\(\int \frac {(a^2+2 a b x+b^2 x^2)^{5/2}}{(d+e x)^2} \, dx\) [298]

Optimal result

Integrand size = 28, antiderivative size = 292 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=-\frac {10 b^2 (b d-a e)^3 x \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}+\frac {(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x) (d+e x)}+\frac {5 b^3 (b d-a e)^2 (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x)}-\frac {5 b^4 (b d-a e) (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^6 (a+b x)}+\frac {b^5 (d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^6 (a+b x)}+\frac {5 b (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^6 (a+b x)} \] Output:

-10*b^2*(-a*e+b*d)^3*x*((b*x+a)^2)^(1/2)/e^5/(b*x+a)+(-a*e+b*d)^5*((b*x+a) 
^2)^(1/2)/e^6/(b*x+a)/(e*x+d)+5*b^3*(-a*e+b*d)^2*(e*x+d)^2*((b*x+a)^2)^(1/ 
2)/e^6/(b*x+a)-5/3*b^4*(-a*e+b*d)*(e*x+d)^3*((b*x+a)^2)^(1/2)/e^6/(b*x+a)+ 
1/4*b^5*(e*x+d)^4*((b*x+a)^2)^(1/2)/e^6/(b*x+a)+5*b*(-a*e+b*d)^4*((b*x+a)^ 
2)^(1/2)*ln(e*x+d)/e^6/(b*x+a)
 

Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=\frac {\sqrt {(a+b x)^2} \left (60 a^4 b d e^4-12 a^5 e^5+120 a^3 b^2 e^3 \left (-d^2+d e x+e^2 x^2\right )+60 a^2 b^3 e^2 \left (2 d^3-4 d^2 e x-3 d e^2 x^2+e^3 x^3\right )+20 a b^4 e \left (-3 d^4+9 d^3 e x+6 d^2 e^2 x^2-2 d e^3 x^3+e^4 x^4\right )+b^5 \left (12 d^5-48 d^4 e x-30 d^3 e^2 x^2+10 d^2 e^3 x^3-5 d e^4 x^4+3 e^5 x^5\right )+60 b (b d-a e)^4 (d+e x) \log (d+e x)\right )}{12 e^6 (a+b x) (d+e x)} \] Input:

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/(d + e*x)^2,x]
 

Output:

(Sqrt[(a + b*x)^2]*(60*a^4*b*d*e^4 - 12*a^5*e^5 + 120*a^3*b^2*e^3*(-d^2 + 
d*e*x + e^2*x^2) + 60*a^2*b^3*e^2*(2*d^3 - 4*d^2*e*x - 3*d*e^2*x^2 + e^3*x 
^3) + 20*a*b^4*e*(-3*d^4 + 9*d^3*e*x + 6*d^2*e^2*x^2 - 2*d*e^3*x^3 + e^4*x 
^4) + b^5*(12*d^5 - 48*d^4*e*x - 30*d^3*e^2*x^2 + 10*d^2*e^3*x^3 - 5*d*e^4 
*x^4 + 3*e^5*x^5) + 60*b*(b*d - a*e)^4*(d + e*x)*Log[d + e*x]))/(12*e^6*(a 
 + b*x)*(d + e*x))
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.54, 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.

\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx\)

\(\Big \downarrow \) 1102

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^5 (a+b x)^5}{(d+e x)^2}dx}{b^5 (a+b x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^5}{(d+e x)^2}dx}{a+b x}\)

\(\Big \downarrow \) 49

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {(d+e x)^3 b^5}{e^5}-\frac {5 (b d-a e) (d+e x)^2 b^4}{e^5}+\frac {10 (b d-a e)^2 (d+e x) b^3}{e^5}-\frac {10 (b d-a e)^3 b^2}{e^5}+\frac {5 (b d-a e)^4 b}{e^5 (d+e x)}+\frac {(a e-b d)^5}{e^5 (d+e x)^2}\right )dx}{a+b x}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {5 b^4 (d+e x)^3 (b d-a e)}{3 e^6}+\frac {5 b^3 (d+e x)^2 (b d-a e)^2}{e^6}-\frac {10 b^2 x (b d-a e)^3}{e^5}+\frac {(b d-a e)^5}{e^6 (d+e x)}+\frac {5 b (b d-a e)^4 \log (d+e x)}{e^6}+\frac {b^5 (d+e x)^4}{4 e^6}\right )}{a+b x}\)

