3.18.38 \(\int (d+e x)^m (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\) [1738]

3.18.38.1 Optimal result
3.18.38.2 Mathematica [A] (verified)
3.18.38.3 Rubi [A] (verified)
3.18.38.4 Maple [B] (verified)
3.18.38.5 Fricas [B] (verification not implemented)
3.18.38.6 Sympy [F]
3.18.38.7 Maxima [A] (verification not implemented)
3.18.38.8 Giac [B] (verification not implemented)
3.18.38.9 Mupad [F(-1)]

3.18.38.1 Optimal result

Integrand size = 28, antiderivative size = 219 \[ \int (d+e x)^m \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=-\frac {(b d-a e)^3 (d+e x)^{1+m} \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (1+m) (a+b x)}+\frac {3 b (b d-a e)^2 (d+e x)^{2+m} \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (2+m) (a+b x)}-\frac {3 b^2 (b d-a e) (d+e x)^{3+m} \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (3+m) (a+b x)}+\frac {b^3 (d+e x)^{4+m} \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (4+m) (a+b x)} \]

output
-(-a*e+b*d)^3*(e*x+d)^(1+m)*((b*x+a)^2)^(1/2)/e^4/(1+m)/(b*x+a)+3*b*(-a*e+ 
b*d)^2*(e*x+d)^(2+m)*((b*x+a)^2)^(1/2)/e^4/(2+m)/(b*x+a)-3*b^2*(-a*e+b*d)* 
(e*x+d)^(3+m)*((b*x+a)^2)^(1/2)/e^4/(3+m)/(b*x+a)+b^3*(e*x+d)^(4+m)*((b*x+ 
a)^2)^(1/2)/e^4/(4+m)/(b*x+a)
 
3.18.38.2 Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.52 \[ \int (d+e x)^m \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {\left ((a+b x)^2\right )^{3/2} (d+e x)^{1+m} \left (-\frac {(b d-a e)^3}{1+m}+\frac {3 b (b d-a e)^2 (d+e x)}{2+m}-\frac {3 b^2 (b d-a e) (d+e x)^2}{3+m}+\frac {b^3 (d+e x)^3}{4+m}\right )}{e^4 (a+b x)^3} \]

input
Integrate[(d + e*x)^m*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
 
output
(((a + b*x)^2)^(3/2)*(d + e*x)^(1 + m)*(-((b*d - a*e)^3/(1 + m)) + (3*b*(b 
*d - a*e)^2*(d + e*x))/(2 + m) - (3*b^2*(b*d - a*e)*(d + e*x)^2)/(3 + m) + 
 (b^3*(d + e*x)^3)/(4 + m)))/(e^4*(a + b*x)^3)
 
3.18.38.3 Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.63, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1102, 27, 53, 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.

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

\(\Big \downarrow \) 1102

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b^3 (a+b x)^3 (d+e x)^mdx}{b^3 (a+b x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int (a+b x)^3 (d+e x)^mdx}{a+b x}\)

\(\Big \downarrow \) 53

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {3 b^2 (b d-a e) (d+e x)^{m+3}}{e^4 (m+3)}-\frac {(b d-a e)^3 (d+e x)^{m+1}}{e^4 (m+1)}+\frac {3 b (b d-a e)^2 (d+e x)^{m+2}}{e^4 (m+2)}+\frac {b^3 (d+e x)^{m+4}}{e^4 (m+4)}\right )}{a+b x}\)

input
Int[(d + e*x)^m*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
 
output
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-(((b*d - a*e)^3*(d + e*x)^(1 + m))/(e^4*( 
1 + m))) + (3*b*(b*d - a*e)^2*(d + e*x)^(2 + m))/(e^4*(2 + m)) - (3*b^2*(b 
*d - a*e)*(d + e*x)^(3 + m))/(e^4*(3 + m)) + (b^3*(d + e*x)^(4 + m))/(e^4* 
(4 + m))))/(a + b*x)
 

3.18.38.3.1 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 53
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, n}, 
x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) 
|| LtQ[9*m + 5*(n + 1), 0] || GtQ[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]
 
3.18.38.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(401\) vs. \(2(175)=350\).

