Integrand size = 35, antiderivative size = 359 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {5 e^2 (8 b B d-A b e-7 a B e) \sqrt {d+e x}}{64 b^4 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (8 b B d-A b e-7 a B e) (d+e x)^{3/2}}{96 b^3 (b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(8 b B d-A b e-7 a B e) (d+e x)^{5/2}}{24 b^2 (b d-a e) (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (d+e x)^{7/2}}{4 b (b d-a e) (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e^3 (8 b B d-A b e-7 a B e) (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 b^{9/2} (b d-a e)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
-5/96*e*(-A*b*e-7*B*a*e+8*B*b*d)*(e*x+d)^(3/2)/b^3/(-a*e+b*d)/(b*x+a)/((b* x+a)^2)^(1/2)-1/24*(-A*b*e-7*B*a*e+8*B*b*d)*(e*x+d)^(5/2)/b^2/(-a*e+b*d)/( b*x+a)^2/((b*x+a)^2)^(1/2)-1/4*(A*b-B*a)*(e*x+d)^(7/2)/b/(-a*e+b*d)/(b*x+a )^3/((b*x+a)^2)^(1/2)-5/64*e^3*(-A*b*e-7*B*a*e+8*B*b*d)*(b*x+a)*arctanh(b^ (1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(9/2)/(-a*e+b*d)^(3/2)/((b*x+a)^2) ^(1/2)-5/64*e^2*(-A*b*e-7*B*a*e+8*B*b*d)*(e*x+d)^(1/2)/b^4/(-a*e+b*d)/((b* x+a)^2)^(1/2)
Time = 2.04 (sec) , antiderivative size = 328, normalized size of antiderivative = 0.91 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {e^3 (a+b x) \left (\frac {\sqrt {b} \sqrt {d+e x} \left (B \left (-105 a^4 e^3+5 a^3 b e^2 (10 d-77 e x)+a^2 b^2 e \left (24 d^2+188 d e x-511 e^2 x^2\right )+8 b^4 d x \left (8 d^2+26 d e x+33 e^2 x^2\right )+a b^3 \left (16 d^3+88 d^2 e x+258 d e^2 x^2-279 e^3 x^3\right )\right )+A b \left (-15 a^3 e^3-5 a^2 b e^2 (2 d+11 e x)-a b^2 e \left (8 d^2+36 d e x+73 e^2 x^2\right )+b^3 \left (48 d^3+136 d^2 e x+118 d e^2 x^2+15 e^3 x^3\right )\right )\right )}{e^3 (-b d+a e) (a+b x)^4}-\frac {15 (8 b B d-A b e-7 a B e) \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{3/2}}\right )}{192 b^{9/2} \sqrt {(a+b x)^2}} \]
(e^3*(a + b*x)*((Sqrt[b]*Sqrt[d + e*x]*(B*(-105*a^4*e^3 + 5*a^3*b*e^2*(10* d - 77*e*x) + a^2*b^2*e*(24*d^2 + 188*d*e*x - 511*e^2*x^2) + 8*b^4*d*x*(8* d^2 + 26*d*e*x + 33*e^2*x^2) + a*b^3*(16*d^3 + 88*d^2*e*x + 258*d*e^2*x^2 - 279*e^3*x^3)) + A*b*(-15*a^3*e^3 - 5*a^2*b*e^2*(2*d + 11*e*x) - a*b^2*e* (8*d^2 + 36*d*e*x + 73*e^2*x^2) + b^3*(48*d^3 + 136*d^2*e*x + 118*d*e^2*x^ 2 + 15*e^3*x^3))))/(e^3*(-(b*d) + a*e)*(a + b*x)^4) - (15*(8*b*B*d - A*b*e - 7*a*B*e)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(-(b*d) + a*e)^(3/2)))/(192*b^(9/2)*Sqrt[(a + b*x)^2])
Time = 0.35 (sec) , antiderivative size = 235, 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, 51, 51, 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.
