Integrand size = 28, antiderivative size = 253 \[ \int \frac {(d+e x)^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {10 e^3 (b d-a e)^2}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(b d-a e)^5}{4 b^6 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (b d-a e)^4}{3 b^6 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e^2 (b d-a e)^3}{b^6 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^5 x (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e^4 (b d-a e) (a+b x) \log (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:
-10*e^3*(-a*e+b*d)^2/b^6/((b*x+a)^2)^(1/2)-1/4*(-a*e+b*d)^5/b^6/(b*x+a)^3/ ((b*x+a)^2)^(1/2)-5/3*e*(-a*e+b*d)^4/b^6/(b*x+a)^2/((b*x+a)^2)^(1/2)-5*e^2 *(-a*e+b*d)^3/b^6/(b*x+a)/((b*x+a)^2)^(1/2)+e^5*x*(b*x+a)/b^5/((b*x+a)^2)^ (1/2)+5*e^4*(-a*e+b*d)*(b*x+a)*ln(b*x+a)/b^6/((b*x+a)^2)^(1/2)
Time = 1.10 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.96 \[ \int \frac {(d+e x)^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {-77 a^5 e^5+a^4 b e^4 (125 d-248 e x)-2 a^3 b^2 e^3 \left (15 d^2-220 d e x+126 e^2 x^2\right )-2 a^2 b^3 e^2 \left (5 d^3+60 d^2 e x-270 d e^2 x^2+24 e^3 x^3\right )+a b^4 e \left (-5 d^4-40 d^3 e x-180 d^2 e^2 x^2+240 d e^3 x^3+48 e^4 x^4\right )-b^5 \left (3 d^5+20 d^4 e x+60 d^3 e^2 x^2+120 d^2 e^3 x^3-12 e^5 x^5\right )-60 e^4 (-b d+a e) (a+b x)^4 \log (a+b x)}{12 b^6 (a+b x)^3 \sqrt {(a+b x)^2}} \] Input:
Integrate[(d + e*x)^5/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
Output:
(-77*a^5*e^5 + a^4*b*e^4*(125*d - 248*e*x) - 2*a^3*b^2*e^3*(15*d^2 - 220*d *e*x + 126*e^2*x^2) - 2*a^2*b^3*e^2*(5*d^3 + 60*d^2*e*x - 270*d*e^2*x^2 + 24*e^3*x^3) + a*b^4*e*(-5*d^4 - 40*d^3*e*x - 180*d^2*e^2*x^2 + 240*d*e^3*x ^3 + 48*e^4*x^4) - b^5*(3*d^5 + 20*d^4*e*x + 60*d^3*e^2*x^2 + 120*d^2*e^3* x^3 - 12*e^5*x^5) - 60*e^4*(-(b*d) + a*e)*(a + b*x)^4*Log[a + b*x])/(12*b^ 6*(a + b*x)^3*Sqrt[(a + b*x)^2])
Time = 0.62 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.62, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1102, 27, 49, 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.
| \(\displaystyle \int \frac {(d+e x)^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1102 |
| \(\displaystyle \frac {b^5 (a+b x) \int \frac {(d+e x)^5}{b^5 (a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {(d+e x)^5}{(a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {(a+b x) \int \left (\frac {e^5}{b^5}+\frac {5 (b d-a e) e^4}{b^5 (a+b x)}+\frac {10 (b d-a e)^2 e^3}{b^5 (a+b x)^2}+\frac {10 (b d-a e)^3 e^2}{b^5 (a+b x)^3}+\frac {5 (b d-a e)^4 e}{b^5 (a+b x)^4}+\frac {(b d-a e)^5}{b^5 (a+b x)^5}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(a+b x) \left (\frac {5 e^4 (b d-a e) \log (a+b x)}{b^6}-\frac {10 e^3 (b d-a e)^2}{b^6 (a+b x)}-\frac {5 e^2 (b d-a e)^3}{b^6 (a+b x)^2}-\frac {5 e (b d-a e)^4}{3 b^6 (a+b x)^3}-\frac {(b d-a e)^5}{4 b^6 (a+b x)^4}+\frac {e^5 x}{b^5}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
Input:
Int[(d + e*x)^5/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
Output:
