3.22.36 \(\int \frac {a+b x}{(d+e x)^{5/2} (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [2136]

3.22.36.1 Optimal result
3.22.36.2 Mathematica [A] (verified)
3.22.36.3 Rubi [A] (verified)
3.22.36.4 Maple [A] (verified)
3.22.36.5 Fricas [B] (verification not implemented)
3.22.36.6 Sympy [F]
3.22.36.7 Maxima [F]
3.22.36.8 Giac [A] (verification not implemented)
3.22.36.9 Mupad [F(-1)]

3.22.36.1 Optimal result

Integrand size = 35, antiderivative size = 212 \[ \int \frac {a+b x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {1}{(b d-a e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (a+b x)}{3 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 b e (a+b x)}{(b d-a e)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 b^{3/2} e (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]

output
-1/(-a*e+b*d)/(e*x+d)^(3/2)/((b*x+a)^2)^(1/2)-5/3*e*(b*x+a)/(-a*e+b*d)^2/( 
e*x+d)^(3/2)/((b*x+a)^2)^(1/2)+5*b^(3/2)*e*(b*x+a)*arctanh(b^(1/2)*(e*x+d) 
^(1/2)/(-a*e+b*d)^(1/2))/(-a*e+b*d)^(7/2)/((b*x+a)^2)^(1/2)-5*b*e*(b*x+a)/ 
(-a*e+b*d)^3/(e*x+d)^(1/2)/((b*x+a)^2)^(1/2)
 
3.22.36.2 Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.68 \[ \int \frac {a+b x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {e (a+b x) \left (\frac {-2 a^2 e^2+2 a b e (7 d+5 e x)+b^2 \left (3 d^2+20 d e x+15 e^2 x^2\right )}{e (-b d+a e)^3 (a+b x) (d+e x)^{3/2}}+\frac {15 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{7/2}}\right )}{3 \sqrt {(a+b x)^2}} \]

input
Integrate[(a + b*x)/((d + e*x)^(5/2)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]
 
output
(e*(a + b*x)*((-2*a^2*e^2 + 2*a*b*e*(7*d + 5*e*x) + b^2*(3*d^2 + 20*d*e*x 
+ 15*e^2*x^2))/(e*(-(b*d) + a*e)^3*(a + b*x)*(d + e*x)^(3/2)) + (15*b^(3/2 
)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(-(b*d) + a*e)^(7/2) 
))/(3*Sqrt[(a + b*x)^2])
 
3.22.36.3 Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.83, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1187, 27, 52, 61, 61, 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.

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

\(\Big \downarrow \) 1187

\(\displaystyle \frac {b^3 (a+b x) \int \frac {1}{b^3 (a+b x)^2 (d+e x)^{5/2}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(a+b x) \int \frac {1}{(a+b x)^2 (d+e x)^{5/2}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {(a+b x) \left (-\frac {5 e \left (\frac {b \left (\frac {2}{\sqrt {d+e x} (b d-a e)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{3/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

input
Int[(a + b*x)/((d + e*x)^(5/2)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]
 
output
((a + b*x)*(-(1/((b*d - a*e)*(a + b*x)*(d + e*x)^(3/2))) - (5*e*(2/(3*(b*d 
 - a*e)*(d + e*x)^(3/2)) + (b*(2/((b*d - a*e)*Sqrt[d + e*x]) - (2*Sqrt[b]* 
ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(b*d - a*e)^(3/2)))/(b*d 
 - a*e)))/(2*(b*d - a*e))))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
 

3.22.36.3.1 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 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 61
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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

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 1187
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]
 
3.22.36.4 Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.14

method result size
default \(\frac {\left (15 \left (e x +d \right )^{\frac {3}{2}} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) b^{3} e x +15 \left (e x +d \right )^{\frac {3}{2}} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{2} e +15 \sqrt {\left (a e -b d \right ) b}\, b^{2} e^{2} x^{2}+10 \sqrt {\left (a e -b d \right ) b}\, a b \,e^{2} x +20 \sqrt {\left (a e -b d \right ) b}\, b^{2} d e x -2 \sqrt {\left (a e -b d \right ) b}\, a^{2} e^{2}+14 \sqrt {\left (a e -b d \right ) b}\, a b d e +3 \sqrt {\left (a e -b d \right ) b}\, b^{2} d^{2}\right ) \left (b x +a \right )^{2}}{3 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} \left (a e -b d \right )^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\) \(242\)

input
int((b*x+a)/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERB 
OSE)
 
