Integrand size = 35, antiderivative size = 281 \[ \int \frac {A+B x}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {-4 b B d+5 A b e-a B e}{4 b (b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A b-a B}{2 b (b d-a e) (a+b x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 e (4 b B d-5 A b e+a B e) (a+b x)}{4 b (b d-a e)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e (4 b B d-5 A b e+a B e) (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {b} (b d-a e)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
3/4*e*(-5*A*b*e+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)^(7/2)/b^(1/2)/((b*x+a)^2)^(1/2)+1/4*(5*A*b*e-B*a*e -4*B*b*d)/b/(-a*e+b*d)^2/(e*x+d)^(1/2)/((b*x+a)^2)^(1/2)+1/2*(-A*b+B*a)/b/ (-a*e+b*d)/(b*x+a)/(e*x+d)^(1/2)/((b*x+a)^2)^(1/2)-3/4*e*(-5*A*b*e+B*a*e+4 *B*b*d)*(b*x+a)/b/(-a*e+b*d)^3/(e*x+d)^(1/2)/((b*x+a)^2)^(1/2)
Time = 0.29 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.77 \[ \int \frac {A+B x}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {e (a+b x)^3 \left (\frac {A \left (-8 a^2 e^2-a b e (9 d+25 e x)+b^2 \left (2 d^2-5 d e x-15 e^2 x^2\right )\right )+B \left (4 b^2 d x (d+3 e x)+a^2 e (13 d+5 e x)+a b \left (2 d^2+21 d e x+3 e^2 x^2\right )\right )}{e (-b d+a e)^3 (a+b x)^2 \sqrt {d+e x}}+\frac {3 (4 b B d-5 A b e+a B e) \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{\sqrt {b} (-b d+a e)^{7/2}}\right )}{4 \left ((a+b x)^2\right )^{3/2}} \]
(e*(a + b*x)^3*((A*(-8*a^2*e^2 - a*b*e*(9*d + 25*e*x) + b^2*(2*d^2 - 5*d*e *x - 15*e^2*x^2)) + B*(4*b^2*d*x*(d + 3*e*x) + a^2*e*(13*d + 5*e*x) + a*b* (2*d^2 + 21*d*e*x + 3*e^2*x^2)))/(e*(-(b*d) + a*e)^3*(a + b*x)^2*Sqrt[d + e*x]) + (3*(4*b*B*d - 5*A*b*e + a*B*e)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt [-(b*d) + a*e]])/(Sqrt[b]*(-(b*d) + a*e)^(7/2))))/(4*((a + b*x)^2)^(3/2))
Time = 0.36 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.76, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1187, 27, 87, 52, 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.
| \(\displaystyle \int \frac {A+B x}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (d+e x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b^3 (a+b x) \int \frac {A+B x}{b^3 (a+b x)^3 (d+e x)^{3/2}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {A+B x}{(a+b x)^3 (d+e x)^{3/2}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(a B e-5 A b e+4 b B d) \int \frac {1}{(a+b x)^2 (d+e x)^{3/2}}dx}{4 b (b d-a e)}-\frac {A b-a B}{2 b (a+b x)^2 \sqrt {d+e x} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(a B e-5 A b e+4 b B d) \left (-\frac {3 e \int \frac {1}{(a+b x) (d+e x)^{3/2}}dx}{2 (b d-a e)}-\frac {1}{(a+b x) \sqrt {d+e x} (b d-a e)}\right )}{4 b (b d-a e)}-\frac {A b-a B}{2 b (a+b x)^2 \sqrt {d+e x} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(a B e-5 A b e+4 b B d) \left (-\frac {3 e \left (\frac {b \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b d-a e}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) \sqrt {d+e x} (b d-a e)}\right )}{4 b (b d-a e)}-\frac {A b-a B}{2 b (a+b x)^2 \sqrt {d+e x} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(a B e-5 A b e+4 b B d) \left (-\frac {3 e \left (\frac {2 b \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{e (b d-a e)}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) \sqrt {d+e x} (b d-a e)}\right )}{4 b (b d-a e)}-\frac {A b-a B}{2 b (a+b x)^2 \sqrt {d+e x} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(a B e-5 A b e+4 b B d) \left (-\frac {3 e \left (\frac {2}{\sqrt {d+e x} (b d-a e)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) \sqrt {d+e x} (b d-a e)}\right )}{4 b (b d-a e)}-\frac {A b-a B}{2 b (a+b x)^2 \sqrt {d+e x} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(-1/2*(A*b - a*B)/(b*(b*d - a*e)*(a + b*x)^2*Sqrt[d + e*x]) + ( (4*b*B*d - 5*A*b*e + a*B*e)*(-(1/((b*d - a*e)*(a + b*x)*Sqrt[d + e*x])) - (3*e*(2/((b*d - a*e)*Sqrt[d + e*x]) - (2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(b*d - a*e)^(3/2)))/(2*(b*d - a*e))))/(4*b*(b*d - a*e))))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.19.71.3.1 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[(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]
Leaf count of result is larger than twice the leaf count of optimal. \(680\) vs. \(2(213)=426\).
Time = 0.29 (sec) , antiderivative size = 681, normalized size of antiderivative = 2.42
| method | result | size |
| default | \(-\frac {\left (15 A \sqrt {e x +d}\, \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) b^{3} e^{2} x^{2}-3 B \sqrt {e x +d}\, \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{2} e^{2} x^{2}-12 B \sqrt {e x +d}\, \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) b^{3} d e \,x^{2}+30 A \sqrt {e x +d}\, \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{2} e^{2} x -6 B \sqrt {e x +d}\, \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} b \,e^{2} x -24 B \sqrt {e x +d}\, \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{2} d e x +15 A \sqrt {e x +d}\, \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} b \,e^{2}+15 A \sqrt {\left (a e -b d \right ) b}\, b^{2} e^{2} x^{2}-3 B \sqrt {e x +d}\, \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{3} e^{2}-12 B \sqrt {e x +d}\, \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} b d e -3 B \sqrt {\left (a e -b d \right ) b}\, a b \,e^{2} x^{2}-12 B \sqrt {\left (a e -b d \right ) b}\, b^{2} d e \,x^{2}+25 A \sqrt {\left (a e -b d \right ) b}\, a b \,e^{2} x +5 A \sqrt {\left (a e -b d \right ) b}\, b^{2} d e x -5 B \sqrt {\left (a e -b d \right ) b}\, a^{2} e^{2} x -21 B \sqrt {\left (a e -b d \right ) b}\, a b d e x -4 B \sqrt {\left (a e -b d \right ) b}\, b^{2} d^{2} x +8 A \sqrt {\left (a e -b d \right ) b}\, a^{2} e^{2}+9 A \sqrt {\left (a e -b d \right ) b}\, a b d e -2 A \sqrt {\left (a e -b d \right ) b}\, b^{2} d^{2}-13 B \sqrt {\left (a e -b d \right ) b}\, a^{2} d e -2 B \sqrt {\left (a e -b d \right ) b}\, a b \,d^{2}\right ) \left (b x +a \right )}{4 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, \left (a e -b d \right )^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\) | \(681\) |
