\(\int (a+b x)^{3/2} (A+B x) (d+e x)^{3/2} \, dx\) [183]

Optimal result

Integrand size = 24, antiderivative size = 294 \[ \int (a+b x)^{3/2} (A+B x) (d+e x)^{3/2} \, dx=\frac {3 (b d-a e)^3 (b B d-2 A b e+a B e) \sqrt {a+b x} \sqrt {d+e x}}{128 b^3 e^3}-\frac {(b d-a e)^2 (b B d-2 A b e+a B e) (a+b x)^{3/2} \sqrt {d+e x}}{64 b^3 e^2}-\frac {(b d-a e) (b B d-2 A b e+a B e) (a+b x)^{5/2} \sqrt {d+e x}}{16 b^3 e}-\frac {(b B d-2 A b e+a B e) (a+b x)^{5/2} (d+e x)^{3/2}}{8 b^2 e}+\frac {B (a+b x)^{5/2} (d+e x)^{5/2}}{5 b e}-\frac {3 (b d-a e)^4 (b B d-2 A b e+a B e) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{128 b^{7/2} e^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.09 \[ \int (a+b x)^{3/2} (A+B x) (d+e x)^{3/2} \, dx=\frac {\sqrt {a+b x} \sqrt {d+e x} \left (15 a^4 B e^4-10 a^3 b e^3 (4 B d+3 A e+B e x)+2 a^2 b^2 e^2 \left (5 A e (11 d+2 e x)+B \left (9 d^2+13 d e x+4 e^2 x^2\right )\right )+2 a b^3 e \left (5 A e \left (11 d^2+44 d e x+24 e^2 x^2\right )+B \left (-20 d^3+13 d^2 e x+136 d e^2 x^2+88 e^3 x^3\right )\right )+b^4 \left (10 A e \left (-3 d^3+2 d^2 e x+24 d e^2 x^2+16 e^3 x^3\right )+B \left (15 d^4-10 d^3 e x+8 d^2 e^2 x^2+176 d e^3 x^3+128 e^4 x^4\right )\right )\right )}{640 b^3 e^3}-\frac {3 (b d-a e)^4 (b B d-2 A b e+a B e) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {e} \sqrt {a+b x}}\right )}{128 b^{7/2} e^{7/2}} \] Input:

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

Output:

(Sqrt[a + b*x]*Sqrt[d + e*x]*(15*a^4*B*e^4 - 10*a^3*b*e^3*(4*B*d + 3*A*e + 
 B*e*x) + 2*a^2*b^2*e^2*(5*A*e*(11*d + 2*e*x) + B*(9*d^2 + 13*d*e*x + 4*e^ 
2*x^2)) + 2*a*b^3*e*(5*A*e*(11*d^2 + 44*d*e*x + 24*e^2*x^2) + B*(-20*d^3 + 
 13*d^2*e*x + 136*d*e^2*x^2 + 88*e^3*x^3)) + b^4*(10*A*e*(-3*d^3 + 2*d^2*e 
*x + 24*d*e^2*x^2 + 16*e^3*x^3) + B*(15*d^4 - 10*d^3*e*x + 8*d^2*e^2*x^2 + 
 176*d*e^3*x^3 + 128*e^4*x^4))))/(640*b^3*e^3) - (3*(b*d - a*e)^4*(b*B*d - 
 2*A*b*e + a*B*e)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/(Sqrt[e]*Sqrt[a + b*x])] 
)/(128*b^(7/2)*e^(7/2))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.86, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {90, 60, 60, 60, 60, 66, 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.

\(\displaystyle \int (a+b x)^{3/2} (A+B x) (d+e x)^{3/2} \, dx\)

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 60

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

rule 66

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1371\) vs. \(2(250)=500\).

Time = 0.27 (sec) , antiderivative size = 1372, normalized size of antiderivative = 4.67

method	result	size
default	\(\text {Expression too large to display}\)	\(1372\)

Input:

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

Output:

1/1280*(b*x+a)^(1/2)*(e*x+d)^(1/2)*(256*B*b^4*e^4*x^4*((e*x+d)*(b*x+a))^(1 
/2)*(b*e)^(1/2)+320*A*b^4*e^4*x^3*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+352* 
B*a*b^3*e^4*x^3*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+352*B*b^4*d*e^3*x^3*(( 
e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+480*A*a*b^3*e^4*x^2*((e*x+d)*(b*x+a))^(1 
/2)*(b*e)^(1/2)+220*A*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)*a*b^3*d^2*e^2-80 
*B*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)*a^3*b*d*e^3+36*B*((e*x+d)*(b*x+a))^ 
(1/2)*(b*e)^(1/2)*a^2*b^2*d^2*e^2-80*B*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2) 
*a*b^3*d^3*e+30*B*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)*a^4*e^4+30*B*((e*x+d 
)*(b*x+a))^(1/2)*(b*e)^(1/2)*b^4*d^4+544*B*a*b^3*d*e^3*x^2*((e*x+d)*(b*x+a 
))^(1/2)*(b*e)^(1/2)+880*A*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)*a*b^3*d*e^3 
*x+52*B*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)*a^2*b^2*d*e^3*x+52*B*((e*x+d)* 
(b*x+a))^(1/2)*(b*e)^(1/2)*a*b^3*d^2*e^2*x+30*A*ln(1/2*(2*b*e*x+2*((e*x+d) 
*(b*x+a))^(1/2)*(b*e)^(1/2)+a*e+d*b)/(b*e)^(1/2))*a^4*b*e^5+30*A*ln(1/2*(2 
*b*e*x+2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+a*e+d*b)/(b*e)^(1/2))*b^5*d^4 
*e+180*A*ln(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+a*e+d*b)/(b 
*e)^(1/2))*a^2*b^3*d^2*e^3-120*A*ln(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/2) 
*(b*e)^(1/2)+a*e+d*b)/(b*e)^(1/2))*a*b^4*d^3*e^2+45*B*ln(1/2*(2*b*e*x+2*(( 
e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+a*e+d*b)/(b*e)^(1/2))*a^4*b*d*e^4-30*B*l 
n(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+a*e+d*b)/(b*e)^(1/2)) 
*a^3*b^2*d^2*e^3-30*B*ln(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 511 vs. \(2 (250) = 500\).

Time = 0.16 (sec) , antiderivative size = 1036, normalized size of antiderivative = 3.52 \[ \int (a+b x)^{3/2} (A+B x) (d+e x)^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/2560*(15*(B*b^5*d^5 - (3*B*a*b^4 + 2*A*b^5)*d^4*e + 2*(B*a^2*b^3 + 4*A 
*a*b^4)*d^3*e^2 + 2*(B*a^3*b^2 - 6*A*a^2*b^3)*d^2*e^3 - (3*B*a^4*b - 8*A*a 
^3*b^2)*d*e^4 + (B*a^5 - 2*A*a^4*b)*e^5)*sqrt(b*e)*log(8*b^2*e^2*x^2 + b^2 
*d^2 + 6*a*b*d*e + a^2*e^2 + 4*(2*b*e*x + b*d + a*e)*sqrt(b*e)*sqrt(b*x + 
a)*sqrt(e*x + d) + 8*(b^2*d*e + a*b*e^2)*x) - 4*(128*B*b^5*e^5*x^4 + 15*B* 
b^5*d^4*e - 10*(4*B*a*b^4 + 3*A*b^5)*d^3*e^2 + 2*(9*B*a^2*b^3 + 55*A*a*b^4 
)*d^2*e^3 - 10*(4*B*a^3*b^2 - 11*A*a^2*b^3)*d*e^4 + 15*(B*a^4*b - 2*A*a^3* 
b^2)*e^5 + 16*(11*B*b^5*d*e^4 + (11*B*a*b^4 + 10*A*b^5)*e^5)*x^3 + 8*(B*b^ 
5*d^2*e^3 + 2*(17*B*a*b^4 + 15*A*b^5)*d*e^4 + (B*a^2*b^3 + 30*A*a*b^4)*e^5 
)*x^2 - 2*(5*B*b^5*d^3*e^2 - (13*B*a*b^4 + 10*A*b^5)*d^2*e^3 - (13*B*a^2*b 
^3 + 220*A*a*b^4)*d*e^4 + 5*(B*a^3*b^2 - 2*A*a^2*b^3)*e^5)*x)*sqrt(b*x + a 
)*sqrt(e*x + d))/(b^4*e^4), 1/1280*(15*(B*b^5*d^5 - (3*B*a*b^4 + 2*A*b^5)* 
d^4*e + 2*(B*a^2*b^3 + 4*A*a*b^4)*d^3*e^2 + 2*(B*a^3*b^2 - 6*A*a^2*b^3)*d^ 
2*e^3 - (3*B*a^4*b - 8*A*a^3*b^2)*d*e^4 + (B*a^5 - 2*A*a^4*b)*e^5)*sqrt(-b 
*e)*arctan(1/2*(2*b*e*x + b*d + a*e)*sqrt(-b*e)*sqrt(b*x + a)*sqrt(e*x + d 
)/(b^2*e^2*x^2 + a*b*d*e + (b^2*d*e + a*b*e^2)*x)) + 2*(128*B*b^5*e^5*x^4 
+ 15*B*b^5*d^4*e - 10*(4*B*a*b^4 + 3*A*b^5)*d^3*e^2 + 2*(9*B*a^2*b^3 + 55* 
A*a*b^4)*d^2*e^3 - 10*(4*B*a^3*b^2 - 11*A*a^2*b^3)*d*e^4 + 15*(B*a^4*b - 2 
*A*a^3*b^2)*e^5 + 16*(11*B*b^5*d*e^4 + (11*B*a*b^4 + 10*A*b^5)*e^5)*x^3 + 
8*(B*b^5*d^2*e^3 + 2*(17*B*a*b^4 + 15*A*b^5)*d*e^4 + (B*a^2*b^3 + 30*A*...
 