output
1/3*(15*(e*x+d)^(3/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*b^3*e*x+ 
15*(e*x+d)^(3/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a*b^2*e+15*(( 
a*e-b*d)*b)^(1/2)*b^2*e^2*x^2+10*((a*e-b*d)*b)^(1/2)*a*b*e^2*x+20*((a*e-b* 
d)*b)^(1/2)*b^2*d*e*x-2*((a*e-b*d)*b)^(1/2)*a^2*e^2+14*((a*e-b*d)*b)^(1/2) 
*a*b*d*e+3*((a*e-b*d)*b)^(1/2)*b^2*d^2)*(b*x+a)^2/((a*e-b*d)*b)^(1/2)/(e*x 
+d)^(3/2)/(a*e-b*d)^3/((b*x+a)^2)^(3/2)
 
3.22.36.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 386 vs. \(2 (150) = 300\).

Time = 0.33 (sec) , antiderivative size = 782, normalized size of antiderivative = 3.69 \[ \int \frac {a+b x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\left [-\frac {15 \, {\left (b^{2} e^{3} x^{3} + a b d^{2} e + {\left (2 \, b^{2} d e^{2} + a b e^{3}\right )} x^{2} + {\left (b^{2} d^{2} e + 2 \, a b d e^{2}\right )} x\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, {\left (b d - a e\right )} \sqrt {e x + d} \sqrt {\frac {b}{b d - a e}}}{b x + a}\right ) + 2 \, {\left (15 \, b^{2} e^{2} x^{2} + 3 \, b^{2} d^{2} + 14 \, a b d e - 2 \, a^{2} e^{2} + 10 \, {\left (2 \, b^{2} d e + a b e^{2}\right )} x\right )} \sqrt {e x + d}}{6 \, {\left (a b^{3} d^{5} - 3 \, a^{2} b^{2} d^{4} e + 3 \, a^{3} b d^{3} e^{2} - a^{4} d^{2} e^{3} + {\left (b^{4} d^{3} e^{2} - 3 \, a b^{3} d^{2} e^{3} + 3 \, a^{2} b^{2} d e^{4} - a^{3} b e^{5}\right )} x^{3} + {\left (2 \, b^{4} d^{4} e - 5 \, a b^{3} d^{3} e^{2} + 3 \, a^{2} b^{2} d^{2} e^{3} + a^{3} b d e^{4} - a^{4} e^{5}\right )} x^{2} + {\left (b^{4} d^{5} - a b^{3} d^{4} e - 3 \, a^{2} b^{2} d^{3} e^{2} + 5 \, a^{3} b d^{2} e^{3} - 2 \, a^{4} d e^{4}\right )} x\right )}}, \frac {15 \, {\left (b^{2} e^{3} x^{3} + a b d^{2} e + {\left (2 \, b^{2} d e^{2} + a b e^{3}\right )} x^{2} + {\left (b^{2} d^{2} e + 2 \, a b d e^{2}\right )} x\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {e x + d} \sqrt {-\frac {b}{b d - a e}}}{b e x + b d}\right ) - {\left (15 \, b^{2} e^{2} x^{2} + 3 \, b^{2} d^{2} + 14 \, a b d e - 2 \, a^{2} e^{2} + 10 \, {\left (2 \, b^{2} d e + a b e^{2}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (a b^{3} d^{5} - 3 \, a^{2} b^{2} d^{4} e + 3 \, a^{3} b d^{3} e^{2} - a^{4} d^{2} e^{3} + {\left (b^{4} d^{3} e^{2} - 3 \, a b^{3} d^{2} e^{3} + 3 \, a^{2} b^{2} d e^{4} - a^{3} b e^{5}\right )} x^{3} + {\left (2 \, b^{4} d^{4} e - 5 \, a b^{3} d^{3} e^{2} + 3 \, a^{2} b^{2} d^{2} e^{3} + a^{3} b d e^{4} - a^{4} e^{5}\right )} x^{2} + {\left (b^{4} d^{5} - a b^{3} d^{4} e - 3 \, a^{2} b^{2} d^{3} e^{2} + 5 \, a^{3} b d^{2} e^{3} - 2 \, a^{4} d e^{4}\right )} x\right )}}\right ] \]

input
integrate((b*x+a)/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm=" 
fricas")
 
