Integrand size = 35, antiderivative size = 251 \[ \int \frac {A+B x}{(d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 (B d-A e) (a+b x)}{5 e (b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) (a+b x)}{3 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b (A b-a B) (a+b x)}{(b d-a e)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b^{3/2} (A b-a B) (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
-2/5*(-A*e+B*d)*(b*x+a)/e/(-a*e+b*d)/(e*x+d)^(5/2)/((b*x+a)^2)^(1/2)+2/3*( A*b-B*a)*(b*x+a)/(-a*e+b*d)^2/(e*x+d)^(3/2)/((b*x+a)^2)^(1/2)-2*b^(3/2)*(A *b-B*a)*(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*x+a)^2)^(1/2)+2*b*(A*b-B*a)*(b*x+a)/(-a*e+b*d)^3/(e*x+d)^(1/2)/ ((b*x+a)^2)^(1/2)
Time = 0.20 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.77 \[ \int \frac {A+B x}{(d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 (a+b x) \left (\frac {-a^2 e^2 (2 B d+3 A e+5 B e x)+a b e \left (A e (11 d+5 e x)+B \left (14 d^2+35 d e x+15 e^2 x^2\right )\right )+b^2 \left (3 B d^3-A e \left (23 d^2+35 d e x+15 e^2 x^2\right )\right )}{e (-b d+a e)^3 (d+e x)^{5/2}}-\frac {15 b^{3/2} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{7/2}}\right )}{15 \sqrt {(a+b x)^2}} \]
(2*(a + b*x)*((-(a^2*e^2*(2*B*d + 3*A*e + 5*B*e*x)) + a*b*e*(A*e*(11*d + 5 *e*x) + B*(14*d^2 + 35*d*e*x + 15*e^2*x^2)) + b^2*(3*B*d^3 - A*e*(23*d^2 + 35*d*e*x + 15*e^2*x^2)))/(e*(-(b*d) + a*e)^3*(d + e*x)^(5/2)) - (15*b^(3/ 2)*(A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(-(b*d) + a*e)^(7/2)))/(15*Sqrt[(a + b*x)^2])
Time = 0.31 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.74, 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, 61, 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.
| \(\displaystyle \int \frac {A+B x}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{7/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b (a+b x) \int \frac {A+B x}{b (a+b x) (d+e x)^{7/2}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {A+B x}{(a+b x) (d+e x)^{7/2}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \int \frac {1}{(a+b x) (d+e x)^{5/2}}dx}{b d-a e}-\frac {2 (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {b \int \frac {1}{(a+b x) (d+e x)^{3/2}}dx}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}-\frac {2 (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {b \left (\frac {b \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b d-a e}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}-\frac {2 (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {b \left (\frac {2 b \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{e (b d-a e)}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}-\frac {2 (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {b \left (\frac {2}{\sqrt {d+e x} (b d-a e)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}-\frac {2 (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*((-2*(B*d - A*e))/(5*e*(b*d - a*e)*(d + e*x)^(5/2)) + ((A*b - a *B)*(2/(3*(b*d - a*e)*(d + e*x)^(3/2)) + (b*(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)))/(b*d - a*e)))/(b*d - a*e)))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.19.65.3.1 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] && 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. \(385\) vs. \(2(187)=374\).
