Integrand size = 33, antiderivative size = 222 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^8} \, dx=\frac {b^2 (A b-a B) (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 (b d-a e)^4 (d+e x)^4}+\frac {(B d-A e) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{7 (b d-a e)^2 (d+e x)^7}+\frac {(2 b B d-9 A b e+7 a B e) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{42 (b d-a e)^3 (d+e x)^6}+\frac {b (2 b B d-51 A b e+49 a B e) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{210 (b d-a e)^4 (d+e x)^5} \] Output:
1/4*b^2*(A*b-B*a)*(b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(-a*e+b*d)^4/(e*x+d) ^4+1/7*(-A*e+B*d)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/(-a*e+b*d)^2/(e*x+d)^7+1/42* (-9*A*b*e+7*B*a*e+2*B*b*d)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/(-a*e+b*d)^3/(e*x+d )^6+1/210*b*(-51*A*b*e+49*B*a*e+2*B*b*d)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/(-a*e +b*d)^4/(e*x+d)^5
Time = 1.11 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^8} \, dx=-\frac {\sqrt {(a+b x)^2} \left (10 a^3 e^3 (6 A e+B (d+7 e x))+6 a^2 b e^2 \left (5 A e (d+7 e x)+2 B \left (d^2+7 d e x+21 e^2 x^2\right )\right )+3 a b^2 e \left (4 A e \left (d^2+7 d e x+21 e^2 x^2\right )+3 B \left (d^3+7 d^2 e x+21 d e^2 x^2+35 e^3 x^3\right )\right )+b^3 \left (3 A e \left (d^3+7 d^2 e x+21 d e^2 x^2+35 e^3 x^3\right )+4 B \left (d^4+7 d^3 e x+21 d^2 e^2 x^2+35 d e^3 x^3+35 e^4 x^4\right )\right )\right )}{420 e^5 (a+b x) (d+e x)^7} \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^8,x]
Output:
-1/420*(Sqrt[(a + b*x)^2]*(10*a^3*e^3*(6*A*e + B*(d + 7*e*x)) + 6*a^2*b*e^ 2*(5*A*e*(d + 7*e*x) + 2*B*(d^2 + 7*d*e*x + 21*e^2*x^2)) + 3*a*b^2*e*(4*A* e*(d^2 + 7*d*e*x + 21*e^2*x^2) + 3*B*(d^3 + 7*d^2*e*x + 21*d*e^2*x^2 + 35* e^3*x^3)) + b^3*(3*A*e*(d^3 + 7*d^2*e*x + 21*d*e^2*x^2 + 35*e^3*x^3) + 4*B *(d^4 + 7*d^3*e*x + 21*d^2*e^2*x^2 + 35*d*e^3*x^3 + 35*e^4*x^4))))/(e^5*(a + b*x)*(d + e*x)^7)
Time = 0.67 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {1187, 27, 86, 2009}
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.
