\(\int (a+b x)^{5/2} (A+B x) \sqrt {d+e x} \, dx\) [195]

Optimal result

Integrand size = 24, antiderivative size = 304 \[ \int (a+b x)^{5/2} (A+B x) \sqrt {d+e x} \, dx=-\frac {(b d-a e)^3 (7 b B d-10 A b e+3 a B e) \sqrt {a+b x} \sqrt {d+e x}}{128 b^2 e^4}+\frac {(b d-a e)^2 (7 b B d-10 A b e+3 a B e) (a+b x)^{3/2} \sqrt {d+e x}}{192 b^2 e^3}-\frac {(b d-a e) (7 b B d-10 A b e+3 a B e) (a+b x)^{5/2} \sqrt {d+e x}}{240 b^2 e^2}-\frac {(7 b B d-10 A b e+3 a B e) (a+b x)^{7/2} \sqrt {d+e x}}{40 b^2 e}+\frac {B (a+b x)^{7/2} (d+e x)^{3/2}}{5 b e}+\frac {(b d-a e)^4 (7 b B d-10 A b e+3 a B e) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{128 b^{5/2} e^{9/2}} \] Output:

-1/128*(-a*e+b*d)^3*(-10*A*b*e+3*B*a*e+7*B*b*d)*(b*x+a)^(1/2)*(e*x+d)^(1/2 
)/b^2/e^4+1/192*(-a*e+b*d)^2*(-10*A*b*e+3*B*a*e+7*B*b*d)*(b*x+a)^(3/2)*(e* 
x+d)^(1/2)/b^2/e^3-1/240*(-a*e+b*d)*(-10*A*b*e+3*B*a*e+7*B*b*d)*(b*x+a)^(5 
/2)*(e*x+d)^(1/2)/b^2/e^2-1/40*(-10*A*b*e+3*B*a*e+7*B*b*d)*(b*x+a)^(7/2)*( 
e*x+d)^(1/2)/b^2/e+1/5*B*(b*x+a)^(7/2)*(e*x+d)^(3/2)/b/e+1/128*(-a*e+b*d)^ 
4*(-10*A*b*e+3*B*a*e+7*B*b*d)*arctanh(e^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(e*x+d 
)^(1/2))/b^(5/2)/e^(9/2)
 

Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.06 \[ \int (a+b x)^{5/2} (A+B x) \sqrt {d+e x} \, dx=\frac {\sqrt {a+b x} \sqrt {d+e x} \left (-45 a^4 B e^4+30 a^3 b e^3 (2 B d+5 A e+B e x)+2 a^2 b^2 e^2 \left (5 A e (73 d+118 e x)+B \left (-173 d^2+109 d e x+372 e^2 x^2\right )\right )+2 a b^3 e \left (5 A e \left (-55 d^2+36 d e x+136 e^2 x^2\right )+B \left (170 d^3-111 d^2 e x+88 d e^2 x^2+504 e^3 x^3\right )\right )+b^4 \left (10 A e \left (15 d^3-10 d^2 e x+8 d e^2 x^2+48 e^3 x^3\right )+B \left (-105 d^4+70 d^3 e x-56 d^2 e^2 x^2+48 d e^3 x^3+384 e^4 x^4\right )\right )\right )}{1920 b^2 e^4}+\frac {(b d-a e)^4 (7 b B d-10 A b e+3 a B e) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {e} \sqrt {a+b x}}\right )}{128 b^{5/2} e^{9/2}} \] Input:

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

Output:

(Sqrt[a + b*x]*Sqrt[d + e*x]*(-45*a^4*B*e^4 + 30*a^3*b*e^3*(2*B*d + 5*A*e 
+ B*e*x) + 2*a^2*b^2*e^2*(5*A*e*(73*d + 118*e*x) + B*(-173*d^2 + 109*d*e*x 
 + 372*e^2*x^2)) + 2*a*b^3*e*(5*A*e*(-55*d^2 + 36*d*e*x + 136*e^2*x^2) + B 
*(170*d^3 - 111*d^2*e*x + 88*d*e^2*x^2 + 504*e^3*x^3)) + b^4*(10*A*e*(15*d 
^3 - 10*d^2*e*x + 8*d*e^2*x^2 + 48*e^3*x^3) + B*(-105*d^4 + 70*d^3*e*x - 5 
6*d^2*e^2*x^2 + 48*d*e^3*x^3 + 384*e^4*x^4))))/(1920*b^2*e^4) + ((b*d - a* 
e)^4*(7*b*B*d - 10*A*b*e + 3*a*B*e)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/(Sqrt[ 
e]*Sqrt[a + b*x])])/(128*b^(5/2)*e^(9/2))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.83, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, 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.

