3} b e^{4}\right )} x\right )} \sqrt {e x + d}}{20 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}\right ] \]
[1/40*(315*(a^2*b^2*d^2*e^2 - 2*a^3*b*d*e^3 + a^4*e^4 + (b^4*d^2*e^2 - 2*a *b^3*d*e^3 + a^2*b^2*e^4)*x^2 + 2*(a*b^3*d^2*e^2 - 2*a^2*b^2*d*e^3 + a^3*b *e^4)*x)*sqrt((b*d - a*e)/b)*log((b*e*x + 2*b*d - a*e - 2*sqrt(e*x + d)*b* sqrt((b*d - a*e)/b))/(b*x + a)) + 2*(8*b^4*e^4*x^4 - 10*b^4*d^4 - 45*a*b^3 *d^3*e + 483*a^2*b^2*d^2*e^2 - 735*a^3*b*d*e^3 + 315*a^4*e^4 + 8*(7*b^4*d* e^3 - 3*a*b^3*e^4)*x^3 + 24*(12*b^4*d^2*e^2 - 17*a*b^3*d*e^3 + 7*a^2*b^2*e ^4)*x^2 - (85*b^4*d^3*e - 831*a*b^3*d^2*e^2 + 1239*a^2*b^2*d*e^3 - 525*a^3 *b*e^4)*x)*sqrt(e*x + d))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5), -1/20*(315*(a^2 *b^2*d^2*e^2 - 2*a^3*b*d*e^3 + a^4*e^4 + (b^4*d^2*e^2 - 2*a*b^3*d*e^3 + a^ 2*b^2*e^4)*x^2 + 2*(a*b^3*d^2*e^2 - 2*a^2*b^2*d*e^3 + a^3*b*e^4)*x)*sqrt(- (b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (8*b^4*e^4*x^4 - 10*b^4*d^4 - 45*a*b^3*d^3*e + 483*a^2*b^2*d^2*e^2 - 735* a^3*b*d*e^3 + 315*a^4*e^4 + 8*(7*b^4*d*e^3 - 3*a*b^3*e^4)*x^3 + 24*(12*b^4 *d^2*e^2 - 17*a*b^3*d*e^3 + 7*a^2*b^2*e^4)*x^2 - (85*b^4*d^3*e - 831*a*b^3 *d^2*e^2 + 1239*a^2*b^2*d*e^3 - 525*a^3*b*e^4)*x)*sqrt(e*x + d))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)]
Timed out. \[ \int \frac {(a+b x) (d+e x)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(a+b x) (d+e x)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]
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. 373 vs. \(2 (143) = 286\).
Time = 0.28 (sec) , antiderivative size = 373, normalized size of antiderivative = 2.13 \[ \int \frac {(a+b x) (d+e x)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {63 \, {\left (b^{3} d^{3} e^{2} - 3 \, a b^{2} d^{2} e^{3} + 3 \, a^{2} b d e^{4} - a^{3} e^{5}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, \sqrt {-b^{2} d + a b e} b^{5}} - \frac {17 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{4} d^{3} e^{2} - 15 \, \sqrt {e x + d} b^{4} d^{4} e^{2} - 51 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{3} d^{2} e^{3} + 60 \, \sqrt {e x + d} a b^{3} d^{3} e^{3} + 51 \, {\left (e x + d\right )}^{\frac {3}{2}} a^{2} b^{2} d e^{4} - 90 \, \sqrt {e x + d} a^{2} b^{2} d^{2} e^{4} - 17 \, {\left (e x + d\right )}^{\frac {3}{2}} a^{3} b e^{5} + 60 \, \sqrt {e x + d} a^{3} b d e^{5} - 15 \, \sqrt {e x + d} a^{4} e^{6}}{4 \, {\left ({\left (e x + d\right )} b - b d + a e\right )}^{2} b^{5}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {5}{2}} b^{12} e^{2} + 5 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{12} d e^{2} + 30 \, \sqrt {e x + d} b^{12} d^{2} e^{2} - 5 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{11} e^{3} - 60 \, \sqrt {e x + d} a b^{11} d e^{3} + 30 \, \sqrt {e x + d} a^{2} b^{10} e^{4}\right )}}{5 \, b^{15}} \]