output
[-1/6*(15*(b^2*e^3*x^3 + a*b*d^2*e + (2*b^2*d*e^2 + a*b*e^3)*x^2 + (b^2*d^ 
2*e + 2*a*b*d*e^2)*x)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e - 2*(b* 
d - a*e)*sqrt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) + 2*(15*b^2*e^2*x^2 
 + 3*b^2*d^2 + 14*a*b*d*e - 2*a^2*e^2 + 10*(2*b^2*d*e + a*b*e^2)*x)*sqrt(e 
*x + d))/(a*b^3*d^5 - 3*a^2*b^2*d^4*e + 3*a^3*b*d^3*e^2 - a^4*d^2*e^3 + (b 
^4*d^3*e^2 - 3*a*b^3*d^2*e^3 + 3*a^2*b^2*d*e^4 - a^3*b*e^5)*x^3 + (2*b^4*d 
^4*e - 5*a*b^3*d^3*e^2 + 3*a^2*b^2*d^2*e^3 + a^3*b*d*e^4 - a^4*e^5)*x^2 + 
(b^4*d^5 - a*b^3*d^4*e - 3*a^2*b^2*d^3*e^2 + 5*a^3*b*d^2*e^3 - 2*a^4*d*e^4 
)*x), 1/3*(15*(b^2*e^3*x^3 + a*b*d^2*e + (2*b^2*d*e^2 + a*b*e^3)*x^2 + (b^ 
2*d^2*e + 2*a*b*d*e^2)*x)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a*e)*sqrt(e* 
x + d)*sqrt(-b/(b*d - a*e))/(b*e*x + b*d)) - (15*b^2*e^2*x^2 + 3*b^2*d^2 + 
 14*a*b*d*e - 2*a^2*e^2 + 10*(2*b^2*d*e + a*b*e^2)*x)*sqrt(e*x + d))/(a*b^ 
3*d^5 - 3*a^2*b^2*d^4*e + 3*a^3*b*d^3*e^2 - a^4*d^2*e^3 + (b^4*d^3*e^2 - 3 
*a*b^3*d^2*e^3 + 3*a^2*b^2*d*e^4 - a^3*b*e^5)*x^3 + (2*b^4*d^4*e - 5*a*b^3 
*d^3*e^2 + 3*a^2*b^2*d^2*e^3 + a^3*b*d*e^4 - a^4*e^5)*x^2 + (b^4*d^5 - a*b 
^3*d^4*e - 3*a^2*b^2*d^3*e^2 + 5*a^3*b*d^2*e^3 - 2*a^4*d*e^4)*x)]
 
3.22.36.6 Sympy [F]

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

input
integrate((b*x+a)/(e*x+d)**(5/2)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)
 
output
Integral((a + b*x)/((d + e*x)**(5/2)*((a + b*x)**2)**(3/2)), x)
 
3.22.36.7 Maxima [F]

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

input
integrate((b*x+a)/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm=" 
maxima")
 
output
integrate((b*x + a)/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*(e*x + d)^(5/2)), x)
 
3.22.36.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.36 \[ \int \frac {a+b x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {5 \, b^{2} e \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\left (b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} \sqrt {-b^{2} d + a b e}} - \frac {\sqrt {e x + d} b^{2} e}{{\left (b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}} - \frac {2 \, {\left (6 \, {\left (e x + d\right )} b e + b d e - a e^{2}\right )}}{3 \, {\left (b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} {\left (e x + d\right )}^{\frac {3}{2}}} \]

input
integrate((b*x+a)/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm=" 
giac")
 
output
-5*b^2*e*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((b^3*d^3*sgn(b*x + 
a) - 3*a*b^2*d^2*e*sgn(b*x + a) + 3*a^2*b*d*e^2*sgn(b*x + a) - a^3*e^3*sgn 
(b*x + a))*sqrt(-b^2*d + a*b*e)) - sqrt(e*x + d)*b^2*e/((b^3*d^3*sgn(b*x + 
 a) - 3*a*b^2*d^2*e*sgn(b*x + a) + 3*a^2*b*d*e^2*sgn(b*x + a) - a^3*e^3*sg 
n(b*x + a))*((e*x + d)*b - b*d + a*e)) - 2/3*(6*(e*x + d)*b*e + b*d*e - a* 
e^2)/((b^3*d^3*sgn(b*x + a) - 3*a*b^2*d^2*e*sgn(b*x + a) + 3*a^2*b*d*e^2*s 
gn(b*x + a) - a^3*e^3*sgn(b*x + a))*(e*x + d)^(3/2))
 
3.22.36.9 Mupad [F(-1)]

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

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