Time = 2.25 (sec) , antiderivative size = 402, normalized size of antiderivative = 1.84

method result size
gosper \(\frac {\left (e x +d \right )^{1+m} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (b^{3} e^{3} m^{3} x^{3}+3 a \,b^{2} e^{3} m^{3} x^{2}+6 b^{3} e^{3} m^{2} x^{3}+3 a^{2} b \,e^{3} m^{3} x +21 a \,b^{2} e^{3} m^{2} x^{2}-3 b^{3} d \,e^{2} m^{2} x^{2}+11 b^{3} e^{3} m \,x^{3}+a^{3} e^{3} m^{3}+24 a^{2} b \,e^{3} m^{2} x -6 a \,b^{2} d \,e^{2} m^{2} x +42 a \,b^{2} e^{3} m \,x^{2}-9 b^{3} d \,e^{2} m \,x^{2}+6 e^{3} x^{3} b^{3}+9 a^{3} e^{3} m^{2}-3 a^{2} b d \,e^{2} m^{2}+57 a^{2} b \,e^{3} m x -30 a \,b^{2} d \,e^{2} m x +24 x^{2} a \,b^{2} e^{3}+6 b^{3} d^{2} e m x -6 x^{2} b^{3} d \,e^{2}+26 a^{3} e^{3} m -21 a^{2} b d \,e^{2} m +36 a^{2} b \,e^{3} x +6 a \,b^{2} d^{2} e m -24 x a \,b^{2} d \,e^{2}+6 b^{3} d^{2} e x +24 a^{3} e^{3}-36 a^{2} b d \,e^{2}+24 a \,b^{2} d^{2} e -6 b^{3} d^{3}\right )}{e^{4} \left (b x +a \right )^{3} \left (m^{4}+10 m^{3}+35 m^{2}+50 m +24\right )}\) \(402\)
risch \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (b^{3} e^{4} m^{3} x^{4}+3 a \,b^{2} e^{4} m^{3} x^{3}+b^{3} d \,e^{3} m^{3} x^{3}+6 b^{3} e^{4} m^{2} x^{4}+3 a^{2} b \,e^{4} m^{3} x^{2}+3 a \,b^{2} d \,e^{3} m^{3} x^{2}+21 a \,b^{2} e^{4} m^{2} x^{3}+3 b^{3} d \,e^{3} m^{2} x^{3}+11 b^{3} e^{4} m \,x^{4}+a^{3} e^{4} m^{3} x +3 a^{2} b d \,e^{3} m^{3} x +24 a^{2} b \,e^{4} m^{2} x^{2}+15 a \,b^{2} d \,e^{3} m^{2} x^{2}+42 a \,b^{2} e^{4} m \,x^{3}-3 b^{3} d^{2} e^{2} m^{2} x^{2}+2 b^{3} d \,e^{3} m \,x^{3}+6 b^{3} x^{4} e^{4}+a^{3} d \,e^{3} m^{3}+9 a^{3} e^{4} m^{2} x +21 a^{2} b d \,e^{3} m^{2} x +57 a^{2} b \,e^{4} m \,x^{2}-6 a \,b^{2} d^{2} e^{2} m^{2} x +12 a \,b^{2} d \,e^{3} m \,x^{2}+24 a \,b^{2} e^{4} x^{3}-3 b^{3} d^{2} e^{2} m \,x^{2}+9 a^{3} d \,e^{3} m^{2}+26 a^{3} e^{4} m x -3 a^{2} b \,d^{2} e^{2} m^{2}+36 a^{2} b d \,e^{3} m x +36 a^{2} b \,e^{4} x^{2}-24 a \,b^{2} d^{2} e^{2} m x +6 b^{3} d^{3} e m x +26 a^{3} d \,e^{3} m +24 e^{4} a^{3} x -21 a^{2} b \,d^{2} e^{2} m +6 a \,b^{2} d^{3} e m +24 a^{3} d \,e^{3}-36 a^{2} b \,d^{2} e^{2}+24 a \,b^{2} d^{3} e -6 b^{3} d^{4}\right ) \left (e x +d \right )^{m}}{\left (b x +a \right ) \left (3+m \right ) \left (4+m \right ) \left (2+m \right ) \left (1+m \right ) e^{4}}\) \(563\)

input
int((e*x+d)^m*(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERBOSE)
 