((a + b*x)*((e^5*x)/b^5 - (b*d - a*e)^5/(4*b^6*(a + b*x)^4) - (5*e*(b*d - a*e)^4)/(3*b^6*(a + b*x)^3) - (5*e^2*(b*d - a*e)^3)/(b^6*(a + b*x)^2) - (1 0*e^3*(b*d - a*e)^2)/(b^6*(a + b*x)) + (5*e^4*(b*d - a*e)*Log[a + b*x])/b^ 6))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
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] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F racPart[p])) Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
Time = 1.84 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.15
| method | result | size |
| risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, e^{5} x}{\left (b x +a \right ) b^{5}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (-10 a^{2} b^{2} e^{5}+20 a \,b^{3} d \,e^{4}-10 d^{2} e^{3} b^{4}\right ) x^{3}-5 b \,e^{2} \left (5 e^{3} a^{3}-9 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) x^{2}-\frac {5 e \left (13 a^{4} e^{4}-22 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}+2 a \,b^{3} d^{3} e +b^{4} d^{4}\right ) x}{3}-\frac {77 e^{5} a^{5}-125 a^{4} b d \,e^{4}+30 a^{3} b^{2} d^{2} e^{3}+10 a^{2} b^{3} d^{3} e^{2}+5 a \,b^{4} d^{4} e +3 b^{5} d^{5}}{12 b}\right )}{\left (b x +a \right )^{5} b^{5}}-\frac {5 \sqrt {\left (b x +a \right )^{2}}\, e^{4} \left (a e -b d \right ) \ln \left (b x +a \right )}{\left (b x +a \right ) b^{6}}\) | \(291\) |
| default | \(-\frac {\left (-240 \ln \left (b x +a \right ) x \,a^{3} b^{2} d \,e^{4}-440 a^{3} b^{2} d \,e^{4} x +3 b^{5} d^{5}-240 \ln \left (b x +a \right ) x^{3} a \,b^{4} d \,e^{4}-60 \ln \left (b x +a \right ) b^{5} d \,e^{4} x^{4}+240 \ln \left (b x +a \right ) x^{3} a^{2} b^{3} e^{5}-125 a^{4} b d \,e^{4}+30 a^{3} b^{2} d^{2} e^{3}+248 a^{4} b \,e^{5} x +20 b^{5} d^{4} e x -360 \ln \left (b x +a \right ) x^{2} a^{2} b^{3} d \,e^{4}+10 a^{2} b^{3} d^{3} e^{2}+5 a \,b^{4} d^{4} e +360 \ln \left (b x +a \right ) x^{2} a^{3} b^{2} e^{5}+120 x \,a^{2} b^{3} d^{2} e^{3}+40 x a \,b^{4} d^{3} e^{2}+60 \ln \left (b x +a \right ) x^{4} a \,b^{4} e^{5}-540 x^{2} a^{2} b^{3} d \,e^{4}+180 x^{2} a \,b^{4} d^{2} e^{3}-240 x^{3} a \,b^{4} d \,e^{4}+240 \ln \left (b x +a \right ) x \,a^{4} b \,e^{5}-60 \ln \left (b x +a \right ) a^{4} b d \,e^{4}-48 x^{4} a \,b^{4} e^{5}+48 x^{3} a^{2} b^{3} e^{5}+120 x^{3} b^{5} d^{2} e^{3}+252 x^{2} a^{3} b^{2} e^{5}+60 x^{2} b^{5} d^{3} e^{2}+77 e^{5} a^{5}-12 x^{5} e^{5} b^{5}+60 \ln \left (b x +a \right ) a^{5} e^{5}\right ) \left (b x +a \right )}{12 b^{6} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(449\) |
Input:
int((e*x+d)^5/(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
((b*x+a)^2)^(1/2)/(b*x+a)*e^5/b^5*x+((b*x+a)^2)^(1/2)/(b*x+a)^5*((-10*a^2* b^2*e^5+20*a*b^3*d*e^4-10*b^4*d^2*e^3)*x^3-5*b*e^2*(5*a^3*e^3-9*a^2*b*d*e^ 2+3*a*b^2*d^2*e+b^3*d^3)*x^2-5/3*e*(13*a^4*e^4-22*a^3*b*d*e^3+6*a^2*b^2*d^ 2*e^2+2*a*b^3*d^3*e+b^4*d^4)*x-1/12*(77*a^5*e^5-125*a^4*b*d*e^4+30*a^3*b^2 *d^2*e^3+10*a^2*b^3*d^3*e^2+5*a*b^4*d^4*e+3*b^5*d^5)/b)/b^5-5*((b*x+a)^2)^ (1/2)/(b*x+a)/b^6*e^4*(a*e-b*d)*ln(b*x+a)
Leaf count of result is larger than twice the leaf count of optimal. 412 vs. \(2 (183) = 366\).
