Integrand size = 26, antiderivative size = 310 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{(d+e x)^5} \, dx=\frac {\left (16 A c^2 d^2-8 b c d (B d+2 A e)+b^2 e (3 B d+5 A e)\right ) (b d+(2 c d-b e) x) \sqrt {b x+c x^2}}{64 d^3 (c d-b e)^3 (d+e x)^2}+\frac {(B d-A e) \left (b x+c x^2\right )^{3/2}}{4 d (c d-b e) (d+e x)^4}-\frac {(5 A e (2 c d-b e)-B d (2 c d+3 b e)) \left (b x+c x^2\right )^{3/2}}{24 d^2 (c d-b e)^2 (d+e x)^3}-\frac {b^2 \left (16 A c^2 d^2-8 b c d (B d+2 A e)+b^2 e (3 B d+5 A e)\right ) \text {arctanh}\left (\frac {b d+(2 c d-b e) x}{2 \sqrt {d} \sqrt {c d-b e} \sqrt {b x+c x^2}}\right )}{128 d^{7/2} (c d-b e)^{7/2}} \]
1/4*(-A*e+B*d)*(c*x^2+b*x)^(3/2)/d/(-b*e+c*d)/(e*x+d)^4-1/24*(5*A*e*(-b*e+ 2*c*d)-B*d*(3*b*e+2*c*d))*(c*x^2+b*x)^(3/2)/d^2/(-b*e+c*d)^2/(e*x+d)^3-1/1 28*b^2*(16*A*c^2*d^2-8*b*c*d*(2*A*e+B*d)+b^2*e*(5*A*e+3*B*d))*arctanh(1/2* (b*d+(-b*e+2*c*d)*x)/d^(1/2)/(-b*e+c*d)^(1/2)/(c*x^2+b*x)^(1/2))/d^(7/2)/( -b*e+c*d)^(7/2)+1/64*(16*A*c^2*d^2-8*b*c*d*(2*A*e+B*d)+b^2*e*(5*A*e+3*B*d) )*(b*d+(-b*e+2*c*d)*x)*(c*x^2+b*x)^(1/2)/d^3/(-b*e+c*d)^3/(e*x+d)^2
Time = 10.55 (sec) , antiderivative size = 279, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{(d+e x)^5} \, dx=\frac {\sqrt {x (b+c x)} \left (48 (-B d+A e) x^{3/2} (b+c x)-\frac {8 (5 A e (-2 c d+b e)+B d (2 c d+3 b e)) x^{3/2} (b+c x) (d+e x)}{d (c d-b e)}+\frac {3 \left (16 A c^2 d^2-8 b c d (B d+2 A e)+b^2 e (3 B d+5 A e)\right ) (d+e x)^2 \left (\sqrt {d} \sqrt {c d-b e} \sqrt {x} \sqrt {b+c x} (-b d-2 c d x+b e x)+b^2 (d+e x)^2 \text {arctanh}\left (\frac {\sqrt {c d-b e} \sqrt {x}}{\sqrt {d} \sqrt {b+c x}}\right )\right )}{d^{5/2} (c d-b e)^{5/2} \sqrt {b+c x}}\right )}{192 d (-c d+b e) \sqrt {x} (d+e x)^4} \]
(Sqrt[x*(b + c*x)]*(48*(-(B*d) + A*e)*x^(3/2)*(b + c*x) - (8*(5*A*e*(-2*c* d + b*e) + B*d*(2*c*d + 3*b*e))*x^(3/2)*(b + c*x)*(d + e*x))/(d*(c*d - b*e )) + (3*(16*A*c^2*d^2 - 8*b*c*d*(B*d + 2*A*e) + b^2*e*(3*B*d + 5*A*e))*(d + e*x)^2*(Sqrt[d]*Sqrt[c*d - b*e]*Sqrt[x]*Sqrt[b + c*x]*(-(b*d) - 2*c*d*x + b*e*x) + b^2*(d + e*x)^2*ArcTanh[(Sqrt[c*d - b*e]*Sqrt[x])/(Sqrt[d]*Sqrt [b + c*x])]))/(d^(5/2)*(c*d - b*e)^(5/2)*Sqrt[b + c*x])))/(192*d*(-(c*d) + b*e)*Sqrt[x]*(d + e*x)^4)
Time = 0.50 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1237, 27, 1228, 1152, 1154, 219}
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.