Input:

Int[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/(d + e*x)^2,x]
 

Output:

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((-10*b^2*(b*d - a*e)^3*x)/e^5 + (b*d - a*e 
)^5/(e^6*(d + e*x)) + (5*b^3*(b*d - a*e)^2*(d + e*x)^2)/e^6 - (5*b^4*(b*d 
- a*e)*(d + e*x)^3)/(3*e^6) + (b^5*(d + e*x)^4)/(4*e^6) + (5*b*(b*d - a*e) 
^4*Log[d + e*x])/e^6))/(a + b*x)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 49

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]
 

rule 1102

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 1.28 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.05

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{2} \left (\frac {1}{4} b^{3} x^{4} e^{3}+\frac {5}{3} a \,b^{2} e^{3} x^{3}-\frac {2}{3} b^{3} d \,e^{2} x^{3}+5 a^{2} b \,e^{3} x^{2}-5 a \,b^{2} d \,e^{2} x^{2}+\frac {3}{2} b^{3} d^{2} e \,x^{2}+10 e^{3} a^{3} x -20 a^{2} b d \,e^{2} x +15 a \,b^{2} d^{2} e x -4 b^{3} d^{3} x \right )}{\left (b x +a \right ) e^{5}}+\frac {5 \sqrt {\left (b x +a \right )^{2}}\, b \left (a^{4} e^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right ) \ln \left (e x +d \right )}{\left (b x +a \right ) e^{6}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (e^{5} a^{5}-5 a^{4} b d \,e^{4}+10 a^{3} b^{2} d^{2} e^{3}-10 a^{2} b^{3} d^{3} e^{2}+5 a \,b^{4} d^{4} e -b^{5} d^{5}\right )}{\left (b x +a \right ) e^{6} \left (e x +d \right )}\)	\(307\)
default	\(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (120 a^{3} b^{2} d \,e^{4} x +60 \ln \left (e x +d \right ) a^{4} b d \,e^{4}+60 \ln \left (e x +d \right ) a^{4} b \,e^{5} x +12 b^{5} d^{5}+60 \ln \left (e x +d \right ) b^{5} d^{4} e x -240 \ln \left (e x +d \right ) a^{3} b^{2} d^{2} e^{3}+360 \ln \left (e x +d \right ) a^{2} b^{3} d^{3} e^{2}-240 \ln \left (e x +d \right ) a \,b^{4} d^{4} e +60 a^{4} b d \,e^{4}-120 a^{3} b^{2} d^{2} e^{3}-240 \ln \left (e x +d \right ) a^{3} b^{2} d \,e^{4} x +360 \ln \left (e x +d \right ) a^{2} b^{3} d^{2} e^{3} x -240 \ln \left (e x +d \right ) a \,b^{4} d^{3} e^{2} x -48 b^{5} d^{4} e x +120 a^{2} b^{3} d^{3} e^{2}-60 a \,b^{4} d^{4} e -240 x \,a^{2} b^{3} d^{2} e^{3}+180 x a \,b^{4} d^{3} e^{2}+60 \ln \left (e x +d \right ) b^{5} d^{5}-180 x^{2} a^{2} b^{3} d \,e^{4}+120 x^{2} a \,b^{4} d^{2} e^{3}-40 x^{3} a \,b^{4} d \,e^{4}+20 x^{4} a \,b^{4} e^{5}-5 x^{4} b^{5} d \,e^{4}+60 x^{3} a^{2} b^{3} e^{5}+10 x^{3} b^{5} d^{2} e^{3}+120 x^{2} a^{3} b^{2} e^{5}-30 x^{2} b^{5} d^{3} e^{2}-12 e^{5} a^{5}+3 x^{5} e^{5} b^{5}\right )}{12 \left (b x +a \right )^{5} e^{6} \left (e x +d \right )}\)	\(456\)

