Integrand size = 28, antiderivative size = 172 \[ \int \frac {(d+e x)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {105 e^3 (b d-a e) \sqrt {d+e x}}{8 b^5}+\frac {35 e^3 (d+e x)^{3/2}}{8 b^4}-\frac {21 e^2 (d+e x)^{5/2}}{8 b^3 (a+b x)}-\frac {3 e (d+e x)^{7/2}}{4 b^2 (a+b x)^2}-\frac {(d+e x)^{9/2}}{3 b (a+b x)^3}-\frac {105 e^3 (b d-a e)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{11/2}} \]
35/8*e^3*(e*x+d)^(3/2)/b^4-21/8*e^2*(e*x+d)^(5/2)/b^3/(b*x+a)-3/4*e*(e*x+d )^(7/2)/b^2/(b*x+a)^2-1/3*(e*x+d)^(9/2)/b/(b*x+a)^3-105/8*e^3*(-a*e+b*d)^( 3/2)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(11/2)+105/8*e^3*(- a*e+b*d)*(e*x+d)^(1/2)/b^5
Time = 0.82 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.21 \[ \int \frac {(d+e x)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {\sqrt {d+e x} \left (315 a^4 e^4-420 a^3 b e^3 (d-2 e x)+63 a^2 b^2 e^2 \left (d^2-18 d e x+11 e^2 x^2\right )+18 a b^3 e \left (d^3+10 d^2 e x-53 d e^2 x^2+8 e^3 x^3\right )+b^4 \left (8 d^4+50 d^3 e x+165 d^2 e^2 x^2-208 d e^3 x^3-16 e^4 x^4\right )\right )}{24 b^5 (a+b x)^3}+\frac {105 e^3 (-b d+a e)^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{8 b^{11/2}} \]
-1/24*(Sqrt[d + e*x]*(315*a^4*e^4 - 420*a^3*b*e^3*(d - 2*e*x) + 63*a^2*b^2 *e^2*(d^2 - 18*d*e*x + 11*e^2*x^2) + 18*a*b^3*e*(d^3 + 10*d^2*e*x - 53*d*e ^2*x^2 + 8*e^3*x^3) + b^4*(8*d^4 + 50*d^3*e*x + 165*d^2*e^2*x^2 - 208*d*e^ 3*x^3 - 16*e^4*x^4)))/(b^5*(a + b*x)^3) + (105*e^3*(-(b*d) + a*e)^(3/2)*Ar cTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(8*b^(11/2))
Time = 0.28 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {1098, 27, 51, 51, 51, 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.
\(\displaystyle \int \frac {(d+e x)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 1098 |
\(\displaystyle b^4 \int \frac {(d+e x)^{9/2}}{b^4 (a+b x)^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(d+e x)^{9/2}}{(a+b x)^4}dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {3 e \int \frac {(d+e x)^{7/2}}{(a+b x)^3}dx}{2 b}-\frac {(d+e x)^{9/2}}{3 b (a+b x)^3}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {3 e \left (\frac {7 e \int \frac {(d+e x)^{5/2}}{(a+b x)^2}dx}{4 b}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}\right )}{2 b}-\frac {(d+e x)^{9/2}}{3 b (a+b x)^3}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {3 e \left (\frac {7 e \left (\frac {5 e \int \frac {(d+e x)^{3/2}}{a+b x}dx}{2 b}-\frac {(d+e x)^{5/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}\right )}{2 b}-\frac {(d+e x)^{9/2}}{3 b (a+b x)^3}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {3 e \left (\frac {7 e \left (\frac {5 e \left (\frac {(b d-a e) \int \frac {\sqrt {d+e x}}{a+b x}dx}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{2 b}-\frac {(d+e x)^{5/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}\right )}{2 b}-\frac {(d+e x)^{9/2}}{3 b (a+b x)^3}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {3 e \left (\frac {7 e \left (\frac {5 e \left (\frac {(b d-a e) \left (\frac {(b d-a e) \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{2 b}-\frac {(d+e x)^{5/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}\right )}{2 b}-\frac {(d+e x)^{9/2}}{3 b (a+b x)^3}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {3 e \left (\frac {7 e \left (\frac {5 e \left (\frac {(b d-a e) \left (\frac {2 (b d-a e) \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b e}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{2 b}-\frac {(d+e x)^{5/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}\right )}{2 b}-\frac {(d+e x)^{9/2}}{3 b (a+b x)^3}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {3 e \left (\frac {7 e \left (\frac {5 e \left (\frac {(b d-a e) \left (\frac {2 \sqrt {d+e x}}{b}-\frac {2 \sqrt {b d-a e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2}}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{2 b}-\frac {(d+e x)^{5/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}\right )}{2 b}-\frac {(d+e x)^{9/2}}{3 b (a+b x)^3}\) |