| \(\displaystyle \int \frac {(A+B x) \sqrt {b x+c x^2}}{(d+e x)^5} \, dx\) |
| \(\Big \downarrow \) 1237 |
| \(\displaystyle \frac {\left (b x+c x^2\right )^{3/2} (B d-A e)}{4 d (d+e x)^4 (c d-b e)}-\frac {\int \frac {(3 b B d-8 A c d+5 A b e-2 c (B d-A e) x) \sqrt {c x^2+b x}}{2 (d+e x)^4}dx}{4 d (c d-b e)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\left (b x+c x^2\right )^{3/2} (B d-A e)}{4 d (d+e x)^4 (c d-b e)}-\frac {\int \frac {(3 b B d-8 A c d+5 A b e-2 c (B d-A e) x) \sqrt {c x^2+b x}}{(d+e x)^4}dx}{8 d (c d-b e)}\) |
| \(\Big \downarrow \) 1228 |
| \(\displaystyle \frac {\left (b x+c x^2\right )^{3/2} (B d-A e)}{4 d (d+e x)^4 (c d-b e)}-\frac {\frac {\left (b x+c x^2\right )^{3/2} (5 A e (2 c d-b e)-B d (3 b e+2 c d))}{3 d (d+e x)^3 (c d-b e)}-\frac {\left (b^2 e (5 A e+3 B d)-8 b c d (2 A e+B d)+16 A c^2 d^2\right ) \int \frac {\sqrt {c x^2+b x}}{(d+e x)^3}dx}{2 d (c d-b e)}}{8 d (c d-b e)}\) |
| \(\Big \downarrow \) 1152 |
| \(\displaystyle \frac {\left (b x+c x^2\right )^{3/2} (B d-A e)}{4 d (d+e x)^4 (c d-b e)}-\frac {\frac {\left (b x+c x^2\right )^{3/2} (5 A e (2 c d-b e)-B d (3 b e+2 c d))}{3 d (d+e x)^3 (c d-b e)}-\frac {\left (b^2 e (5 A e+3 B d)-8 b c d (2 A e+B d)+16 A c^2 d^2\right ) \left (\frac {\sqrt {b x+c x^2} (x (2 c d-b e)+b d)}{4 d (d+e x)^2 (c d-b e)}-\frac {b^2 \int \frac {1}{(d+e x) \sqrt {c x^2+b x}}dx}{8 d (c d-b e)}\right )}{2 d (c d-b e)}}{8 d (c d-b e)}\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle \frac {\left (b x+c x^2\right )^{3/2} (B d-A e)}{4 d (d+e x)^4 (c d-b e)}-\frac {\frac {\left (b x+c x^2\right )^{3/2} (5 A e (2 c d-b e)-B d (3 b e+2 c d))}{3 d (d+e x)^3 (c d-b e)}-\frac {\left (b^2 e (5 A e+3 B d)-8 b c d (2 A e+B d)+16 A c^2 d^2\right ) \left (\frac {b^2 \int \frac {1}{4 d (c d-b e)-\frac {(b d+(2 c d-b e) x)^2}{c x^2+b x}}d\left (-\frac {b d+(2 c d-b e) x}{\sqrt {c x^2+b x}}\right )}{4 d (c d-b e)}+\frac {\sqrt {b x+c x^2} (x (2 c d-b e)+b d)}{4 d (d+e x)^2 (c d-b e)}\right )}{2 d (c d-b e)}}{8 d (c d-b e)}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\left (b x+c x^2\right )^{3/2} (B d-A e)}{4 d (d+e x)^4 (c d-b e)}-\frac {\frac {\left (b x+c x^2\right )^{3/2} (5 A e (2 c d-b e)-B d (3 b e+2 c d))}{3 d (d+e x)^3 (c d-b e)}-\frac {\left (b^2 e (5 A e+3 B d)-8 b c d (2 A e+B d)+16 A c^2 d^2\right ) \left (\frac {\sqrt {b x+c x^2} (x (2 c d-b e)+b d)}{4 d (d+e x)^2 (c d-b e)}-\frac {b^2 \text {arctanh}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{8 d^{3/2} (c d-b e)^{3/2}}\right )}{2 d (c d-b e)}}{8 d (c d-b e)}\) |
((B*d - A*e)*(b*x + c*x^2)^(3/2))/(4*d*(c*d - b*e)*(d + e*x)^4) - (((5*A*e *(2*c*d - b*e) - B*d*(2*c*d + 3*b*e))*(b*x + c*x^2)^(3/2))/(3*d*(c*d - b*e )*(d + e*x)^3) - ((16*A*c^2*d^2 - 8*b*c*d*(B*d + 2*A*e) + b^2*e*(3*B*d + 5 *A*e))*(((b*d + (2*c*d - b*e)*x)*Sqrt[b*x + c*x^2])/(4*d*(c*d - b*e)*(d + e*x)^2) - (b^2*ArcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b*e]* Sqrt[b*x + c*x^2])])/(8*d^(3/2)*(c*d - b*e)^(3/2))))/(2*d*(c*d - b*e)))/(8 *d*(c*d - b*e))
