\(\int \frac {(d+e x)^{5/2}}{(a^2+2 a b x+b^2 x^2)^3} \, dx\) [194]

Optimal result

Integrand size = 28, antiderivative size = 198 \[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {e^2 \sqrt {d+e x}}{16 b^3 (a+b x)^3}-\frac {e^3 \sqrt {d+e x}}{64 b^3 (b d-a e) (a+b x)^2}+\frac {3 e^4 \sqrt {d+e x}}{128 b^3 (b d-a e)^2 (a+b x)}-\frac {e (d+e x)^{3/2}}{8 b^2 (a+b x)^4}-\frac {(d+e x)^{5/2}}{5 b (a+b x)^5}-\frac {3 e^5 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{128 b^{7/2} (b d-a e)^{5/2}} \] Output:

-1/16*e^2*(e*x+d)^(1/2)/b^3/(b*x+a)^3-1/64*e^3*(e*x+d)^(1/2)/b^3/(-a*e+b*d 
)/(b*x+a)^2+3/128*e^4*(e*x+d)^(1/2)/b^3/(-a*e+b*d)^2/(b*x+a)-1/8*e*(e*x+d) 
^(3/2)/b^2/(b*x+a)^4-1/5*(e*x+d)^(5/2)/b/(b*x+a)^5-3/128*e^5*arctanh(b^(1/ 
2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(7/2)/(-a*e+b*d)^(5/2)
 

Mathematica [A] (verified)

Time = 1.38 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.12 \[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {\sqrt {d+e x} \left (15 a^4 e^4+10 a^3 b e^3 (d+7 e x)+2 a^2 b^2 e^2 \left (4 d^2+23 d e x+64 e^2 x^2\right )-2 a b^3 e \left (88 d^3+256 d^2 e x+233 d e^2 x^2+35 e^3 x^3\right )+b^4 \left (128 d^4+336 d^3 e x+248 d^2 e^2 x^2+10 d e^3 x^3-15 e^4 x^4\right )\right )}{640 b^3 (b d-a e)^2 (a+b x)^5}+\frac {3 e^5 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{128 b^{7/2} (-b d+a e)^{5/2}} \] Input:

Integrate[(d + e*x)^(5/2)/(a^2 + 2*a*b*x + b^2*x^2)^3,x]
 

Output:

-1/640*(Sqrt[d + e*x]*(15*a^4*e^4 + 10*a^3*b*e^3*(d + 7*e*x) + 2*a^2*b^2*e 
^2*(4*d^2 + 23*d*e*x + 64*e^2*x^2) - 2*a*b^3*e*(88*d^3 + 256*d^2*e*x + 233 
*d*e^2*x^2 + 35*e^3*x^3) + b^4*(128*d^4 + 336*d^3*e*x + 248*d^2*e^2*x^2 + 
10*d*e^3*x^3 - 15*e^4*x^4)))/(b^3*(b*d - a*e)^2*(a + b*x)^5) + (3*e^5*ArcT 
an[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(128*b^(7/2)*(-(b*d) + a*e 
)^(5/2))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {1098, 27, 51, 51, 51, 52, 52, 73, 221}

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.

\(\displaystyle \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx\)

