Integrand size = 26, antiderivative size = 331 \[ \int \frac {A+B x}{(d+e x)^4 \sqrt {b x+c x^2}} \, dx=\frac {(B d-A e) \sqrt {b x+c x^2}}{3 d (c d-b e) (d+e x)^3}-\frac {(5 A e (2 c d-b e)-B d (4 c d+b e)) \sqrt {b x+c x^2}}{12 d^2 (c d-b e)^2 (d+e x)^2}+\frac {\left (B d \left (8 c^2 d^2+10 b c d e-3 b^2 e^2\right )-A e \left (44 c^2 d^2-44 b c d e+15 b^2 e^2\right )\right ) \sqrt {b x+c x^2}}{24 d^3 (c d-b e)^3 (d+e x)}+\frac {\left (16 A c^3 d^3-8 b c^2 d^2 (B d+3 A e)-b^3 e^2 (B d+5 A e)+2 b^2 c d e (2 B d+9 A e)\right ) \text {arctanh}\left (\frac {b d+(2 c d-b e) x}{2 \sqrt {d} \sqrt {c d-b e} \sqrt {b x+c x^2}}\right )}{16 d^{7/2} (c d-b e)^{7/2}} \]
1/16*(16*A*c^3*d^3-8*b*c^2*d^2*(3*A*e+B*d)-b^3*e^2*(5*A*e+B*d)+2*b^2*c*d*e *(9*A*e+2*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)+1/3*(-A*e+B*d)*(c*x^2+b*x)^(1/ 2)/d/(-b*e+c*d)/(e*x+d)^3-1/12*(5*A*e*(-b*e+2*c*d)-B*d*(b*e+4*c*d))*(c*x^2 +b*x)^(1/2)/d^2/(-b*e+c*d)^2/(e*x+d)^2+1/24*(B*d*(-3*b^2*e^2+10*b*c*d*e+8* c^2*d^2)-A*e*(15*b^2*e^2-44*b*c*d*e+44*c^2*d^2))*(c*x^2+b*x)^(1/2)/d^3/(-b *e+c*d)^3/(e*x+d)
Time = 11.06 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.04 \[ \int \frac {A+B x}{(d+e x)^4 \sqrt {b x+c x^2}} \, dx=-\frac {\sqrt {x} \left (8 (B d-A e) \sqrt {x} (b+c x)+\frac {\sqrt {b+c x} (d+e x) \left (2 d^{3/2} (c d-b e)^{3/2} (5 A e (-2 c d+b e)+B d (4 c d+b e)) \sqrt {x} \sqrt {b+c x}-(d+e x) \left (-\sqrt {d} \sqrt {c d-b e} \left (A e \left (-44 c^2 d^2+44 b c d e-15 b^2 e^2\right )+B d \left (8 c^2 d^2+10 b c d e-3 b^2 e^2\right )\right ) \sqrt {x} \sqrt {b+c x}-3 \left (16 A c^3 d^3-8 b c^2 d^2 (B d+3 A e)-b^3 e^2 (B d+5 A e)+2 b^2 c d e (2 B d+9 A e)\right ) (d+e x) \text {arctanh}\left (\frac {\sqrt {c d-b e} \sqrt {x}}{\sqrt {d} \sqrt {b+c x}}\right )\right )\right )}{d^{5/2} (c d-b e)^{5/2}}\right )}{24 d (-c d+b e) \sqrt {x (b+c x)} (d+e x)^3} \]
-1/24*(Sqrt[x]*(8*(B*d - A*e)*Sqrt[x]*(b + c*x) + (Sqrt[b + c*x]*(d + e*x) *(2*d^(3/2)*(c*d - b*e)^(3/2)*(5*A*e*(-2*c*d + b*e) + B*d*(4*c*d + b*e))*S qrt[x]*Sqrt[b + c*x] - (d + e*x)*(-(Sqrt[d]*Sqrt[c*d - b*e]*(A*e*(-44*c^2* d^2 + 44*b*c*d*e - 15*b^2*e^2) + B*d*(8*c^2*d^2 + 10*b*c*d*e - 3*b^2*e^2)) *Sqrt[x]*Sqrt[b + c*x]) - 3*(16*A*c^3*d^3 - 8*b*c^2*d^2*(B*d + 3*A*e) - b^ 3*e^2*(B*d + 5*A*e) + 2*b^2*c*d*e*(2*B*d + 9*A*e))*(d + e*x)*ArcTanh[(Sqrt [c*d - b*e]*Sqrt[x])/(Sqrt[d]*Sqrt[b + c*x])])))/(d^(5/2)*(c*d - b*e)^(5/2 ))))/(d*(-(c*d) + b*e)*Sqrt[x*(b + c*x)]*(d + e*x)^3)
Time = 0.66 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {1237, 27, 1237, 27, 1228, 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.
