Integrand size = 33, antiderivative size = 424 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^3} \, dx=-\frac {10 b^2 (b d-a e)^2 (2 b B d-A b e-a B e) x \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x)}-\frac {(b d-a e)^5 (B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^7 (a+b x) (d+e x)^2}+\frac {(b d-a e)^4 (6 b B d-5 A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^7 (a+b x) (d+e x)}+\frac {5 b^3 (b d-a e) (3 b B d-A b e-2 a B e) (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^7 (a+b x)}-\frac {b^4 (6 b B d-A b e-5 a B e) (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^7 (a+b x)}+\frac {b^5 B (d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^7 (a+b x)}+\frac {5 b (b d-a e)^3 (3 b B d-2 A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^7 (a+b x)} \] Output:
-10*b^2*(-a*e+b*d)^2*(-A*b*e-B*a*e+2*B*b*d)*x*((b*x+a)^2)^(1/2)/e^6/(b*x+a )-1/2*(-a*e+b*d)^5*(-A*e+B*d)*((b*x+a)^2)^(1/2)/e^7/(b*x+a)/(e*x+d)^2+(-a* e+b*d)^4*(-5*A*b*e-B*a*e+6*B*b*d)*((b*x+a)^2)^(1/2)/e^7/(b*x+a)/(e*x+d)+5/ 2*b^3*(-a*e+b*d)*(-A*b*e-2*B*a*e+3*B*b*d)*(e*x+d)^2*((b*x+a)^2)^(1/2)/e^7/ (b*x+a)-1/3*b^4*(-A*b*e-5*B*a*e+6*B*b*d)*(e*x+d)^3*((b*x+a)^2)^(1/2)/e^7/( b*x+a)+1/4*b^5*B*(e*x+d)^4*((b*x+a)^2)^(1/2)/e^7/(b*x+a)+5*b*(-a*e+b*d)^3* (-2*A*b*e-B*a*e+3*B*b*d)*((b*x+a)^2)^(1/2)*ln(e*x+d)/e^7/(b*x+a)
Time = 1.31 (sec) , antiderivative size = 501, normalized size of antiderivative = 1.18 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^3} \, dx=\frac {\sqrt {(a+b x)^2} \left (-6 a^5 e^5 (A e+B (d+2 e x))-30 a^4 b e^4 (A e (d+2 e x)-B d (3 d+4 e x))+60 a^3 b^2 e^3 \left (A d e (3 d+4 e x)+B \left (-5 d^3-4 d^2 e x+4 d e^2 x^2+2 e^3 x^3\right )\right )+60 a^2 b^3 e^2 \left (A e \left (-5 d^3-4 d^2 e x+4 d e^2 x^2+2 e^3 x^3\right )+B \left (7 d^4+2 d^3 e x-11 d^2 e^2 x^2-4 d e^3 x^3+e^4 x^4\right )\right )+10 a b^4 e \left (3 A e \left (7 d^4+2 d^3 e x-11 d^2 e^2 x^2-4 d e^3 x^3+e^4 x^4\right )+B \left (-27 d^5+6 d^4 e x+63 d^3 e^2 x^2+20 d^2 e^3 x^3-5 d e^4 x^4+2 e^5 x^5\right )\right )+b^5 \left (2 A e \left (-27 d^5+6 d^4 e x+63 d^3 e^2 x^2+20 d^2 e^3 x^3-5 d e^4 x^4+2 e^5 x^5\right )+3 B \left (22 d^6-16 d^5 e x-68 d^4 e^2 x^2-20 d^3 e^3 x^3+5 d^2 e^4 x^4-2 d e^5 x^5+e^6 x^6\right )\right )+60 b (b d-a e)^3 (3 b B d-2 A b e-a B e) (d+e x)^2 \log (d+e x)\right )}{12 e^7 (a+b x) (d+e x)^2} \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/(d + e*x)^3,x]
Output:
(Sqrt[(a + b*x)^2]*(-6*a^5*e^5*(A*e + B*(d + 2*e*x)) - 30*a^4*b*e^4*(A*e*( d + 2*e*x) - B*d*(3*d + 4*e*x)) + 60*a^3*b^2*e^3*(A*d*e*(3*d + 4*e*x) + B* (-5*d^3 - 4*d^2*e*x + 4*d*e^2*x^2 + 2*e^3*x^3)) + 60*a^2*b^3*e^2*(A*e*(-5* d^3 - 4*d^2*e*x + 4*d*e^2*x^2 + 2*e^3*x^3) + B*(7*d^4 + 2*d^3*e*x - 11*d^2 *e^2*x^2 - 4*d*e^3*x^3 + e^4*x^4)) + 10*a*b^4*e*(3*A*e*(7*d^4 + 2*d^3*e*x - 11*d^2*e^2*x^2 - 4*d*e^3*x^3 + e^4*x^4) + B*(-27*d^5 + 6*d^4*e*x + 63*d^ 3*e^2*x^2 + 20*d^2*e^3*x^3 - 5*d*e^4*x^4 + 2*e^5*x^5)) + b^5*(2*A*e*(-27*d ^5 + 6*d^4*e*x + 63*d^3*e^2*x^2 + 20*d^2*e^3*x^3 - 5*d*e^4*x^4 + 2*e^5*x^5 ) + 3*B*(22*d^6 - 16*d^5*e*x - 68*d^4*e^2*x^2 - 20*d^3*e^3*x^3 + 5*d^2*e^4 *x^4 - 2*d*e^5*x^5 + e^6*x^6)) + 60*b*(b*d - a*e)^3*(3*b*B*d - 2*A*b*e - a *B*e)*(d + e*x)^2*Log[d + e*x]))/(12*e^7*(a + b*x)*(d + e*x)^2)
Time = 1.03 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.62, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2} (A+B x)}{(d+e x)^3} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^5 (a+b x)^5 (A+B x)}{(d+e x)^3}dx}{b^5 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^5 (A+B x)}{(d+e x)^3}dx}{a+b x}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {B (d+e x)^3 b^5}{e^6}+\frac {(-6 b B d+A b e+5 a B e) (d+e x)^2 b^4}{e^6}-\frac {5 (b d-a e) (-3 b B d+A b e+2 a B e) (d+e x) b^3}{e^6}+\frac {10 (b d-a e)^2 (-2 b B d+A b e+a B e) b^2}{e^6}-\frac {5 (b d-a e)^3 (-3 b B d+2 A b e+a B e) b}{e^6 (d+e x)}+\frac {(a e-b d)^4 (-6 b B d+5 A b e+a B e)}{e^6 (d+e x)^2}+\frac {(a e-b d)^5 (A e-B d)}{e^6 (d+e x)^3}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {b^4 (d+e x)^3 (-5 a B e-A b e+6 b B d)}{3 e^7}+\frac {5 b^3 (d+e x)^2 (b d-a e) (-2 a B e-A b e+3 b B d)}{2 e^7}-\frac {10 b^2 x (b d-a e)^2 (-a B e-A b e+2 b B d)}{e^6}+\frac {(b d-a e)^4 (-a B e-5 A b e+6 b B d)}{e^7 (d+e x)}-\frac {(b d-a e)^5 (B d-A e)}{2 e^7 (d+e x)^2}+\frac {5 b (b d-a e)^3 \log (d+e x) (-a B e-2 A b e+3 b B d)}{e^7}+\frac {b^5 B (d+e x)^4}{4 e^7}\right )}{a+b x}\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/(d + e*x)^3,x]
Output:
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((-10*b^2*(b*d - a*e)^2*(2*b*B*d - A*b*e - a*B*e)*x)/e^6 - ((b*d - a*e)^5*(B*d - A*e))/(2*e^7*(d + e*x)^2) + ((b*d - a*e)^4*(6*b*B*d - 5*A*b*e - a*B*e))/(e^7*(d + e*x)) + (5*b^3*(b*d - a*e)*( 3*b*B*d - A*b*e - 2*a*B*e)*(d + e*x)^2)/(2*e^7) - (b^4*(6*b*B*d - A*b*e - 5*a*B*e)*(d + e*x)^3)/(3*e^7) + (b^5*B*(d + e*x)^4)/(4*e^7) + (5*b*(b*d - a*e)^3*(3*b*B*d - 2*A*b*e - a*B*e)*Log[d + e*x])/e^7))/(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 = 1.66 (sec) , antiderivative size = 661, normalized size of antiderivative = 1.56