\(\Big \downarrow \) 1098

\(\displaystyle b^6 \int \frac {(d+e x)^{5/2}}{b^6 (a+b x)^6}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {(d+e x)^{5/2}}{(a+b x)^6}dx\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {e \int \frac {(d+e x)^{3/2}}{(a+b x)^5}dx}{2 b}-\frac {(d+e x)^{5/2}}{5 b (a+b x)^5}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {e \left (\frac {3 e \int \frac {\sqrt {d+e x}}{(a+b x)^4}dx}{8 b}-\frac {(d+e x)^{3/2}}{4 b (a+b x)^4}\right )}{2 b}-\frac {(d+e x)^{5/2}}{5 b (a+b x)^5}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {e \left (\frac {3 e \left (\frac {e \int \frac {1}{(a+b x)^3 \sqrt {d+e x}}dx}{6 b}-\frac {\sqrt {d+e x}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(d+e x)^{3/2}}{4 b (a+b x)^4}\right )}{2 b}-\frac {(d+e x)^{5/2}}{5 b (a+b x)^5}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {e \left (\frac {3 e \left (\frac {e \left (-\frac {3 e \int \frac {1}{(a+b x)^2 \sqrt {d+e x}}dx}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{6 b}-\frac {\sqrt {d+e x}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(d+e x)^{3/2}}{4 b (a+b x)^4}\right )}{2 b}-\frac {(d+e x)^{5/2}}{5 b (a+b x)^5}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {e \left (\frac {3 e \left (\frac {e \left (-\frac {3 e \left (-\frac {e \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{2 (b d-a e)}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{6 b}-\frac {\sqrt {d+e x}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(d+e x)^{3/2}}{4 b (a+b x)^4}\right )}{2 b}-\frac {(d+e x)^{5/2}}{5 b (a+b x)^5}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {e \left (\frac {3 e \left (\frac {e \left (-\frac {3 e \left (-\frac {\int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b d-a e}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{6 b}-\frac {\sqrt {d+e x}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(d+e x)^{3/2}}{4 b (a+b x)^4}\right )}{2 b}-\frac {(d+e x)^{5/2}}{5 b (a+b x)^5}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {e \left (\frac {3 e \left (\frac {e \left (-\frac {3 e \left (\frac {e \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{3/2}}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{6 b}-\frac {\sqrt {d+e x}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(d+e x)^{3/2}}{4 b (a+b x)^4}\right )}{2 b}-\frac {(d+e x)^{5/2}}{5 b (a+b x)^5}\)

Input:

Int[(d + e*x)^(5/2)/(a^2 + 2*a*b*x + b^2*x^2)^3,x]
 

Output:

-1/5*(d + e*x)^(5/2)/(b*(a + b*x)^5) + (e*(-1/4*(d + e*x)^(3/2)/(b*(a + b* 
x)^4) + (3*e*(-1/3*Sqrt[d + e*x]/(b*(a + b*x)^3) + (e*(-1/2*Sqrt[d + e*x]/ 
((b*d - a*e)*(a + b*x)^2) - (3*e*(-(Sqrt[d + e*x]/((b*d - a*e)*(a + b*x))) 
 + (e*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(Sqrt[b]*(b*d - a* 
e)^(3/2))))/(4*(b*d - a*e))))/(6*b)))/(8*b)))/(2*b)
 

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 1098

Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ 
Symbol] :> Simp[1/c^p   Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[ 
{a, b, c, d, e, m}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
 

Maple [A] (verified)