| \(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x)}{(d+e x)^8} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3 (A+B x)}{(d+e x)^8}dx}{b^3 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^8}dx}{a+b x}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {B b^3}{e^4 (d+e x)^4}+\frac {(-4 b B d+A b e+3 a B e) b^2}{e^4 (d+e x)^5}-\frac {3 (b d-a e) (-2 b B d+A b e+a B e) b}{e^4 (d+e x)^6}+\frac {(a e-b d)^2 (-4 b B d+3 A b e+a B e)}{e^4 (d+e x)^7}+\frac {(a e-b d)^3 (A e-B d)}{e^4 (d+e x)^8}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b^2 (-3 a B e-A b e+4 b B d)}{4 e^5 (d+e x)^4}-\frac {3 b (b d-a e) (-a B e-A b e+2 b B d)}{5 e^5 (d+e x)^5}+\frac {(b d-a e)^2 (-a B e-3 A b e+4 b B d)}{6 e^5 (d+e x)^6}-\frac {(b d-a e)^3 (B d-A e)}{7 e^5 (d+e x)^7}-\frac {b^3 B}{3 e^5 (d+e x)^3}\right )}{a+b x}\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^8,x]
Output:
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-1/7*((b*d - a*e)^3*(B*d - A*e))/(e^5*(d + e*x)^7) + ((b*d - a*e)^2*(4*b*B*d - 3*A*b*e - a*B*e))/(6*e^5*(d + e*x)^6) - (3*b*(b*d - a*e)*(2*b*B*d - A*b*e - a*B*e))/(5*e^5*(d + e*x)^5) + (b^2* (4*b*B*d - A*b*e - 3*a*B*e))/(4*e^5*(d + e*x)^4) - (b^3*B)/(3*e^5*(d + e*x )^3)))/(a + b*x)
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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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]
Time = 2.62 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.29
| method | result | size |
| risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B \,b^{3} x^{4}}{3 e}-\frac {b^{2} \left (3 A b e +9 B a e +4 B b d \right ) x^{3}}{12 e^{2}}-\frac {b \left (12 A a b \,e^{2}+3 A \,b^{2} d e +12 B \,e^{2} a^{2}+9 B a b d e +4 B \,b^{2} d^{2}\right ) x^{2}}{20 e^{3}}-\frac {\left (30 A \,a^{2} b \,e^{3}+12 A a \,b^{2} d \,e^{2}+3 A \,b^{3} d^{2} e +10 B \,e^{3} a^{3}+12 B \,a^{2} b d \,e^{2}+9 B a \,b^{2} d^{2} e +4 B \,b^{3} d^{3}\right ) x}{60 e^{4}}-\frac {60 A \,a^{3} e^{4}+30 A \,a^{2} b d \,e^{3}+12 A a \,b^{2} d^{2} e^{2}+3 A \,b^{3} d^{3} e +10 B \,a^{3} d \,e^{3}+12 B \,a^{2} b \,d^{2} e^{2}+9 B a \,b^{2} d^{3} e +4 B \,b^{3} d^{4}}{420 e^{5}}\right )}{\left (b x +a \right ) \left (e x +d \right )^{7}}\) | \(286\) |
| gosper | \(-\frac {\left (140 B \,b^{3} e^{4} x^{4}+105 A \,b^{3} e^{4} x^{3}+315 B a \,b^{2} e^{4} x^{3}+140 B \,b^{3} d \,e^{3} x^{3}+252 A a \,b^{2} e^{4} x^{2}+63 A \,b^{3} d \,e^{3} x^{2}+252 B \,a^{2} b \,e^{4} x^{2}+189 B a \,b^{2} d \,e^{3} x^{2}+84 B \,b^{3} d^{2} e^{2} x^{2}+210 A \,a^{2} b \,e^{4} x +84 A a \,b^{2} d \,e^{3} x +21 A \,b^{3} d^{2} e^{2} x +70 B \,a^{3} e^{4} x +84 B \,a^{2} b d \,e^{3} x +63 B a \,b^{2} d^{2} e^{2} x +28 B \,b^{3} d^{3} e x +60 A \,a^{3} e^{4}+30 A \,a^{2} b d \,e^{3}+12 A a \,b^{2} d^{2} e^{2}+3 A \,b^{3} d^{3} e +10 B \,a^{3} d \,e^{3}+12 B \,a^{2} b \,d^{2} e^{2}+9 B a \,b^{2} d^{3} e +4 B \,b^{3} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{420 e^{5} \left (e x +d \right )^{7} \left (b x +a \right )^{3}}\) | \(317\) |