Time = 0.25 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.54
| method | result | size |
| default | \(-\frac {2 \left (b x +a \right ) \left (15 A \left (e x +d \right )^{\frac {5}{2}} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) b^{3} e -15 B \left (e x +d \right )^{\frac {5}{2}} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{2} e +15 A \sqrt {\left (a e -b d \right ) b}\, b^{2} e^{3} x^{2}-15 B \sqrt {\left (a e -b d \right ) b}\, a b \,e^{3} x^{2}-5 A \sqrt {\left (a e -b d \right ) b}\, a b \,e^{3} x +35 A \sqrt {\left (a e -b d \right ) b}\, b^{2} d \,e^{2} x +5 B \sqrt {\left (a e -b d \right ) b}\, a^{2} e^{3} x -35 B \sqrt {\left (a e -b d \right ) b}\, a b d \,e^{2} x +3 A \sqrt {\left (a e -b d \right ) b}\, a^{2} e^{3}-11 A \sqrt {\left (a e -b d \right ) b}\, a b d \,e^{2}+23 A \sqrt {\left (a e -b d \right ) b}\, b^{2} d^{2} e +2 B \sqrt {\left (a e -b d \right ) b}\, a^{2} d \,e^{2}-14 B \sqrt {\left (a e -b d \right ) b}\, a b \,d^{2} e -3 B \sqrt {\left (a e -b d \right ) b}\, b^{2} d^{3}\right )}{15 \sqrt {\left (b x +a \right )^{2}}\, e \left (a e -b d \right )^{3} \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}}\) | \(386\) |
-2/15*(b*x+a)*(15*A*(e*x+d)^(5/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/ 2))*b^3*e-15*B*(e*x+d)^(5/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a *b^2*e+15*A*((a*e-b*d)*b)^(1/2)*b^2*e^3*x^2-15*B*((a*e-b*d)*b)^(1/2)*a*b*e ^3*x^2-5*A*((a*e-b*d)*b)^(1/2)*a*b*e^3*x+35*A*((a*e-b*d)*b)^(1/2)*b^2*d*e^ 2*x+5*B*((a*e-b*d)*b)^(1/2)*a^2*e^3*x-35*B*((a*e-b*d)*b)^(1/2)*a*b*d*e^2*x +3*A*((a*e-b*d)*b)^(1/2)*a^2*e^3-11*A*((a*e-b*d)*b)^(1/2)*a*b*d*e^2+23*A*( (a*e-b*d)*b)^(1/2)*b^2*d^2*e+2*B*((a*e-b*d)*b)^(1/2)*a^2*d*e^2-14*B*((a*e- b*d)*b)^(1/2)*a*b*d^2*e-3*B*((a*e-b*d)*b)^(1/2)*b^2*d^3)/((b*x+a)^2)^(1/2) /e/(a*e-b*d)^3/(e*x+d)^(5/2)/((a*e-b*d)*b)^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 446 vs. \(2 (187) = 374\).
Time = 0.68 (sec) , antiderivative size = 902, normalized size of antiderivative = 3.59 \[ \int \frac {A+B x}{(d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\left [\frac {15 \, {\left ({\left (B a b - A b^{2}\right )} e^{4} x^{3} + 3 \, {\left (B a b - A b^{2}\right )} d e^{3} x^{2} + 3 \, {\left (B a b - A b^{2}\right )} d^{2} e^{2} x + {\left (B a b - A b^{2}\right )} d^{3} e\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, {\left (b d - a e\right )} \sqrt {e x + d} \sqrt {\frac {b}{b d - a e}}}{b x + a}\right ) - 2 \, {\left (3 \, B b^{2} d^{3} - 3 \, A a^{2} e^{3} + 15 \, {\left (B a b - A b^{2}\right )} e^{3} x^{2} + {\left (14 \, B a b - 23 \, A b^{2}\right )} d^{2} e - {\left (2 \, B a^{2} - 11 \, A a b\right )} d e^{2} + 5 \, {\left (7 \, {\left (B a b - A b^{2}\right )} d e^{2} - {\left (B a^{2} - A a b\right )} e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (b^{3} d^{6} e - 3 \, a b^{2} d^{5} e^{2} + 3 \, a^{2} b d^{4} e^{3} - a^{3} d^{3} e^{4} + {\left (b^{3} d^{3} e^{4} - 3 \, a b^{2} d^{2} e^{5} + 3 \, a^{2} b d e^{6} - a^{3} e^{7}\right )} x^{3} + 3 \, {\left (b^{3} d^{4} e^{3} - 3 \, a b^{2} d^{3} e^{4} + 3 \, a^{2} b d^{2} e^{5} - a^{3} d e^{6}\right )} x^{2} + 3 \, {\left (b^{3} d^{5} e^{2} - 3 \, a b^{2} d^{4} e^{3} + 3 \, a^{2} b d^{3} e^{4} - a^{3} d^{2} e^{5}\right )} x\right )}}, \frac {2 \, {\left (15 \, {\left ({\left (B a b - A b^{2}\right )} e^{4} x^{3} + 3 \, {\left (B a b - A b^{2}\right )} d e^{3} x^{2} + 3 \, {\left (B a b - A b^{2}\right )} d^{2} e^{2} x + {\left (B a b - A b^{2}\right )} d^{3} e\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {e x + d} \sqrt {-\frac {b}{b d - a e}}}{b e x + b d}\right ) - {\left (3 \, B b^{2} d^{3} - 3 \, A a^{2} e^{3} + 15 \, {\left (B a b - A b^{2}\right )} e^{3} x^{2} + {\left (14 \, B a b - 23 \, A b^{2}\right )} d^{2} e - {\left (2 \, B a^{2} - 11 \, A a b\right )} d e^{2} + 5 \, {\left (7 \, {\left (B a b - A b^{2}\right )} d e^{2} - {\left (B a^{2} - A a b\right )} e^{3}\right )} x\right )} \sqrt {e x + d}\right )}}{15 \, {\left (b^{3} d^{6} e - 3 \, a b^{2} d^{5} e^{2} + 3 \, a^{2} b d^{4} e^{3} - a^{3} d^{3} e^{4} + {\left (b^{3} d^{3} e^{4} - 3 \, a b^{2} d^{2} e^{5} + 3 \, a^{2} b d e^{6} - a^{3} e^{7}\right )} x^{3} + 3 \, {\left (b^{3} d^{4} e^{3} - 3 \, a b^{2} d^{3} e^{4} + 3 \, a^{2} b d^{2} e^{5} - a^{3} d e^{6}\right )} x^{2} + 3 \, {\left (b^{3} d^{5} e^{2} - 3 \, a b^{2} d^{4} e^{3} + 3 \, a^{2} b d^{3} e^{4} - a^{3} d^{2} e^{5}\right )} x\right )}}\right ] \]
[1/15*(15*((B*a*b - A*b^2)*e^4*x^3 + 3*(B*a*b - A*b^2)*d*e^3*x^2 + 3*(B*a* b - A*b^2)*d^2*e^2*x + (B*a*b - A*b^2)*d^3*e)*sqrt(b/(b*d - a*e))*log((b*e *x + 2*b*d - a*e + 2*(b*d - a*e)*sqrt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) - 2*(3*B*b^2*d^3 - 3*A*a^2*e^3 + 15*(B*a*b - A*b^2)*e^3*x^2 + (14*B*a *b - 23*A*b^2)*d^2*e - (2*B*a^2 - 11*A*a*b)*d*e^2 + 5*(7*(B*a*b - A*b^2)*d *e^2 - (B*a^2 - A*a*b)*e^3)*x)*sqrt(e*x + d))/(b^3*d^6*e - 3*a*b^2*d^5*e^2 + 3*a^2*b*d^4*e^3 - a^3*d^3*e^4 + (b^3*d^3*e^4 - 3*a*b^2*d^2*e^5 + 3*a^2* b*d*e^6 - a^3*e^7)*x^3 + 3*(b^3*d^4*e^3 - 3*a*b^2*d^3*e^4 + 3*a^2*b*d^2*e^ 5 - a^3*d*e^6)*x^2 + 3*(b^3*d^5*e^2 - 3*a*b^2*d^4*e^3 + 3*a^2*b*d^3*e^4 - a^3*d^2*e^5)*x), 2/15*(15*((B*a*b - A*b^2)*e^4*x^3 + 3*(B*a*b - A*b^2)*d*e ^3*x^2 + 3*(B*a*b - A*b^2)*d^2*e^2*x + (B*a*b - A*b^2)*d^3*e)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a*e)*sqrt(e*x + d)*sqrt(-b/(b*d - a*e))/(b*e*x + b *d)) - (3*B*b^2*d^3 - 3*A*a^2*e^3 + 15*(B*a*b - A*b^2)*e^3*x^2 + (14*B*a*b - 23*A*b^2)*d^2*e - (2*B*a^2 - 11*A*a*b)*d*e^2 + 5*(7*(B*a*b - A*b^2)*d*e ^2 - (B*a^2 - A*a*b)*e^3)*x)*sqrt(e*x + d))/(b^3*d^6*e - 3*a*b^2*d^5*e^2 + 3*a^2*b*d^4*e^3 - a^3*d^3*e^4 + (b^3*d^3*e^4 - 3*a*b^2*d^2*e^5 + 3*a^2*b* d*e^6 - a^3*e^7)*x^3 + 3*(b^3*d^4*e^3 - 3*a*b^2*d^3*e^4 + 3*a^2*b*d^2*e^5 - a^3*d*e^6)*x^2 + 3*(b^3*d^5*e^2 - 3*a*b^2*d^4*e^3 + 3*a^2*b*d^3*e^4 - a^ 3*d^2*e^5)*x)]