Input:

int((b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^2,x,method=_RETURNVERBOSE)
 

Output:

((b*x+a)^2)^(1/2)/(b*x+a)*b^2/e^5*(1/4*b^3*x^4*e^3+5/3*a*b^2*e^3*x^3-2/3*b 
^3*d*e^2*x^3+5*a^2*b*e^3*x^2-5*a*b^2*d*e^2*x^2+3/2*b^3*d^2*e*x^2+10*e^3*a^ 
3*x-20*a^2*b*d*e^2*x+15*a*b^2*d^2*e*x-4*b^3*d^3*x)+5*((b*x+a)^2)^(1/2)/(b* 
x+a)*b/e^6*(a^4*e^4-4*a^3*b*d*e^3+6*a^2*b^2*d^2*e^2-4*a*b^3*d^3*e+b^4*d^4) 
*ln(e*x+d)-((b*x+a)^2)^(1/2)/(b*x+a)/e^6*(a^5*e^5-5*a^4*b*d*e^4+10*a^3*b^2 
*d^2*e^3-10*a^2*b^3*d^3*e^2+5*a*b^4*d^4*e-b^5*d^5)/(e*x+d)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=\frac {3 \, b^{5} e^{5} x^{5} + 12 \, b^{5} d^{5} - 60 \, a b^{4} d^{4} e + 120 \, a^{2} b^{3} d^{3} e^{2} - 120 \, a^{3} b^{2} d^{2} e^{3} + 60 \, a^{4} b d e^{4} - 12 \, a^{5} e^{5} - 5 \, {\left (b^{5} d e^{4} - 4 \, a b^{4} e^{5}\right )} x^{4} + 10 \, {\left (b^{5} d^{2} e^{3} - 4 \, a b^{4} d e^{4} + 6 \, a^{2} b^{3} e^{5}\right )} x^{3} - 30 \, {\left (b^{5} d^{3} e^{2} - 4 \, a b^{4} d^{2} e^{3} + 6 \, a^{2} b^{3} d e^{4} - 4 \, a^{3} b^{2} e^{5}\right )} x^{2} - 12 \, {\left (4 \, b^{5} d^{4} e - 15 \, a b^{4} d^{3} e^{2} + 20 \, a^{2} b^{3} d^{2} e^{3} - 10 \, a^{3} b^{2} d e^{4}\right )} x + 60 \, {\left (b^{5} d^{5} - 4 \, a b^{4} d^{4} e + 6 \, a^{2} b^{3} d^{3} e^{2} - 4 \, a^{3} b^{2} d^{2} e^{3} + a^{4} b d e^{4} + {\left (b^{5} d^{4} e - 4 \, a b^{4} d^{3} e^{2} + 6 \, a^{2} b^{3} d^{2} e^{3} - 4 \, a^{3} b^{2} d e^{4} + a^{4} b e^{5}\right )} x\right )} \log \left (e x + d\right )}{12 \, {\left (e^{7} x + d e^{6}\right )}} \] Input:

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^2,x, algorithm="fricas")
 

Output:

1/12*(3*b^5*e^5*x^5 + 12*b^5*d^5 - 60*a*b^4*d^4*e + 120*a^2*b^3*d^3*e^2 - 
120*a^3*b^2*d^2*e^3 + 60*a^4*b*d*e^4 - 12*a^5*e^5 - 5*(b^5*d*e^4 - 4*a*b^4 
*e^5)*x^4 + 10*(b^5*d^2*e^3 - 4*a*b^4*d*e^4 + 6*a^2*b^3*e^5)*x^3 - 30*(b^5 
*d^3*e^2 - 4*a*b^4*d^2*e^3 + 6*a^2*b^3*d*e^4 - 4*a^3*b^2*e^5)*x^2 - 12*(4* 
b^5*d^4*e - 15*a*b^4*d^3*e^2 + 20*a^2*b^3*d^2*e^3 - 10*a^3*b^2*d*e^4)*x + 
60*(b^5*d^5 - 4*a*b^4*d^4*e + 6*a^2*b^3*d^3*e^2 - 4*a^3*b^2*d^2*e^3 + a^4* 
b*d*e^4 + (b^5*d^4*e - 4*a*b^4*d^3*e^2 + 6*a^2*b^3*d^2*e^3 - 4*a^3*b^2*d*e 
^4 + a^4*b*e^5)*x)*log(e*x + d))/(e^7*x + d*e^6)
 

