Integrand size = 33, antiderivative size = 214 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=\frac {(b d-a e) (2 b B d+5 A b e-7 a B e) \sqrt {d+e x}}{b^4}+\frac {(2 b B d+5 A b e-7 a B e) (d+e x)^{3/2}}{3 b^3}+\frac {(2 b B d+5 A b e-7 a B e) (d+e x)^{5/2}}{5 b^2 (b d-a e)}-\frac {(A b-a B) (d+e x)^{7/2}}{b (b d-a e) (a+b x)}-\frac {(b d-a e)^{3/2} (2 b B d+5 A b e-7 a B e) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{9/2}} \]
1/3*(5*A*b*e-7*B*a*e+2*B*b*d)*(e*x+d)^(3/2)/b^3+1/5*(5*A*b*e-7*B*a*e+2*B*b *d)*(e*x+d)^(5/2)/b^2/(-a*e+b*d)-(A*b-B*a)*(e*x+d)^(7/2)/b/(-a*e+b*d)/(b*x +a)-(-a*e+b*d)^(3/2)*(5*A*b*e-7*B*a*e+2*B*b*d)*arctanh(b^(1/2)*(e*x+d)^(1/ 2)/(-a*e+b*d)^(1/2))/b^(9/2)+(-a*e+b*d)*(5*A*b*e-7*B*a*e+2*B*b*d)*(e*x+d)^ (1/2)/b^4
Time = 0.13 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.98 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=\frac {\sqrt {d+e x} \left (-5 A b \left (15 a^2 e^2+10 a b e (-2 d+e x)+b^2 \left (3 d^2-14 d e x-2 e^2 x^2\right )\right )+B \left (105 a^3 e^2+10 a^2 b e (-17 d+7 e x)+a b^2 \left (61 d^2-118 d e x-14 e^2 x^2\right )+2 b^3 x \left (23 d^2+11 d e x+3 e^2 x^2\right )\right )\right )}{15 b^4 (a+b x)}+\frac {(-b d+a e)^{3/2} (2 b B d+5 A b e-7 a B e) \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{b^{9/2}} \]
(Sqrt[d + e*x]*(-5*A*b*(15*a^2*e^2 + 10*a*b*e*(-2*d + e*x) + b^2*(3*d^2 - 14*d*e*x - 2*e^2*x^2)) + B*(105*a^3*e^2 + 10*a^2*b*e*(-17*d + 7*e*x) + a*b ^2*(61*d^2 - 118*d*e*x - 14*e^2*x^2) + 2*b^3*x*(23*d^2 + 11*d*e*x + 3*e^2* x^2))))/(15*b^4*(a + b*x)) + ((-(b*d) + a*e)^(3/2)*(2*b*B*d + 5*A*b*e - 7* a*B*e)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/b^(9/2)
Time = 0.34 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.90, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {1184, 27, 87, 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.
\(\displaystyle \int \frac {(A+B x) (d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle b^2 \int \frac {(A+B x) (d+e x)^{5/2}}{b^2 (a+b x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(A+B x) (d+e x)^{5/2}}{(a+b x)^2}dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {(-7 a B e+5 A b e+2 b B d) \int \frac {(d+e x)^{5/2}}{a+b x}dx}{2 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{b (a+b x) (b d-a e)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(-7 a B e+5 A b e+2 b B d) \left (\frac {(b d-a e) \int \frac {(d+e x)^{3/2}}{a+b x}dx}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\right )}{2 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{b (a+b x) (b d-a e)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(-7 a B e+5 A b e+2 b B d) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \int \frac {\sqrt {d+e x}}{a+b x}dx}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\right )}{2 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{b (a+b x) (b d-a e)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(-7 