| \(\displaystyle \int \frac {A+B x}{\sqrt {b x+c x^2} (d+e x)^4} \, dx\) |
| \(\Big \downarrow \) 1237 |
| \(\displaystyle \frac {\sqrt {b x+c x^2} (B d-A e)}{3 d (d+e x)^3 (c d-b e)}-\frac {\int \frac {b B d-6 A c d+5 A b e-4 c (B d-A e) x}{2 (d+e x)^3 \sqrt {c x^2+b x}}dx}{3 d (c d-b e)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {b x+c x^2} (B d-A e)}{3 d (d+e x)^3 (c d-b e)}-\frac {\int \frac {b B d-6 A c d+5 A b e-4 c (B d-A e) x}{(d+e x)^3 \sqrt {c x^2+b x}}dx}{6 d (c d-b e)}\) |
| \(\Big \downarrow \) 1237 |
| \(\displaystyle \frac {\sqrt {b x+c x^2} (B d-A e)}{3 d (d+e x)^3 (c d-b e)}-\frac {\frac {\sqrt {b x+c x^2} (5 A e (2 c d-b e)-B d (b e+4 c d))}{2 d (d+e x)^2 (c d-b e)}-\frac {\int \frac {3 e (B d+5 A e) b^2-2 c d (4 B d+17 A e) b+24 A c^2 d^2-2 c (5 A e (2 c d-b e)-B d (4 c d+b e)) x}{2 (d+e x)^2 \sqrt {c x^2+b x}}dx}{2 d (c d-b e)}}{6 d (c d-b e)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {b x+c x^2} (B d-A e)}{3 d (d+e x)^3 (c d-b e)}-\frac {\frac {\sqrt {b x+c x^2} (5 A e (2 c d-b e)-B d (b e+4 c d))}{2 d (d+e x)^2 (c d-b e)}-\frac {\int \frac {3 e (B d+5 A e) b^2-2 c d (4 B d+17 A e) b+24 A c^2 d^2-2 c (5 A e (2 c d-b e)-B d (4 c d+b e)) x}{(d+e x)^2 \sqrt {c x^2+b x}}dx}{4 d (c d-b e)}}{6 d (c d-b e)}\) |
| \(\Big \downarrow \) 1228 |
| \(\displaystyle \frac {\sqrt {b x+c x^2} (B d-A e)}{3 d (d+e x)^3 (c d-b e)}-\frac {\frac {\sqrt {b x+c x^2} (5 A e (2 c d-b e)-B d (b e+4 c d))}{2 d (d+e x)^2 (c d-b e)}-\frac {\frac {3 \left (b^3 \left (-e^2\right ) (5 A e+B d)+2 b^2 c d e (9 A e+2 B d)-8 b c^2 d^2 (3 A e+B d)+16 A c^3 d^3\right ) \int \frac {1}{(d+e x) \sqrt {c x^2+b x}}dx}{2 d (c d-b e)}+\frac {\sqrt {b x+c x^2} \left (B d \left (-3 b^2 e^2+10 b c d e+8 c^2 d^2\right )-A e \left (15 b^2 e^2-44 b c d e+44 c^2 d^2\right )\right )}{d (d+e x) (c d-b e)}}{4 d (c d-b e)}}{6 d (c d-b e)}\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle \frac {\sqrt {b x+c x^2} (B d-A e)}{3 d (d+e x)^3 (c d-b e)}-\frac {\frac {\sqrt {b x+c x^2} (5 A e (2 c d-b e)-B d (b e+4 c d))}{2 d (d+e x)^2 (c d-b e)}-\frac {\frac {\sqrt {b x+c x^2} \left (B d \left (-3 b^2 e^2+10 b c d e+8 c^2 d^2\right )-A e \left (15 b^2 e^2-44 b c d e+44 c^2 d^2\right )\right )}{d (d+e x) (c d-b e)}-\frac {3 \left (b^3 \left (-e^2\right ) (5 A e+B d)+2 b^2 c d e (9 A e+2 B d)-8 b c^2 d^2 (3 A e+B d)+16 A c^3 d^3\right ) \int \frac {1}{4 d (c d-b e)-\frac {(b d+(2 c d-b e) x)^2}{c x^2+b x}}d\left (-\frac {b d+(2 c d-b e) x}{\sqrt {c x^2+b x}}\right )}{d (c d-b e)}}{4 d (c d-b e)}}{6 d (c d-b e)}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\sqrt {b x+c x^2} (B d-A e)}{3 d (d+e x)^3 (c d-b e)}-\frac {\frac {\sqrt {b x+c x^2} (5 A e (2 c d-b e)-B d (b e+4 c d))}{2 d (d+e x)^2 (c d-b e)}-\frac {\frac {3 \left (b^3 \left (-e^2\right ) (5 A e+B d)+2 b^2 c d e (9 A e+2 B d)-8 b c^2 d^2 (3 A e+B d)+16 A c^3 d^3\right ) \text {arctanh}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{2 d^{3/2} (c d-b e)^{3/2}}+\frac {\sqrt {b x+c x^2} \left (B d \left (-3 b^2 e^2+10 b c d e+8 c^2 d^2\right )-A e \left (15 b^2 e^2-44 b c d e+44 c^2 d^2\right )\right )}{d (d+e x) (c d-b e)}}{4 d (c d-b e)}}{6 d (c d-b e)}\) |
((B*d - A*e)*Sqrt[b*x + c*x^2])/(3*d*(c*d - b*e)*(d + e*x)^3) - (((5*A*e*( 2*c*d - b*e) - B*d*(4*c*d + b*e))*Sqrt[b*x + c*x^2])/(2*d*(c*d - b*e)*(d + e*x)^2) - (((B*d*(8*c^2*d^2 + 10*b*c*d*e - 3*b^2*e^2) - A*e*(44*c^2*d^2 - 44*b*c*d*e + 15*b^2*e^2))*Sqrt[b*x + c*x^2])/(d*(c*d - b*e)*(d + e*x)) + (3*(16*A*c^3*d^3 - 8*b*c^2*d^2*(B*d + 3*A*e) - b^3*e^2*(B*d + 5*A*e) + 2*b ^2*c*d*e*(2*B*d + 9*A*e))*ArcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqrt[d]*Sqrt[ c*d - b*e]*Sqrt[b*x + c*x^2])])/(2*d^(3/2)*(c*d - b*e)^(3/2)))/(4*d*(c*d - b*e)))/(6*d*(c*d - b*e))
3.12.98.3.1 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[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 = 0.67 (sec) , antiderivative size = 294, normalized size of antiderivative = 0.89
| method | result | size |
| pseudoelliptic | \(-\frac {5 \left (-\frac {11 \sqrt {d \left (b e -c d \right )}\, \left (\left (-\frac {B \,d^{3}}{11}+e \left (A +\frac {8 B x}{33}\right ) d^{2}+\frac {40 x \left (\frac {3 B x}{40}+A \right ) e^{2} d}{33}+\frac {5 A \,e^{3} x^{2}}{11}\right ) e^{2} b^{2}-\frac {30 c d e \left (-\frac {2 B \,d^{3}}{15}+e \left (\frac {7 B x}{45}+A \right ) d^{2}+\frac {59 \left (\frac {5 B x}{59}+A \right ) x \,e^{2} d}{45}+\frac {22 A \,e^{3} x^{2}}{45}\right ) b}{11}+\frac {24 c^{2} d^{2} \left (-\frac {B \,d^{3}}{3}+e \left (-\frac {B x}{3}+A \right ) d^{2}+\frac {3 x \,e^{2} \left (-\frac {2 B x}{27}+A \right ) d}{2}+\frac {11 A \,e^{3} x^{2}}{18}\right )}{11}\right ) \sqrt {x \left (c x +b \right )}}{5}+\arctan \left (\frac {\sqrt {x \left (c x +b \right )}\, d}{x \sqrt {d \left (b e -c d \right )}}\right ) \left (e x +d \right )^{3} \left (e^{2} \left (A e +\frac {B d}{5}\right ) b^{3}-\frac {18 c \left (A e +\frac {2 B d}{9}\right ) d e \,b^{2}}{5}+\frac {24 c^{2} \left (A e +\frac {B d}{3}\right ) d^{2} b}{5}-\frac {16 A \,c^{3} d^{3}}{5}\right )\right )}{8 \sqrt {d \left (b e -c d \right )}\, \left (e x +d \right )^{3} \left (b e -c d \right )^{3} d^{3}}\) | \(294\) |