| \(\displaystyle \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b^5 (a+b x) \int \frac {(A+B x) (d+e x)^{5/2}}{b^5 (a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {(A+B x) (d+e x)^{5/2}}{(a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(-7 a B e-A b e+8 b B d) \int \frac {(d+e x)^{5/2}}{(a+b x)^4}dx}{8 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{4 b (a+b x)^4 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(-7 a B e-A b e+8 b B d) \left (\frac {5 e \int \frac {(d+e x)^{3/2}}{(a+b x)^3}dx}{6 b}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}\right )}{8 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{4 b (a+b x)^4 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(-7 a B e-A b e+8 b B d) \left (\frac {5 e \left (\frac {3 e \int \frac {\sqrt {d+e x}}{(a+b x)^2}dx}{4 b}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}\right )}{8 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{4 b (a+b x)^4 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(-7 a B e-A b e+8 b B d) \left (\frac {5 e \left (\frac {3 e \left (\frac {e \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{2 b}-\frac {\sqrt {d+e x}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}\right )}{8 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{4 b (a+b x)^4 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(-7 a B e-A b e+8 b B d) \left (\frac {5 e \left (\frac {3 e \left (\frac {\int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b}-\frac {\sqrt {d+e x}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}\right )}{8 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{4 b (a+b x)^4 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(-7 a B e-A b e+8 b B d) \left (\frac {5 e \left (\frac {3 e \left (-\frac {e \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2} \sqrt {b d-a e}}-\frac {\sqrt {d+e x}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}\right )}{8 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{4 b (a+b x)^4 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(-1/4*((A*b - a*B)*(d + e*x)^(7/2))/(b*(b*d - a*e)*(a + b*x)^4) + ((8*b*B*d - A*b*e - 7*a*B*e)*(-1/3*(d + e*x)^(5/2)/(b*(a + b*x)^3) + (5 *e*(-1/2*(d + e*x)^(3/2)/(b*(a + b*x)^2) + (3*e*(-(Sqrt[d + e*x]/(b*(a + b *x))) - (e*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(b^(3/2)*Sqrt [b*d - a*e])))/(4*b)))/(6*b)))/(8*b*(b*d - a*e))))/Sqrt[a^2 + 2*a*b*x + b^ 2*x^2]
3.19.77.3.1 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] :> 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. \(1272\) vs. \(2(276)=552\).
Time = 0.30 (sec) , antiderivative size = 1273, normalized size of antiderivative = 3.55
1/192*(90*A*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^2*b^3*e^5*x^2+10 5*B*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^5*e^5+73*A*(e*x+d)^(5/2) *((a*e-b*d)*b)^(1/2)*b^4*d*e+15*A*(e*x+d)^(7/2)*((a*e-b*d)*b)^(1/2)*b^4*e+ 264*B*(e*x+d)^(7/2)*((a*e-b*d)*b)^(1/2)*b^4*d-584*B*(e*x+d)^(5/2)*((a*e-b* d)*b)^(1/2)*b^4*d^2+630*B*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^3* b^2*e^5*x^2+60*A*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^3*b^2*e^5*x -511*B*(e*x+d)^(5/2)*((a*e-b*d)*b)^(1/2)*a^2*b^2*e^2+420*B*arctan(b*(e*x+d )^(1/2)/((a*e-b*d)*b)^(1/2))*a^4*b*e^5*x-55*A*(e*x+d)^(3/2)*((a*e-b*d)*b)^ (1/2)*a^2*b^2*e^3-55*A*(e*x+d)^(3/2)*((a*e-b*d)*b)^(1/2)*b^4*d^2*e-720*B*a rctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^2*b^3*d*e^4*x^2+1095*B*(e*x+d )^(5/2)*((a*e-b*d)*b)^(1/2)*a*b^3*d*e+435*B*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1 /2)*a^3*b*d*e^3-675*B*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*a^2*b^2*d^2*e^2+46 5*B*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2)*a*b^3*d^3*e-480*B*arctan(b*(e*x+d)^( 1/2)/((a*e-b*d)*b)^(1/2))*a*b^4*d*e^4*x^3+105*B*arctan(b*(e*x+d)^(1/2)/((a *e-b*d)*b)^(1/2))*a*b^4*e^5*x^4-120*B*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b) ^(1/2))*b^5*d*e^4*x^4+60*A*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a*b ^4*e^5*x^3+420*B*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^2*b^3*e^5*x ^3-279*B*(e*x+d)^(7/2)*((a*e-b*d)*b)^(1/2)*a*b^3*e-73*A*(e*x+d)^(5/2)*((a* e-b*d)*b)^(1/2)*a*b^3*e^2-385*B*(e*x+d)^(3/2)*((a*e-b*d)*b)^(1/2)*a^3*b*e^ 3-120*B*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^4*b*d*e^4-15*A*(e...
Leaf count of result is larger than twice the leaf count of optimal. 767 vs. \(2 (276) = 552\).