-1/4*(15*A*(e*x+d)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*b^3*e ^2*x^2-3*B*(e*x+d)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a*b^2 *e^2*x^2-12*B*(e*x+d)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*b^ 3*d*e*x^2+30*A*(e*x+d)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a *b^2*e^2*x-6*B*(e*x+d)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a ^2*b*e^2*x-24*B*(e*x+d)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))* a*b^2*d*e*x+15*A*(e*x+d)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)) *a^2*b*e^2+15*A*((a*e-b*d)*b)^(1/2)*b^2*e^2*x^2-3*B*(e*x+d)^(1/2)*arctan(b *(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^3*e^2-12*B*(e*x+d)^(1/2)*arctan(b*(e *x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^2*b*d*e-3*B*((a*e-b*d)*b)^(1/2)*a*b*e^2 *x^2-12*B*((a*e-b*d)*b)^(1/2)*b^2*d*e*x^2+25*A*((a*e-b*d)*b)^(1/2)*a*b*e^2 *x+5*A*((a*e-b*d)*b)^(1/2)*b^2*d*e*x-5*B*((a*e-b*d)*b)^(1/2)*a^2*e^2*x-21* B*((a*e-b*d)*b)^(1/2)*a*b*d*e*x-4*B*((a*e-b*d)*b)^(1/2)*b^2*d^2*x+8*A*((a* e-b*d)*b)^(1/2)*a^2*e^2+9*A*((a*e-b*d)*b)^(1/2)*a*b*d*e-2*A*((a*e-b*d)*b)^ (1/2)*b^2*d^2-13*B*((a*e-b*d)*b)^(1/2)*a^2*d*e-2*B*((a*e-b*d)*b)^(1/2)*a*b *d^2)*(b*x+a)/((a*e-b*d)*b)^(1/2)/(e*x+d)^(1/2)/(a*e-b*d)^3/((b*x+a)^2)^(3 /2)
Leaf count of result is larger than twice the leaf count of optimal. 698 vs. \(2 (212) = 424\).
Time = 0.43 (sec) , antiderivative size = 1410, normalized size of antiderivative = 5.02 \[ \int \frac {A+B x}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\text {Too large to display} \]
[1/8*(3*(4*B*a^2*b*d^2*e + (B*a^3 - 5*A*a^2*b)*d*e^2 + (4*B*b^3*d*e^2 + (B *a*b^2 - 5*A*b^3)*e^3)*x^3 + (4*B*b^3*d^2*e + (9*B*a*b^2 - 5*A*b^3)*d*e^2 + 2*(B*a^2*b - 5*A*a*b^2)*e^3)*x^2 + (8*B*a*b^2*d^2*e + 2*(3*B*a^2*b - 5*A *a*b^2)*d*e^2 + (B*a^3 - 5*A*a^2*b)*e^3)*x)*sqrt(b^2*d - a*b*e)*log((b*e*x + 2*b*d - a*e + 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(8*A* a^3*b*e^3 + 2*(B*a*b^3 + A*b^4)*d^3 + 11*(B*a^2*b^2 - A*a*b^3)*d^2*e - (13 *B*a^3*b - A*a^2*b^2)*d*e^2 + 3*(4*B*b^4*d^2*e - (3*B*a*b^3 + 5*A*b^4)*d*e ^2 - (B*a^2*b^2 - 5*A*a*b^3)*e^3)*x^2 + (4*B*b^4*d^3 + (17*B*a*b^3 - 5*A*b ^4)*d^2*e - 4*(4*B*a^2*b^2 + 5*A*a*b^3)*d*e^2 - 5*(B*a^3*b - 5*A*a^2*b^2)* e^3)*x)*sqrt(e*x + d))/(a^2*b^5*d^5 - 4*a^3*b^4*d^4*e + 6*a^4*b^3*d^3*e^2 - 4*a^5*b^2*d^2*e^3 + a^6*b*d*e^4 + (b^7*d^4*e - 4*a*b^6*d^3*e^2 + 6*a^2*b ^5*d^2*e^3 - 4*a^3*b^4*d*e^4 + a^4*b^3*e^5)*x^3 + (b^7*d^5 - 2*a*b^6*d^4*e - 2*a^2*b^5*d^3*e^2 + 8*a^3*b^4*d^2*e^3 - 7*a^4*b^3*d*e^4 + 2*a^5*b^2*e^5 )*x^2 + (2*a*b^6*d^5 - 7*a^2*b^5*d^4*e + 8*a^3*b^4*d^3*e^2 - 2*a^4*b^3*d^2 *e^3 - 2*a^5*b^2*d*e^4 + a^6*b*e^5)*x), -1/4*(3*(4*B*a^2*b*d^2*e + (B*a^3 - 5*A*a^2*b)*d*e^2 + (4*B*b^3*d*e^2 + (B*a*b^2 - 5*A*b^3)*e^3)*x^3 + (4*B* b^3*d^2*e + (9*B*a*b^2 - 5*A*b^3)*d*e^2 + 2*(B*a^2*b - 5*A*a*b^2)*e^3)*x^2 + (8*B*a*b^2*d^2*e + 2*(3*B*a^2*b - 5*A*a*b^2)*d*e^2 + (B*a^3 - 5*A*a^2*b )*e^3)*x)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/( b*e*x + b*d)) + (8*A*a^3*b*e^3 + 2*(B*a*b^3 + A*b^4)*d^3 + 11*(B*a^2*b^...