| default | \(\text {Expression too large to display}\) | \(1297\) |
-5/8/(d*(b*e-c*d))^(1/2)*(-11/5*(d*(b*e-c*d))^(1/2)*((-1/11*B*d^3+e*(A+8/3 3*B*x)*d^2+40/33*x*(3/40*B*x+A)*e^2*d+5/11*A*e^3*x^2)*e^2*b^2-30/11*c*d*e* (-2/15*B*d^3+e*(7/45*B*x+A)*d^2+59/45*(5/59*B*x+A)*x*e^2*d+22/45*A*e^3*x^2 )*b+24/11*c^2*d^2*(-1/3*B*d^3+e*(-1/3*B*x+A)*d^2+3/2*x*e^2*(-2/27*B*x+A)*d +11/18*A*e^3*x^2))*(x*(c*x+b))^(1/2)+arctan((x*(c*x+b))^(1/2)/x*d/(d*(b*e- c*d))^(1/2))*(e*x+d)^3*(e^2*(A*e+1/5*B*d)*b^3-18/5*c*(A*e+2/9*B*d)*d*e*b^2 +24/5*c^2*(A*e+1/3*B*d)*d^2*b-16/5*A*c^3*d^3))/(e*x+d)^3/(b*e-c*d)^3/d^3
Leaf count of result is larger than twice the leaf count of optimal. 885 vs. \(2 (304) = 608\).
Time = 0.50 (sec) , antiderivative size = 1781, normalized size of antiderivative = 5.38 \[ \int \frac {A+B x}{(d+e x)^4 \sqrt {b x+c x^2}} \, dx=\text {Too large to display} \]
[-1/48*(3*(5*A*b^3*d^3*e^3 + 8*(B*b*c^2 - 2*A*c^3)*d^6 - 4*(B*b^2*c - 6*A* b*c^2)*d^5*e + (B*b^3 - 18*A*b^2*c)*d^4*e^2 + (5*A*b^3*e^6 + 8*(B*b*c^2 - 2*A*c^3)*d^3*e^3 - 4*(B*b^2*c - 6*A*b*c^2)*d^2*e^4 + (B*b^3 - 18*A*b^2*c)* d*e^5)*x^3 + 3*(5*A*b^3*d*e^5 + 8*(B*b*c^2 - 2*A*c^3)*d^4*e^2 - 4*(B*b^2*c - 6*A*b*c^2)*d^3*e^3 + (B*b^3 - 18*A*b^2*c)*d^2*e^4)*x^2 + 3*(5*A*b^3*d^2 *e^4 + 8*(B*b*c^2 - 2*A*c^3)*d^5*e - 4*(B*b^2*c - 6*A*b*c^2)*d^4*e^2 + (B* b^3 - 18*A*b^2*c)*d^3*e^3)*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*(24*B*c^3*d^7 + 33*A*b^3*d^3*e^4 - 36*(B*b*c^2 + 2*A*c^3)*d^6*e + 3*(5*B*b^2*c + 54*A*b *c^2)*d^5*e^2 - 3*(B*b^3 + 41*A*b^2*c)*d^4*e^3 + (8*B*c^3*d^5*e^2 + 15*A*b ^3*d*e^6 + 2*(B*b*c^2 - 22*A*c^3)*d^4*e^3 - (13*B*b^2*c - 88*A*b*c^2)*d^3* e^4 + (3*B*b^3 - 59*A*b^2*c)*d^2*e^5)*x^2 + 2*(12*B*c^3*d^6*e + 20*A*b^3*d ^2*e^5 - (5*B*b*c^2 + 54*A*c^3)*d^5*e^2 - (11*B*b^2*c - 113*A*b*c^2)*d^4*e ^3 + (4*B*b^3 - 79*A*b^2*c)*d^3*e^4)*x)*sqrt(c*x^2 + b*x))/(c^4*d^11 - 4*b *c^3*d^10*e + 6*b^2*c^2*d^9*e^2 - 4*b^3*c*d^8*e^3 + b^4*d^7*e^4 + (c^4*d^8 *e^3 - 4*b*c^3*d^7*e^4 + 6*b^2*c^2*d^6*e^5 - 4*b^3*c*d^5*e^6 + b^4*d^4*e^7 )*x^3 + 3*(c^4*d^9*e^2 - 4*b*c^3*d^8*e^3 + 6*b^2*c^2*d^7*e^4 - 4*b^3*c*d^6 *e^5 + b^4*d^5*e^6)*x^2 + 3*(c^4*d^10*e - 4*b*c^3*d^9*e^2 + 6*b^2*c^2*d^8* e^3 - 4*b^3*c*d^7*e^4 + b^4*d^6*e^5)*x), -1/24*(3*(5*A*b^3*d^3*e^3 + 8*(B* b*c^2 - 2*A*c^3)*d^6 - 4*(B*b^2*c - 6*A*b*c^2)*d^5*e + (B*b^3 - 18*A*b^...