Time = 3.56 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(2 e^{5} \left (\frac {\frac {3 b \left (e x +d \right )^{\frac {9}{2}}}{256 \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right )}+\frac {7 \left (e x +d \right )^{\frac {7}{2}}}{128 \left (a e -b d \right )}-\frac {\left (e x +d \right )^{\frac {5}{2}}}{10 b}-\frac {7 \left (a e -b d \right ) \left (e x +d \right )^{\frac {3}{2}}}{128 b^{2}}-\frac {3 \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {e x +d}}{256 b^{3}}}{\left (b \left (e x +d \right )+a e -b d \right )^{5}}+\frac {3 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )}{256 b^{3} \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {b \left (a e -b d \right )}}\right )\)	\(205\)
default	\(2 e^{5} \left (\frac {\frac {3 b \left (e x +d \right )^{\frac {9}{2}}}{256 \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right )}+\frac {7 \left (e x +d \right )^{\frac {7}{2}}}{128 \left (a e -b d \right )}-\frac {\left (e x +d \right )^{\frac {5}{2}}}{10 b}-\frac {7 \left (a e -b d \right ) \left (e x +d \right )^{\frac {3}{2}}}{128 b^{2}}-\frac {3 \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {e x +d}}{256 b^{3}}}{\left (b \left (e x +d \right )+a e -b d \right )^{5}}+\frac {3 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )}{256 b^{3} \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {b \left (a e -b d \right )}}\right )\)	\(205\)
pseudoelliptic	\(-\frac {3 \left (\sqrt {b \left (a e -b d \right )}\, \left (\left (-b^{4} x^{4}-\frac {14}{3} a \,b^{3} x^{3}+\frac {128}{15} a^{2} b^{2} x^{2}+\frac {14}{3} a^{3} b x +a^{4}\right ) e^{4}+\frac {2 b d \left (b^{3} x^{3}-\frac {233}{5} a \,b^{2} x^{2}+\frac {23}{5} a^{2} b x +a^{3}\right ) e^{3}}{3}+\frac {8 b^{2} d^{2} \left (31 b^{2} x^{2}-64 a b x +a^{2}\right ) e^{2}}{15}-\frac {176 \left (-\frac {21 b x}{11}+a \right ) b^{3} d^{3} e}{15}+\frac {128 b^{4} d^{4}}{15}\right ) \sqrt {e x +d}-e^{5} \left (b x +a \right )^{5} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )\right )}{128 \sqrt {b \left (a e -b d \right )}\, \left (b x +a \right )^{5} \left (a e -b d \right )^{2} b^{3}}\)	\(219\)

Input:

int((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^3,x,method=_RETURNVERBOSE)
 

Output:

2*e^5*((3/256*b/(a^2*e^2-2*a*b*d*e+b^2*d^2)*(e*x+d)^(9/2)+7/128/(a*e-b*d)* 
(e*x+d)^(7/2)-1/10/b*(e*x+d)^(5/2)-7/128*(a*e-b*d)/b^2*(e*x+d)^(3/2)-3/256 
/b^3*(a^2*e^2-2*a*b*d*e+b^2*d^2)*(e*x+d)^(1/2))/(b*(e*x+d)+a*e-b*d)^5+3/25 
6/b^3/(a^2*e^2-2*a*b*d*e+b^2*d^2)/(b*(a*e-b*d))^(1/2)*arctan(b*(e*x+d)^(1/ 
2)/(b*(a*e-b*d))^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 662 vs. \(2 (166) = 332\).

Time = 0.14 (sec) , antiderivative size = 1337, normalized size of antiderivative = 6.75 \[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\text {Too large to display} \] Input:

integrate((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")
 

Output:

[1/1280*(15*(b^5*e^5*x^5 + 5*a*b^4*e^5*x^4 + 10*a^2*b^3*e^5*x^3 + 10*a^3*b 
^2*e^5*x^2 + 5*a^4*b*e^5*x + a^5*e^5)*sqrt(b^2*d - a*b*e)*log((b*e*x + 2*b 
*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(128*b^6*d^ 
5 - 304*a*b^5*d^4*e + 184*a^2*b^4*d^3*e^2 + 2*a^3*b^3*d^2*e^3 + 5*a^4*b^2* 
d*e^4 - 15*a^5*b*e^5 - 15*(b^6*d*e^4 - a*b^5*e^5)*x^4 + 10*(b^6*d^2*e^3 - 
8*a*b^5*d*e^4 + 7*a^2*b^4*e^5)*x^3 + 2*(124*b^6*d^3*e^2 - 357*a*b^5*d^2*e^ 
3 + 297*a^2*b^4*d*e^4 - 64*a^3*b^3*e^5)*x^2 + 2*(168*b^6*d^4*e - 424*a*b^5 
*d^3*e^2 + 279*a^2*b^4*d^2*e^3 + 12*a^3*b^3*d*e^4 - 35*a^4*b^2*e^5)*x)*sqr 
t(e*x + d))/(a^5*b^7*d^3 - 3*a^6*b^6*d^2*e + 3*a^7*b^5*d*e^2 - a^8*b^4*e^3 
 + (b^12*d^3 - 3*a*b^11*d^2*e + 3*a^2*b^10*d*e^2 - a^3*b^9*e^3)*x^5 + 5*(a 
*b^11*d^3 - 3*a^2*b^10*d^2*e + 3*a^3*b^9*d*e^2 - a^4*b^8*e^3)*x^4 + 10*(a^ 
2*b^10*d^3 - 3*a^3*b^9*d^2*e + 3*a^4*b^8*d*e^2 - a^5*b^7*e^3)*x^3 + 10*(a^ 
3*b^9*d^3 - 3*a^4*b^8*d^2*e + 3*a^5*b^7*d*e^2 - a^6*b^6*e^3)*x^2 + 5*(a^4* 
b^8*d^3 - 3*a^5*b^7*d^2*e + 3*a^6*b^6*d*e^2 - a^7*b^5*e^3)*x), 1/640*(15*( 
b^5*e^5*x^5 + 5*a*b^4*e^5*x^4 + 10*a^2*b^3*e^5*x^3 + 10*a^3*b^2*e^5*x^2 + 
5*a^4*b*e^5*x + a^5*e^5)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)* 
sqrt(e*x + d)/(b*e*x + b*d)) - (128*b^6*d^5 - 304*a*b^5*d^4*e + 184*a^2*b^ 
4*d^3*e^2 + 2*a^3*b^3*d^2*e^3 + 5*a^4*b^2*d*e^4 - 15*a^5*b*e^5 - 15*(b^6*d 
*e^4 - a*b^5*e^5)*x^4 + 10*(b^6*d^2*e^3 - 8*a*b^5*d*e^4 + 7*a^2*b^4*e^5)*x 
^3 + 2*(124*b^6*d^3*e^2 - 357*a*b^5*d^2*e^3 + 297*a^2*b^4*d*e^4 - 64*a^...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\text {Timed out} \] Input:

integrate((e*x+d)**(5/2)/(b**2*x**2+2*a*b*x+a**2)**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^3,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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (166) = 332\).

Time = 0.18 (sec) , antiderivative size = 380, normalized size of antiderivative = 1.92 \[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {3 \, e^{5} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{128 \, {\left (b^{5} d^{2} - 2 \, a b^{4} d e + a^{2} b^{3} e^{2}\right )} \sqrt {-b^{2} d + a b e}} + \frac {15 \, {\left (e x + d\right )}^{\frac {9}{2}} b^{4} e^{5} - 70 \, {\left (e x + d\right )}^{\frac {7}{2}} b^{4} d e^{5} - 128 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{4} d^{2} e^{5} + 70 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{4} d^{3} e^{5} - 15 \, \sqrt {e x + d} b^{4} d^{4} e^{5} + 70 \, {\left (e x + d\right )}^{\frac {7}{2}} a b^{3} e^{6} + 256 \, {\left (e x + d\right )}^{\frac {5}{2}} a b^{3} d e^{6} - 210 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{3} d^{2} e^{6} + 60 \, \sqrt {e x + d} a b^{3} d^{3} e^{6} - 128 \, {\left (e x + d\right )}^{\frac {5}{2}} a^{2} b^{2} e^{7} + 210 \, {\left (e x + d\right )}^{\frac {3}{2}} a^{2} b^{2} d e^{7} - 90 \, \sqrt {e x + d} a^{2} b^{2} d^{2} e^{7} - 70 \, {\left (e x + d\right )}^{\frac {3}{2}} a^{3} b e^{8} + 60 \, \sqrt {e x + d} a^{3} b d e^{8} - 15 \, \sqrt {e x + d} a^{4} e^{9}}{640 \, {\left (b^{5} d^{2} - 2 \, a b^{4} d e + a^{2} b^{3} e^{2}\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{5}} \] Input:

integrate((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="giac")
 