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{2} \left (\frac {1}{4} b^{3} B \,x^{4} e^{3}+\frac {1}{3} A \,b^{3} e^{3} x^{3}+\frac {5}{3} B a \,b^{2} e^{3} x^{3}-B \,b^{3} d \,e^{2} x^{3}+\frac {5}{2} A a \,b^{2} e^{3} x^{2}-\frac {3}{2} A \,b^{3} d \,e^{2} x^{2}+5 B \,a^{2} b \,e^{3} x^{2}-\frac {15}{2} B a \,b^{2} d \,e^{2} x^{2}+3 B \,b^{3} d^{2} e \,x^{2}+10 A \,a^{2} b \,e^{3} x -15 A a \,b^{2} d \,e^{2} x +6 A \,b^{3} d^{2} e x +10 B \,e^{3} a^{3} x -30 B \,a^{2} b d \,e^{2} x +30 B a \,b^{2} d^{2} e x -10 B \,b^{3} d^{3} x \right )}{\left (b x +a \right ) e^{6}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (-5 A \,a^{4} b \,e^{5}+20 A \,a^{3} b^{2} d \,e^{4}-30 A \,a^{2} b^{3} d^{2} e^{3}+20 A a \,b^{4} d^{3} e^{2}-5 A \,b^{5} d^{4} e -B \,a^{5} e^{5}+10 B \,a^{4} b d \,e^{4}-30 B \,a^{3} b^{2} d^{2} e^{3}+40 B \,a^{2} b^{3} d^{3} e^{2}-25 B a \,b^{4} d^{4} e +6 B \,b^{5} d^{5}\right ) x -\frac {A \,a^{5} e^{6}+5 A \,a^{4} b d \,e^{5}-30 A \,a^{3} b^{2} d^{2} e^{4}+50 A \,a^{2} b^{3} d^{3} e^{3}-35 A a \,b^{4} d^{4} e^{2}+9 A \,b^{5} d^{5} e +B \,a^{5} d \,e^{5}-15 B \,a^{4} b \,d^{2} e^{4}+50 B \,a^{3} b^{2} d^{3} e^{3}-70 B \,a^{2} b^{3} d^{4} e^{2}+45 B a \,b^{4} d^{5} e -11 b^{5} B \,d^{6}}{2 e}\right )}{\left (b x +a \right ) e^{6} \left (e x +d \right )^{2}}+\frac {5 \sqrt {\left (b x +a \right )^{2}}\, b \left (2 A \,a^{3} b \,e^{4}-6 A \,a^{2} b^{2} d \,e^{3}+6 A a \,b^{3} d^{2} e^{2}-2 A \,b^{4} d^{3} e +B \,e^{4} a^{4}-6 B \,a^{3} b d \,e^{3}+12 B \,a^{2} b^{2} d^{2} e^{2}-10 B a \,b^{3} d^{3} e +3 B \,b^{4} d^{4}\right ) \ln \left (e x +d \right )}{\left (b x +a \right ) e^{7}}\) | \(661\) |
default | \(\text {Expression too large to display}\) | \(1205\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^3,x,method=_RETURNVERBOSE)
Output:
((b*x+a)^2)^(1/2)/(b*x+a)*b^2/e^6*(1/4*b^3*B*x^4*e^3+1/3*A*b^3*e^3*x^3+5/3 *B*a*b^2*e^3*x^3-B*b^3*d*e^2*x^3+5/2*A*a*b^2*e^3*x^2-3/2*A*b^3*d*e^2*x^2+5 *B*a^2*b*e^3*x^2-15/2*B*a*b^2*d*e^2*x^2+3*B*b^3*d^2*e*x^2+10*A*a^2*b*e^3*x -15*A*a*b^2*d*e^2*x+6*A*b^3*d^2*e*x+10*B*e^3*a^3*x-30*B*a^2*b*d*e^2*x+30*B *a*b^2*d^2*e*x-10*B*b^3*d^3*x)+((b*x+a)^2)^(1/2)/(b*x+a)*((-5*A*a^4*b*e^5+ 20*A*a^3*b^2*d*e^4-30*A*a^2*b^3*d^2*e^3+20*A*a*b^4*d^3*e^2-5*A*b^5*d^4*e-B *a^5*e^5+10*B*a^4*b*d*e^4-30*B*a^3*b^2*d^2*e^3+40*B*a^2*b^3*d^3*e^2-25*B*a *b^4*d^4*e+6*B*b^5*d^5)*x-1/2*(A*a^5*e^6+5*A*a^4*b*d*e^5-30*A*a^3*b^2*d^2* e^4+50*A*a^2*b^3*d^3*e^3-35*A*a*b^4*d^4*e^2+9*A*b^5*d^5*e+B*a^5*d*e^5-15*B *a^4*b*d^2*e^4+50*B*a^3*b^2*d^3*e^3-70*B*a^2*b^3*d^4*e^2+45*B*a*b^4*d^5*e- 11*B*b^5*d^6)/e)/e^6/(e*x+d)^2+5*((b*x+a)^2)^(1/2)/(b*x+a)*b/e^7*(2*A*a^3* b*e^4-6*A*a^2*b^2*d*e^3+6*A*a*b^3*d^2*e^2-2*A*b^4*d^3*e+B*a^4*e^4-6*B*a^3* b*d*e^3+12*B*a^2*b^2*d^2*e^2-10*B*a*b^3*d^3*e+3*B*b^4*d^4)*ln(e*x+d)
Leaf count of result is larger than twice the leaf count of optimal. 871 vs. \(2 (339) = 678\).