output
1/e^4*(e*x+d)^(1+m)/(b*x+a)^3*((b*x+a)^2)^(3/2)/(m^4+10*m^3+35*m^2+50*m+24 
)*(b^3*e^3*m^3*x^3+3*a*b^2*e^3*m^3*x^2+6*b^3*e^3*m^2*x^3+3*a^2*b*e^3*m^3*x 
+21*a*b^2*e^3*m^2*x^2-3*b^3*d*e^2*m^2*x^2+11*b^3*e^3*m*x^3+a^3*e^3*m^3+24* 
a^2*b*e^3*m^2*x-6*a*b^2*d*e^2*m^2*x+42*a*b^2*e^3*m*x^2-9*b^3*d*e^2*m*x^2+6 
*b^3*e^3*x^3+9*a^3*e^3*m^2-3*a^2*b*d*e^2*m^2+57*a^2*b*e^3*m*x-30*a*b^2*d*e 
^2*m*x+24*a*b^2*e^3*x^2+6*b^3*d^2*e*m*x-6*b^3*d*e^2*x^2+26*a^3*e^3*m-21*a^ 
2*b*d*e^2*m+36*a^2*b*e^3*x+6*a*b^2*d^2*e*m-24*a*b^2*d*e^2*x+6*b^3*d^2*e*x+ 
24*a^3*e^3-36*a^2*b*d*e^2+24*a*b^2*d^2*e-6*b^3*d^3)
 
3.18.38.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 496 vs. \(2 (175) = 350\).

Time = 0.58 (sec) , antiderivative size = 496, normalized size of antiderivative = 2.26 \[ \int (d+e x)^m \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {{\left (a^{3} d e^{3} m^{3} - 6 \, b^{3} d^{4} + 24 \, a b^{2} d^{3} e - 36 \, a^{2} b d^{2} e^{2} + 24 \, a^{3} d e^{3} + {\left (b^{3} e^{4} m^{3} + 6 \, b^{3} e^{4} m^{2} + 11 \, b^{3} e^{4} m + 6 \, b^{3} e^{4}\right )} x^{4} + {\left (24 \, a b^{2} e^{4} + {\left (b^{3} d e^{3} + 3 \, a b^{2} e^{4}\right )} m^{3} + 3 \, {\left (b^{3} d e^{3} + 7 \, a b^{2} e^{4}\right )} m^{2} + 2 \, {\left (b^{3} d e^{3} + 21 \, a b^{2} e^{4}\right )} m\right )} x^{3} - 3 \, {\left (a^{2} b d^{2} e^{2} - 3 \, a^{3} d e^{3}\right )} m^{2} + 3 \, {\left (12 \, a^{2} b e^{4} + {\left (a b^{2} d e^{3} + a^{2} b e^{4}\right )} m^{3} - {\left (b^{3} d^{2} e^{2} - 5 \, a b^{2} d e^{3} - 8 \, a^{2} b e^{4}\right )} m^{2} - {\left (b^{3} d^{2} e^{2} - 4 \, a b^{2} d e^{3} - 19 \, a^{2} b e^{4}\right )} m\right )} x^{2} + {\left (6 \, a b^{2} d^{3} e - 21 \, a^{2} b d^{2} e^{2} + 26 \, a^{3} d e^{3}\right )} m + {\left (24 \, a^{3} e^{4} + {\left (3 \, a^{2} b d e^{3} + a^{3} e^{4}\right )} m^{3} - 3 \, {\left (2 \, a b^{2} d^{2} e^{2} - 7 \, a^{2} b d e^{3} - 3 \, a^{3} e^{4}\right )} m^{2} + 2 \, {\left (3 \, b^{3} d^{3} e - 12 \, a b^{2} d^{2} e^{2} + 18 \, a^{2} b d e^{3} + 13 \, a^{3} e^{4}\right )} m\right )} x\right )} {\left (e x + d\right )}^{m}}{e^{4} m^{4} + 10 \, e^{4} m^{3} + 35 \, e^{4} m^{2} + 50 \, e^{4} m + 24 \, e^{4}} \]

input
integrate((e*x+d)^m*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")
 