Output:

3/128*e^5*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((b^5*d^2 - 2*a*b^4 
*d*e + a^2*b^3*e^2)*sqrt(-b^2*d + a*b*e)) + 1/640*(15*(e*x + d)^(9/2)*b^4* 
e^5 - 70*(e*x + d)^(7/2)*b^4*d*e^5 - 128*(e*x + d)^(5/2)*b^4*d^2*e^5 + 70* 
(e*x + d)^(3/2)*b^4*d^3*e^5 - 15*sqrt(e*x + d)*b^4*d^4*e^5 + 70*(e*x + d)^ 
(7/2)*a*b^3*e^6 + 256*(e*x + d)^(5/2)*a*b^3*d*e^6 - 210*(e*x + d)^(3/2)*a* 
b^3*d^2*e^6 + 60*sqrt(e*x + d)*a*b^3*d^3*e^6 - 128*(e*x + d)^(5/2)*a^2*b^2 
*e^7 + 210*(e*x + d)^(3/2)*a^2*b^2*d*e^7 - 90*sqrt(e*x + d)*a^2*b^2*d^2*e^ 
7 - 70*(e*x + d)^(3/2)*a^3*b*e^8 + 60*sqrt(e*x + d)*a^3*b*d*e^8 - 15*sqrt( 
e*x + d)*a^4*e^9)/((b^5*d^2 - 2*a*b^4*d*e + a^2*b^3*e^2)*((e*x + d)*b - b* 
d + a*e)^5)
 

Mupad [B] (verification not implemented)

Time = 9.53 (sec) , antiderivative size = 411, normalized size of antiderivative = 2.08 \[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {3\,e^5\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )}{128\,b^{7/2}\,{\left (a\,e-b\,d\right )}^{5/2}}-\frac {\frac {e^5\,{\left (d+e\,x\right )}^{5/2}}{5\,b}-\frac {7\,e^5\,{\left (d+e\,x\right )}^{7/2}}{64\,\left (a\,e-b\,d\right )}+\frac {3\,e^5\,\sqrt {d+e\,x}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}{128\,b^3}+\frac {7\,e^5\,\left (a\,e-b\,d\right )\,{\left (d+e\,x\right )}^{3/2}}{64\,b^2}-\frac {3\,b\,e^5\,{\left (d+e\,x\right )}^{9/2}}{128\,{\left (a\,e-b\,d\right )}^2}}{\left (d+e\,x\right )\,\left (5\,a^4\,b\,e^4-20\,a^3\,b^2\,d\,e^3+30\,a^2\,b^3\,d^2\,e^2-20\,a\,b^4\,d^3\,e+5\,b^5\,d^4\right )-{\left (d+e\,x\right )}^2\,\left (-10\,a^3\,b^2\,e^3+30\,a^2\,b^3\,d\,e^2-30\,a\,b^4\,d^2\,e+10\,b^5\,d^3\right )+b^5\,{\left (d+e\,x\right )}^5-\left (5\,b^5\,d-5\,a\,b^4\,e\right )\,{\left (d+e\,x\right )}^4+a^5\,e^5-b^5\,d^5+{\left (d+e\,x\right )}^3\,\left (10\,a^2\,b^3\,e^2-20\,a\,b^4\,d\,e+10\,b^5\,d^2\right )-10\,a^2\,b^3\,d^3\,e^2+10\,a^3\,b^2\,d^2\,e^3+5\,a\,b^4\,d^4\,e-5\,a^4\,b\,d\,e^4} \] Input:

int((d + e*x)^(5/2)/(a^2 + b^2*x^2 + 2*a*b*x)^3,x)
 

Output:

(3*e^5*atan((b^(1/2)*(d + e*x)^(1/2))/(a*e - b*d)^(1/2)))/(128*b^(7/2)*(a* 
e - b*d)^(5/2)) - ((e^5*(d + e*x)^(5/2))/(5*b) - (7*e^5*(d + e*x)^(7/2))/( 
64*(a*e - b*d)) + (3*e^5*(d + e*x)^(1/2)*(a^2*e^2 + b^2*d^2 - 2*a*b*d*e))/ 
(128*b^3) + (7*e^5*(a*e - b*d)*(d + e*x)^(3/2))/(64*b^2) - (3*b*e^5*(d + e 
*x)^(9/2))/(128*(a*e - b*d)^2))/((d + e*x)*(5*b^5*d^4 + 5*a^4*b*e^4 - 20*a 
^3*b^2*d*e^3 + 30*a^2*b^3*d^2*e^2 - 20*a*b^4*d^3*e) - (d + e*x)^2*(10*b^5* 
d^3 - 10*a^3*b^2*e^3 + 30*a^2*b^3*d*e^2 - 30*a*b^4*d^2*e) + b^5*(d + e*x)^ 
5 - (5*b^5*d - 5*a*b^4*e)*(d + e*x)^4 + a^5*e^5 - b^5*d^5 + (d + e*x)^3*(1 
0*b^5*d^2 + 10*a^2*b^3*e^2 - 20*a*b^4*d*e) - 10*a^2*b^3*d^3*e^2 + 10*a^3*b 
^2*d^2*e^3 + 5*a*b^4*d^4*e - 5*a^4*b*d*e^4)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 973, normalized size of antiderivative = 4.91 \[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx =\text {Too large to display} \] Input:

int((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^3,x)
 

Output:

(15*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d 
)))*a**5*e**5 + 75*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b) 
*sqrt(a*e - b*d)))*a**4*b*e**5*x + 150*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt( 
d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*a**3*b**2*e**5*x**2 + 150*sqrt(b)*s 
qrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*a**2*b**3 
*e**5*x**3 + 75*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sq 
rt(a*e - b*d)))*a*b**4*e**5*x**4 + 15*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d 
 + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*b**5*e**5*x**5 - 15*sqrt(d + e*x)*a* 
*5*b*e**5 + 5*sqrt(d + e*x)*a**4*b**2*d*e**4 - 70*sqrt(d + e*x)*a**4*b**2* 
e**5*x + 2*sqrt(d + e*x)*a**3*b**3*d**2*e**3 + 24*sqrt(d + e*x)*a**3*b**3* 
d*e**4*x - 128*sqrt(d + e*x)*a**3*b**3*e**5*x**2 + 184*sqrt(d + e*x)*a**2* 
b**4*d**3*e**2 + 558*sqrt(d + e*x)*a**2*b**4*d**2*e**3*x + 594*sqrt(d + e* 
x)*a**2*b**4*d*e**4*x**2 + 70*sqrt(d + e*x)*a**2*b**4*e**5*x**3 - 304*sqrt 
(d + e*x)*a*b**5*d**4*e - 848*sqrt(d + e*x)*a*b**5*d**3*e**2*x - 714*sqrt( 
d + e*x)*a*b**5*d**2*e**3*x**2 - 80*sqrt(d + e*x)*a*b**5*d*e**4*x**3 + 15* 
sqrt(d + e*x)*a*b**5*e**5*x**4 + 128*sqrt(d + e*x)*b**6*d**5 + 336*sqrt(d 
+ e*x)*b**6*d**4*e*x + 248*sqrt(d + e*x)*b**6*d**3*e**2*x**2 + 10*sqrt(d + 
 e*x)*b**6*d**2*e**3*x**3 - 15*sqrt(d + e*x)*b**6*d*e**4*x**4)/(640*b**4*( 
a**8*e**3 - 3*a**7*b*d*e**2 + 5*a**7*b*e**3*x + 3*a**6*b**2*d**2*e - 15*a* 
*6*b**2*d*e**2*x + 10*a**6*b**2*e**3*x**2 - a**5*b**3*d**3 + 15*a**5*b*...