Integrand size = 35, antiderivative size = 341 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {5 e (4 b B d+3 A b e-7 a B e) (a+b x) \sqrt {d+e x}}{4 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e (4 b B d+3 A b e-7 a B e) (a+b x) (d+e x)^{3/2}}{12 b^3 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(4 b B d+3 A b e-7 a B e) (d+e x)^{5/2}}{4 b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (d+e x)^{7/2}}{2 b (b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e \sqrt {b d-a e} (4 b B d+3 A b e-7 a B e) (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
5/12*e*(3*A*b*e-7*B*a*e+4*B*b*d)*(b*x+a)*(e*x+d)^(3/2)/b^3/(-a*e+b*d)/((b* x+a)^2)^(1/2)-1/4*(3*A*b*e-7*B*a*e+4*B*b*d)*(e*x+d)^(5/2)/b^2/(-a*e+b*d)/( (b*x+a)^2)^(1/2)-1/2*(A*b-B*a)*(e*x+d)^(7/2)/b/(-a*e+b*d)/(b*x+a)/((b*x+a) ^2)^(1/2)-5/4*e*(3*A*b*e-7*B*a*e+4*B*b*d)*(b*x+a)*arctanh(b^(1/2)*(e*x+d)^ (1/2)/(-a*e+b*d)^(1/2))*(-a*e+b*d)^(1/2)/b^(9/2)/((b*x+a)^2)^(1/2)+5/4*e*( 3*A*b*e-7*B*a*e+4*B*b*d)*(b*x+a)*(e*x+d)^(1/2)/b^4/((b*x+a)^2)^(1/2)
Time = 0.16 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.69 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {e (a+b x)^3 \left (\frac {\sqrt {b} \sqrt {d+e x} \left (-3 A b \left (-15 a^2 e^2+5 a b e (d-5 e x)+b^2 \left (2 d^2+9 d e x-8 e^2 x^2\right )\right )+B \left (-105 a^3 e^2+5 a^2 b e (19 d-35 e x)+a b^2 \left (-6 d^2+163 d e x-56 e^2 x^2\right )+4 b^3 x \left (-3 d^2+14 d e x+2 e^2 x^2\right )\right )\right )}{e (a+b x)^2}-15 \sqrt {-b d+a e} (4 b B d+3 A b e-7 a B e) \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )\right )}{12 b^{9/2} \left ((a+b x)^2\right )^{3/2}} \]
(e*(a + b*x)^3*((Sqrt[b]*Sqrt[d + e*x]*(-3*A*b*(-15*a^2*e^2 + 5*a*b*e*(d - 5*e*x) + b^2*(2*d^2 + 9*d*e*x - 8*e^2*x^2)) + B*(-105*a^3*e^2 + 5*a^2*b*e *(19*d - 35*e*x) + a*b^2*(-6*d^2 + 163*d*e*x - 56*e^2*x^2) + 4*b^3*x*(-3*d ^2 + 14*d*e*x + 2*e^2*x^2))))/(e*(a + b*x)^2) - 15*Sqrt[-(b*d) + a*e]*(4*b *B*d + 3*A*b*e - 7*a*B*e)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e ]]))/(12*b^(9/2)*((a + b*x)^2)^(3/2))
Time = 0.37 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.65, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {1187, 27, 87, 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.
| \(\displaystyle \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b^3 (a+b x) \int \frac {(A+B x) (d+e x)^{5/2}}{b^3 (a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {(A+B x) (d+e x)^{5/2}}{(a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(-7 a B e+3 A b e+4 b B d) \int \frac {(d+e x)^{5/2}}{(a+b x)^2}dx}{4 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{2 b (a+b x)^2 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(-7 a B e+3 A b e+4 b B d) \left (\frac {5 e \int \frac {(d+e x)^{3/2}}{a+b x}dx}{2 b}-\frac {(d+e x)^{5/2}}{b (a+b x)}\right )}{4 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{2 b (a+b x)^2 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(-7 a B e+3 A b e+4 b B d) \left (\frac {5 e \left (\frac {(b d-a e) \int \frac {\sqrt {d+e x}}{a+b x}dx}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{2 b}-\frac {(d+e x)^{5/2}}{b (a+b x)}\right )}{4 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{2 b (a+b x)^2 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(-7 a B e+3 A b e+4 b B d) \left (\frac {5 e \left (\frac {(b d-a e) \left (\frac {(b d-a e) \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{2 b}-\frac {(d+e x)^{5/2}}{b (a+b x)}\right )}{4 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{2 b (a+b x)^2 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(-7 a B e+3 A b e+4 b B d) \left (\frac {5 e \left (\frac {(b d-a e) \left (\frac {2 (b d-a