Time = 0.10 (sec) , antiderivative size = 412, normalized size of antiderivative = 1.63 \[ \int \frac {(d+e x)^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {12 \, b^{5} e^{5} x^{5} + 48 \, a b^{4} e^{5} x^{4} - 3 \, b^{5} d^{5} - 5 \, a b^{4} d^{4} e - 10 \, a^{2} b^{3} d^{3} e^{2} - 30 \, a^{3} b^{2} d^{2} e^{3} + 125 \, a^{4} b d e^{4} - 77 \, a^{5} e^{5} - 24 \, {\left (5 \, b^{5} d^{2} e^{3} - 10 \, a b^{4} d e^{4} + 2 \, a^{2} b^{3} e^{5}\right )} x^{3} - 12 \, {\left (5 \, b^{5} d^{3} e^{2} + 15 \, a b^{4} d^{2} e^{3} - 45 \, a^{2} b^{3} d e^{4} + 21 \, a^{3} b^{2} e^{5}\right )} x^{2} - 4 \, {\left (5 \, b^{5} d^{4} e + 10 \, a b^{4} d^{3} e^{2} + 30 \, a^{2} b^{3} d^{2} e^{3} - 110 \, a^{3} b^{2} d e^{4} + 62 \, a^{4} b e^{5}\right )} x + 60 \, {\left (a^{4} b d e^{4} - a^{5} e^{5} + {\left (b^{5} d e^{4} - a b^{4} e^{5}\right )} x^{4} + 4 \, {\left (a b^{4} d e^{4} - a^{2} b^{3} e^{5}\right )} x^{3} + 6 \, {\left (a^{2} b^{3} d e^{4} - a^{3} b^{2} e^{5}\right )} x^{2} + 4 \, {\left (a^{3} b^{2} d e^{4} - a^{4} b e^{5}\right )} x\right )} \log \left (b x + a\right )}{12 \, {\left (b^{10} x^{4} + 4 \, a b^{9} x^{3} + 6 \, a^{2} b^{8} x^{2} + 4 \, a^{3} b^{7} x + a^{4} b^{6}\right )}} \] Input:
integrate((e*x+d)^5/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")
Output:
1/12*(12*b^5*e^5*x^5 + 48*a*b^4*e^5*x^4 - 3*b^5*d^5 - 5*a*b^4*d^4*e - 10*a ^2*b^3*d^3*e^2 - 30*a^3*b^2*d^2*e^3 + 125*a^4*b*d*e^4 - 77*a^5*e^5 - 24*(5 *b^5*d^2*e^3 - 10*a*b^4*d*e^4 + 2*a^2*b^3*e^5)*x^3 - 12*(5*b^5*d^3*e^2 + 1 5*a*b^4*d^2*e^3 - 45*a^2*b^3*d*e^4 + 21*a^3*b^2*e^5)*x^2 - 4*(5*b^5*d^4*e + 10*a*b^4*d^3*e^2 + 30*a^2*b^3*d^2*e^3 - 110*a^3*b^2*d*e^4 + 62*a^4*b*e^5 )*x + 60*(a^4*b*d*e^4 - a^5*e^5 + (b^5*d*e^4 - a*b^4*e^5)*x^4 + 4*(a*b^4*d *e^4 - a^2*b^3*e^5)*x^3 + 6*(a^2*b^3*d*e^4 - a^3*b^2*e^5)*x^2 + 4*(a^3*b^2 *d*e^4 - a^4*b*e^5)*x)*log(b*x + a))/(b^10*x^4 + 4*a*b^9*x^3 + 6*a^2*b^8*x ^2 + 4*a^3*b^7*x + a^4*b^6)
\[ \int \frac {(d+e x)^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (d + e x\right )^{5}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((e*x+d)**5/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)
Output:
Integral((d + e*x)**5/((a + b*x)**2)**(5/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 443 vs. \(2 (183) = 366\).