3.12.70.3.1 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_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(-(d + e*x)^(m + 1))*(d*b - 2*a*e + (2*c*d - b*e)*x)*((a + b *x + c*x^2)^p/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[p*((b^2 - 4*a *c)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2))) Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[m + 2*p + 2, 0] && GtQ[p, 0]
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Simp[(b*(e *f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^ (m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x ] && EqQ[Simplify[m + 2*p + 3], 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*f - d*g)*(d + e*x)^(m + 1)*((a + b* x + c*x^2)^(p + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((m + 1) *(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[ (c*d*f - f*b*e + a*e*g)*(m + 1) + b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && LtQ[m, -1 ] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Time = 0.63 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.15
| method | result | size |
| pseudoelliptic | \(-\frac {5 \left (\left (\frac {8 \left (2 A \,c^{2}-B b c \right ) d^{2}}{5}-\frac {16 \left (A c -\frac {3 B b}{16}\right ) e b d}{5}+A \,b^{2} e^{2}\right ) \left (e x +d \right )^{4} b^{2} \arctan \left (\frac {\sqrt {x \left (c x +b \right )}\, d}{x \sqrt {d \left (b e -c d \right )}}\right )+\sqrt {d \left (b e -c d \right )}\, \sqrt {x \left (c x +b \right )}\, \left (\frac {16 c \left (-\frac {B \,b^{2}}{2}+c \left (\frac {B x}{3}+A \right ) b +2 c^{2} x \left (\frac {2 B x}{3}+A \right )\right ) d^{5}}{5}-\frac {16 \left (-\frac {3 B \,b^{3}}{16}+c \left (\frac {47 B x}{24}+A \right ) b^{2}+\frac {11 c^{2} \left (\frac {21 B x}{22}+A \right ) x b}{3}-\frac {4 c^{3} x^{2} \left (\frac {B x}{4}+A \right )}{3}\right ) e \,d^{4}}{5}+\left (\left (\frac {11 B x}{5}+A \right ) b^{3}+\frac {66 c x \left (\frac {46 B x}{99}+A \right ) b^{2}}{5}-\frac {104 c^{2} \left (\frac {5 B x}{13}+A \right ) x^{2} b}{15}+\frac {16 A \,c^{3} x^{3}}{15}\right ) e^{2} d^{3}-\frac {73 \left (\left (\frac {33 B x}{73}+A \right ) b^{2}-\frac {140 c x \left (\frac {9 B x}{70}+A \right ) b}{73}+\frac {24 A \,c^{2} x^{2}}{73}\right ) x \,e^{3} b \,d^{2}}{15}-\frac {11 x^{2} e^{4} \left (\left (\frac {9 B x}{55}+A \right ) b -\frac {38 A c x}{55}\right ) b^{2} d}{3}-A \,b^{3} e^{5} x^{3}\right )\right )}{64 \sqrt {d \left (b e -c d \right )}\, \left (e x +d \right )^{4} \left (b e -c d \right )^{3} d^{3}}\) | \(355\) |
| default | \(\text {Expression too large to display}\) | \(3278\) |
-5/64/(d*(b*e-c*d))^(1/2)*((8/5*(2*A*c^2-B*b*c)*d^2-16/5*(A*c-3/16*B*b)*e* b*d+A*b^2*e^2)*(e*x+d)^4*b^2*arctan((x*(c*x+b))^(1/2)/x*d/(d*(b*e-c*d))^(1 /2))+(d*(b*e-c*d))^(1/2)*(x*(c*x+b))^(1/2)*(16/5*c*(-1/2*B*b^2+c*(1/3*B*x+ A)*b+2*c^2*x*(2/3*B*x+A))*d^5-16/5*(-3/16*B*b^3+c*(47/24*B*x+A)*b^2+11/3*c ^2*(21/22*B*x+A)*x*b-4/3*c^3*x^2*(1/4*B*x+A))*e*d^4+((11/5*B*x+A)*b^3+66/5 *c*x*(46/99*B*x+A)*b^2-104/15*c^2*(5/13*B*x+A)*x^2*b+16/15*A*c^3*x^3)*e^2* d^3-73/15*((33/73*B*x+A)*b^2-140/73*c*x*(9/70*B*x+A)*b+24/73*A*c^2*x^2)*x* e^3*b*d^2-11/3*x^2*e^4*((9/55*B*x+A)*b-38/55*A*c*x)*b^2*d-A*b^3*e^5*x^3))/ (e*x+d)^4/(b*e-c*d)^3/d^3
Leaf count of result is larger than twice the leaf count of optimal. 1089 vs. \(2 (283) = 566\).
