\(\int \frac {(A+B x) (d+e x)^{3/2}}{\sqrt {a+b x}} \, dx\) [206]

Optimal result

Integrand size = 24, antiderivative size = 193 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{\sqrt {a+b x}} \, dx=-\frac {(b d-a e) (b B d-6 A b e+5 a B e) \sqrt {a+b x} \sqrt {d+e x}}{8 b^3 e}-\frac {(b B d-6 A b e+5 a B e) \sqrt {a+b x} (d+e x)^{3/2}}{12 b^2 e}+\frac {B \sqrt {a+b x} (d+e x)^{5/2}}{3 b e}-\frac {(b d-a e)^2 (b B d-6 A b e+5 a B e) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{8 b^{7/2} e^{3/2}} \] Output:

-1/8*(-a*e+b*d)*(-6*A*b*e+5*B*a*e+B*b*d)*(b*x+a)^(1/2)*(e*x+d)^(1/2)/b^3/e 
-1/12*(-6*A*b*e+5*B*a*e+B*b*d)*(b*x+a)^(1/2)*(e*x+d)^(3/2)/b^2/e+1/3*B*(b* 
x+a)^(1/2)*(e*x+d)^(5/2)/b/e-1/8*(-a*e+b*d)^2*(-6*A*b*e+5*B*a*e+B*b*d)*arc 
tanh(e^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(e*x+d)^(1/2))/b^(7/2)/e^(3/2)
 

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.85 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{\sqrt {a+b x}} \, dx=\frac {\sqrt {a+b x} \sqrt {d+e x} \left (15 a^2 B e^2-2 a b e (11 B d+9 A e+5 B e x)+b^2 \left (6 A e (5 d+2 e x)+B \left (3 d^2+14 d e x+8 e^2 x^2\right )\right )\right )}{24 b^3 e}-\frac {(b d-a e)^2 (b B d-6 A b e+5 a B e) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{8 b^{7/2} e^{3/2}} \] Input:

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

Output:

(Sqrt[a + b*x]*Sqrt[d + e*x]*(15*a^2*B*e^2 - 2*a*b*e*(11*B*d + 9*A*e + 5*B 
*e*x) + b^2*(6*A*e*(5*d + 2*e*x) + B*(3*d^2 + 14*d*e*x + 8*e^2*x^2))))/(24 
*b^3*e) - ((b*d - a*e)^2*(b*B*d - 6*A*b*e + 5*a*B*e)*ArcTanh[(Sqrt[e]*Sqrt 
[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(8*b^(7/2)*e^(3/2))
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.87, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {90, 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.

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(B*Sqrt[a + b*x]*(d + e*x)^(5/2))/(3*b*e) - ((b*B*d - 6*A*b*e + 5*a*B*e)*( 
(Sqrt[a + b*x]*(d + e*x)^(3/2))/(2*b) + (3*(b*d - a*e)*((Sqrt[a + b*x]*Sqr 
t[d + e*x])/b + ((b*d - a*e)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt 
[d + e*x])])/(b^(3/2)*Sqrt[e])))/(4*b)))/(6*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. \(635\) vs. \(2(161)=322\).

Time = 0.26 (sec) , antiderivative size = 636, normalized size of antiderivative = 3.30

method	result	size
default	\(\frac {\sqrt {e x +d}\, \sqrt {b x +a}\, \left (16 B \,b^{2} e^{2} x^{2} \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+18 A \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) a^{2} b \,e^{3}-36 A \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) a \,b^{2} d \,e^{2}+18 A \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) b^{3} d^{2} e +24 A \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}\, b^{2} e^{2} x -15 B \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) a^{3} e^{3}+27 B \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) a^{2} b d \,e^{2}-9 B \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) a \,b^{2} d^{2} e -3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) b^{3} d^{3}-20 B \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}\, a b \,e^{2} x +28 B \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}\, b^{2} d e x -36 A \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}\, a b \,e^{2}+60 A \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}\, b^{2} d e +30 B \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}\, a^{2} e^{2}-44 B \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}\, a b d e +6 B \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}\, b^{2} d^{2}\right )}{48 b^{3} e \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}}\)	\(636\)

Input:

int((B*x+A)*(e*x+d)^(3/2)/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/48*(e*x+d)^(1/2)*(b*x+a)^(1/2)*(16*B*b^2*e^2*x^2*((e*x+d)*(b*x+a))^(1/2) 
*(b*e)^(1/2)+18*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*e^3-36*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^2*d*e^2+18*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^3*d^2*e+24*A*((e 
*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)*b^2*e^2*x-15*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*e^3+27*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^2*b*d*e 
^2-9*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*b^2*d^2*e-3*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))*b^3*d^3-20*B*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2 
)*a*b*e^2*x+28*B*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)*b^2*d*e*x-36*A*((e*x+ 
d)*(b*x+a))^(1/2)*(b*e)^(1/2)*a*b*e^2+60*A*((e*x+d)*(b*x+a))^(1/2)*(b*e)^( 
1/2)*b^2*d*e+30*B*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)*a^2*e^2-44*B*((e*x+d 
)*(b*x+a))^(1/2)*(b*e)^(1/2)*a*b*d*e+6*B*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/ 
2)*b^2*d^2)/b^3/e/((e*x+d)*(b*x+a))^(1/2)/(b*e)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 542, normalized size of antiderivative = 2.81 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{\sqrt {a+b x}} \, dx=\left [-\frac {3 \, {\left (B b^{3} d^{3} + 3 \, {\left (B a b^{2} - 2 \, A b^{3}\right )} d^{2} e - 3 \, {\left (3 \, B a^{2} b - 4 \, A a b^{2}\right )} d e^{2} + {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} e^{3}\right )} \sqrt {b e} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b e x + b d + a e\right )} \sqrt {b e} \sqrt {b x + a} \sqrt {e x + d} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} x\right ) - 4 \, {\left (8 \, B b^{3} e^{3} x^{2} + 3 \, B b^{3} d^{2} e - 2 \, {\left (11 \, B a b^{2} - 15 \, A b^{3}\right )} d e^{2} + 3 \, {\left (5 \, B a^{2} b - 6 \, A a b^{2}\right )} e^{3} + 2 \, {\left (7 \, B b^{3} d e^{2} - {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} e^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{96 \, b^{4} e^{2}}, \frac {3 \, {\left (B b^{3} d^{3} + 3 \, {\left (B a b^{2} - 2 \, A b^{3}\right )} d^{2} e - 3 \, {\left (3 \, B a^{2} b - 4 \, A a b^{2}\right )} d e^{2} + {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} e^{3}\right )} \sqrt {-b e} \arctan \left (\frac {{\left (2 \, b e x + b d + a e\right )} \sqrt {-b e} \sqrt {b x + a} \sqrt {e x + d}}{2 \, {\left (b^{2} e^{2} x^{2} + a b d e + {\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right ) + 2 \, {\left (8 \, B b^{3} e^{3} x^{2} + 3 \, B b^{3} d^{2} e - 2 \, {\left (11 \, B a b^{2} - 15 \, A b^{3}\right )} d e^{2} + 3 \, {\left (5 \, B a^{2} b - 6 \, A a b^{2}\right )} e^{3} + 2 \, {\left (7 \, B b^{3} d e^{2} - {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} e^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{48 \, b^{4} e^{2}}\right ] \] Input:

integrate((B*x+A)*(e*x+d)^(3/2)/(b*x+a)^(1/2),x, algorithm="fricas")
 

Output:

[-1/96*(3*(B*b^3*d^3 + 3*(B*a*b^2 - 2*A*b^3)*d^2*e - 3*(3*B*a^2*b - 4*A*a* 
b^2)*d*e^2 + (5*B*a^3 - 6*A*a^2*b)*e^3)*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*(8*B*b^3*e^3*x^2 + 3*B*b^3* 
d^2*e - 2*(11*B*a*b^2 - 15*A*b^3)*d*e^2 + 3*(5*B*a^2*b - 6*A*a*b^2)*e^3 + 
2*(7*B*b^3*d*e^2 - (5*B*a*b^2 - 6*A*b^3)*e^3)*x)*sqrt(b*x + a)*sqrt(e*x + 
d))/(b^4*e^2), 1/48*(3*(B*b^3*d^3 + 3*(B*a*b^2 - 2*A*b^3)*d^2*e - 3*(3*B*a 
^2*b - 4*A*a*b^2)*d*e^2 + (5*B*a^3 - 6*A*a^2*b)*e^3)*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*(8*B*b^3*e^3*x^2 + 3*B*b^3*d^2*e 
- 2*(11*B*a*b^2 - 15*A*b^3)*d*e^2 + 3*(5*B*a^2*b - 6*A*a*b^2)*e^3 + 2*(7*B 
*b^3*d*e^2 - (5*B*a*b^2 - 6*A*b^3)*e^3)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b 
^4*e^2)]
 

Sympy [F]

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

integrate((B*x+A)*(e*x+d)**(3/2)/(b*x+a)**(1/2),x)
 

Output:

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

Maxima [F(-2)]

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

integrate((B*x+A)*(e*x+d)^(3/2)/(b*x+a)^(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(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. 559 vs. \(2 (161) = 322\).