a B e+5 A b e+2 b B d) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\right )}{2 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{b (a+b x) (b d-a e)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(-7 a B e+5 A b e+2 b B d) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {2 (b d-a e) \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b e}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\right )}{2 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{b (a+b x) (b d-a e)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(-7 a B e+5 A b e+2 b B d) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {2 \sqrt {d+e x}}{b}-\frac {2 \sqrt {b d-a e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2}}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\right )}{2 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{b (a+b x) (b d-a e)}\) |
-(((A*b - a*B)*(d + e*x)^(7/2))/(b*(b*d - a*e)*(a + b*x))) + ((2*b*B*d + 5 *A*b*e - 7*a*B*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]*ArcT anh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/b^(3/2)))/b))/b))/(2*b*(b*d - a*e))
3.19.9.3.1 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 + 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)*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 = 0.37 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.99
method | result | size |
risch | \(-\frac {2 \left (-3 B \,b^{2} e^{2} x^{2}-5 A \,b^{2} e^{2} x +10 B a b \,e^{2} x -11 B \,b^{2} d e x +30 A a b \,e^{2}-35 A \,b^{2} d e -45 a^{2} B \,e^{2}+70 B a b d e -23 B \,b^{2} d^{2}\right ) \sqrt {e x +d}}{15 b^{4}}+\frac {\left (2 e^{2} a^{2}-4 a b d e +2 b^{2} d^{2}\right ) \left (\frac {\left (-\frac {1}{2} A b e +\frac {1}{2} B a e \right ) \sqrt {e x +d}}{b \left (e x +d \right )+a e -b d}+\frac {\left (5 A b e -7 B a e +2 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \sqrt {\left (a e -b d \right ) b}}\right )}{b^{4}}\) | \(212\) |
pseudoelliptic | \(\frac {-5 \left (\frac {\left (\frac {2 \left (-\frac {23}{3} x \,d^{2}-x^{3} e^{2}-\frac {11}{3} d e \,x^{2}\right ) b^{3}}{5}-\frac {61 \left (-\frac {14}{61} e^{2} x^{2}-\frac {118}{61} d e x +d^{2}\right ) a \,b^{2}}{15}+\frac {34 \left (-\frac {7 e x}{17}+d \right ) e \,a^{2} b}{3}-7 a^{3} e^{2}\right ) B}{5}+b \left (\frac {\left (-\frac {14}{3} d e x +d^{2}-\frac {2}{3} e^{2} x^{2}\right ) b^{2}}{5}-\frac {4 \left (-\frac {e x}{2}+d \right ) e a b}{3}+e^{2} a^{2}\right ) A \right ) \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}+5 \left (a e -b d \right )^{2} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) \left (\frac {\left (-7 a e +2 b d \right ) B}{5}+A b e \right ) \left (b x +a \right )}{b^{4} \left (b x +a \right ) \sqrt {\left (a e -b d \right ) b}}\) | \(222\) |