Time = 0.53 (sec) , antiderivative size = 1547, normalized size of antiderivative = 4.31 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\text {Too large to display} \]
[1/384*(15*(8*B*a^4*b*d*e^3 - (7*B*a^5 + A*a^4*b)*e^4 + (8*B*b^5*d*e^3 - ( 7*B*a*b^4 + A*b^5)*e^4)*x^4 + 4*(8*B*a*b^4*d*e^3 - (7*B*a^2*b^3 + A*a*b^4) *e^4)*x^3 + 6*(8*B*a^2*b^3*d*e^3 - (7*B*a^3*b^2 + A*a^2*b^3)*e^4)*x^2 + 4* (8*B*a^3*b^2*d*e^3 - (7*B*a^4*b + A*a^3*b^2)*e^4)*x)*sqrt(b^2*d - a*b*e)*l og((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(16*(B*a*b^5 + 3*A*b^6)*d^4 + 8*(B*a^2*b^4 - 7*A*a*b^5)*d^3*e + 2*(13* B*a^3*b^3 - A*a^2*b^4)*d^2*e^2 - 5*(31*B*a^4*b^2 + A*a^3*b^3)*d*e^3 + 15*( 7*B*a^5*b + A*a^4*b^2)*e^4 + 3*(88*B*b^6*d^2*e^2 - (181*B*a*b^5 - 5*A*b^6) *d*e^3 + (93*B*a^2*b^4 - 5*A*a*b^5)*e^4)*x^3 + (208*B*b^6*d^3*e + 2*(25*B* a*b^5 + 59*A*b^6)*d^2*e^2 - (769*B*a^2*b^4 + 191*A*a*b^5)*d*e^3 + 73*(7*B* a^3*b^3 + A*a^2*b^4)*e^4)*x^2 + (64*B*b^6*d^4 + 8*(3*B*a*b^5 + 17*A*b^6)*d ^3*e + 4*(25*B*a^2*b^4 - 43*A*a*b^5)*d^2*e^2 - (573*B*a^3*b^3 + 19*A*a^2*b ^4)*d*e^3 + 55*(7*B*a^4*b^2 + A*a^3*b^3)*e^4)*x)*sqrt(e*x + d))/(a^4*b^7*d ^2 - 2*a^5*b^6*d*e + a^6*b^5*e^2 + (b^11*d^2 - 2*a*b^10*d*e + a^2*b^9*e^2) *x^4 + 4*(a*b^10*d^2 - 2*a^2*b^9*d*e + a^3*b^8*e^2)*x^3 + 6*(a^2*b^9*d^2 - 2*a^3*b^8*d*e + a^4*b^7*e^2)*x^2 + 4*(a^3*b^8*d^2 - 2*a^4*b^7*d*e + a^5*b ^6*e^2)*x), 1/192*(15*(8*B*a^4*b*d*e^3 - (7*B*a^5 + A*a^4*b)*e^4 + (8*B*b^ 5*d*e^3 - (7*B*a*b^4 + A*b^5)*e^4)*x^4 + 4*(8*B*a*b^4*d*e^3 - (7*B*a^2*b^3 + A*a*b^4)*e^4)*x^3 + 6*(8*B*a^2*b^3*d*e^3 - (7*B*a^3*b^2 + A*a^2*b^3)*e^ 4)*x^2 + 4*(8*B*a^3*b^2*d*e^3 - (7*B*a^4*b + A*a^3*b^2)*e^4)*x)*sqrt(-b...
Timed out. \[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{\frac {5}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 579 vs. \(2 (276) = 552\).