Time = 0.10 (sec) , antiderivative size = 871, normalized size of antiderivative = 2.05 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^3} \, dx =\text {Too large to display} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^3,x, algorithm="fric as")
Output:
1/12*(3*B*b^5*e^6*x^6 + 66*B*b^5*d^6 - 6*A*a^5*e^6 - 54*(5*B*a*b^4 + A*b^5 )*d^5*e + 210*(2*B*a^2*b^3 + A*a*b^4)*d^4*e^2 - 300*(B*a^3*b^2 + A*a^2*b^3 )*d^3*e^3 + 90*(B*a^4*b + 2*A*a^3*b^2)*d^2*e^4 - 6*(B*a^5 + 5*A*a^4*b)*d*e ^5 - 2*(3*B*b^5*d*e^5 - 2*(5*B*a*b^4 + A*b^5)*e^6)*x^5 + 5*(3*B*b^5*d^2*e^ 4 - 2*(5*B*a*b^4 + A*b^5)*d*e^5 + 6*(2*B*a^2*b^3 + A*a*b^4)*e^6)*x^4 - 20* (3*B*b^5*d^3*e^3 - 2*(5*B*a*b^4 + A*b^5)*d^2*e^4 + 6*(2*B*a^2*b^3 + A*a*b^ 4)*d*e^5 - 6*(B*a^3*b^2 + A*a^2*b^3)*e^6)*x^3 - 6*(34*B*b^5*d^4*e^2 - 21*( 5*B*a*b^4 + A*b^5)*d^3*e^3 + 55*(2*B*a^2*b^3 + A*a*b^4)*d^2*e^4 - 40*(B*a^ 3*b^2 + A*a^2*b^3)*d*e^5)*x^2 - 12*(4*B*b^5*d^5*e - (5*B*a*b^4 + A*b^5)*d^ 4*e^2 - 5*(2*B*a^2*b^3 + A*a*b^4)*d^3*e^3 + 20*(B*a^3*b^2 + A*a^2*b^3)*d^2 *e^4 - 10*(B*a^4*b + 2*A*a^3*b^2)*d*e^5 + (B*a^5 + 5*A*a^4*b)*e^6)*x + 60* (3*B*b^5*d^6 - 2*(5*B*a*b^4 + A*b^5)*d^5*e + 6*(2*B*a^2*b^3 + A*a*b^4)*d^4 *e^2 - 6*(B*a^3*b^2 + A*a^2*b^3)*d^3*e^3 + (B*a^4*b + 2*A*a^3*b^2)*d^2*e^4 + (3*B*b^5*d^4*e^2 - 2*(5*B*a*b^4 + A*b^5)*d^3*e^3 + 6*(2*B*a^2*b^3 + A*a *b^4)*d^2*e^4 - 6*(B*a^3*b^2 + A*a^2*b^3)*d*e^5 + (B*a^4*b + 2*A*a^3*b^2)* e^6)*x^2 + 2*(3*B*b^5*d^5*e - 2*(5*B*a*b^4 + A*b^5)*d^4*e^2 + 6*(2*B*a^2*b ^3 + A*a*b^4)*d^3*e^3 - 6*(B*a^3*b^2 + A*a^2*b^3)*d^2*e^4 + (B*a^4*b + 2*A *a^3*b^2)*d*e^5)*x)*log(e*x + d))/(e^9*x^2 + 2*d*e^8*x + d^2*e^7)
\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^3} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{\left (d + e x\right )^{3}}\, dx \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(5/2)/(e*x+d)**3,x)
Output:
Integral((A + B*x)*((a + b*x)**2)**(5/2)/(d + e*x)**3, x)
Exception generated. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^3,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. 924 vs. \(2 (339) = 678\).