-1/3*(d + e*x)^(9/2)/(b*(a + b*x)^3) + (3*e*(-1/2*(d + e*x)^(7/2)/(b*(a + b*x)^2) + (7*e*(-((d + e*x)^(5/2)/(b*(a + b*x))) + (5*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[ (Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/b^(3/2)))/b))/(2*b)))/(4*b)))/(2 *b)
3.17.55.3.1 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_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[ {a, b, c, d, e, m}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 2.68 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.03
method | result | size |
risch | \(-\frac {2 e^{3} \left (-b e x +12 a e -13 b d \right ) \sqrt {e x +d}}{3 b^{5}}+\frac {\left (2 a^{2} e^{2}-4 a b d e +2 b^{2} d^{2}\right ) e^{3} \left (\frac {-\frac {55 b^{2} \left (e x +d \right )^{\frac {5}{2}}}{16}-\frac {35 \left (a e -b d \right ) b \left (e x +d \right )^{\frac {3}{2}}}{6}+\left (-\frac {41}{16} a^{2} e^{2}+\frac {41}{8} a b d e -\frac {41}{16} b^{2} d^{2}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{3}}+\frac {105 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{16 \sqrt {\left (a e -b d \right ) b}}\right )}{b^{5}}\) | \(178\) |
pseudoelliptic | \(\frac {\frac {105 e^{3} \left (b x +a \right )^{3} \left (a e -b d \right )^{2} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8}-\frac {105 \sqrt {e x +d}\, \left (\left (-\frac {16}{315} e^{4} x^{4}-\frac {208}{315} d \,e^{3} x^{3}+\frac {11}{21} d^{2} e^{2} x^{2}+\frac {10}{63} d^{3} e x +\frac {8}{315} d^{4}\right ) b^{4}+\frac {2 a e \left (8 e^{3} x^{3}-53 d \,e^{2} x^{2}+10 d^{2} e x +d^{3}\right ) b^{3}}{35}+\frac {a^{2} e^{2} \left (11 x^{2} e^{2}-18 d e x +d^{2}\right ) b^{2}}{5}-\frac {4 a^{3} e^{3} \left (-2 e x +d \right ) b}{3}+e^{4} a^{4}\right ) \sqrt {\left (a e -b d \right ) b}}{8}}{b^{5} \left (b x +a \right )^{3} \sqrt {\left (a e -b d \right ) b}}\) | \(221\) |
derivativedivides | \(2 e^{3} \left (-\frac {-\frac {\left (e x +d \right )^{\frac {3}{2}} b}{3}+4 a e \sqrt {e x +d}-4 d b \sqrt {e x +d}}{b^{5}}+\frac {\frac {\left (-\frac {55}{16} a^{2} b^{2} e^{2}+\frac {55}{8} a \,b^{3} d e -\frac {55}{16} b^{4} d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}-\frac {35 b \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}}{6}+\left (-\frac {41}{16} e^{4} a^{4}+\frac {41}{4} b \,e^{3} d \,a^{3}-\frac {123}{8} b^{2} e^{2} d^{2} a^{2}+\frac {41}{4} a \,b^{3} d^{3} e -\frac {41}{16} b^{4} d^{4}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{3}}+\frac {105 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{16 \sqrt {\left (a e -b d \right ) b}}}{b^{5}}\right )\) | \(267\) |
default | \(2 e^{3} \left (-\frac {-\frac {\left (e x +d \right )^{\frac {3}{2}} b}{3}+4 a e \sqrt {e x +d}-4 d b \sqrt {e x +d}}{b^{5}}+\frac {\frac {\left (-\frac {55}{16} a^{2} b^{2} e^{2}+\frac {55}{8} a \,b^{3} d e -\frac {55}{16} b^{4} d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}-\frac {35 b \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}}{6}+\left (-\frac {41}{16} e^{4} a^{4}+\frac {41}{4} b \,e^{3} d \,a^{3}-\frac {123}{8} b^{2} e^{2} d^{2} a^{2}+\frac {41}{4} a \,b^{3} d^{3} e -\frac {41}{16} b^{4} d^{4}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{3}}+\frac {105 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{16 \sqrt {\left (a e -b d \right ) b}}}{b^{5}}\right )\) | \(267\) |
-2/3*e^3*(-b*e*x+12*a*e-13*b*d)*(e*x+d)^(1/2)/b^5+1/b^5*(2*a^2*e^2-4*a*b*d *e+2*b^2*d^2)*e^3*((-55/16*b^2*(e*x+d)^(5/2)-35/6*(a*e-b*d)*b*(e*x+d)^(3/2 )+(-41/16*a^2*e^2+41/8*a*b*d*e-41/16*b^2*d^2)*(e*x+d)^(1/2))/(b*(e*x+d)+a* e-b*d)^3+105/16/((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. \(2 (140) = 280\).