\[ \int \frac {A+B x}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {A + B x}{\left (d + e x\right )^{\frac {3}{2}} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {A+B x}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {B x + A}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (e x + d\right )}^{\frac {3}{2}}} \,d x } \]
Time = 0.31 (sec) , antiderivative size = 412, normalized size of antiderivative = 1.47 \[ \int \frac {A+B x}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {3 \, {\left (4 \, B b d e + B a e^{2} - 5 \, A b e^{2}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, {\left (b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} \sqrt {-b^{2} d + a b e}} - \frac {2 \, {\left (B d e - A e^{2}\right )}}{{\left (b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} \sqrt {e x + d}} - \frac {4 \, {\left (e x + d\right )}^{\frac {3}{2}} B b^{2} d e - 4 \, \sqrt {e x + d} B b^{2} d^{2} e + 3 \, {\left (e x + d\right )}^{\frac {3}{2}} B a b e^{2} - 7 \, {\left (e x + d\right )}^{\frac {3}{2}} A b^{2} e^{2} - \sqrt {e x + d} B a b d e^{2} + 9 \, \sqrt {e x + d} A b^{2} d e^{2} + 5 \, \sqrt {e x + d} B a^{2} e^{3} - 9 \, \sqrt {e x + d} A a b e^{3}}{4 \, {\left (b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{2}} \]
-3/4*(4*B*b*d*e + B*a*e^2 - 5*A*b*e^2)*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((b^3*d^3*sgn(b*x + a) - 3*a*b^2*d^2*e*sgn(b*x + a) + 3*a^2*b*d* e^2*sgn(b*x + a) - a^3*e^3*sgn(b*x + a))*sqrt(-b^2*d + a*b*e)) - 2*(B*d*e - A*e^2)/((b^3*d^3*sgn(b*x + a) - 3*a*b^2*d^2*e*sgn(b*x + a) + 3*a^2*b*d*e ^2*sgn(b*x + a) - a^3*e^3*sgn(b*x + a))*sqrt(e*x + d)) - 1/4*(4*(e*x + d)^ (3/2)*B*b^2*d*e - 4*sqrt(e*x + d)*B*b^2*d^2*e + 3*(e*x + d)^(3/2)*B*a*b*e^ 2 - 7*(e*x + d)^(3/2)*A*b^2*e^2 - sqrt(e*x + d)*B*a*b*d*e^2 + 9*sqrt(e*x + d)*A*b^2*d*e^2 + 5*sqrt(e*x + d)*B*a^2*e^3 - 9*sqrt(e*x + d)*A*a*b*e^3)/( (b^3*d^3*sgn(b*x + a) - 3*a*b^2*d^2*e*sgn(b*x + a) + 3*a^2*b*d*e^2*sgn(b*x + a) - a^3*e^3*sgn(b*x + a))*((e*x + d)*b - b*d + a*e)^2)
Timed out. \[ \int \frac {A+B x}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {A+B\,x}{{\left (d+e\,x\right )}^{3/2}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]