\[ \int \frac {A+B x}{(d+e x)^4 \sqrt {b x+c x^2}} \, dx=\int \frac {A + B x}{\sqrt {x \left (b + c x\right )} \left (d + e x\right )^{4}}\, dx \]
Exception generated. \[ \int \frac {A+B x}{(d+e x)^4 \sqrt {b x+c x^2}} \, dx=\text {Exception raised: ValueError} \]
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. 1693 vs. \(2 (304) = 608\).
Time = 0.30 (sec) , antiderivative size = 1693, normalized size of antiderivative = 5.11 \[ \int \frac {A+B x}{(d+e x)^4 \sqrt {b x+c x^2}} \, dx=\text {Too large to display} \]
-1/8*(8*B*b*c^2*d^3 - 16*A*c^3*d^3 - 4*B*b^2*c*d^2*e + 24*A*b*c^2*d^2*e + B*b^3*d*e^2 - 18*A*b^2*c*d*e^2 + 5*A*b^3*e^3)*arctan(-((sqrt(c)*x - sqrt(c *x^2 + b*x))*e + sqrt(c)*d)/sqrt(-c*d^2 + b*d*e))/((c^3*d^6 - 3*b*c^2*d^5* e + 3*b^2*c*d^4*e^2 - b^3*d^3*e^3)*sqrt(-c*d^2 + b*d*e)) + 1/24*(24*(sqrt( c)*x - sqrt(c*x^2 + b*x))^5*B*b*c^2*d^3*e^3 - 48*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*A*c^3*d^3*e^3 - 12*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b^2*c*d^2* e^4 + 72*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*A*b*c^2*d^2*e^4 + 3*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b^3*d*e^5 - 54*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5* A*b^2*c*d*e^5 + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*A*b^3*e^6 + 120*(sqrt (c)*x - sqrt(c*x^2 + b*x))^4*B*b*c^(5/2)*d^4*e^2 - 240*(sqrt(c)*x - sqrt(c *x^2 + b*x))^4*A*c^(7/2)*d^4*e^2 - 60*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B* b^2*c^(3/2)*d^3*e^3 + 360*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*A*b*c^(5/2)*d^ 3*e^3 + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b^3*sqrt(c)*d^2*e^4 - 270*( sqrt(c)*x - sqrt(c*x^2 + b*x))^4*A*b^2*c^(3/2)*d^2*e^4 + 75*(sqrt(c)*x - s qrt(c*x^2 + b*x))^4*A*b^3*sqrt(c)*d*e^5 + 64*(sqrt(c)*x - sqrt(c*x^2 + b*x ))^3*B*c^4*d^6 - 16*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b*c^3*d^5*e - 352* (sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*c^4*d^5*e + 168*(sqrt(c)*x - sqrt(c*x^ 2 + b*x))^3*B*b^2*c^2*d^4*e^2 + 400*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*b* c^3*d^4*e^2 - 74*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^3*c*d^3*e^3 - 204*( sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*b^2*c^2*d^3*e^3 + 8*(sqrt(c)*x - sqr...
Timed out. \[ \int \frac {A+B x}{(d+e x)^4 \sqrt {b x+c x^2}} \, dx=\int \frac {A+B\,x}{\sqrt {c\,x^2+b\,x}\,{\left (d+e\,x\right )}^4} \,d x \]