Time = 0.31 (sec) , antiderivative size = 730, normalized size of antiderivative = 4.24 \[ \int \frac {(d+e x)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\left [-\frac {315 \, {\left (a^{3} b d e^{3} - a^{4} e^{4} + {\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 3 \, {\left (a b^{3} d e^{3} - a^{2} b^{2} e^{4}\right )} x^{2} + 3 \, {\left (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 (16 \, b^{4} e^{4} x^{4} - 8 \, b^{4} d^{4} - 18 \, a b^{3} d^{3} e - 63 \, a^{2} b^{2} d^{2} e^{2} + 420 \, a^{3} b d e^{3} - 315 \, a^{4} e^{4} + 16 \, {\left (13 \, b^{4} d e^{3} - 9 \, a b^{3} e^{4}\right )} x^{3} - 3 \, {\left (55 \, b^{4} d^{2} e^{2} - 318 \, a b^{3} d e^{3} + 231 \, a^{2} b^{2} e^{4}\right )} x^{2} - 2 \, {\left (25 \, b^{4} d^{3} e + 90 \, a b^{3} d^{2} e^{2} - 567 \, a^{2} b^{2} d e^{3} + 420 \, a^{3} b e^{4}\right )} x\right )} \sqrt {e x + d}}{48 \, {\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}}, -\frac {315 \, {\left (a^{3} b d e^{3} - a^{4} e^{4} + {\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 3 \, {\left (a b^{3} d e^{3} - a^{2} b^{2} e^{4}\right )} x^{2} + 3 \, {\left (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 (16 \, b^{4} e^{4} x^{4} - 8 \, b^{4} d^{4} - 18 \, a b^{3} d^{3} e - 63 \, a^{2} b^{2} d^{2} e^{2} + 420 \, a^{3} b d e^{3} - 315 \, a^{4} e^{4} + 16 \, {\left (13 \, b^{4} d e^{3} - 9 \, a b^{3} e^{4}\right )} x^{3} - 3 \, {\left (55 \, b^{4} d^{2} e^{2} - 318 \, a b^{3} d e^{3} + 231 \, a^{2} b^{2} e^{4}\right )} x^{2} - 2 \, {\left (25 \, b^{4} d^{3} e + 90 \, a b^{3} d^{2} e^{2} - 567 \, a^{2} b^{2} d e^{3} + 420 \, a^{3} b e^{4}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}}\right ] \]
[-1/48*(315*(a^3*b*d*e^3 - a^4*e^4 + (b^4*d*e^3 - a*b^3*e^4)*x^3 + 3*(a*b^ 3*d*e^3 - a^2*b^2*e^4)*x^2 + 3*(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*(16*b^4*e^4*x^4 - 8*b^4*d^4 - 18*a*b^3*d^3*e - 63*a^2*b^2*d ^2*e^2 + 420*a^3*b*d*e^3 - 315*a^4*e^4 + 16*(13*b^4*d*e^3 - 9*a*b^3*e^4)*x ^3 - 3*(55*b^4*d^2*e^2 - 318*a*b^3*d*e^3 + 231*a^2*b^2*e^4)*x^2 - 2*(25*b^ 4*d^3*e + 90*a*b^3*d^2*e^2 - 567*a^2*b^2*d*e^3 + 420*a^3*b*e^4)*x)*sqrt(e* x + d))/(b^8*x^3 + 3*a*b^7*x^2 + 3*a^2*b^6*x + a^3*b^5), -1/24*(315*(a^3*b *d*e^3 - a^4*e^4 + (b^4*d*e^3 - a*b^3*e^4)*x^3 + 3*(a*b^3*d*e^3 - a^2*b^2* e^4)*x^2 + 3*(a^2*b^2*d*e^3 - a^3*b*e^4)*x)*sqrt(-(b*d - a*e)/b)*arctan(-s qrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (16*b^4*e^4*x^4 - 8*b^4 *d^4 - 18*a*b^3*d^3*e - 63*a^2*b^2*d^2*e^2 + 420*a^3*b*d*e^3 - 315*a^4*e^4 + 16*(13*b^4*d*e^3 - 9*a*b^3*e^4)*x^3 - 3*(55*b^4*d^2*e^2 - 318*a*b^3*d*e ^3 + 231*a^2*b^2*e^4)*x^2 - 2*(25*b^4*d^3*e + 90*a*b^3*d^2*e^2 - 567*a^2*b ^2*d*e^3 + 420*a^3*b*e^4)*x)*sqrt(e*x + d))/(b^8*x^3 + 3*a*b^7*x^2 + 3*a^2 *b^6*x + a^3*b^5)]
Timed out. \[ \int \frac {(d+e x)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(d+e x)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]
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. 358 vs. \(2 (140) = 280\).
Time = 0.27 (sec) , antiderivative size = 358, normalized size of antiderivative = 2.08 \[ \int \frac {(d+e x)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {105 \, {\left (b^{2} d^{2} e^{3} - 2 \, a b d e^{4} + a^{2} e^{5}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{8 \, \sqrt {-b^{2} d + a b e} b^{5}} - \frac {165 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{4} d^{2} e^{3} - 280 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{4} d^{3} e^{3} + 123 \, \sqrt {e x + d} b^{4} d^{4} e^{3} - 330 \, {\left (e x + d\right )}^{\frac {5}{2}} a b^{3} d e^{4} + 840 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{3} d^{2} e^{4} - 492 \, \sqrt {e x + d} a b^{3} d^{3} e^{4} + 165 \, {\left (e x + d\right )}^{\frac {5}{2}} a^{2} b^{2} e^{5} - 840 \, {\left (e x + d\right )}^{\frac {3}{2}} a^{2} b^{2} d e^{5} + 738 \, \sqrt {e x + d} a^{2} b^{2} d^{2} e^{5} + 280 \, {\left (e x + d\right )}^{\frac {3}{2}} a^{3} b e^{6} - 492 \, \sqrt {e x + d} a^{3} b d e^{6} + 123 \, \sqrt {e x + d} a^{4} e^{7}}{24 \, {\left ({\left (e x + d\right )} b - b d + a e\right )}^{3} b^{5}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} b^{8} e^{3} + 12 \, \sqrt {e x + d} b^{8} d e^{3} - 12 \, \sqrt {e x + d} a b^{7} e^{4}\right )}}{3 \, b^{12}} \]