| default | \(-\frac {\left (140 B \,b^{3} e^{4} x^{4}+105 A \,b^{3} e^{4} x^{3}+315 B a \,b^{2} e^{4} x^{3}+140 B \,b^{3} d \,e^{3} x^{3}+252 A a \,b^{2} e^{4} x^{2}+63 A \,b^{3} d \,e^{3} x^{2}+252 B \,a^{2} b \,e^{4} x^{2}+189 B a \,b^{2} d \,e^{3} x^{2}+84 B \,b^{3} d^{2} e^{2} x^{2}+210 A \,a^{2} b \,e^{4} x +84 A a \,b^{2} d \,e^{3} x +21 A \,b^{3} d^{2} e^{2} x +70 B \,a^{3} e^{4} x +84 B \,a^{2} b d \,e^{3} x +63 B a \,b^{2} d^{2} e^{2} x +28 B \,b^{3} d^{3} e x +60 A \,a^{3} e^{4}+30 A \,a^{2} b d \,e^{3}+12 A a \,b^{2} d^{2} e^{2}+3 A \,b^{3} d^{3} e +10 B \,a^{3} d \,e^{3}+12 B \,a^{2} b \,d^{2} e^{2}+9 B a \,b^{2} d^{3} e +4 B \,b^{3} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{420 e^{5} \left (e x +d \right )^{7} \left (b x +a \right )^{3}}\) | \(317\) |
| orering | \(-\frac {\left (140 B \,b^{3} e^{4} x^{4}+105 A \,b^{3} e^{4} x^{3}+315 B a \,b^{2} e^{4} x^{3}+140 B \,b^{3} d \,e^{3} x^{3}+252 A a \,b^{2} e^{4} x^{2}+63 A \,b^{3} d \,e^{3} x^{2}+252 B \,a^{2} b \,e^{4} x^{2}+189 B a \,b^{2} d \,e^{3} x^{2}+84 B \,b^{3} d^{2} e^{2} x^{2}+210 A \,a^{2} b \,e^{4} x +84 A a \,b^{2} d \,e^{3} x +21 A \,b^{3} d^{2} e^{2} x +70 B \,a^{3} e^{4} x +84 B \,a^{2} b d \,e^{3} x +63 B a \,b^{2} d^{2} e^{2} x +28 B \,b^{3} d^{3} e x +60 A \,a^{3} e^{4}+30 A \,a^{2} b d \,e^{3}+12 A a \,b^{2} d^{2} e^{2}+3 A \,b^{3} d^{3} e +10 B \,a^{3} d \,e^{3}+12 B \,a^{2} b \,d^{2} e^{2}+9 B a \,b^{2} d^{3} e +4 B \,b^{3} d^{4}\right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {3}{2}}}{420 e^{5} \left (b x +a \right )^{3} \left (e x +d \right )^{7}}\) | \(326\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^8,x,method=_RETURNVERBOSE)
Output:
((b*x+a)^2)^(1/2)/(b*x+a)*(-1/3*B*b^3/e*x^4-1/12*b^2/e^2*(3*A*b*e+9*B*a*e+ 4*B*b*d)*x^3-1/20*b/e^3*(12*A*a*b*e^2+3*A*b^2*d*e+12*B*a^2*e^2+9*B*a*b*d*e +4*B*b^2*d^2)*x^2-1/60/e^4*(30*A*a^2*b*e^3+12*A*a*b^2*d*e^2+3*A*b^3*d^2*e+ 10*B*a^3*e^3+12*B*a^2*b*d*e^2+9*B*a*b^2*d^2*e+4*B*b^3*d^3)*x-1/420/e^5*(60 *A*a^3*e^4+30*A*a^2*b*d*e^3+12*A*a*b^2*d^2*e^2+3*A*b^3*d^3*e+10*B*a^3*d*e^ 3+12*B*a^2*b*d^2*e^2+9*B*a*b^2*d^3*e+4*B*b^3*d^4))/(e*x+d)^7
Time = 0.08 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.50 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^8} \, dx=-\frac {140 \, B b^{3} e^{4} x^{4} + 4 \, B b^{3} d^{4} + 60 \, A a^{3} e^{4} + 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 12 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} + 10 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} + 35 \, {\left (4 \, B b^{3} d e^{3} + 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 21 \, {\left (4 \, B b^{3} d^{2} e^{2} + 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 12 \, {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 7 \, {\left (4 \, B b^{3} d^{3} e + 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 12 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} + 10 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x}{420 \, {\left (e^{12} x^{7} + 7 \, d e^{11} x^{6} + 21 \, d^{2} e^{10} x^{5} + 35 \, d^{3} e^{9} x^{4} + 35 \, d^{4} e^{8} x^{3} + 21 \, d^{5} e^{7} x^{2} + 7 \, d^{6} e^{6} x + d^{7} e^{5}\right )}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^8,x, algorithm="fric as")
Output:
-1/420*(140*B*b^3*e^4*x^4 + 4*B*b^3*d^4 + 60*A*a^3*e^4 + 3*(3*B*a*b^2 + A* b^3)*d^3*e + 12*(B*a^2*b + A*a*b^2)*d^2*e^2 + 10*(B*a^3 + 3*A*a^2*b)*d*e^3 + 35*(4*B*b^3*d*e^3 + 3*(3*B*a*b^2 + A*b^3)*e^4)*x^3 + 21*(4*B*b^3*d^2*e^ 2 + 3*(3*B*a*b^2 + A*b^3)*d*e^3 + 12*(B*a^2*b + A*a*b^2)*e^4)*x^2 + 7*(4*B *b^3*d^3*e + 3*(3*B*a*b^2 + A*b^3)*d^2*e^2 + 12*(B*a^2*b + A*a*b^2)*d*e^3 + 10*(B*a^3 + 3*A*a^2*b)*e^4)*x)/(e^12*x^7 + 7*d*e^11*x^6 + 21*d^2*e^10*x^ 5 + 35*d^3*e^9*x^4 + 35*d^4*e^8*x^3 + 21*d^5*e^7*x^2 + 7*d^6*e^6*x + d^7*e ^5)
\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^8} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{8}}\, dx \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**8,x)
Output:
Integral((A + B*x)*((a + b*x)**2)**(3/2)/(d + e*x)**8, x)
Exception generated. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^8} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^8,x, algorithm="maxi ma")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 532 vs. \(2 (206) = 412\).
Time = 0.18 (sec) , antiderivative size = 532, normalized size of antiderivative = 2.40 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^8} \, dx=\frac {{\left (4 \, B b^{7} d - 7 \, B a b^{6} e + 3 \, A b^{7} e\right )} \mathrm {sgn}\left (b x + a\right )}{420 \, {\left (b^{4} d^{4} e^{5} - 4 \, a b^{3} d^{3} e^{6} + 6 \, a^{2} b^{2} d^{2} e^{7} - 4 \, a^{3} b d e^{8} + a^{4} e^{9}\right )}} - \frac {140 \, B b^{3} e^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + 140 \, B b^{3} d e^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 315 \, B a b^{2} e^{4} x^{3} \mathrm {sgn}\left (b x + a\right ) + 105 \, A b^{3} e^{4} x^{3} \mathrm {sgn}\left (b x + a\right ) + 84 \, B b^{3} d^{2} e^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 189 \, B a b^{2} d e^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 63 \, A b^{3} d e^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 252 \, B a^{2} b e^{4} x^{2} \mathrm {sgn}\left (b x + a\right ) + 252 \, A a b^{2} e^{4} x^{2} \mathrm {sgn}\left (b x + a\right ) + 28 \, B b^{3} d^{3} e x \mathrm {sgn}\left (b x + a\right ) + 63 \, B a b^{2} d^{2} e^{2} x \mathrm {sgn}\left (b x + a\right ) + 21 \, A b^{3} d^{2} e^{2} x \mathrm {sgn}\left (b x + a\right ) + 84 \, B a^{2} b d e^{3} x \mathrm {sgn}\left (b x + a\right ) + 84 \, A a b^{2} d e^{3} x \mathrm {sgn}\left (b x + a\right ) + 70 \, B a^{3} e^{4} x \mathrm {sgn}\left (b x + a\right ) + 210 \, A a^{2} b e^{4} x \mathrm {sgn}\left (b x + a\right ) + 4 \, B b^{3} d^{4} \mathrm {sgn}\left (b x + a\right ) + 9 \, B a b^{2} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 3 \, A b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 12 \, B a^{2} b d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 12 \, A a b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 10 \, B a^{3} d e^{3} \mathrm {sgn}\left (b x + a\right ) + 30 \, A a^{2} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + 60 \, A a^{3} e^{4} \mathrm {sgn}\left (b x + a\right )}{420 \, {\left (e x + d\right )}^{7} e^{5}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^8,x, algorithm="giac ")
Output:
1/420*(4*B*b^7*d - 7*B*a*b^6*e + 3*A*b^7*e)*sgn(b*x + a)/(b^4*d^4*e^5 - 4* a*b^3*d^3*e^6 + 6*a^2*b^2*d^2*e^7 - 4*a^3*b*d*e^8 + a^4*e^9) - 1/420*(140* B*b^3*e^4*x^4*sgn(b*x + a) + 140*B*b^3*d*e^3*x^3*sgn(b*x + a) + 315*B*a*b^ 2*e^4*x^3*sgn(b*x + a) + 105*A*b^3*e^4*x^3*sgn(b*x + a) + 84*B*b^3*d^2*e^2 *x^2*sgn(b*x + a) + 189*B*a*b^2*d*e^3*x^2*sgn(b*x + a) + 63*A*b^3*d*e^3*x^ 2*sgn(b*x + a) + 252*B*a^2*b*e^4*x^2*sgn(b*x + a) + 252*A*a*b^2*e^4*x^2*sg n(b*x + a) + 28*B*b^3*d^3*e*x*sgn(b*x + a) + 63*B*a*b^2*d^2*e^2*x*sgn(b*x + a) + 21*A*b^3*d^2*e^2*x*sgn(b*x + a) + 84*B*a^2*b*d*e^3*x*sgn(b*x + a) + 84*A*a*b^2*d*e^3*x*sgn(b*x + a) + 70*B*a^3*e^4*x*sgn(b*x + a) + 210*A*a^2 *b*e^4*x*sgn(b*x + a) + 4*B*b^3*d^4*sgn(b*x + a) + 9*B*a*b^2*d^3*e*sgn(b*x + a) + 3*A*b^3*d^3*e*sgn(b*x + a) + 12*B*a^2*b*d^2*e^2*sgn(b*x + a) + 12* A*a*b^2*d^2*e^2*sgn(b*x + a) + 10*B*a^3*d*e^3*sgn(b*x + a) + 30*A*a^2*b*d* e^3*sgn(b*x + a) + 60*A*a^3*e^4*sgn(b*x + a))/((e*x + d)^7*e^5)
Time = 11.21 (sec) , antiderivative size = 577, normalized size of antiderivative = 2.60 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^8} \, dx=-\frac {\left (\frac {A\,b^3\,e-3\,B\,b^3\,d+3\,B\,a\,b^2\,e}{4\,e^5}-\frac {B\,b^3\,d}{4\,e^5}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4}-\frac {\left (\frac {A\,a^3}{7\,e}-\frac {d\,\left (\frac {B\,a^3+3\,A\,b\,a^2}{7\,e}+\frac {d\,\left (\frac {d\,\left (\frac {A\,b^3+3\,B\,a\,b^2}{7\,e}-\frac {B\,b^3\,d}{7\,e^2}\right )}{e}-\frac {3\,a\,b\,\left (A\,b+B\,a\right )}{7\,e}\right )}{e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^7}-\frac {\left (\frac {B\,a^3\,e^3-3\,B\,a^2\,b\,d\,e^2+3\,A\,a^2\,b\,e^3+3\,B\,a\,b^2\,d^2\,e-3\,A\,a\,b^2\,d\,e^2-B\,b^3\,d^3+A\,b^3\,d^2\,e}{6\,e^5}-\frac {d\,\left (\frac {3\,B\,a^2\,b\,e^3-3\,B\,a\,b^2\,d\,e^2+3\,A\,a\,b^2\,e^3+B\,b^3\,d^2\,e-A\,b^3\,d\,e^2}{6\,e^5}-\frac {d\,\left (\frac {b^2\,\left (A\,b\,e+3\,B\,a\,e-B\,b\,d\right )}{6\,e^3}-\frac {B\,b^3\,d}{6\,e^3}\right )}{e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^6}-\frac {\left (\frac {3\,B\,a^2\,b\,e^2-6\,B\,a\,b^2\,d\,e+3\,A\,a\,b^2\,e^2+3\,B\,b^3\,d^2-2\,A\,b^3\,d\,e}{5\,e^5}-\frac {d\,\left (\frac {b^2\,\left (A\,b\,e+3\,B\,a\,e-2\,B\,b\,d\right )}{5\,e^4}-\frac {B\,b^3\,d}{5\,e^4}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5}-\frac {B\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{3\,e^5\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^3} \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(d + e*x)^8,x)
Output:
- (((A*b^3*e - 3*B*b^3*d + 3*B*a*b^2*e)/(4*e^5) - (B*b^3*d)/(4*e^5))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^4) - (((A*a^3)/(7*e) - (d *((B*a^3 + 3*A*a^2*b)/(7*e) + (d*((d*((A*b^3 + 3*B*a*b^2)/(7*e) - (B*b^3*d )/(7*e^2)))/e - (3*a*b*(A*b + B*a))/(7*e)))/e))/e)*(a^2 + b^2*x^2 + 2*a*b* x)^(1/2))/((a + b*x)*(d + e*x)^7) - (((B*a^3*e^3 - B*b^3*d^3 + 3*A*a^2*b*e ^3 + A*b^3*d^2*e - 3*A*a*b^2*d*e^2 + 3*B*a*b^2*d^2*e - 3*B*a^2*b*d*e^2)/(6 *e^5) - (d*((3*A*a*b^2*e^3 + 3*B*a^2*b*e^3 - A*b^3*d*e^2 + B*b^3*d^2*e - 3 *B*a*b^2*d*e^2)/(6*e^5) - (d*((b^2*(A*b*e + 3*B*a*e - B*b*d))/(6*e^3) - (B *b^3*d)/(6*e^3)))/e))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^6) - (((3*B*b^3*d^2 - 2*A*b^3*d*e + 3*A*a*b^2*e^2 + 3*B*a^2*b*e^2 - 6 *B*a*b^2*d*e)/(5*e^5) - (d*((b^2*(A*b*e + 3*B*a*e - 2*B*b*d))/(5*e^4) - (B *b^3*d)/(5*e^4)))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x) ^5) - (B*b^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(3*e^5*(a + b*x)*(d + e*x)^3 )
Time = 0.20 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.13 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^8} \, dx=\frac {-35 b^{4} e^{4} x^{4}-105 a \,b^{3} e^{4} x^{3}-35 b^{4} d \,e^{3} x^{3}-126 a^{2} b^{2} e^{4} x^{2}-63 a \,b^{3} d \,e^{3} x^{2}-21 b^{4} d^{2} e^{2} x^{2}-70 a^{3} b \,e^{4} x -42 a^{2} b^{2} d \,e^{3} x -21 a \,b^{3} d^{2} e^{2} x -7 b^{4} d^{3} e x -15 a^{4} e^{4}-10 a^{3} b d \,e^{3}-6 a^{2} b^{2} d^{2} e^{2}-3 a \,b^{3} d^{3} e -b^{4} d^{4}}{105 e^{5} \left (e^{7} x^{7}+7 d \,e^{6} x^{6}+21 d^{2} e^{5} x^{5}+35 d^{3} e^{4} x^{4}+35 d^{4} e^{3} x^{3}+21 d^{5} e^{2} x^{2}+7 d^{6} e x +d^{7}\right )} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^8,x)
Output:
( - 15*a**4*e**4 - 10*a**3*b*d*e**3 - 70*a**3*b*e**4*x - 6*a**2*b**2*d**2* e**2 - 42*a**2*b**2*d*e**3*x - 126*a**2*b**2*e**4*x**2 - 3*a*b**3*d**3*e - 21*a*b**3*d**2*e**2*x - 63*a*b**3*d*e**3*x**2 - 105*a*b**3*e**4*x**3 - b* *4*d**4 - 7*b**4*d**3*e*x - 21*b**4*d**2*e**2*x**2 - 35*b**4*d*e**3*x**3 - 35*b**4*e**4*x**4)/(105*e**5*(d**7 + 7*d**6*e*x + 21*d**5*e**2*x**2 + 35* d**4*e**3*x**3 + 35*d**3*e**4*x**4 + 21*d**2*e**5*x**5 + 7*d*e**6*x**6 + e **7*x**7))