derivativedivides | \(-\frac {2 \left (-\frac {B \,b^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {A \,b^{2} e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {2 B a b e \left (e x +d \right )^{\frac {3}{2}}}{3}-\frac {B \,b^{2} d \left (e x +d \right )^{\frac {3}{2}}}{3}+2 A a b \,e^{2} \sqrt {e x +d}-2 A \,b^{2} d e \sqrt {e x +d}-3 a^{2} e^{2} B \sqrt {e x +d}+4 B a b d e \sqrt {e x +d}-b^{2} d^{2} B \sqrt {e x +d}\right )}{b^{4}}+\frac {\frac {2 \left (-\frac {1}{2} A \,a^{2} b \,e^{3}+A a \,b^{2} d \,e^{2}-\frac {1}{2} A \,b^{3} d^{2} e +\frac {1}{2} B \,e^{3} a^{3}-B \,a^{2} b d \,e^{2}+\frac {1}{2} B a \,b^{2} d^{2} e \right ) \sqrt {e x +d}}{b \left (e x +d \right )+a e -b d}+\frac {\left (5 A \,a^{2} b \,e^{3}-10 A a \,b^{2} d \,e^{2}+5 A \,b^{3} d^{2} e -7 B \,e^{3} a^{3}+16 B \,a^{2} b d \,e^{2}-11 B a \,b^{2} d^{2} e +2 B \,b^{3} d^{3}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}}}{b^{4}}\) | \(339\) |
default | \(-\frac {2 \left (-\frac {B \,b^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {A \,b^{2} e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {2 B a b e \left (e x +d \right )^{\frac {3}{2}}}{3}-\frac {B \,b^{2} d \left (e x +d \right )^{\frac {3}{2}}}{3}+2 A a b \,e^{2} \sqrt {e x +d}-2 A \,b^{2} d e \sqrt {e x +d}-3 a^{2} e^{2} B \sqrt {e x +d}+4 B a b d e \sqrt {e x +d}-b^{2} d^{2} B \sqrt {e x +d}\right )}{b^{4}}+\frac {\frac {2 \left (-\frac {1}{2} A \,a^{2} b \,e^{3}+A a \,b^{2} d \,e^{2}-\frac {1}{2} A \,b^{3} d^{2} e +\frac {1}{2} B \,e^{3} a^{3}-B \,a^{2} b d \,e^{2}+\frac {1}{2} B a \,b^{2} d^{2} e \right ) \sqrt {e x +d}}{b \left (e x +d \right )+a e -b d}+\frac {\left (5 A \,a^{2} b \,e^{3}-10 A a \,b^{2} d \,e^{2}+5 A \,b^{3} d^{2} e -7 B \,e^{3} a^{3}+16 B \,a^{2} b d \,e^{2}-11 B a \,b^{2} d^{2} e +2 B \,b^{3} d^{3}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}}}{b^{4}}\) | \(339\) |
-2/15*(-3*B*b^2*e^2*x^2-5*A*b^2*e^2*x+10*B*a*b*e^2*x-11*B*b^2*d*e*x+30*A*a *b*e^2-35*A*b^2*d*e-45*B*a^2*e^2+70*B*a*b*d*e-23*B*b^2*d^2)*(e*x+d)^(1/2)/ b^4+1/b^4*(2*a^2*e^2-4*a*b*d*e+2*b^2*d^2)*((-1/2*A*b*e+1/2*B*a*e)*(e*x+d)^ (1/2)/(b*(e*x+d)+a*e-b*d)+1/2*(5*A*b*e-7*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)))
Time = 0.41 (sec) , antiderivative size = 666, normalized size of antiderivative = 3.11 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=\left [-\frac {15 \, {\left (2 \, B a b^{2} d^{2} - {\left (9 \, B a^{2} b - 5 \, A a b^{2}\right )} d e + {\left (7 \, B a^{3} - 5 \, A a^{2} b\right )} e^{2} + {\left (2 \, B b^{3} d^{2} - {\left (9 \, B a b^{2} - 5 \, A b^{3}\right )} d e + {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} e^{2}\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 (6 \, B b^{3} e^{2} x^{3} + {\left (61 \, B a b^{2} - 15 \, A b^{3}\right )} d^{2} - 10 \, {\left (17 \, B a^{2} b - 10 \, A a b^{2}\right )} d e + 15 \, {\left (7 \, B a^{3} - 5 \, A a^{2} b\right )} e^{2} + 2 \, {\left (11 \, B b^{3} d e - {\left (7 \, B a b^{2} - 5 \, A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (23 \, B b^{3} d^{2} - {\left (59 \, B a b^{2} - 35 \, A b^{3}\right )} d e + 5 \, {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{30 \, {\left (b^{5} x + a b^{4}\right )}}, -\frac {15 \, {\left (2 \, B a b^{2} d^{2} - {\left (9 \, B a^{2} b - 5 \, A a b^{2}\right )} d e + {\left (7 \, B a^{3} - 5 \, A a^{2} b\right )} e^{2} + {\left (2 \, B b^{3} d^{2} - {\left (9 \, B a b^{2} - 5 \, A b^{3}\right )} d e + {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} e^{2}\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 (6 \, B b^{3} e^{2} x^{3} + {\left (61 \, B a b^{2} - 15 \, A b^{3}\right )} d^{2} - 10 \, {\left (17 \, B a^{2} b - 10 \, A a b^{2}\right )} d e + 15 \, {\left (7 \, B a^{3} - 5 \, A a^{2} b\right )} e^{2} + 2 \, {\left (11 \, B b^{3} d e - {\left (7 \, B a b^{2} - 5 \, A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (23 \, B b^{3} d^{2} - {\left (59 \, B a b^{2} - 35 \, A b^{3}\right )} d e + 5 \, {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (b^{5} x + a b^{4}\right )}}\right ] \]
[-1/30*(15*(2*B*a*b^2*d^2 - (9*B*a^2*b - 5*A*a*b^2)*d*e + (7*B*a^3 - 5*A*a ^2*b)*e^2 + (2*B*b^3*d^2 - (9*B*a*b^2 - 5*A*b^3)*d*e + (7*B*a^2*b - 5*A*a* b^2)*e^2)*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*(6*B*b^3*e^2*x^3 + (61*B*a*b^2 - 1 5*A*b^3)*d^2 - 10*(17*B*a^2*b - 10*A*a*b^2)*d*e + 15*(7*B*a^3 - 5*A*a^2*b) *e^2 + 2*(11*B*b^3*d*e - (7*B*a*b^2 - 5*A*b^3)*e^2)*x^2 + 2*(23*B*b^3*d^2 - (59*B*a*b^2 - 35*A*b^3)*d*e + 5*(7*B*a^2*b - 5*A*a*b^2)*e^2)*x)*sqrt(e*x + d))/(b^5*x + a*b^4), -1/15*(15*(2*B*a*b^2*d^2 - (9*B*a^2*b - 5*A*a*b^2) *d*e + (7*B*a^3 - 5*A*a^2*b)*e^2 + (2*B*b^3*d^2 - (9*B*a*b^2 - 5*A*b^3)*d* e + (7*B*a^2*b - 5*A*a*b^2)*e^2)*x)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (6*B*b^3*e^2*x^3 + (61*B*a*b^2 - 15*A*b^3)*d^2 - 10*(17*B*a^2*b - 10*A*a*b^2)*d*e + 15*(7*B*a^3 - 5*A*a^2 *b)*e^2 + 2*(11*B*b^3*d*e - (7*B*a*b^2 - 5*A*b^3)*e^2)*x^2 + 2*(23*B*b^3*d ^2 - (59*B*a*b^2 - 35*A*b^3)*d*e + 5*(7*B*a^2*b - 5*A*a*b^2)*e^2)*x)*sqrt( e*x + d))/(b^5*x + a*b^4)]
Timed out. \[ \int \frac {(A+B x) (d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(A+B x) (d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.31 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.79 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=\frac {{\left (2 \, B b^{3} d^{3} - 11 \, B a b^{2} d^{2} e + 5 \, A b^{3} d^{2} e + 16 \, B a^{2} b d e^{2} - 10 \, A a b^{2} d e^{2} - 7 \, B a^{3} e^{3} + 5 \, A a^{2} b e^{3}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{4}} + \frac {\sqrt {e x + d} B a b^{2} d^{2} e - \sqrt {e x + d} A b^{3} d^{2} e - 2 \, \sqrt {e x + d} B a^{2} b d e^{2} + 2 \, \sqrt {e x + d} A a b^{2} d e^{2} + \sqrt {e x + d} B a^{3} e^{3} - \sqrt {e x + d} A a^{2} b e^{3}}{{\left ({\left (e x + d\right )} b - b d + a e\right )} b^{4}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} B b^{8} + 5 \, {\left (e x + d\right )}^{\frac {3}{2}} B b^{8} d + 15 \, \sqrt {e x + d} B b^{8} d^{2} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} B a b^{7} e + 5 \, {\left (e x + d\right )}^{\frac {3}{2}} A b^{8} e - 60 \, \sqrt {e x + d} B a b^{7} d e + 30 \, \sqrt {e x + d} A b^{8} d e + 45 \, \sqrt {e x + d} B a^{2} b^{6} e^{2} - 30 \, \sqrt {e x + d} A a b^{7} e^{2}\right )}}{15 \, b^{10}} \]