Time = 0.33 (sec) , antiderivative size = 579, normalized size of antiderivative = 1.61 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {5 \, {\left (8 \, B b d e^{3} - 7 \, B a e^{4} - A b e^{4}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{64 \, {\left (b^{5} d \mathrm {sgn}\left (b x + a\right ) - a b^{4} e \mathrm {sgn}\left (b x + a\right )\right )} \sqrt {-b^{2} d + a b e}} - \frac {264 \, {\left (e x + d\right )}^{\frac {7}{2}} B b^{4} d e^{3} - 584 \, {\left (e x + d\right )}^{\frac {5}{2}} B b^{4} d^{2} e^{3} + 440 \, {\left (e x + d\right )}^{\frac {3}{2}} B b^{4} d^{3} e^{3} - 120 \, \sqrt {e x + d} B b^{4} d^{4} e^{3} - 279 \, {\left (e x + d\right )}^{\frac {7}{2}} B a b^{3} e^{4} + 15 \, {\left (e x + d\right )}^{\frac {7}{2}} A b^{4} e^{4} + 1095 \, {\left (e x + d\right )}^{\frac {5}{2}} B a b^{3} d e^{4} + 73 \, {\left (e x + d\right )}^{\frac {5}{2}} A b^{4} d e^{4} - 1265 \, {\left (e x + d\right )}^{\frac {3}{2}} B a b^{3} d^{2} e^{4} - 55 \, {\left (e x + d\right )}^{\frac {3}{2}} A b^{4} d^{2} e^{4} + 465 \, \sqrt {e x + d} B a b^{3} d^{3} e^{4} + 15 \, \sqrt {e x + d} A b^{4} d^{3} e^{4} - 511 \, {\left (e x + d\right )}^{\frac {5}{2}} B a^{2} b^{2} e^{5} - 73 \, {\left (e x + d\right )}^{\frac {5}{2}} A a b^{3} e^{5} + 1210 \, {\left (e x + d\right )}^{\frac {3}{2}} B a^{2} b^{2} d e^{5} + 110 \, {\left (e x + d\right )}^{\frac {3}{2}} A a b^{3} d e^{5} - 675 \, \sqrt {e x + d} B a^{2} b^{2} d^{2} e^{5} - 45 \, \sqrt {e x + d} A a b^{3} d^{2} e^{5} - 385 \, {\left (e x + d\right )}^{\frac {3}{2}} B a^{3} b e^{6} - 55 \, {\left (e x + d\right )}^{\frac {3}{2}} A a^{2} b^{2} e^{6} + 435 \, \sqrt {e x + d} B a^{3} b d e^{6} + 45 \, \sqrt {e x + d} A a^{2} b^{2} d e^{6} - 105 \, \sqrt {e x + d} B a^{4} e^{7} - 15 \, \sqrt {e x + d} A a^{3} b e^{7}}{192 \, {\left (b^{5} d \mathrm {sgn}\left (b x + a\right ) - a b^{4} e \mathrm {sgn}\left (b x + a\right )\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{4}} \]
5/64*(8*B*b*d*e^3 - 7*B*a*e^4 - A*b*e^4)*arctan(sqrt(e*x + d)*b/sqrt(-b^2* d + a*b*e))/((b^5*d*sgn(b*x + a) - a*b^4*e*sgn(b*x + a))*sqrt(-b^2*d + a*b *e)) - 1/192*(264*(e*x + d)^(7/2)*B*b^4*d*e^3 - 584*(e*x + d)^(5/2)*B*b^4* d^2*e^3 + 440*(e*x + d)^(3/2)*B*b^4*d^3*e^3 - 120*sqrt(e*x + d)*B*b^4*d^4* e^3 - 279*(e*x + d)^(7/2)*B*a*b^3*e^4 + 15*(e*x + d)^(7/2)*A*b^4*e^4 + 109 5*(e*x + d)^(5/2)*B*a*b^3*d*e^4 + 73*(e*x + d)^(5/2)*A*b^4*d*e^4 - 1265*(e *x + d)^(3/2)*B*a*b^3*d^2*e^4 - 55*(e*x + d)^(3/2)*A*b^4*d^2*e^4 + 465*sqr t(e*x + d)*B*a*b^3*d^3*e^4 + 15*sqrt(e*x + d)*A*b^4*d^3*e^4 - 511*(e*x + d )^(5/2)*B*a^2*b^2*e^5 - 73*(e*x + d)^(5/2)*A*a*b^3*e^5 + 1210*(e*x + d)^(3 /2)*B*a^2*b^2*d*e^5 + 110*(e*x + d)^(3/2)*A*a*b^3*d*e^5 - 675*sqrt(e*x + d )*B*a^2*b^2*d^2*e^5 - 45*sqrt(e*x + d)*A*a*b^3*d^2*e^5 - 385*(e*x + d)^(3/ 2)*B*a^3*b*e^6 - 55*(e*x + d)^(3/2)*A*a^2*b^2*e^6 + 435*sqrt(e*x + d)*B*a^ 3*b*d*e^6 + 45*sqrt(e*x + d)*A*a^2*b^2*d*e^6 - 105*sqrt(e*x + d)*B*a^4*e^7 - 15*sqrt(e*x + d)*A*a^3*b*e^7)/((b^5*d*sgn(b*x + a) - a*b^4*e*sgn(b*x + a))*((e*x + d)*b - b*d + a*e)^4)
Timed out. \[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^{5/2}}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]