105/8*(b^2*d^2*e^3 - 2*a*b*d*e^4 + a^2*e^5)*arctan(sqrt(e*x + d)*b/sqrt(-b ^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^5) - 1/24*(165*(e*x + d)^(5/2)*b^4* d^2*e^3 - 280*(e*x + d)^(3/2)*b^4*d^3*e^3 + 123*sqrt(e*x + d)*b^4*d^4*e^3 - 330*(e*x + d)^(5/2)*a*b^3*d*e^4 + 840*(e*x + d)^(3/2)*a*b^3*d^2*e^4 - 49 2*sqrt(e*x + d)*a*b^3*d^3*e^4 + 165*(e*x + d)^(5/2)*a^2*b^2*e^5 - 840*(e*x + d)^(3/2)*a^2*b^2*d*e^5 + 738*sqrt(e*x + d)*a^2*b^2*d^2*e^5 + 280*(e*x + d)^(3/2)*a^3*b*e^6 - 492*sqrt(e*x + d)*a^3*b*d*e^6 + 123*sqrt(e*x + d)*a^ 4*e^7)/(((e*x + d)*b - b*d + a*e)^3*b^5) + 2/3*((e*x + d)^(3/2)*b^8*e^3 + 12*sqrt(e*x + d)*b^8*d*e^3 - 12*sqrt(e*x + d)*a*b^7*e^4)/b^12
Time = 9.49 (sec) , antiderivative size = 388, normalized size of antiderivative = 2.26 \[ \int \frac {(d+e x)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {2\,e^3\,{\left (d+e\,x\right )}^{3/2}}{3\,b^4}-\frac {{\left (d+e\,x\right )}^{5/2}\,\left (\frac {55\,a^2\,b^2\,e^5}{8}-\frac {55\,a\,b^3\,d\,e^4}{4}+\frac {55\,b^4\,d^2\,e^3}{8}\right )+{\left (d+e\,x\right )}^{3/2}\,\left (\frac {35\,a^3\,b\,e^6}{3}-35\,a^2\,b^2\,d\,e^5+35\,a\,b^3\,d^2\,e^4-\frac {35\,b^4\,d^3\,e^3}{3}\right )+\sqrt {d+e\,x}\,\left (\frac {41\,a^4\,e^7}{8}-\frac {41\,a^3\,b\,d\,e^6}{2}+\frac {123\,a^2\,b^2\,d^2\,e^5}{4}-\frac {41\,a\,b^3\,d^3\,e^4}{2}+\frac {41\,b^4\,d^4\,e^3}{8}\right )}{b^8\,{\left (d+e\,x\right )}^3-\left (3\,b^8\,d-3\,a\,b^7\,e\right )\,{\left (d+e\,x\right )}^2+\left (d+e\,x\right )\,\left (3\,a^2\,b^6\,e^2-6\,a\,b^7\,d\,e+3\,b^8\,d^2\right )-b^8\,d^3+a^3\,b^5\,e^3-3\,a^2\,b^6\,d\,e^2+3\,a\,b^7\,d^2\,e}+\frac {2\,e^3\,\left (4\,b^4\,d-4\,a\,b^3\,e\right )\,\sqrt {d+e\,x}}{b^8}+\frac {105\,e^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,e^3\,{\left (a\,e-b\,d\right )}^{3/2}\,\sqrt {d+e\,x}}{a^2\,e^5-2\,a\,b\,d\,e^4+b^2\,d^2\,e^3}\right )\,{\left (a\,e-b\,d\right )}^{3/2}}{8\,b^{11/2}} \]
(2*e^3*(d + e*x)^(3/2))/(3*b^4) - ((d + e*x)^(5/2)*((55*a^2*b^2*e^5)/8 + ( 55*b^4*d^2*e^3)/8 - (55*a*b^3*d*e^4)/4) + (d + e*x)^(3/2)*((35*a^3*b*e^6)/ 3 - (35*b^4*d^3*e^3)/3 + 35*a*b^3*d^2*e^4 - 35*a^2*b^2*d*e^5) + (d + e*x)^ (1/2)*((41*a^4*e^7)/8 + (41*b^4*d^4*e^3)/8 - (41*a*b^3*d^3*e^4)/2 + (123*a ^2*b^2*d^2*e^5)/4 - (41*a^3*b*d*e^6)/2))/(b^8*(d + e*x)^3 - (3*b^8*d - 3*a *b^7*e)*(d + e*x)^2 + (d + e*x)*(3*b^8*d^2 + 3*a^2*b^6*e^2 - 6*a*b^7*d*e) - b^8*d^3 + a^3*b^5*e^3 - 3*a^2*b^6*d*e^2 + 3*a*b^7*d^2*e) + (2*e^3*(4*b^4 *d - 4*a*b^3*e)*(d + e*x)^(1/2))/b^8 + (105*e^3*atan((b^(1/2)*e^3*(a*e - b *d)^(3/2)*(d + e*x)^(1/2))/(a^2*e^5 + b^2*d^2*e^3 - 2*a*b*d*e^4))*(a*e - b *d)^(3/2))/(8*b^(11/2))