\(\int \frac {(a^2+2 a b x+b^2 x^2)^{5/2}}{(d+e x)^5} \, dx\) [301]

Optimal result

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

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

Mathematica [A] (verified)

Time = 1.12 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^5} \, dx=-\frac {\sqrt {(a+b x)^2} \left (3 a^5 e^5+5 a^4 b e^4 (d+4 e x)+10 a^3 b^2 e^3 \left (d^2+4 d e x+6 e^2 x^2\right )+30 a^2 b^3 e^2 \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )-5 a b^4 d e \left (25 d^3+88 d^2 e x+108 d e^2 x^2+48 e^3 x^3\right )+b^5 \left (77 d^5+248 d^4 e x+252 d^3 e^2 x^2+48 d^2 e^3 x^3-48 d e^4 x^4-12 e^5 x^5\right )+60 b^4 (b d-a e) (d+e x)^4 \log (d+e x)\right )}{12 e^6 (a+b x) (d+e x)^4} \] Input:

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

Output:

-1/12*(Sqrt[(a + b*x)^2]*(3*a^5*e^5 + 5*a^4*b*e^4*(d + 4*e*x) + 10*a^3*b^2 
*e^3*(d^2 + 4*d*e*x + 6*e^2*x^2) + 30*a^2*b^3*e^2*(d^3 + 4*d^2*e*x + 6*d*e 
^2*x^2 + 4*e^3*x^3) - 5*a*b^4*d*e*(25*d^3 + 88*d^2*e*x + 108*d*e^2*x^2 + 4 
8*e^3*x^3) + b^5*(77*d^5 + 248*d^4*e*x + 252*d^3*e^2*x^2 + 48*d^2*e^3*x^3 
- 48*d*e^4*x^4 - 12*e^5*x^5) + 60*b^4*(b*d - a*e)*(d + e*x)^4*Log[d + e*x] 
))/(e^6*(a + b*x)*(d + e*x)^4)
 

Rubi [A] (verified)

Time = 0.59 (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.

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

\(\Big \downarrow \) 1102

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 49

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((b^5*x)/e^5 + (b*d - a*e)^5/(4*e^6*(d + e* 
x)^4) - (5*b*(b*d - a*e)^4)/(3*e^6*(d + e*x)^3) + (5*b^2*(b*d - a*e)^3)/(e 
^6*(d + e*x)^2) - (10*b^3*(b*d - a*e)^2)/(e^6*(d + e*x)) - (5*b^4*(b*d - a 
*e)*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 = 1.86 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.02

method	result	size
risch	\(\frac {b^{5} x \sqrt {\left (b x +a \right )^{2}}}{e^{5} \left (b x +a \right )}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (-10 a^{2} b^{3} e^{4}+20 a \,b^{4} d \,e^{3}-10 b^{5} d^{2} e^{2}\right ) x^{3}-5 b^{2} e \left (e^{3} a^{3}+3 a^{2} b d \,e^{2}-9 a \,b^{2} d^{2} e +5 b^{3} d^{3}\right ) x^{2}-\frac {5 b \left (a^{4} e^{4}+2 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-22 a \,b^{3} d^{3} e +13 b^{4} d^{4}\right ) x}{3}-\frac {3 e^{5} a^{5}+5 a^{4} b d \,e^{4}+10 a^{3} b^{2} d^{2} e^{3}+30 a^{2} b^{3} d^{3} e^{2}-125 a \,b^{4} d^{4} e +77 b^{5} d^{5}}{12 e}\right )}{\left (b x +a \right ) e^{5} \left (e x +d \right )^{4}}+\frac {5 \sqrt {\left (b x +a \right )^{2}}\, b^{4} \left (a e -b d \right ) \ln \left (e x +d \right )}{\left (b x +a \right ) e^{6}}\)	\(298\)
default	\(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (-360 \ln \left (e x +d \right ) b^{5} d^{3} e^{2} x^{2}-240 \ln \left (e x +d \right ) b^{5} d^{2} e^{3} x^{3}+360 \ln \left (e x +d \right ) a \,b^{4} d^{2} e^{3} x^{2}-40 a^{3} b^{2} d \,e^{4} x -77 b^{5} d^{5}-240 \ln \left (e x +d \right ) b^{5} d^{4} e x +60 \ln \left (e x +d \right ) a \,b^{4} d^{4} e +60 \ln \left (e x +d \right ) a \,b^{4} e^{5} x^{4}-60 \ln \left (e x +d \right ) b^{5} d \,e^{4} x^{4}+240 \ln \left (e x +d \right ) a \,b^{4} d \,e^{4} x^{3}-5 a^{4} b d \,e^{4}-10 a^{3} b^{2} d^{2} e^{3}+240 \ln \left (e x +d \right ) a \,b^{4} d^{3} e^{2} x -20 a^{4} b \,e^{5} x -248 b^{5} d^{4} e x -30 a^{2} b^{3} d^{3} e^{2}+125 a \,b^{4} d^{4} e -120 x \,a^{2} b^{3} d^{2} e^{3}+440 x a \,b^{4} d^{3} e^{2}-60 \ln \left (e x +d \right ) b^{5} d^{5}-180 x^{2} a^{2} b^{3} d \,e^{4}+540 x^{2} a \,b^{4} d^{2} e^{3}+240 x^{3} a \,b^{4} d \,e^{4}+48 x^{4} b^{5} d \,e^{4}-120 x^{3} a^{2} b^{3} e^{5}-48 x^{3} b^{5} d^{2} e^{3}-60 x^{2} a^{3} b^{2} e^{5}-252 x^{2} b^{5} d^{3} e^{2}-3 e^{5} a^{5}+12 x^{5} e^{5} b^{5}\right )}{12 \left (b x +a \right )^{5} e^{6} \left (e x +d \right )^{4}}\)	\(458\)