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{\left (d + e x\right )^{2}}\, dx \] Input:

integrate((b**2*x**2+2*a*b*x+a**2)**(5/2)/(e*x+d)**2,x)
 

Output:

Integral(((a + b*x)**2)**(5/2)/(d + e*x)**2, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^2,x, algorithm="maxima")
 

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
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.36 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=\frac {5 \, {\left (b^{5} d^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{4} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{3} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b^{2} d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} b e^{4} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | e x + d \right |}\right )}{e^{6}} + \frac {b^{5} d^{5} \mathrm {sgn}\left (b x + a\right ) - 5 \, a b^{4} d^{4} e \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b^{3} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 10 \, a^{3} b^{2} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} b d e^{4} \mathrm {sgn}\left (b x + a\right ) - a^{5} e^{5} \mathrm {sgn}\left (b x + a\right )}{{\left (e x + d\right )} e^{6}} + \frac {3 \, b^{5} e^{6} x^{4} \mathrm {sgn}\left (b x + a\right ) - 8 \, b^{5} d e^{5} x^{3} \mathrm {sgn}\left (b x + a\right ) + 20 \, a b^{4} e^{6} x^{3} \mathrm {sgn}\left (b x + a\right ) + 18 \, b^{5} d^{2} e^{4} x^{2} \mathrm {sgn}\left (b x + a\right ) - 60 \, a b^{4} d e^{5} x^{2} \mathrm {sgn}\left (b x + a\right ) + 60 \, a^{2} b^{3} e^{6} x^{2} \mathrm {sgn}\left (b x + a\right ) - 48 \, b^{5} d^{3} e^{3} x \mathrm {sgn}\left (b x + a\right ) + 180 \, a b^{4} d^{2} e^{4} x \mathrm {sgn}\left (b x + a\right ) - 240 \, a^{2} b^{3} d e^{5} x \mathrm {sgn}\left (b x + a\right ) + 120 \, a^{3} b^{2} e^{6} x \mathrm {sgn}\left (b x + a\right )}{12 \, e^{8}} \] Input:

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^2,x, algorithm="giac")
 

Output:

5*(b^5*d^4*sgn(b*x + a) - 4*a*b^4*d^3*e*sgn(b*x + a) + 6*a^2*b^3*d^2*e^2*s 
gn(b*x + a) - 4*a^3*b^2*d*e^3*sgn(b*x + a) + a^4*b*e^4*sgn(b*x + a))*log(a 
bs(e*x + d))/e^6 + (b^5*d^5*sgn(b*x + a) - 5*a*b^4*d^4*e*sgn(b*x + a) + 10 
*a^2*b^3*d^3*e^2*sgn(b*x + a) - 10*a^3*b^2*d^2*e^3*sgn(b*x + a) + 5*a^4*b* 
d*e^4*sgn(b*x + a) - a^5*e^5*sgn(b*x + a))/((e*x + d)*e^6) + 1/12*(3*b^5*e 
^6*x^4*sgn(b*x + a) - 8*b^5*d*e^5*x^3*sgn(b*x + a) + 20*a*b^4*e^6*x^3*sgn( 
b*x + a) + 18*b^5*d^2*e^4*x^2*sgn(b*x + a) - 60*a*b^4*d*e^5*x^2*sgn(b*x + 
a) + 60*a^2*b^3*e^6*x^2*sgn(b*x + a) - 48*b^5*d^3*e^3*x*sgn(b*x + a) + 180 
*a*b^4*d^2*e^4*x*sgn(b*x + a) - 240*a^2*b^3*d*e^5*x*sgn(b*x + a) + 120*a^3 
*b^2*e^6*x*sgn(b*x + a))/e^8
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}}{{\left (d+e\,x\right )}^2} \,d x \] Input:

int((a^2 + b^2*x^2 + 2*a*b*x)^(5/2)/(d + e*x)^2,x)
 