Time = 0.07 (sec) , antiderivative size = 443, normalized size of antiderivative = 1.75 \[ \int \frac {(d+e x)^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {1}{12} \, e^{5} {\left (\frac {12 \, b^{5} x^{5} + 48 \, a b^{4} x^{4} - 48 \, a^{2} b^{3} x^{3} - 252 \, a^{3} b^{2} x^{2} - 248 \, a^{4} b x - 77 \, a^{5}}{b^{10} x^{4} + 4 \, a b^{9} x^{3} + 6 \, a^{2} b^{8} x^{2} + 4 \, a^{3} b^{7} x + a^{4} b^{6}} - \frac {60 \, a \log \left (b x + a\right )}{b^{6}}\right )} + \frac {5}{12} \, d e^{4} {\left (\frac {48 \, a b^{3} x^{3} + 108 \, a^{2} b^{2} x^{2} + 88 \, a^{3} b x + 25 \, a^{4}}{b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}} + \frac {12 \, \log \left (b x + a\right )}{b^{5}}\right )} - \frac {5}{6} \, d^{2} e^{3} {\left (\frac {12 \, x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} + \frac {8 \, a^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} + \frac {6 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {3 \, a^{3}}{b^{8} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {5}{12} \, d^{4} e {\left (\frac {4}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {3 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {5}{6} \, d^{3} e^{2} {\left (\frac {6}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {3 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {d^{5}}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} \] Input:
integrate((e*x+d)^5/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")
Output:
1/12*e^5*((12*b^5*x^5 + 48*a*b^4*x^4 - 48*a^2*b^3*x^3 - 252*a^3*b^2*x^2 - 248*a^4*b*x - 77*a^5)/(b^10*x^4 + 4*a*b^9*x^3 + 6*a^2*b^8*x^2 + 4*a^3*b^7* x + a^4*b^6) - 60*a*log(b*x + a)/b^6) + 5/12*d*e^4*((48*a*b^3*x^3 + 108*a^ 2*b^2*x^2 + 88*a^3*b*x + 25*a^4)/(b^9*x^4 + 4*a*b^8*x^3 + 6*a^2*b^7*x^2 + 4*a^3*b^6*x + a^4*b^5) + 12*log(b*x + a)/b^5) - 5/6*d^2*e^3*(12*x^2/((b^2* x^2 + 2*a*b*x + a^2)^(3/2)*b^2) + 8*a^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b ^4) + 6*a/(b^6*(x + a/b)^2) - 8*a^2/(b^7*(x + a/b)^3) - 3*a^3/(b^8*(x + a/ b)^4)) - 5/12*d^4*e*(4/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 3*a/(b^6*(x + a/b)^4)) - 5/6*d^3*e^2*(6/(b^5*(x + a/b)^2) - 8*a/(b^6*(x + a/b)^3) + 3 *a^2/(b^7*(x + a/b)^4)) - 1/4*d^5/(b^5*(x + a/b)^4)
Time = 0.13 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.11 \[ \int \frac {(d+e x)^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {e^{5} x}{b^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {5 \, {\left (b d e^{4} - a e^{5}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{6} \mathrm {sgn}\left (b x + a\right )} - \frac {3 \, b^{5} d^{5} + 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} + 30 \, a^{3} b^{2} d^{2} e^{3} - 125 \, a^{4} b d e^{4} + 77 \, a^{5} e^{5} + 120 \, {\left (b^{5} d^{2} e^{3} - 2 \, a b^{4} d e^{4} + a^{2} b^{3} e^{5}\right )} x^{3} + 60 \, {\left (b^{5} d^{3} e^{2} + 3 \, a b^{4} d^{2} e^{3} - 9 \, a^{2} b^{3} d e^{4} + 5 \, a^{3} b^{2} e^{5}\right )} x^{2} + 20 \, {\left (b^{5} d^{4} e + 2 \, a b^{4} d^{3} e^{2} + 6 \, a^{2} b^{3} d^{2} e^{3} - 22 \, a^{3} b^{2} d e^{4} + 13 \, a^{4} b e^{5}\right )} x}{12 \, {\left (b x + a\right )}^{4} b^{6} \mathrm {sgn}\left (b x + a\right )} \] Input:
integrate((e*x+d)^5/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")
Output:
e^5*x/(b^5*sgn(b*x + a)) + 5*(b*d*e^4 - a*e^5)*log(abs(b*x + a))/(b^6*sgn( b*x + a)) - 1/12*(3*b^5*d^5 + 5*a*b^4*d^4*e + 10*a^2*b^3*d^3*e^2 + 30*a^3* b^2*d^2*e^3 - 125*a^4*b*d*e^4 + 77*a^5*e^5 + 120*(b^5*d^2*e^3 - 2*a*b^4*d* e^4 + a^2*b^3*e^5)*x^3 + 60*(b^5*d^3*e^2 + 3*a*b^4*d^2*e^3 - 9*a^2*b^3*d*e ^4 + 5*a^3*b^2*e^5)*x^2 + 20*(b^5*d^4*e + 2*a*b^4*d^3*e^2 + 6*a^2*b^3*d^2* e^3 - 22*a^3*b^2*d*e^4 + 13*a^4*b*e^5)*x)/((b*x + a)^4*b^6*sgn(b*x + a))
Timed out. \[ \int \frac {(d+e x)^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^5}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \] Input:
int((d + e*x)^5/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)
Output:
int((d + e*x)^5/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2), x)
Time = 0.22 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.72 \[ \int \frac {(d+e x)^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {-3 a \,b^{5} d^{5}+60 \,\mathrm {log}\left (b x +a \right ) a^{5} b d \,e^{4}-240 \,\mathrm {log}\left (b x +a \right ) a^{5} b \,e^{5} x +240 \,\mathrm {log}\left (b x +a \right ) a^{4} b^{2} d \,e^{4} x +65 a^{5} b d \,e^{4}-5 a^{2} b^{4} d^{4} e -10 a^{3} b^{3} d^{3} e^{2}-60 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{4} e^{5} x^{4}-200 a^{5} b \,e^{5} x -180 a^{4} b^{2} e^{5} x^{2}+60 a^{2} b^{4} e^{5} x^{4}+12 a \,b^{5} e^{5} x^{5}+30 b^{6} d^{2} e^{3} x^{4}-60 \,\mathrm {log}\left (b x +a \right ) a^{6} e^{5}-65 a^{6} e^{5}-360 \,\mathrm {log}\left (b x +a \right ) a^{4} b^{2} e^{5} x^{2}-240 \,\mathrm {log}\left (b x +a \right ) a^{3} b^{3} e^{5} x^{3}+200 a^{4} b^{2} d \,e^{4} x +180 a^{3} b^{3} d \,e^{4} x^{2}-40 a^{2} b^{4} d^{3} e^{2} x -20 a \,b^{5} d^{4} e x -60 a \,b^{5} d^{3} e^{2} x^{2}-60 a \,b^{5} d \,e^{4} x^{4}+360 \,\mathrm {log}\left (b x +a \right ) a^{3} b^{3} d \,e^{4} x^{2}+240 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{4} d \,e^{4} x^{3}+60 \,\mathrm {log}\left (b x +a \right ) a \,b^{5} d \,e^{4} x^{4}}{12 a \,b^{6} \left (b^{4} x^{4}+4 a \,b^{3} x^{3}+6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}\right )} \] Input:
int((e*x+d)^5/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)
Output:
( - 60*log(a + b*x)*a**6*e**5 + 60*log(a + b*x)*a**5*b*d*e**4 - 240*log(a + b*x)*a**5*b*e**5*x + 240*log(a + b*x)*a**4*b**2*d*e**4*x - 360*log(a + b *x)*a**4*b**2*e**5*x**2 + 360*log(a + b*x)*a**3*b**3*d*e**4*x**2 - 240*log (a + b*x)*a**3*b**3*e**5*x**3 + 240*log(a + b*x)*a**2*b**4*d*e**4*x**3 - 6 0*log(a + b*x)*a**2*b**4*e**5*x**4 + 60*log(a + b*x)*a*b**5*d*e**4*x**4 - 65*a**6*e**5 + 65*a**5*b*d*e**4 - 200*a**5*b*e**5*x + 200*a**4*b**2*d*e**4 *x - 180*a**4*b**2*e**5*x**2 - 10*a**3*b**3*d**3*e**2 + 180*a**3*b**3*d*e* *4*x**2 - 5*a**2*b**4*d**4*e - 40*a**2*b**4*d**3*e**2*x + 60*a**2*b**4*e** 5*x**4 - 3*a*b**5*d**5 - 20*a*b**5*d**4*e*x - 60*a*b**5*d**3*e**2*x**2 - 6 0*a*b**5*d*e**4*x**4 + 12*a*b**5*e**5*x**5 + 30*b**6*d**2*e**3*x**4)/(12*a *b**6*(a**4 + 4*a**3*b*x + 6*a**2*b**2*x**2 + 4*a*b**3*x**3 + b**4*x**4))