Input:

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

Output:

b^5*x*((b*x+a)^2)^(1/2)/e^5/(b*x+a)+((b*x+a)^2)^(1/2)/(b*x+a)*((-10*a^2*b^ 
3*e^4+20*a*b^4*d*e^3-10*b^5*d^2*e^2)*x^3-5*b^2*e*(a^3*e^3+3*a^2*b*d*e^2-9* 
a*b^2*d^2*e+5*b^3*d^3)*x^2-5/3*b*(a^4*e^4+2*a^3*b*d*e^3+6*a^2*b^2*d^2*e^2- 
22*a*b^3*d^3*e+13*b^4*d^4)*x-1/12*(3*a^5*e^5+5*a^4*b*d*e^4+10*a^3*b^2*d^2* 
e^3+30*a^2*b^3*d^3*e^2-125*a*b^4*d^4*e+77*b^5*d^5)/e)/e^5/(e*x+d)^4+5*((b* 
x+a)^2)^(1/2)/(b*x+a)*b^4/e^6*(a*e-b*d)*ln(e*x+d)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 412, normalized size of antiderivative = 1.41 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^5} \, dx=\frac {12 \, b^{5} e^{5} x^{5} + 48 \, b^{5} d e^{4} x^{4} - 77 \, b^{5} d^{5} + 125 \, a b^{4} d^{4} e - 30 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} - 5 \, a^{4} b d e^{4} - 3 \, a^{5} e^{5} - 24 \, {\left (2 \, b^{5} d^{2} e^{3} - 10 \, a b^{4} d e^{4} + 5 \, a^{2} b^{3} e^{5}\right )} x^{3} - 12 \, {\left (21 \, b^{5} d^{3} e^{2} - 45 \, a b^{4} d^{2} e^{3} + 15 \, a^{2} b^{3} d e^{4} + 5 \, a^{3} b^{2} e^{5}\right )} x^{2} - 4 \, {\left (62 \, b^{5} d^{4} e - 110 \, a b^{4} d^{3} e^{2} + 30 \, a^{2} b^{3} d^{2} e^{3} + 10 \, a^{3} b^{2} d e^{4} + 5 \, a^{4} b e^{5}\right )} x - 60 \, {\left (b^{5} d^{5} - a b^{4} d^{4} e + {\left (b^{5} d e^{4} - a b^{4} e^{5}\right )} x^{4} + 4 \, {\left (b^{5} d^{2} e^{3} - a b^{4} d e^{4}\right )} x^{3} + 6 \, {\left (b^{5} d^{3} e^{2} - a b^{4} d^{2} e^{3}\right )} x^{2} + 4 \, {\left (b^{5} d^{4} e - a b^{4} d^{3} e^{2}\right )} x\right )} \log \left (e x + d\right )}{12 \, {\left (e^{10} x^{4} + 4 \, d e^{9} x^{3} + 6 \, d^{2} e^{8} x^{2} + 4 \, d^{3} e^{7} x + d^{4} e^{6}\right )}} \] Input:

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

Output:

1/12*(12*b^5*e^5*x^5 + 48*b^5*d*e^4*x^4 - 77*b^5*d^5 + 125*a*b^4*d^4*e - 3 
0*a^2*b^3*d^3*e^2 - 10*a^3*b^2*d^2*e^3 - 5*a^4*b*d*e^4 - 3*a^5*e^5 - 24*(2 
*b^5*d^2*e^3 - 10*a*b^4*d*e^4 + 5*a^2*b^3*e^5)*x^3 - 12*(21*b^5*d^3*e^2 - 
45*a*b^4*d^2*e^3 + 15*a^2*b^3*d*e^4 + 5*a^3*b^2*e^5)*x^2 - 4*(62*b^5*d^4*e 
 - 110*a*b^4*d^3*e^2 + 30*a^2*b^3*d^2*e^3 + 10*a^3*b^2*d*e^4 + 5*a^4*b*e^5 
)*x - 60*(b^5*d^5 - a*b^4*d^4*e + (b^5*d*e^4 - a*b^4*e^5)*x^4 + 4*(b^5*d^2 
*e^3 - a*b^4*d*e^4)*x^3 + 6*(b^5*d^3*e^2 - a*b^4*d^2*e^3)*x^2 + 4*(b^5*d^4 
*e - a*b^4*d^3*e^2)*x)*log(e*x + d))/(e^10*x^4 + 4*d*e^9*x^3 + 6*d^2*e^8*x 
^2 + 4*d^3*e^7*x + d^4*e^6)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^5,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 = 0.42 (sec) , antiderivative size = 383, normalized size of antiderivative = 1.31 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^5} \, dx=\frac {b^{5} x \mathrm {sgn}\left (b x + a\right )}{e^{5}} - \frac {5 \, {\left (b^{5} d \mathrm {sgn}\left (b x + a\right ) - a b^{4} e \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | e x + d \right |}\right )}{e^{6}} - \frac {77 \, b^{5} d^{5} \mathrm {sgn}\left (b x + a\right ) - 125 \, a b^{4} d^{4} e \mathrm {sgn}\left (b x + a\right ) + 30 \, a^{2} b^{3} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{3} b^{2} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} b d e^{4} \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{5} e^{5} \mathrm {sgn}\left (b x + a\right ) + 120 \, {\left (b^{5} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b^{4} d e^{4} \mathrm {sgn}\left (b x + a\right ) + a^{2} b^{3} e^{5} \mathrm {sgn}\left (b x + a\right )\right )} x^{3} + 60 \, {\left (5 \, b^{5} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 9 \, a b^{4} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b^{3} d e^{4} \mathrm {sgn}\left (b x + a\right ) + a^{3} b^{2} e^{5} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 20 \, {\left (13 \, b^{5} d^{4} e \mathrm {sgn}\left (b x + a\right ) - 22 \, a b^{4} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{3} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{3} b^{2} d e^{4} \mathrm {sgn}\left (b x + a\right ) + a^{4} b e^{5} \mathrm {sgn}\left (b x + a\right )\right )} x}{12 \, {\left (e x + d\right )}^{4} e^{6}} \] Input:

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

Output:

b^5*x*sgn(b*x + a)/e^5 - 5*(b^5*d*sgn(b*x + a) - a*b^4*e*sgn(b*x + a))*log 
(abs(e*x + d))/e^6 - 1/12*(77*b^5*d^5*sgn(b*x + a) - 125*a*b^4*d^4*e*sgn(b 
*x + a) + 30*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) + 3*a^5*e^5*sgn(b*x + a) + 120*(b^5*d^2*e^3 
*sgn(b*x + a) - 2*a*b^4*d*e^4*sgn(b*x + a) + a^2*b^3*e^5*sgn(b*x + a))*x^3 
 + 60*(5*b^5*d^3*e^2*sgn(b*x + a) - 9*a*b^4*d^2*e^3*sgn(b*x + a) + 3*a^2*b 
^3*d*e^4*sgn(b*x + a) + a^3*b^2*e^5*sgn(b*x + a))*x^2 + 20*(13*b^5*d^4*e*s 
gn(b*x + a) - 22*a*b^4*d^3*e^2*sgn(b*x + a) + 6*a^2*b^3*d^2*e^3*sgn(b*x + 
a) + 2*a^3*b^2*d*e^4*sgn(b*x + a) + a^4*b*e^5*sgn(b*x + a))*x)/((e*x + d)^ 
4*e^6)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.49 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^5} \, dx=\frac {240 \,\mathrm {log}\left (e x +d \right ) a \,b^{4} d^{2} e^{4} x^{3}-60 \,\mathrm {log}\left (e x +d \right ) b^{5} d^{2} e^{4} x^{4}-65 b^{5} d^{6}+30 a^{2} b^{3} e^{6} x^{4}-3 a^{5} d \,e^{5}+60 \,\mathrm {log}\left (e x +d \right ) a \,b^{4} d \,e^{5} x^{4}-360 \,\mathrm {log}\left (e x +d \right ) b^{5} d^{4} e^{2} x^{2}+60 \,\mathrm {log}\left (e x +d \right ) a \,b^{4} d^{5} e -240 \,\mathrm {log}\left (e x +d \right ) b^{5} d^{5} e x -20 a^{4} b d \,e^{5} x -40 a^{3} b^{2} d^{2} e^{4} x -60 a^{3} b^{2} d \,e^{5} x^{2}+200 a \,b^{4} d^{4} e^{2} x +180 a \,b^{4} d^{3} e^{3} x^{2}-60 a \,b^{4} d \,e^{5} x^{4}-5 a^{4} b \,d^{2} e^{4}-10 a^{3} b^{2} d^{3} e^{3}+65 a \,b^{4} d^{5} e -200 b^{5} d^{5} e x -180 b^{5} d^{4} e^{2} x^{2}+60 b^{5} d^{2} e^{4} x^{4}+12 b^{5} d \,e^{5} x^{5}+240 \,\mathrm {log}\left (e x +d \right ) a \,b^{4} d^{4} e^{2} x +360 \,\mathrm {log}\left (e x +d \right ) a \,b^{4} d^{3} e^{3} x^{2}-60 \,\mathrm {log}\left (e x +d \right ) b^{5} d^{6}-240 \,\mathrm {log}\left (e x +d \right ) b^{5} d^{3} e^{3} x^{3}}{12 d \,e^{6} \left (e^{4} x^{4}+4 d \,e^{3} x^{3}+6 d^{2} e^{2} x^{2}+4 d^{3} e x +d^{4}\right )} \] Input:

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

Output:

(60*log(d + e*x)*a*b**4*d**5*e + 240*log(d + e*x)*a*b**4*d**4*e**2*x + 360 
*log(d + e*x)*a*b**4*d**3*e**3*x**2 + 240*log(d + e*x)*a*b**4*d**2*e**4*x* 
*3 + 60*log(d + e*x)*a*b**4*d*e**5*x**4 - 60*log(d + e*x)*b**5*d**6 - 240* 
log(d + e*x)*b**5*d**5*e*x - 360*log(d + e*x)*b**5*d**4*e**2*x**2 - 240*lo 
g(d + e*x)*b**5*d**3*e**3*x**3 - 60*log(d + e*x)*b**5*d**2*e**4*x**4 - 3*a 
**5*d*e**5 - 5*a**4*b*d**2*e**4 - 20*a**4*b*d*e**5*x - 10*a**3*b**2*d**3*e 
**3 - 40*a**3*b**2*d**2*e**4*x - 60*a**3*b**2*d*e**5*x**2 + 30*a**2*b**3*e 
**6*x**4 + 65*a*b**4*d**5*e + 200*a*b**4*d**4*e**2*x + 180*a*b**4*d**3*e** 
3*x**2 - 60*a*b**4*d*e**5*x**4 - 65*b**5*d**6 - 200*b**5*d**5*e*x - 180*b* 
*5*d**4*e**2*x**2 + 60*b**5*d**2*e**4*x**4 + 12*b**5*d*e**5*x**5)/(12*d*e* 
*6*(d**4 + 4*d**3*e*x + 6*d**2*e**2*x**2 + 4*d*e**3*x**3 + e**4*x**4))