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

\(\Big \downarrow \) 90

\(\displaystyle \frac {B (a+b x)^{7/2} (d+e x)^{3/2}}{5 b e}-\frac {(3 a B e-10 A b e+7 b B d) \int (a+b x)^{5/2} \sqrt {d+e x}dx}{10 b e}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {B (a+b x)^{7/2} (d+e x)^{3/2}}{5 b e}-\frac {(3 a B e-10 A b e+7 b B d) \left (\frac {(b d-a e) \int \frac {(a+b x)^{5/2}}{\sqrt {d+e x}}dx}{8 b}+\frac {(a+b x)^{7/2} \sqrt {d+e x}}{4 b}\right )}{10 b e}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {B (a+b x)^{7/2} (d+e x)^{3/2}}{5 b e}-\frac {(3 a B e-10 A b e+7 b B d) \left (\frac {(b d-a e) \left (\frac {(a+b x)^{5/2} \sqrt {d+e x}}{3 e}-\frac {5 (b d-a e) \int \frac {(a+b x)^{3/2}}{\sqrt {d+e x}}dx}{6 e}\right )}{8 b}+\frac {(a+b x)^{7/2} \sqrt {d+e x}}{4 b}\right )}{10 b e}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {B (a+b x)^{7/2} (d+e x)^{3/2}}{5 b e}-\frac {(3 a B e-10 A b e+7 b B d) \left (\frac {(b d-a e) \left (\frac {(a+b x)^{5/2} \sqrt {d+e x}}{3 e}-\frac {5 (b d-a e) \left (\frac {(a+b x)^{3/2} \sqrt {d+e x}}{2 e}-\frac {3 (b d-a e) \int \frac {\sqrt {a+b x}}{\sqrt {d+e x}}dx}{4 e}\right )}{6 e}\right )}{8 b}+\frac {(a+b x)^{7/2} \sqrt {d+e x}}{4 b}\right )}{10 b e}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {B (a+b x)^{7/2} (d+e x)^{3/2}}{5 b e}-\frac {(3 a B e-10 A b e+7 b B d) \left (\frac {(b d-a e) \left (\frac {(a+b x)^{5/2} \sqrt {d+e x}}{3 e}-\frac {5 (b d-a e) \left (\frac {(a+b x)^{3/2} \sqrt {d+e x}}{2 e}-\frac {3 (b d-a e) \left (\frac {\sqrt {a+b x} \sqrt {d+e x}}{e}-\frac {(b d-a e) \int \frac {1}{\sqrt {a+b x} \sqrt {d+e x}}dx}{2 e}\right )}{4 e}\right )}{6 e}\right )}{8 b}+\frac {(a+b x)^{7/2} \sqrt {d+e x}}{4 b}\right )}{10 b e}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {B (a+b x)^{7/2} (d+e x)^{3/2}}{5 b e}-\frac {(3 a B e-10 A b e+7 b B d) \left (\frac {(b d-a e) \left (\frac {(a+b x)^{5/2} \sqrt {d+e x}}{3 e}-\frac {5 (b d-a e) \left (\frac {(a+b x)^{3/2} \sqrt {d+e x}}{2 e}-\frac {3 (b d-a e) \left (\frac {\sqrt {a+b x} \sqrt {d+e x}}{e}-\frac {(b d-a e) \int \frac {1}{b-\frac {e (a+b x)}{d+e x}}d\frac {\sqrt {a+b x}}{\sqrt {d+e x}}}{e}\right )}{4 e}\right )}{6 e}\right )}{8 b}+\frac {(a+b x)^{7/2} \sqrt {d+e x}}{4 b}\right )}{10 b e}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {B (a+b x)^{7/2} (d+e x)^{3/2}}{5 b e}-\frac {(3 a B e-10 A b e+7 b B d) \left (\frac {(b d-a e) \left (\frac {(a+b x)^{5/2} \sqrt {d+e x}}{3 e}-\frac {5 (b d-a e) \left (\frac {(a+b x)^{3/2} \sqrt {d+e x}}{2 e}-\frac {3 (b d-a e) \left (\frac {\sqrt {a+b x} \sqrt {d+e x}}{e}-\frac {(b d-a e) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{\sqrt {b} e^{3/2}}\right )}{4 e}\right )}{6 e}\right )}{8 b}+\frac {(a+b x)^{7/2} \sqrt {d+e x}}{4 b}\right )}{10 b e}\)