Time = 0.19 (sec) , antiderivative size = 924, normalized size of antiderivative = 2.18 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^3} \, dx=\text {Too large to display} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^3,x, algorithm="giac ")
Output:
5*(3*B*b^5*d^4*sgn(b*x + a) - 10*B*a*b^4*d^3*e*sgn(b*x + a) - 2*A*b^5*d^3* e*sgn(b*x + a) + 12*B*a^2*b^3*d^2*e^2*sgn(b*x + a) + 6*A*a*b^4*d^2*e^2*sgn (b*x + a) - 6*B*a^3*b^2*d*e^3*sgn(b*x + a) - 6*A*a^2*b^3*d*e^3*sgn(b*x + a ) + B*a^4*b*e^4*sgn(b*x + a) + 2*A*a^3*b^2*e^4*sgn(b*x + a))*log(abs(e*x + d))/e^7 + 1/2*(11*B*b^5*d^6*sgn(b*x + a) - 45*B*a*b^4*d^5*e*sgn(b*x + a) - 9*A*b^5*d^5*e*sgn(b*x + a) + 70*B*a^2*b^3*d^4*e^2*sgn(b*x + a) + 35*A*a* b^4*d^4*e^2*sgn(b*x + a) - 50*B*a^3*b^2*d^3*e^3*sgn(b*x + a) - 50*A*a^2*b^ 3*d^3*e^3*sgn(b*x + a) + 15*B*a^4*b*d^2*e^4*sgn(b*x + a) + 30*A*a^3*b^2*d^ 2*e^4*sgn(b*x + a) - B*a^5*d*e^5*sgn(b*x + a) - 5*A*a^4*b*d*e^5*sgn(b*x + a) - A*a^5*e^6*sgn(b*x + a) + 2*(6*B*b^5*d^5*e*sgn(b*x + a) - 25*B*a*b^4*d ^4*e^2*sgn(b*x + a) - 5*A*b^5*d^4*e^2*sgn(b*x + a) + 40*B*a^2*b^3*d^3*e^3* sgn(b*x + a) + 20*A*a*b^4*d^3*e^3*sgn(b*x + a) - 30*B*a^3*b^2*d^2*e^4*sgn( b*x + a) - 30*A*a^2*b^3*d^2*e^4*sgn(b*x + a) + 10*B*a^4*b*d*e^5*sgn(b*x + a) + 20*A*a^3*b^2*d*e^5*sgn(b*x + a) - B*a^5*e^6*sgn(b*x + a) - 5*A*a^4*b* e^6*sgn(b*x + a))*x)/((e*x + d)^2*e^7) + 1/12*(3*B*b^5*e^9*x^4*sgn(b*x + a ) - 12*B*b^5*d*e^8*x^3*sgn(b*x + a) + 20*B*a*b^4*e^9*x^3*sgn(b*x + a) + 4* A*b^5*e^9*x^3*sgn(b*x + a) + 36*B*b^5*d^2*e^7*x^2*sgn(b*x + a) - 90*B*a*b^ 4*d*e^8*x^2*sgn(b*x + a) - 18*A*b^5*d*e^8*x^2*sgn(b*x + a) + 60*B*a^2*b^3* e^9*x^2*sgn(b*x + a) + 30*A*a*b^4*e^9*x^2*sgn(b*x + a) - 120*B*b^5*d^3*e^6 *x*sgn(b*x + a) + 360*B*a*b^4*d^2*e^7*x*sgn(b*x + a) + 72*A*b^5*d^2*e^7...