(2*B*b^3*d^3 - 11*B*a*b^2*d^2*e + 5*A*b^3*d^2*e + 16*B*a^2*b*d*e^2 - 10*A* a*b^2*d*e^2 - 7*B*a^3*e^3 + 5*A*a^2*b*e^3)*arctan(sqrt(e*x + d)*b/sqrt(-b^ 2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^4) + (sqrt(e*x + d)*B*a*b^2*d^2*e - sqrt(e*x + d)*A*b^3*d^2*e - 2*sqrt(e*x + d)*B*a^2*b*d*e^2 + 2*sqrt(e*x + d )*A*a*b^2*d*e^2 + sqrt(e*x + d)*B*a^3*e^3 - sqrt(e*x + d)*A*a^2*b*e^3)/((( e*x + d)*b - b*d + a*e)*b^4) + 2/15*(3*(e*x + d)^(5/2)*B*b^8 + 5*(e*x + d) ^(3/2)*B*b^8*d + 15*sqrt(e*x + d)*B*b^8*d^2 - 10*(e*x + d)^(3/2)*B*a*b^7*e + 5*(e*x + d)^(3/2)*A*b^8*e - 60*sqrt(e*x + d)*B*a*b^7*d*e + 30*sqrt(e*x + d)*A*b^8*d*e + 45*sqrt(e*x + d)*B*a^2*b^6*e^2 - 30*sqrt(e*x + d)*A*a*b^7 *e^2)/b^10
Time = 0.14 (sec) , antiderivative size = 363, normalized size of antiderivative = 1.70 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=\left (\frac {2\,A\,e-2\,B\,d}{3\,b^2}+\frac {2\,B\,\left (2\,b^2\,d-2\,a\,b\,e\right )}{3\,b^4}\right )\,{\left (d+e\,x\right )}^{3/2}+\left (\frac {\left (2\,b^2\,d-2\,a\,b\,e\right )\,\left (\frac {2\,A\,e-2\,B\,d}{b^2}+\frac {2\,B\,\left (2\,b^2\,d-2\,a\,b\,e\right )}{b^4}\right )}{b^2}-\frac {2\,B\,{\left (a\,e-b\,d\right )}^2}{b^4}\right )\,\sqrt {d+e\,x}+\frac {\sqrt {d+e\,x}\,\left (B\,a^3\,e^3-2\,B\,a^2\,b\,d\,e^2-A\,a^2\,b\,e^3+B\,a\,b^2\,d^2\,e+2\,A\,a\,b^2\,d\,e^2-A\,b^3\,d^2\,e\right )}{b^5\,\left (d+e\,x\right )-b^5\,d+a\,b^4\,e}+\frac {2\,B\,{\left (d+e\,x\right )}^{5/2}}{5\,b^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,{\left (a\,e-b\,d\right )}^{3/2}\,\sqrt {d+e\,x}\,\left (5\,A\,b\,e-7\,B\,a\,e+2\,B\,b\,d\right )}{-7\,B\,a^3\,e^3+16\,B\,a^2\,b\,d\,e^2+5\,A\,a^2\,b\,e^3-11\,B\,a\,b^2\,d^2\,e-10\,A\,a\,b^2\,d\,e^2+2\,B\,b^3\,d^3+5\,A\,b^3\,d^2\,e}\right )\,{\left (a\,e-b\,d\right )}^{3/2}\,\left (5\,A\,b\,e-7\,B\,a\,e+2\,B\,b\,d\right )}{b^{9/2}} \]
((2*A*e - 2*B*d)/(3*b^2) + (2*B*(2*b^2*d - 2*a*b*e))/(3*b^4))*(d + e*x)^(3 /2) + (((2*b^2*d - 2*a*b*e)*((2*A*e - 2*B*d)/b^2 + (2*B*(2*b^2*d - 2*a*b*e ))/b^4))/b^2 - (2*B*(a*e - b*d)^2)/b^4)*(d + e*x)^(1/2) + ((d + e*x)^(1/2) *(B*a^3*e^3 - A*a^2*b*e^3 - A*b^3*d^2*e + 2*A*a*b^2*d*e^2 + B*a*b^2*d^2*e - 2*B*a^2*b*d*e^2))/(b^5*(d + e*x) - b^5*d + a*b^4*e) + (2*B*(d + e*x)^(5/ 2))/(5*b^2) + (atan((b^(1/2)*(a*e - b*d)^(3/2)*(d + e*x)^(1/2)*(5*A*b*e - 7*B*a*e + 2*B*b*d))/(2*B*b^3*d^3 - 7*B*a^3*e^3 + 5*A*a^2*b*e^3 + 5*A*b^3*d ^2*e - 10*A*a*b^2*d*e^2 - 11*B*a*b^2*d^2*e + 16*B*a^2*b*d*e^2))*(a*e - b*d )^(3/2)*(5*A*b*e - 7*B*a*e + 2*B*b*d))/b^(9/2)