e) \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b e}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{2 b}-\frac {(d+e x)^{5/2}}{b (a+b x)}\right )}{4 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{2 b (a+b x)^2 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(-7 a B e+3 A b e+4 b B d) \left (\frac {5 e \left (\frac {(b d-a e) \left (\frac {2 \sqrt {d+e x}}{b}-\frac {2 \sqrt {b d-a e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2}}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{2 b}-\frac {(d+e x)^{5/2}}{b (a+b x)}\right )}{4 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{2 b (a+b x)^2 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(-1/2*((A*b - a*B)*(d + e*x)^(7/2))/(b*(b*d - a*e)*(a + b*x)^2) + ((4*b*B*d + 3*A*b*e - 7*a*B*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*(b*d - a*e))))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.19.67.3.1 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)*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[(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 = 0.34 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.69
| method | result | size |
| risch | \(\frac {2 e \left (B b e x +3 A b e -9 B a e +7 B b d \right ) \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{3 b^{4} \left (b x +a \right )}-\frac {\left (2 a e -2 b d \right ) e \left (\frac {\left (-\frac {9}{8} A \,b^{2} e +\frac {13}{8} B e b a -\frac {1}{2} B \,b^{2} d \right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {7}{8} A a b \,e^{2}+\frac {7}{8} A \,b^{2} d e +\frac {11}{8} a^{2} B \,e^{2}-\frac {15}{8} B a b d e +\frac {1}{2} B \,b^{2} d^{2}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{2}}+\frac {5 \left (3 A b e -7 B a e +4 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \sqrt {\left (a e -b d \right ) b}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b^{4} \left (b x +a \right )}\) | \(234\) |
| default | \(\text {Expression too large to display}\) | \(1150\) |
2/3*e*(B*b*e*x+3*A*b*e-9*B*a*e+7*B*b*d)*(e*x+d)^(1/2)/b^4*((b*x+a)^2)^(1/2 )/(b*x+a)-1/b^4*(2*a*e-2*b*d)*e*(((-9/8*A*b^2*e+13/8*B*e*b*a-1/2*B*b^2*d)* (e*x+d)^(3/2)+(-7/8*A*a*b*e^2+7/8*A*b^2*d*e+11/8*a^2*B*e^2-15/8*B*a*b*d*e+ 1/2*B*b^2*d^2)*(e*x+d)^(1/2))/(b*(e*x+d)+a*e-b*d)^2+5/8*(3*A*b*e-7*B*a*e+4 *B*b*d)/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)))*( (b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.32 (sec) , antiderivative size = 680, normalized size of antiderivative = 1.99 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\left [-\frac {15 \, {\left (4 \, B a^{2} b d e - {\left (7 \, B a^{3} - 3 \, A a^{2} b\right )} e^{2} + {\left (4 \, B b^{3} d e - {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (4 \, B a b^{2} d e - {\left (7 \, B a^{2} b - 3 \, 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 (8 \, B b^{3} e^{2} x^{3} - 6 \, {\left (B a b^{2} + A b^{3}\right )} d^{2} + 5 \, {\left (19 \, B a^{2} b - 3 \, A a b^{2}\right )} d e - 15 \, {\left (7 \, B a^{3} - 3 \, A a^{2} b\right )} e^{2} + 8 \, {\left (7 \, B b^{3} d e - {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} e^{2}\right )} x^{2} - {\left (12 \, B b^{3} d^{2} - {\left (163 \, B a b^{2} - 27 \, A b^{3}\right )} d e + 25 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, -\frac {15 \, {\left (4 \, B a^{2} b d e - {\left (7 \, B a^{3} - 3 \, A a^{2} b\right )} e^{2} + {\left (4 \, B b^{3} d e - {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (4 \, B a b^{2} d e - {\left (7 \, B a^{2} b - 3 \, 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 (8 \, B b^{3} e^{2} x^{3} - 6 \, {\left (B a b^{2} + A b^{3}\right )} d^{2} + 5 \, {\left (19 \, B a^{2} b - 3 \, A a b^{2}\right )} d e - 15 \, {\left (7 \, B a^{3} - 3 \, A a^{2} b\right )} e^{2} + 8 \, {\left (7 \, B b^{3} d e - {\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} e^{2}\right )} x^{2} - {\left (12 \, B b^{3} d^{2} - {\left (163 \, B a b^{2} - 27 \, A b^{3}\right )} d e + 25 \, {\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{12 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \]