Output:

int((a^2 + b^2*x^2 + 2*a*b*x)^(5/2)/(d + e*x)^2, x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 415, normalized size of antiderivative = 1.42 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=\frac {60 \,\mathrm {log}\left (e x +d \right ) a^{4} b \,d^{2} e^{4}-240 \,\mathrm {log}\left (e x +d \right ) a^{3} b^{2} d^{3} e^{3}+360 \,\mathrm {log}\left (e x +d \right ) a^{2} b^{3} d^{4} e^{2}-240 \,\mathrm {log}\left (e x +d \right ) a \,b^{4} d^{5} e +60 \,\mathrm {log}\left (e x +d \right ) b^{5} d^{5} e x -60 a^{4} b d \,e^{5} x +240 a^{3} b^{2} d^{2} e^{4} x +120 a^{3} b^{2} d \,e^{5} x^{2}-360 a^{2} b^{3} d^{3} e^{3} x -180 a^{2} b^{3} d^{2} e^{4} x^{2}+60 a^{2} b^{3} d \,e^{5} x^{3}+240 a \,b^{4} d^{4} e^{2} x +120 a \,b^{4} d^{3} e^{3} x^{2}-40 a \,b^{4} d^{2} e^{4} x^{3}+20 a \,b^{4} d \,e^{5} x^{4}-60 b^{5} d^{5} e x -30 b^{5} d^{4} e^{2} x^{2}+10 b^{5} d^{3} e^{3} x^{3}-5 b^{5} d^{2} e^{4} x^{4}+3 b^{5} d \,e^{5} x^{5}+60 \,\mathrm {log}\left (e x +d \right ) a^{4} b d \,e^{5} x -240 \,\mathrm {log}\left (e x +d \right ) a^{3} b^{2} d^{2} e^{4} x +360 \,\mathrm {log}\left (e x +d \right ) a^{2} b^{3} d^{3} e^{3} x -240 \,\mathrm {log}\left (e x +d \right ) a \,b^{4} d^{4} e^{2} x +60 \,\mathrm {log}\left (e x +d \right ) b^{5} d^{6}+12 a^{5} e^{6} x}{12 d \,e^{6} \left (e x +d \right )} \] Input:

int((b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^2,x)
 

Output:

(60*log(d + e*x)*a**4*b*d**2*e**4 + 60*log(d + e*x)*a**4*b*d*e**5*x - 240* 
log(d + e*x)*a**3*b**2*d**3*e**3 - 240*log(d + e*x)*a**3*b**2*d**2*e**4*x 
+ 360*log(d + e*x)*a**2*b**3*d**4*e**2 + 360*log(d + e*x)*a**2*b**3*d**3*e 
**3*x - 240*log(d + e*x)*a*b**4*d**5*e - 240*log(d + e*x)*a*b**4*d**4*e**2 
*x + 60*log(d + e*x)*b**5*d**6 + 60*log(d + e*x)*b**5*d**5*e*x + 12*a**5*e 
**6*x - 60*a**4*b*d*e**5*x + 240*a**3*b**2*d**2*e**4*x + 120*a**3*b**2*d*e 
**5*x**2 - 360*a**2*b**3*d**3*e**3*x - 180*a**2*b**3*d**2*e**4*x**2 + 60*a 
**2*b**3*d*e**5*x**3 + 240*a*b**4*d**4*e**2*x + 120*a*b**4*d**3*e**3*x**2 
- 40*a*b**4*d**2*e**4*x**3 + 20*a*b**4*d*e**5*x**4 - 60*b**5*d**5*e*x - 30 
*b**5*d**4*e**2*x**2 + 10*b**5*d**3*e**3*x**3 - 5*b**5*d**2*e**4*x**4 + 3* 
b**5*d*e**5*x**5)/(12*d*e**6*(d + e*x))