Timed out. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^3} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}}{{\left (d+e\,x\right )}^3} \,d x \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2))/(d + e*x)^3,x)
Output:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2))/(d + e*x)^3, x)
Time = 0.26 (sec) , antiderivative size = 625, normalized size of antiderivative = 1.47 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^3} \, dx=\frac {-60 b^{6} d^{5} e^{2} x^{2}-20 b^{6} d^{4} e^{3} x^{3}+5 b^{6} d^{3} e^{4} x^{4}-2 b^{6} d^{2} e^{5} x^{5}+b^{6} d \,e^{6} x^{6}+60 \,\mathrm {log}\left (e x +d \right ) a^{4} b^{2} d \,e^{6} x^{2}-240 \,\mathrm {log}\left (e x +d \right ) a^{3} b^{3} d^{2} e^{5} x^{2}+360 \,\mathrm {log}\left (e x +d \right ) a^{2} b^{4} d^{3} e^{4} x^{2}-240 \,\mathrm {log}\left (e x +d \right ) a \,b^{5} d^{4} e^{3} x^{2}+120 \,\mathrm {log}\left (e x +d \right ) a^{4} b^{2} d^{2} e^{5} x -480 \,\mathrm {log}\left (e x +d \right ) a^{3} b^{3} d^{3} e^{4} x +720 \,\mathrm {log}\left (e x +d \right ) a^{2} b^{4} d^{4} e^{3} x -480 \,\mathrm {log}\left (e x +d \right ) a \,b^{5} d^{5} e^{2} x +80 a \,b^{5} d^{3} e^{4} x^{3}-20 a \,b^{5} d^{2} e^{5} x^{4}+8 a \,b^{5} d \,e^{6} x^{5}-2 a^{6} d \,e^{6}+60 \,\mathrm {log}\left (e x +d \right ) b^{6} d^{7}+60 \,\mathrm {log}\left (e x +d \right ) a^{4} b^{2} d^{3} e^{4}-240 \,\mathrm {log}\left (e x +d \right ) a^{3} b^{3} d^{4} e^{3}+360 \,\mathrm {log}\left (e x +d \right ) a^{2} b^{4} d^{5} e^{2}-240 \,\mathrm {log}\left (e x +d \right ) a \,b^{5} d^{6} e +120 \,\mathrm {log}\left (e x +d \right ) b^{6} d^{6} e x -60 a^{4} b^{2} d \,e^{6} x^{2}+240 a^{3} b^{3} d^{2} e^{5} x^{2}+80 a^{3} b^{3} d \,e^{6} x^{3}-360 a^{2} b^{4} d^{3} e^{4} x^{2}-120 a^{2} b^{4} d^{2} e^{5} x^{3}+30 a^{2} b^{4} d \,e^{6} x^{4}+240 a \,b^{5} d^{4} e^{3} x^{2}+12 a^{5} b \,e^{7} x^{2}+30 a^{4} b^{2} d^{3} e^{4}-120 a^{3} b^{3} d^{4} e^{3}+180 a^{2} b^{4} d^{5} e^{2}-120 a \,b^{5} d^{6} e +30 b^{6} d^{7}+60 \,\mathrm {log}\left (e x +d \right ) b^{6} d^{5} e^{2} x^{2}}{4 d \,e^{7} \left (e^{2} x^{2}+2 d e x +d^{2}\right )} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^3,x)
Output:
(60*log(d + e*x)*a**4*b**2*d**3*e**4 + 120*log(d + e*x)*a**4*b**2*d**2*e** 5*x + 60*log(d + e*x)*a**4*b**2*d*e**6*x**2 - 240*log(d + e*x)*a**3*b**3*d **4*e**3 - 480*log(d + e*x)*a**3*b**3*d**3*e**4*x - 240*log(d + e*x)*a**3* b**3*d**2*e**5*x**2 + 360*log(d + e*x)*a**2*b**4*d**5*e**2 + 720*log(d + e *x)*a**2*b**4*d**4*e**3*x + 360*log(d + e*x)*a**2*b**4*d**3*e**4*x**2 - 24 0*log(d + e*x)*a*b**5*d**6*e - 480*log(d + e*x)*a*b**5*d**5*e**2*x - 240*l og(d + e*x)*a*b**5*d**4*e**3*x**2 + 60*log(d + e*x)*b**6*d**7 + 120*log(d + e*x)*b**6*d**6*e*x + 60*log(d + e*x)*b**6*d**5*e**2*x**2 - 2*a**6*d*e**6 + 12*a**5*b*e**7*x**2 + 30*a**4*b**2*d**3*e**4 - 60*a**4*b**2*d*e**6*x**2 - 120*a**3*b**3*d**4*e**3 + 240*a**3*b**3*d**2*e**5*x**2 + 80*a**3*b**3*d *e**6*x**3 + 180*a**2*b**4*d**5*e**2 - 360*a**2*b**4*d**3*e**4*x**2 - 120* a**2*b**4*d**2*e**5*x**3 + 30*a**2*b**4*d*e**6*x**4 - 120*a*b**5*d**6*e + 240*a*b**5*d**4*e**3*x**2 + 80*a*b**5*d**3*e**4*x**3 - 20*a*b**5*d**2*e**5 *x**4 + 8*a*b**5*d*e**6*x**5 + 30*b**6*d**7 - 60*b**6*d**5*e**2*x**2 - 20* b**6*d**4*e**3*x**3 + 5*b**6*d**3*e**4*x**4 - 2*b**6*d**2*e**5*x**5 + b**6 *d*e**6*x**6)/(4*d*e**7*(d**2 + 2*d*e*x + e**2*x**2))