output
(a^3*d*e^3*m^3 - 6*b^3*d^4 + 24*a*b^2*d^3*e - 36*a^2*b*d^2*e^2 + 24*a^3*d* 
e^3 + (b^3*e^4*m^3 + 6*b^3*e^4*m^2 + 11*b^3*e^4*m + 6*b^3*e^4)*x^4 + (24*a 
*b^2*e^4 + (b^3*d*e^3 + 3*a*b^2*e^4)*m^3 + 3*(b^3*d*e^3 + 7*a*b^2*e^4)*m^2 
 + 2*(b^3*d*e^3 + 21*a*b^2*e^4)*m)*x^3 - 3*(a^2*b*d^2*e^2 - 3*a^3*d*e^3)*m 
^2 + 3*(12*a^2*b*e^4 + (a*b^2*d*e^3 + a^2*b*e^4)*m^3 - (b^3*d^2*e^2 - 5*a* 
b^2*d*e^3 - 8*a^2*b*e^4)*m^2 - (b^3*d^2*e^2 - 4*a*b^2*d*e^3 - 19*a^2*b*e^4 
)*m)*x^2 + (6*a*b^2*d^3*e - 21*a^2*b*d^2*e^2 + 26*a^3*d*e^3)*m + (24*a^3*e 
^4 + (3*a^2*b*d*e^3 + a^3*e^4)*m^3 - 3*(2*a*b^2*d^2*e^2 - 7*a^2*b*d*e^3 - 
3*a^3*e^4)*m^2 + 2*(3*b^3*d^3*e - 12*a*b^2*d^2*e^2 + 18*a^2*b*d*e^3 + 13*a 
^3*e^4)*m)*x)*(e*x + d)^m/(e^4*m^4 + 10*e^4*m^3 + 35*e^4*m^2 + 50*e^4*m + 
24*e^4)
 
3.18.38.6 Sympy [F]

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

input
integrate((e*x+d)**m*(b**2*x**2+2*a*b*x+a**2)**(3/2),x)
 
output
Integral((d + e*x)**m*((a + b*x)**2)**(3/2), x)
 
3.18.38.7 Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.39 \[ \int (d+e x)^m \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {{\left ({\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} b^{3} e^{4} x^{4} - 3 \, {\left (m^{2} + 7 \, m + 12\right )} a^{2} b d^{2} e^{2} + {\left (m^{3} + 9 \, m^{2} + 26 \, m + 24\right )} a^{3} d e^{3} + 6 \, a b^{2} d^{3} e {\left (m + 4\right )} - 6 \, b^{3} d^{4} + {\left ({\left (m^{3} + 3 \, m^{2} + 2 \, m\right )} b^{3} d e^{3} + 3 \, {\left (m^{3} + 7 \, m^{2} + 14 \, m + 8\right )} a b^{2} e^{4}\right )} x^{3} - 3 \, {\left ({\left (m^{2} + m\right )} b^{3} d^{2} e^{2} - {\left (m^{3} + 5 \, m^{2} + 4 \, m\right )} a b^{2} d e^{3} - {\left (m^{3} + 8 \, m^{2} + 19 \, m + 12\right )} a^{2} b e^{4}\right )} x^{2} - {\left (6 \, {\left (m^{2} + 4 \, m\right )} a b^{2} d^{2} e^{2} - 3 \, {\left (m^{3} + 7 \, m^{2} + 12 \, m\right )} a^{2} b d e^{3} - {\left (m^{3} + 9 \, m^{2} + 26 \, m + 24\right )} a^{3} e^{4} - 6 \, b^{3} d^{3} e m\right )} x\right )} {\left (e x + d\right )}^{m}}{{\left (m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24\right )} e^{4}} \]

input
integrate((e*x+d)^m*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")
 
output
((m^3 + 6*m^2 + 11*m + 6)*b^3*e^4*x^4 - 3*(m^2 + 7*m + 12)*a^2*b*d^2*e^2 + 
 (m^3 + 9*m^2 + 26*m + 24)*a^3*d*e^3 + 6*a*b^2*d^3*e*(m + 4) - 6*b^3*d^4 + 
 ((m^3 + 3*m^2 + 2*m)*b^3*d*e^3 + 3*(m^3 + 7*m^2 + 14*m + 8)*a*b^2*e^4)*x^ 
3 - 3*((m^2 + m)*b^3*d^2*e^2 - (m^3 + 5*m^2 + 4*m)*a*b^2*d*e^3 - (m^3 + 8* 
m^2 + 19*m + 12)*a^2*b*e^4)*x^2 - (6*(m^2 + 4*m)*a*b^2*d^2*e^2 - 3*(m^3 + 
7*m^2 + 12*m)*a^2*b*d*e^3 - (m^3 + 9*m^2 + 26*m + 24)*a^3*e^4 - 6*b^3*d^3* 
e*m)*x)*(e*x + d)^m/((m^4 + 10*m^3 + 35*m^2 + 50*m + 24)*e^4)
 
3.18.38.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1073 vs. \(2 (175) = 350\).