63/4*(b^3*d^3*e^2 - 3*a*b^2*d^2*e^3 + 3*a^2*b*d*e^4 - a^3*e^5)*arctan(sqrt (e*x + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^5) - 1/4*(17*(e* x + d)^(3/2)*b^4*d^3*e^2 - 15*sqrt(e*x + d)*b^4*d^4*e^2 - 51*(e*x + d)^(3/ 2)*a*b^3*d^2*e^3 + 60*sqrt(e*x + d)*a*b^3*d^3*e^3 + 51*(e*x + d)^(3/2)*a^2 *b^2*d*e^4 - 90*sqrt(e*x + d)*a^2*b^2*d^2*e^4 - 17*(e*x + d)^(3/2)*a^3*b*e ^5 + 60*sqrt(e*x + d)*a^3*b*d*e^5 - 15*sqrt(e*x + d)*a^4*e^6)/(((e*x + d)* b - b*d + a*e)^2*b^5) + 2/5*((e*x + d)^(5/2)*b^12*e^2 + 5*(e*x + d)^(3/2)* b^12*d*e^2 + 30*sqrt(e*x + d)*b^12*d^2*e^2 - 5*(e*x + d)^(3/2)*a*b^11*e^3 - 60*sqrt(e*x + d)*a*b^11*d*e^3 + 30*sqrt(e*x + d)*a^2*b^10*e^4)/b^15
Time = 0.15 (sec) , antiderivative size = 361, normalized size of antiderivative = 2.06 \[ \int \frac {(a+b x) (d+e x)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\left (\frac {2\,e^2\,{\left (3\,b^3\,d-3\,a\,b^2\,e\right )}^2}{b^9}-\frac {6\,e^2\,{\left (a\,e-b\,d\right )}^2}{b^5}\right )\,\sqrt {d+e\,x}+\frac {{\left (d+e\,x\right )}^{3/2}\,\left (\frac {17\,a^3\,b\,e^5}{4}-\frac {51\,a^2\,b^2\,d\,e^4}{4}+\frac {51\,a\,b^3\,d^2\,e^3}{4}-\frac {17\,b^4\,d^3\,e^2}{4}\right )+\sqrt {d+e\,x}\,\left (\frac {15\,a^4\,e^6}{4}-15\,a^3\,b\,d\,e^5+\frac {45\,a^2\,b^2\,d^2\,e^4}{2}-15\,a\,b^3\,d^3\,e^3+\frac {15\,b^4\,d^4\,e^2}{4}\right )}{b^7\,{\left (d+e\,x\right )}^2-\left (2\,b^7\,d-2\,a\,b^6\,e\right )\,\left (d+e\,x\right )+b^7\,d^2+a^2\,b^5\,e^2-2\,a\,b^6\,d\,e}+\frac {2\,e^2\,{\left (d+e\,x\right )}^{5/2}}{5\,b^3}+\frac {2\,e^2\,\left (3\,b^3\,d-3\,a\,b^2\,e\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,b^6}-\frac {63\,e^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,e^2\,{\left (a\,e-b\,d\right )}^{5/2}\,\sqrt {d+e\,x}}{a^3\,e^5-3\,a^2\,b\,d\,e^4+3\,a\,b^2\,d^2\,e^3-b^3\,d^3\,e^2}\right )\,{\left (a\,e-b\,d\right )}^{5/2}}{4\,b^{11/2}} \]
((2*e^2*(3*b^3*d - 3*a*b^2*e)^2)/b^9 - (6*e^2*(a*e - b*d)^2)/b^5)*(d + e*x )^(1/2) + ((d + e*x)^(3/2)*((17*a^3*b*e^5)/4 - (17*b^4*d^3*e^2)/4 + (51*a* b^3*d^2*e^3)/4 - (51*a^2*b^2*d*e^4)/4) + (d + e*x)^(1/2)*((15*a^4*e^6)/4 + (15*b^4*d^4*e^2)/4 - 15*a*b^3*d^3*e^3 + (45*a^2*b^2*d^2*e^4)/2 - 15*a^3*b *d*e^5))/(b^7*(d + e*x)^2 - (2*b^7*d - 2*a*b^6*e)*(d + e*x) + b^7*d^2 + a^ 2*b^5*e^2 - 2*a*b^6*d*e) + (2*e^2*(d + e*x)^(5/2))/(5*b^3) + (2*e^2*(3*b^3 *d - 3*a*b^2*e)*(d + e*x)^(3/2))/(3*b^6) - (63*e^2*atan((b^(1/2)*e^2*(a*e - b*d)^(5/2)*(d + e*x)^(1/2))/(a^3*e^5 - b^3*d^3*e^2 + 3*a*b^2*d^2*e^3 - 3 *a^2*b*d*e^4))*(a*e - b*d)^(5/2))/(4*b^(11/2))