[-1/24*(15*(4*B*a^2*b*d*e - (7*B*a^3 - 3*A*a^2*b)*e^2 + (4*B*b^3*d*e - (7* B*a*b^2 - 3*A*b^3)*e^2)*x^2 + 2*(4*B*a*b^2*d*e - (7*B*a^2*b - 3*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*sq rt((b*d - a*e)/b))/(b*x + a)) - 2*(8*B*b^3*e^2*x^3 - 6*(B*a*b^2 + A*b^3)*d ^2 + 5*(19*B*a^2*b - 3*A*a*b^2)*d*e - 15*(7*B*a^3 - 3*A*a^2*b)*e^2 + 8*(7* B*b^3*d*e - (7*B*a*b^2 - 3*A*b^3)*e^2)*x^2 - (12*B*b^3*d^2 - (163*B*a*b^2 - 27*A*b^3)*d*e + 25*(7*B*a^2*b - 3*A*a*b^2)*e^2)*x)*sqrt(e*x + d))/(b^6*x ^2 + 2*a*b^5*x + a^2*b^4), -1/12*(15*(4*B*a^2*b*d*e - (7*B*a^3 - 3*A*a^2*b )*e^2 + (4*B*b^3*d*e - (7*B*a*b^2 - 3*A*b^3)*e^2)*x^2 + 2*(4*B*a*b^2*d*e - (7*B*a^2*b - 3*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)) - (8*B*b^3*e^2*x^3 - 6*(B*a*b^2 + A* b^3)*d^2 + 5*(19*B*a^2*b - 3*A*a*b^2)*d*e - 15*(7*B*a^3 - 3*A*a^2*b)*e^2 + 8*(7*B*b^3*d*e - (7*B*a*b^2 - 3*A*b^3)*e^2)*x^2 - (12*B*b^3*d^2 - (163*B* a*b^2 - 27*A*b^3)*d*e + 25*(7*B*a^2*b - 3*A*a*b^2)*e^2)*x)*sqrt(e*x + d))/ (b^6*x^2 + 2*a*b^5*x + a^2*b^4)]
Timed out. \[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{\frac {5}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}} \,d x } \]
Time = 0.30 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.21 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {5 \, {\left (4 \, B b^{2} d^{2} e - 11 \, B a b d e^{2} + 3 \, A b^{2} d e^{2} + 7 \, B a^{2} e^{3} - 3 \, A a b e^{3}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, \sqrt {-b^{2} d + a b e} b^{4} \mathrm {sgn}\left (b x + a\right )} - \frac {4 \, {\left (e x + d\right )}^{\frac {3}{2}} B b^{3} d^{2} e - 4 \, \sqrt {e x + d} B b^{3} d^{3} e - 17 \, {\left (e x + d\right )}^{\frac {3}{2}} B a b^{2} d e^{2} + 9 \, {\left (e x + d\right )}^{\frac {3}{2}} A b^{3} d e^{2} + 19 \, \sqrt {e x + d} B a b^{2} d^{2} e^{2} - 7 \, \sqrt {e x + d} A b^{3} d^{2} e^{2} + 13 \, {\left (e x + d\right )}^{\frac {3}{2}} B a^{2} b e^{3} - 9 \, {\left (e x + d\right )}^{\frac {3}{2}} A a b^{2} e^{3} - 26 \, \sqrt {e x + d} B a^{2} b d e^{3} + 14 \, \sqrt {e x + d} A a b^{2} d e^{3} + 11 \, \sqrt {e x + d} B a^{3} e^{4} - 7 \, \sqrt {e x + d} A a^{2} b e^{4}}{4 \, {\left ({\left (e x + d\right )} b - b d + a e\right )}^{2} b^{4} \mathrm {sgn}\left (b x + a\right )} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} B b^{6} e + 6 \, \sqrt {e x + d} B b^{6} d e - 9 \, \sqrt {e x + d} B a b^{5} e^{2} + 3 \, \sqrt {e x + d} A b^{6} e^{2}\right )}}{3 \, b^{9} \mathrm {sgn}\left (b x + a\right )} \]
5/4*(4*B*b^2*d^2*e - 11*B*a*b*d*e^2 + 3*A*b^2*d*e^2 + 7*B*a^2*e^3 - 3*A*a* 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*sgn(b*x + a)) - 1/4*(4*(e*x + d)^(3/2)*B*b^3*d^2*e - 4*sqrt(e*x + d)*B *b^3*d^3*e - 17*(e*x + d)^(3/2)*B*a*b^2*d*e^2 + 9*(e*x + d)^(3/2)*A*b^3*d* e^2 + 19*sqrt(e*x + d)*B*a*b^2*d^2*e^2 - 7*sqrt(e*x + d)*A*b^3*d^2*e^2 + 1 3*(e*x + d)^(3/2)*B*a^2*b*e^3 - 9*(e*x + d)^(3/2)*A*a*b^2*e^3 - 26*sqrt(e* x + d)*B*a^2*b*d*e^3 + 14*sqrt(e*x + d)*A*a*b^2*d*e^3 + 11*sqrt(e*x + d)*B *a^3*e^4 - 7*sqrt(e*x + d)*A*a^2*b*e^4)/(((e*x + d)*b - b*d + a*e)^2*b^4*s gn(b*x + a)) + 2/3*((e*x + d)^(3/2)*B*b^6*e + 6*sqrt(e*x + d)*B*b^6*d*e - 9*sqrt(e*x + d)*B*a*b^5*e^2 + 3*sqrt(e*x + d)*A*b^6*e^2)/(b^9*sgn(b*x + a) )
Timed out. \[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^{5/2}}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]