Time = 0.31 (sec) , antiderivative size = 1073, normalized size of antiderivative = 4.90 \[ \int (d+e x)^m \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\text {Too large to display} \]

input
integrate((e*x+d)^m*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")
 
output
((e*x + d)^m*b^3*e^4*m^3*x^4*sgn(b*x + a) + (e*x + d)^m*b^3*d*e^3*m^3*x^3* 
sgn(b*x + a) + 3*(e*x + d)^m*a*b^2*e^4*m^3*x^3*sgn(b*x + a) + 6*(e*x + d)^ 
m*b^3*e^4*m^2*x^4*sgn(b*x + a) + 3*(e*x + d)^m*a*b^2*d*e^3*m^3*x^2*sgn(b*x 
 + a) + 3*(e*x + d)^m*a^2*b*e^4*m^3*x^2*sgn(b*x + a) + 3*(e*x + d)^m*b^3*d 
*e^3*m^2*x^3*sgn(b*x + a) + 21*(e*x + d)^m*a*b^2*e^4*m^2*x^3*sgn(b*x + a) 
+ 11*(e*x + d)^m*b^3*e^4*m*x^4*sgn(b*x + a) + 3*(e*x + d)^m*a^2*b*d*e^3*m^ 
3*x*sgn(b*x + a) + (e*x + d)^m*a^3*e^4*m^3*x*sgn(b*x + a) - 3*(e*x + d)^m* 
b^3*d^2*e^2*m^2*x^2*sgn(b*x + a) + 15*(e*x + d)^m*a*b^2*d*e^3*m^2*x^2*sgn( 
b*x + a) + 24*(e*x + d)^m*a^2*b*e^4*m^2*x^2*sgn(b*x + a) + 2*(e*x + d)^m*b 
^3*d*e^3*m*x^3*sgn(b*x + a) + 42*(e*x + d)^m*a*b^2*e^4*m*x^3*sgn(b*x + a) 
+ 6*(e*x + d)^m*b^3*e^4*x^4*sgn(b*x + a) + (e*x + d)^m*a^3*d*e^3*m^3*sgn(b 
*x + a) - 6*(e*x + d)^m*a*b^2*d^2*e^2*m^2*x*sgn(b*x + a) + 21*(e*x + d)^m* 
a^2*b*d*e^3*m^2*x*sgn(b*x + a) + 9*(e*x + d)^m*a^3*e^4*m^2*x*sgn(b*x + a) 
- 3*(e*x + d)^m*b^3*d^2*e^2*m*x^2*sgn(b*x + a) + 12*(e*x + d)^m*a*b^2*d*e^ 
3*m*x^2*sgn(b*x + a) + 57*(e*x + d)^m*a^2*b*e^4*m*x^2*sgn(b*x + a) + 24*(e 
*x + d)^m*a*b^2*e^4*x^3*sgn(b*x + a) - 3*(e*x + d)^m*a^2*b*d^2*e^2*m^2*sgn 
(b*x + a) + 9*(e*x + d)^m*a^3*d*e^3*m^2*sgn(b*x + a) + 6*(e*x + d)^m*b^3*d 
^3*e*m*x*sgn(b*x + a) - 24*(e*x + d)^m*a*b^2*d^2*e^2*m*x*sgn(b*x + a) + 36 
*(e*x + d)^m*a^2*b*d*e^3*m*x*sgn(b*x + a) + 26*(e*x + d)^m*a^3*e^4*m*x*sgn 
(b*x + a) + 36*(e*x + d)^m*a^2*b*e^4*x^2*sgn(b*x + a) + 6*(e*x + d)^m*a...
 
3.18.38.9 Mupad [F(-1)]

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

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