Time = 0.46 (sec) , antiderivative size = 2190, normalized size of antiderivative = 7.06 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{(d+e x)^5} \, dx=\text {Too large to display} \]
[-1/384*(3*(5*A*b^4*d^4*e^2 - 8*(B*b^3*c - 2*A*b^2*c^2)*d^6 + (3*B*b^4 - 1 6*A*b^3*c)*d^5*e + (5*A*b^4*e^6 - 8*(B*b^3*c - 2*A*b^2*c^2)*d^2*e^4 + (3*B *b^4 - 16*A*b^3*c)*d*e^5)*x^4 + 4*(5*A*b^4*d*e^5 - 8*(B*b^3*c - 2*A*b^2*c^ 2)*d^3*e^3 + (3*B*b^4 - 16*A*b^3*c)*d^2*e^4)*x^3 + 6*(5*A*b^4*d^2*e^4 - 8* (B*b^3*c - 2*A*b^2*c^2)*d^4*e^2 + (3*B*b^4 - 16*A*b^3*c)*d^3*e^3)*x^2 + 4* (5*A*b^4*d^3*e^3 - 8*(B*b^3*c - 2*A*b^2*c^2)*d^5*e + (3*B*b^4 - 16*A*b^3*c )*d^4*e^2)*x)*sqrt(c*d^2 - b*d*e)*log((b*d + (2*c*d - b*e)*x + 2*sqrt(c*d^ 2 - b*d*e)*sqrt(c*x^2 + b*x))/(e*x + d)) + 2*(15*A*b^4*d^4*e^3 + 24*(B*b^2 *c^2 - 2*A*b*c^3)*d^7 - 3*(11*B*b^3*c - 32*A*b^2*c^2)*d^6*e + 9*(B*b^4 - 7 *A*b^3*c)*d^5*e^2 - (16*B*c^4*d^6*e + 15*A*b^4*d*e^6 - 8*(7*B*b*c^3 - 2*A* c^4)*d^5*e^2 + 2*(29*B*b^2*c^2 - 20*A*b*c^3)*d^4*e^3 - (27*B*b^3*c - 62*A* b^2*c^2)*d^3*e^4 + (9*B*b^4 - 53*A*b^3*c)*d^2*e^5)*x^3 - (64*B*c^4*d^7 + 5 5*A*b^4*d^2*e^5 - 8*(29*B*b*c^3 - 8*A*c^4)*d^6*e + 4*(65*B*b^2*c^2 - 42*A* b*c^3)*d^5*e^2 - (125*B*b^3*c - 244*A*b^2*c^2)*d^4*e^3 + 3*(11*B*b^4 - 65* A*b^3*c)*d^3*e^4)*x^2 - (73*A*b^4*d^3*e^4 + 16*(B*b*c^3 + 6*A*c^4)*d^7 - 2 *(55*B*b^2*c^2 + 136*A*b*c^3)*d^6*e + (127*B*b^3*c + 374*A*b^2*c^2)*d^5*e^ 2 - (33*B*b^4 + 271*A*b^3*c)*d^4*e^3)*x)*sqrt(c*x^2 + b*x))/(c^4*d^12 - 4* b*c^3*d^11*e + 6*b^2*c^2*d^10*e^2 - 4*b^3*c*d^9*e^3 + b^4*d^8*e^4 + (c^4*d ^8*e^4 - 4*b*c^3*d^7*e^5 + 6*b^2*c^2*d^6*e^6 - 4*b^3*c*d^5*e^7 + b^4*d^4*e ^8)*x^4 + 4*(c^4*d^9*e^3 - 4*b*c^3*d^8*e^4 + 6*b^2*c^2*d^7*e^5 - 4*b^3*...
\[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{(d+e x)^5} \, dx=\int \frac {\sqrt {x \left (b + c x\right )} \left (A + B x\right )}{\left (d + e x\right )^{5}}\, dx \]
Exception generated. \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{(d+e x)^5} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b*e-c*d>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 1841 vs. \(2 (283) = 566\).