Input:

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

Output:

(B*(a + b*x)^(7/2)*(d + e*x)^(3/2))/(5*b*e) - ((7*b*B*d - 10*A*b*e + 3*a*B 
*e)*(((a + b*x)^(7/2)*Sqrt[d + e*x])/(4*b) + ((b*d - a*e)*(((a + b*x)^(5/2 
)*Sqrt[d + e*x])/(3*e) - (5*(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)*ArcT 
anh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(Sqrt[b]*e^(3/2))))/ 
(4*e)))/(6*e)))/(8*b)))/(10*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. \(2(260)=520\).

Time = 0.26 (sec) , antiderivative size = 1372, normalized size of antiderivative = 4.51

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

Input:

int((b*x+a)^(5/2)*(B*x+A)*(e*x+d)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/3840*(b*x+a)^(1/2)*(e*x+d)^(1/2)*(-768*B*b^4*e^4*x^4*((e*x+d)*(b*x+a))^ 
(1/2)*(b*e)^(1/2)-960*A*b^4*e^4*x^3*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)-20 
16*B*a*b^3*e^4*x^3*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)-96*B*b^4*d*e^3*x^3* 
((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)-2720*A*a*b^3*e^4*x^2*((e*x+d)*(b*x+a)) 
^(1/2)*(b*e)^(1/2)+1100*A*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)*a*b^3*d^2*e^ 
2-120*B*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)*a^3*b*d*e^3+692*B*((e*x+d)*(b* 
x+a))^(1/2)*(b*e)^(1/2)*a^2*b^2*d^2*e^2-680*B*((e*x+d)*(b*x+a))^(1/2)*(b*e 
)^(1/2)*a*b^3*d^3*e+90*B*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)*a^4*e^4+210*B 
*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)*b^4*d^4-352*B*a*b^3*d*e^3*x^2*((e*x+d 
)*(b*x+a))^(1/2)*(b*e)^(1/2)-720*A*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)*a*b 
^3*d*e^3*x-436*B*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)*a^2*b^2*d*e^3*x+444*B 
*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)*a*b^3*d^2*e^2*x+150*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+150 
*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+900*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^3-600*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+75*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+150*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^3*b^2*d^2*e^3-450*B*ln(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a))...
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 1046, normalized size of antiderivative = 3.44 \[ \int (a+b x)^{5/2} (A+B x) \sqrt {d+e x} \, dx=\text {Too large to display} \] Input:

integrate((b*x+a)^(5/2)*(B*x+A)*(e*x+d)^(1/2),x, algorithm="fricas")
 

Output:

[-1/7680*(15*(7*B*b^5*d^5 - 5*(5*B*a*b^4 + 2*A*b^5)*d^4*e + 10*(3*B*a^2*b^ 
3 + 4*A*a*b^4)*d^3*e^2 - 10*(B*a^3*b^2 + 6*A*a^2*b^3)*d^2*e^3 - 5*(B*a^4*b 
 - 8*A*a^3*b^2)*d*e^4 + (3*B*a^5 - 10*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*(384*B*b^5*e^5* 
x^4 - 105*B*b^5*d^4*e + 10*(34*B*a*b^4 + 15*A*b^5)*d^3*e^2 - 2*(173*B*a^2* 
b^3 + 275*A*a*b^4)*d^2*e^3 + 10*(6*B*a^3*b^2 + 73*A*a^2*b^3)*d*e^4 - 15*(3 
*B*a^4*b - 10*A*a^3*b^2)*e^5 + 48*(B*b^5*d*e^4 + (21*B*a*b^4 + 10*A*b^5)*e 
^5)*x^3 - 8*(7*B*b^5*d^2*e^3 - 2*(11*B*a*b^4 + 5*A*b^5)*d*e^4 - (93*B*a^2* 
b^3 + 170*A*a*b^4)*e^5)*x^2 + 2*(35*B*b^5*d^3*e^2 - (111*B*a*b^4 + 50*A*b^ 
5)*d^2*e^3 + (109*B*a^2*b^3 + 180*A*a*b^4)*d*e^4 + 5*(3*B*a^3*b^2 + 118*A* 
a^2*b^3)*e^5)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b^3*e^5), -1/3840*(15*(7*B* 
b^5*d^5 - 5*(5*B*a*b^4 + 2*A*b^5)*d^4*e + 10*(3*B*a^2*b^3 + 4*A*a*b^4)*d^3 
*e^2 - 10*(B*a^3*b^2 + 6*A*a^2*b^3)*d^2*e^3 - 5*(B*a^4*b - 8*A*a^3*b^2)*d* 
e^4 + (3*B*a^5 - 10*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*(384*B*b^5*e^5*x^4 - 105*B*b^5*d^4*e + 10*(34*B*a*b^ 
4 + 15*A*b^5)*d^3*e^2 - 2*(173*B*a^2*b^3 + 275*A*a*b^4)*d^2*e^3 + 10*(6*B* 
a^3*b^2 + 73*A*a^2*b^3)*d*e^4 - 15*(3*B*a^4*b - 10*A*a^3*b^2)*e^5 + 48*(B* 
b^5*d*e^4 + (21*B*a*b^4 + 10*A*b^5)*e^5)*x^3 - 8*(7*B*b^5*d^2*e^3 - 2*(...
 

Sympy [F]

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

integrate((b*x+a)**(5/2)*(B*x+A)*(e*x+d)**(1/2),x)
 

Output:

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

Maxima [F(-2)]

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

integrate((b*x+a)^(5/2)*(B*x+A)*(e*x+d)^(1/2),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. 1672 vs. \(2 (260) = 520\).

Time = 0.34 (sec) , antiderivative size = 1672, normalized size of antiderivative = 5.50 \[ \int (a+b x)^{5/2} (A+B x) \sqrt {d+e x} \, dx=\text {Too large to display} \] Input:

integrate((b*x+a)^(5/2)*(B*x+A)*(e*x+d)^(1/2),x, algorithm="giac")
 

Output:

1/1920*(30*(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*a*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^3*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*b*abs(b) + 480*(sqrt(b^2*d + (b*x + 
a)*b*e - a*b*e)*(2*b*x + 2*a + (b*d*e - 5*a*e^2)/e^2)*sqrt(b*x + a) + (b^3 
*d^2 + 2*a*b^2*d*e - 3*a^2*b*e^2)*log(abs(-sqrt(b*e)*sqrt(b*x + a) + sqrt( 
b^2*d + (b*x + a)*b*e - a*b*e)))/(sqrt(b*e)*e))*B*a^3*abs(b)/b^3 + 1440*(s 
qrt(b^2*d + (b*x + a)*b*e - a*b*e)*(2*b*x + 2*a + (b*d*e - 5*a*e^2)/e^2...
 

Mupad [F(-1)]

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

int((A + B*x)*(a + b*x)^(5/2)*(d + e*x)^(1/2),x)
 

Output:

int((A + B*x)*(a + b*x)^(5/2)*(d + e*x)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 651, normalized size of antiderivative = 2.14 \[ \int (a+b x)^{5/2} (A+B x) \sqrt {d+e x} \, dx=\frac {105 \sqrt {e x +d}\, \sqrt {b x +a}\, a^{4} b \,e^{5}+790 \sqrt {e x +d}\, \sqrt {b x +a}\, a^{3} b^{2} d \,e^{4}+1210 \sqrt {e x +d}\, \sqrt {b x +a}\, a^{3} b^{2} e^{5} x -896 \sqrt {e x +d}\, \sqrt {b x +a}\, a^{2} b^{3} d^{2} e^{3}+578 \sqrt {e x +d}\, \sqrt {b x +a}\, a^{2} b^{3} d \,e^{4} x +2104 \sqrt {e x +d}\, \sqrt {b x +a}\, a^{2} b^{3} e^{5} x^{2}+490 \sqrt {e x +d}\, \sqrt {b x +a}\, a \,b^{4} d^{3} e^{2}-322 \sqrt {e x +d}\, \sqrt {b x +a}\, a \,b^{4} d^{2} e^{3} x +256 \sqrt {e x +d}\, \sqrt {b x +a}\, a \,b^{4} d \,e^{4} x^{2}+1488 \sqrt {e x +d}\, \sqrt {b x +a}\, a \,b^{4} e^{5} x^{3}-105 \sqrt {e x +d}\, \sqrt {b x +a}\, b^{5} d^{4} e +70 \sqrt {e x +d}\, \sqrt {b x +a}\, b^{5} d^{3} e^{2} x -56 \sqrt {e x +d}\, \sqrt {b x +a}\, b^{5} d^{2} e^{3} x^{2}+48 \sqrt {e x +d}\, \sqrt {b x +a}\, b^{5} d \,e^{4} x^{3}+384 \sqrt {e x +d}\, \sqrt {b x +a}\, b^{5} e^{5} x^{4}-105 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) a^{5} e^{5}+525 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) a^{4} b d \,e^{4}-1050 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) a^{3} b^{2} d^{2} e^{3}+1050 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) a^{2} b^{3} d^{3} e^{2}-525 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) a \,b^{4} d^{4} e +105 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) b^{5} d^{5}}{1920 b^{2} e^{5}} \] Input:

int((b*x+a)^(5/2)*(B*x+A)*(e*x+d)^(1/2),x)
 

Output:

(105*sqrt(d + e*x)*sqrt(a + b*x)*a**4*b*e**5 + 790*sqrt(d + e*x)*sqrt(a + 
b*x)*a**3*b**2*d*e**4 + 1210*sqrt(d + e*x)*sqrt(a + b*x)*a**3*b**2*e**5*x 
- 896*sqrt(d + e*x)*sqrt(a + b*x)*a**2*b**3*d**2*e**3 + 578*sqrt(d + e*x)* 
sqrt(a + b*x)*a**2*b**3*d*e**4*x + 2104*sqrt(d + e*x)*sqrt(a + b*x)*a**2*b 
**3*e**5*x**2 + 490*sqrt(d + e*x)*sqrt(a + b*x)*a*b**4*d**3*e**2 - 322*sqr 
t(d + e*x)*sqrt(a + b*x)*a*b**4*d**2*e**3*x + 256*sqrt(d + e*x)*sqrt(a + b 
*x)*a*b**4*d*e**4*x**2 + 1488*sqrt(d + e*x)*sqrt(a + b*x)*a*b**4*e**5*x**3 
 - 105*sqrt(d + e*x)*sqrt(a + b*x)*b**5*d**4*e + 70*sqrt(d + e*x)*sqrt(a + 
 b*x)*b**5*d**3*e**2*x - 56*sqrt(d + e*x)*sqrt(a + b*x)*b**5*d**2*e**3*x** 
2 + 48*sqrt(d + e*x)*sqrt(a + b*x)*b**5*d*e**4*x**3 + 384*sqrt(d + e*x)*sq 
rt(a + b*x)*b**5*e**5*x**4 - 105*sqrt(e)*sqrt(b)*log((sqrt(e)*sqrt(a + b*x 
) + sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*a**5*e**5 + 525*sqrt(e)*sqrt(b 
)*log((sqrt(e)*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*a** 
4*b*d*e**4 - 1050*sqrt(e)*sqrt(b)*log((sqrt(e)*sqrt(a + b*x) + sqrt(b)*sqr 
t(d + e*x))/sqrt(a*e - b*d))*a**3*b**2*d**2*e**3 + 1050*sqrt(e)*sqrt(b)*lo 
g((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 - 525*sqrt(e)*sqrt(b)*log((sqrt(e)*sqrt(a + b*x) + sqrt(b)*sq 
rt(d + e*x))/sqrt(a*e - b*d))*a*b**4*d**4*e + 105*sqrt(e)*sqrt(b)*log((sqr 
t(e)*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*b**5*d**5)/(1 
920*b**2*e**5)