Timed out. \[ \int \frac {A+B x}{(d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\text {Timed out} \]
\[ \int \frac {A+B x}{(d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {B x + A}{\sqrt {{\left (b x + a\right )}^{2}} {\left (e x + d\right )}^{\frac {7}{2}}} \,d x } \]
Time = 0.30 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.43 \[ \int \frac {A+B x}{(d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 \, {\left (B a b^{2} \mathrm {sgn}\left (b x + a\right ) - A b^{3} \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt {-b^{2} d + a b e}} - \frac {2 \, {\left (3 \, B b^{2} d^{3} \mathrm {sgn}\left (b x + a\right ) + 15 \, {\left (e x + d\right )}^{2} B a b e \mathrm {sgn}\left (b x + a\right ) - 15 \, {\left (e x + d\right )}^{2} A b^{2} e \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (e x + d\right )} B a b d e \mathrm {sgn}\left (b x + a\right ) - 5 \, {\left (e x + d\right )} A b^{2} d e \mathrm {sgn}\left (b x + a\right ) - 6 \, B a b d^{2} e \mathrm {sgn}\left (b x + a\right ) - 3 \, A b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 5 \, {\left (e x + d\right )} B a^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (e x + d\right )} A a b e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, B a^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, A a b d e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, A a^{2} e^{3} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, {\left (b^{3} d^{3} e - 3 \, a b^{2} d^{2} e^{2} + 3 \, a^{2} b d e^{3} - a^{3} e^{4}\right )} {\left (e x + d\right )}^{\frac {5}{2}}} \]
-2*(B*a*b^2*sgn(b*x + a) - A*b^3*sgn(b*x + a))*arctan(sqrt(e*x + d)*b/sqrt (-b^2*d + a*b*e))/((b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*sqr t(-b^2*d + a*b*e)) - 2/15*(3*B*b^2*d^3*sgn(b*x + a) + 15*(e*x + d)^2*B*a*b *e*sgn(b*x + a) - 15*(e*x + d)^2*A*b^2*e*sgn(b*x + a) + 5*(e*x + d)*B*a*b* d*e*sgn(b*x + a) - 5*(e*x + d)*A*b^2*d*e*sgn(b*x + a) - 6*B*a*b*d^2*e*sgn( b*x + a) - 3*A*b^2*d^2*e*sgn(b*x + a) - 5*(e*x + d)*B*a^2*e^2*sgn(b*x + a) + 5*(e*x + d)*A*a*b*e^2*sgn(b*x + a) + 3*B*a^2*d*e^2*sgn(b*x + a) + 6*A*a *b*d*e^2*sgn(b*x + a) - 3*A*a^2*e^3*sgn(b*x + a))/((b^3*d^3*e - 3*a*b^2*d^ 2*e^2 + 3*a^2*b*d*e^3 - a^3*e^4)*(e*x + d)^(5/2))
Timed out. \[ \int \frac {A+B x}{(d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {A+B\,x}{\sqrt {{\left (a+b\,x\right )}^2}\,{\left (d+e\,x\right )}^{7/2}} \,d x \]