Time = 0.67 (sec) , antiderivative size = 1841, normalized size of antiderivative = 5.94 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{(d+e x)^5} \, dx=\text {Too large to display} \]
1/384*((24*B*b^3*c*d^2*e^4*log(abs(2*c*d*e - b*e^2 - 2*sqrt(c*d^2 - b*d*e) *sqrt(c)*abs(e))) - 48*A*b^2*c^2*d^2*e^4*log(abs(2*c*d*e - b*e^2 - 2*sqrt( c*d^2 - b*d*e)*sqrt(c)*abs(e))) - 9*B*b^4*d*e^5*log(abs(2*c*d*e - b*e^2 - 2*sqrt(c*d^2 - b*d*e)*sqrt(c)*abs(e))) + 48*A*b^3*c*d*e^5*log(abs(2*c*d*e - b*e^2 - 2*sqrt(c*d^2 - b*d*e)*sqrt(c)*abs(e))) - 15*A*b^4*e^6*log(abs(2* c*d*e - b*e^2 - 2*sqrt(c*d^2 - b*d*e)*sqrt(c)*abs(e))) - 32*sqrt(c*d^2 - b *d*e)*B*c^(7/2)*d^4*abs(e) + 80*sqrt(c*d^2 - b*d*e)*B*b*c^(5/2)*d^3*e*abs( e) - 32*sqrt(c*d^2 - b*d*e)*A*c^(7/2)*d^3*e*abs(e) - 36*sqrt(c*d^2 - b*d*e )*B*b^2*c^(3/2)*d^2*e^2*abs(e) + 48*sqrt(c*d^2 - b*d*e)*A*b*c^(5/2)*d^2*e^ 2*abs(e) + 18*sqrt(c*d^2 - b*d*e)*B*b^3*sqrt(c)*d*e^3*abs(e) - 76*sqrt(c*d ^2 - b*d*e)*A*b^2*c^(3/2)*d*e^3*abs(e) + 30*sqrt(c*d^2 - b*d*e)*A*b^3*sqrt (c)*e^4*abs(e))*sgn(1/(e*x + d))*sgn(e)/(sqrt(c*d^2 - b*d*e)*c^3*d^6*e^5*a bs(e) - 3*sqrt(c*d^2 - b*d*e)*b*c^2*d^5*e^6*abs(e) + 3*sqrt(c*d^2 - b*d*e) *b^2*c*d^4*e^7*abs(e) - sqrt(c*d^2 - b*d*e)*b^3*d^3*e^8*abs(e)) + 2*sqrt(c - 2*c*d/(e*x + d) + c*d^2/(e*x + d)^2 + b*e/(e*x + d) - b*d*e/(e*x + d)^2 )*((16*B*c^3*d^4*e^9*sgn(1/(e*x + d))*sgn(e) - 40*B*b*c^2*d^3*e^10*sgn(1/( e*x + d))*sgn(e) + 16*A*c^3*d^3*e^10*sgn(1/(e*x + d))*sgn(e) + 18*B*b^2*c* d^2*e^11*sgn(1/(e*x + d))*sgn(e) - 24*A*b*c^2*d^2*e^11*sgn(1/(e*x + d))*sg n(e) - 9*B*b^3*d*e^12*sgn(1/(e*x + d))*sgn(e) + 38*A*b^2*c*d*e^12*sgn(1/(e *x + d))*sgn(e) - 15*A*b^3*e^13*sgn(1/(e*x + d))*sgn(e))/(c^3*d^6*e^14 ...
Timed out. \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{(d+e x)^5} \, dx=\int \frac {\sqrt {c\,x^2+b\,x}\,\left (A+B\,x\right )}{{\left (d+e\,x\right )}^5} \,d x \]