Sympy [F]

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

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

Output:

Integral((A + B*x)*(a + b*x)**(3/2)*(d + e*x)**(3/2), x)
 

Maxima [F(-2)]

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

integrate((b*x+a)^(3/2)*(B*x+A)*(e*x+d)^(3/2),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(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2484 vs. \(2 (250) = 500\).

Time = 0.45 (sec) , antiderivative size = 2484, normalized size of antiderivative = 8.45 \[ \int (a+b x)^{3/2} (A+B x) (d+e x)^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/1920*(10*(sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*(2*(b*x + a)*(4*(b*x + a)* 
(6*(b*x + a)/b^3 + (b^12*d*e^5 - 25*a*b^11*e^6)/(b^14*e^6)) - (5*b^13*d^2* 
e^4 + 14*a*b^12*d*e^5 - 163*a^2*b^11*e^6)/(b^14*e^6)) + 3*(5*b^14*d^3*e^3 
+ 9*a*b^13*d^2*e^4 + 15*a^2*b^12*d*e^5 - 93*a^3*b^11*e^6)/(b^14*e^6))*sqrt 
(b*x + a) + 3*(5*b^4*d^4 + 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 + 20*a^3*b*d* 
e^3 - 35*a^4*e^4)*log(abs(-sqrt(b*e)*sqrt(b*x + a) + sqrt(b^2*d + (b*x + a 
)*b*e - a*b*e)))/(sqrt(b*e)*b^2*e^3))*B*d*abs(b) - 1920*((b^2*d - a*b*e)*l 
og(abs(-sqrt(b*e)*sqrt(b*x + a) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/sq 
rt(b*e) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a))*A*a^2*d*abs(b 
)/b^2 + 10*(sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*(2*(b*x + a)*(4*(b*x + a)* 
(6*(b*x + a)/b^3 + (b^12*d*e^5 - 25*a*b^11*e^6)/(b^14*e^6)) - (5*b^13*d^2* 
e^4 + 14*a*b^12*d*e^5 - 163*a^2*b^11*e^6)/(b^14*e^6)) + 3*(5*b^14*d^3*e^3 
+ 9*a*b^13*d^2*e^4 + 15*a^2*b^12*d*e^5 - 93*a^3*b^11*e^6)/(b^14*e^6))*sqrt 
(b*x + a) + 3*(5*b^4*d^4 + 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 + 20*a^3*b*d* 
e^3 - 35*a^4*e^4)*log(abs(-sqrt(b*e)*sqrt(b*x + a) + sqrt(b^2*d + (b*x + a 
)*b*e - a*b*e)))/(sqrt(b*e)*b^2*e^3))*A*e*abs(b) + 20*(sqrt(b^2*d + (b*x + 
 a)*b*e - a*b*e)*(2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a)/b^3 + (b^12*d*e^5 
- 25*a*b^11*e^6)/(b^14*e^6)) - (5*b^13*d^2*e^4 + 14*a*b^12*d*e^5 - 163*a^2 
*b^11*e^6)/(b^14*e^6)) + 3*(5*b^14*d^3*e^3 + 9*a*b^13*d^2*e^4 + 15*a^2*b^1 
2*d*e^5 - 93*a^3*b^11*e^6)/(b^14*e^6))*sqrt(b*x + a) + 3*(5*b^4*d^4 + 4...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 651, normalized size of antiderivative = 2.21 \[ \int (a+b x)^{3/2} (A+B x) (d+e x)^{3/2} \, dx=\frac {-15 \sqrt {e x +d}\, \sqrt {b x +a}\, a^{4} b \,e^{5}+70 \sqrt {e x +d}\, \sqrt {b x +a}\, a^{3} b^{2} d \,e^{4}+10 \sqrt {e x +d}\, \sqrt {b x +a}\, a^{3} b^{2} e^{5} x +128 \sqrt {e x +d}\, \sqrt {b x +a}\, a^{2} b^{3} d^{2} e^{3}+466 \sqrt {e x +d}\, \sqrt {b x +a}\, a^{2} b^{3} d \,e^{4} x +248 \sqrt {e x +d}\, \sqrt {b x +a}\, a^{2} b^{3} e^{5} x^{2}-70 \sqrt {e x +d}\, \sqrt {b x +a}\, a \,b^{4} d^{3} e^{2}+46 \sqrt {e x +d}\, \sqrt {b x +a}\, a \,b^{4} d^{2} e^{3} x +512 \sqrt {e x +d}\, \sqrt {b x +a}\, a \,b^{4} d \,e^{4} x^{2}+336 \sqrt {e x +d}\, \sqrt {b x +a}\, a \,b^{4} e^{5} x^{3}+15 \sqrt {e x +d}\, \sqrt {b x +a}\, b^{5} d^{4} e -10 \sqrt {e x +d}\, \sqrt {b x +a}\, b^{5} d^{3} e^{2} x +8 \sqrt {e x +d}\, \sqrt {b x +a}\, b^{5} d^{2} e^{3} x^{2}+176 \sqrt {e x +d}\, \sqrt {b x +a}\, b^{5} d \,e^{4} x^{3}+128 \sqrt {e x +d}\, \sqrt {b x +a}\, b^{5} e^{5} x^{4}+15 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) a^{5} e^{5}-75 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) a^{4} b d \,e^{4}+150 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) a^{3} b^{2} d^{2} e^{3}-150 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) a^{2} b^{3} d^{3} e^{2}+75 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) a \,b^{4} d^{4} e -15 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) b^{5} d^{5}}{640 b^{3} e^{4}} \] Input:

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

Output:

( - 15*sqrt(d + e*x)*sqrt(a + b*x)*a**4*b*e**5 + 70*sqrt(d + e*x)*sqrt(a + 
 b*x)*a**3*b**2*d*e**4 + 10*sqrt(d + e*x)*sqrt(a + b*x)*a**3*b**2*e**5*x + 
 128*sqrt(d + e*x)*sqrt(a + b*x)*a**2*b**3*d**2*e**3 + 466*sqrt(d + e*x)*s 
qrt(a + b*x)*a**2*b**3*d*e**4*x + 248*sqrt(d + e*x)*sqrt(a + b*x)*a**2*b** 
3*e**5*x**2 - 70*sqrt(d + e*x)*sqrt(a + b*x)*a*b**4*d**3*e**2 + 46*sqrt(d 
+ e*x)*sqrt(a + b*x)*a*b**4*d**2*e**3*x + 512*sqrt(d + e*x)*sqrt(a + b*x)* 
a*b**4*d*e**4*x**2 + 336*sqrt(d + e*x)*sqrt(a + b*x)*a*b**4*e**5*x**3 + 15 
*sqrt(d + e*x)*sqrt(a + b*x)*b**5*d**4*e - 10*sqrt(d + e*x)*sqrt(a + b*x)* 
b**5*d**3*e**2*x + 8*sqrt(d + e*x)*sqrt(a + b*x)*b**5*d**2*e**3*x**2 + 176 
*sqrt(d + e*x)*sqrt(a + b*x)*b**5*d*e**4*x**3 + 128*sqrt(d + e*x)*sqrt(a + 
 b*x)*b**5*e**5*x**4 + 15*sqrt(e)*sqrt(b)*log((sqrt(e)*sqrt(a + b*x) + sqr 
t(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*a**5*e**5 - 75*sqrt(e)*sqrt(b)*log((s 
qrt(e)*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*a**4*b*d*e* 
*4 + 150*sqrt(e)*sqrt(b)*log((sqrt(e)*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x 
))/sqrt(a*e - b*d))*a**3*b**2*d**2*e**3 - 150*sqrt(e)*sqrt(b)*log((sqrt(e) 
*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*a**2*b**3*d**3*e* 
*2 + 75*sqrt(e)*sqrt(b)*log((sqrt(e)*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x) 
)/sqrt(a*e - b*d))*a*b**4*d**4*e - 15*sqrt(e)*sqrt(b)*log((sqrt(e)*sqrt(a 
+ b*x) + sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*b**5*d**5)/(640*b**3*e**4 
)