Time = 0.20 (sec) , antiderivative size = 559, normalized size of antiderivative = 2.90 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{\sqrt {a+b x}} \, dx=-\frac {\frac {24 \, {\left (\frac {{\left (b^{2} d - a b e\right )} \log \left ({\left | -\sqrt {b e} \sqrt {b x + a} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b e}} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \sqrt {b x + a}\right )} A d {\left | b \right |}}{b^{2}} - \frac {6 \, {\left (\sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} {\left (2 \, b x + 2 \, a + \frac {b d e - 5 \, a e^{2}}{e^{2}}\right )} \sqrt {b x + a} + \frac {{\left (b^{3} d^{2} + 2 \, a b^{2} d e - 3 \, a^{2} b e^{2}\right )} \log \left ({\left | -\sqrt {b e} \sqrt {b x + a} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b e} e}\right )} B d {\left | b \right |}}{b^{3}} - \frac {6 \, {\left (\sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} {\left (2 \, b x + 2 \, a + \frac {b d e - 5 \, a e^{2}}{e^{2}}\right )} \sqrt {b x + a} + \frac {{\left (b^{3} d^{2} + 2 \, a b^{2} d e - 3 \, a^{2} b e^{2}\right )} \log \left ({\left | -\sqrt {b e} \sqrt {b x + a} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b e} e}\right )} A e {\left | b \right |}}{b^{3}} - \frac {{\left (\sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} {\left (2 \, {\left (4 \, b x + 4 \, a + \frac {b d e^{3} - 13 \, a e^{4}}{e^{4}}\right )} {\left (b x + a\right )} - \frac {3 \, {\left (b^{2} d^{2} e^{2} + 2 \, a b d e^{3} - 11 \, a^{2} e^{4}\right )}}{e^{4}}\right )} \sqrt {b x + a} - \frac {3 \, {\left (b^{4} d^{3} + a b^{3} d^{2} e + 3 \, a^{2} b^{2} d e^{2} - 5 \, a^{3} b e^{3}\right )} \log \left ({\left | -\sqrt {b e} \sqrt {b x + a} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b e} e^{2}}\right )} B e {\left | b \right |}}{b^{4}}}{24 \, b} \] Input:

integrate((B*x+A)*(e*x+d)^(3/2)/(b*x+a)^(1/2),x, algorithm="giac")
 

Output:

-1/24*(24*((b^2*d - a*b*e)*log(abs(-sqrt(b*e)*sqrt(b*x + a) + sqrt(b^2*d + 
 (b*x + a)*b*e - a*b*e)))/sqrt(b*e) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e)* 
sqrt(b*x + a))*A*d*abs(b)/b^2 - 6*(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*d*abs(b)/b^3 - 6*(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))*A*e*abs(b)/b^3 - (sqrt(b^2*d + 
(b*x + a)*b*e - a*b*e)*(2*(4*b*x + 4*a + (b*d*e^3 - 13*a*e^4)/e^4)*(b*x + 
a) - 3*(b^2*d^2*e^2 + 2*a*b*d*e^3 - 11*a^2*e^4)/e^4)*sqrt(b*x + a) - 3*(b^ 
4*d^3 + a*b^3*d^2*e + 3*a^2*b^2*d*e^2 - 5*a^3*b*e^3)*log(abs(-sqrt(b*e)*sq 
rt(b*x + a) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/(sqrt(b*e)*e^2))*B*e*a 
bs(b)/b^4)/b
 

Mupad [F(-1)]

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

int(((A + B*x)*(d + e*x)^(3/2))/(a + b*x)^(1/2),x)
 

Output:

int(((A + B*x)*(d + e*x)^(3/2))/(a + b*x)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.66 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{\sqrt {a+b x}} \, dx=\frac {-3 \sqrt {e x +d}\, \sqrt {b x +a}\, a^{2} b \,e^{3}+8 \sqrt {e x +d}\, \sqrt {b x +a}\, a \,b^{2} d \,e^{2}+2 \sqrt {e x +d}\, \sqrt {b x +a}\, a \,b^{2} e^{3} x +3 \sqrt {e x +d}\, \sqrt {b x +a}\, b^{3} d^{2} e +14 \sqrt {e x +d}\, \sqrt {b x +a}\, b^{3} d \,e^{2} x +8 \sqrt {e x +d}\, \sqrt {b x +a}\, b^{3} e^{3} x^{2}+3 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) a^{3} e^{3}-9 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) a^{2} b d \,e^{2}+9 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) a \,b^{2} d^{2} e -3 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) b^{3} d^{3}}{24 b^{3} e^{2}} \] Input:

int((B*x+A)*(e*x+d)^(3/2)/(b*x+a)^(1/2),x)
 

Output:

( - 3*sqrt(d + e*x)*sqrt(a + b*x)*a**2*b*e**3 + 8*sqrt(d + e*x)*sqrt(a + b 
*x)*a*b**2*d*e**2 + 2*sqrt(d + e*x)*sqrt(a + b*x)*a*b**2*e**3*x + 3*sqrt(d 
 + e*x)*sqrt(a + b*x)*b**3*d**2*e + 14*sqrt(d + e*x)*sqrt(a + b*x)*b**3*d* 
e**2*x + 8*sqrt(d + e*x)*sqrt(a + b*x)*b**3*e**3*x**2 + 3*sqrt(e)*sqrt(b)* 
log((sqrt(e)*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*a**3* 
e**3 - 9*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*d*e**2 + 9*sqrt(e)*sqrt(b)*log((sqrt(e)*sqrt(a 
+ b*x) + sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*a*b**2*d**2*e - 3*sqrt(e) 
*sqrt(b)*log((sqrt(e)*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b* 
d))*b**3*d**3)/(24*b**3*e**2)