\(\int \sqrt {x} (a+b x)^{5/2} (A+B x) \, dx\) [307]

Optimal result

Integrand size = 20, antiderivative size = 187 \[ \int \sqrt {x} (a+b x)^{5/2} (A+B x) \, dx=\frac {a^3 (10 A b-3 a B) \sqrt {x} \sqrt {a+b x}}{128 b^2}+\frac {59 a^2 (10 A b-3 a B) x^{3/2} \sqrt {a+b x}}{960 b}+\frac {17}{240} a (10 A b-3 a B) x^{5/2} \sqrt {a+b x}+\frac {1}{40} b (10 A b-3 a B) x^{7/2} \sqrt {a+b x}+\frac {B x^{3/2} (a+b x)^{7/2}}{5 b}-\frac {a^4 (10 A b-3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{5/2}} \] Output:

1/128*a^3*(10*A*b-3*B*a)*x^(1/2)*(b*x+a)^(1/2)/b^2+59/960*a^2*(10*A*b-3*B* 
a)*x^(3/2)*(b*x+a)^(1/2)/b+17/240*a*(10*A*b-3*B*a)*x^(5/2)*(b*x+a)^(1/2)+1 
/40*b*(10*A*b-3*B*a)*x^(7/2)*(b*x+a)^(1/2)+1/5*B*x^(3/2)*(b*x+a)^(7/2)/b-1 
/128*a^4*(10*A*b-3*B*a)*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.94 \[ \int \sqrt {x} (a+b x)^{5/2} (A+B x) \, dx=\frac {\sqrt {b} \sqrt {x} \sqrt {a+b x} \left (-45 a^4 B+30 a^3 b (5 A+B x)+96 b^4 x^3 (5 A+4 B x)+16 a b^3 x^2 (85 A+63 B x)+4 a^2 b^2 x (295 A+186 B x)\right )+300 a^4 A b \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )+90 a^5 B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{1920 b^{5/2}} \] Input:

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

Output:

(Sqrt[b]*Sqrt[x]*Sqrt[a + b*x]*(-45*a^4*B + 30*a^3*b*(5*A + B*x) + 96*b^4* 
x^3*(5*A + 4*B*x) + 16*a*b^3*x^2*(85*A + 63*B*x) + 4*a^2*b^2*x*(295*A + 18 
6*B*x)) + 300*a^4*A*b*ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt[a] - Sqrt[a + b*x])] 
 + 90*a^5*B*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])])/(1920*b 
^(5/2))
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.85, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {90, 60, 60, 60, 60, 65, 219}

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.

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 65

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(B*x^(3/2)*(a + b*x)^(7/2))/(5*b) + ((10*A*b - 3*a*B)*((x^(3/2)*(a + b*x)^ 
(5/2))/4 + (5*a*((x^(3/2)*(a + b*x)^(3/2))/3 + (a*((x^(3/2)*Sqrt[a + b*x]) 
/2 + (a*((Sqrt[x]*Sqrt[a + b*x])/b - (a*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + 
 b*x]])/b^(3/2)))/4))/2))/8))/(10*b)
 

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 65

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2   Sub 
st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d 
}, x] &&  !GtQ[c, 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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.85

method	result	size
risch	\(\frac {\left (384 B \,x^{4} b^{4}+480 A \,x^{3} b^{4}+1008 B \,x^{3} a \,b^{3}+1360 A \,x^{2} a \,b^{3}+744 B \,x^{2} a^{2} b^{2}+1180 A x \,a^{2} b^{2}+30 B x \,a^{3} b +150 A \,a^{3} b -45 B \,a^{4}\right ) \sqrt {x}\, \sqrt {b x +a}}{1920 b^{2}}-\frac {a^{4} \left (10 A b -3 B a \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{256 b^{\frac {5}{2}} \sqrt {x}\, \sqrt {b x +a}}\)	\(159\)
default	\(-\frac {\sqrt {x}\, \sqrt {b x +a}\, \left (-768 B \,b^{\frac {9}{2}} x^{4} \sqrt {x \left (b x +a \right )}-960 A \,b^{\frac {9}{2}} x^{3} \sqrt {x \left (b x +a \right )}-2016 B a \,b^{\frac {7}{2}} x^{3} \sqrt {x \left (b x +a \right )}-2720 A a \,b^{\frac {7}{2}} x^{2} \sqrt {x \left (b x +a \right )}-1488 B \,a^{2} b^{\frac {5}{2}} x^{2} \sqrt {x \left (b x +a \right )}-2360 A \sqrt {x \left (b x +a \right )}\, b^{\frac {5}{2}} a^{2} x -60 B \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a^{3} x +150 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{4} b -300 A \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a^{3}-45 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{5}+90 B \sqrt {x \left (b x +a \right )}\, \sqrt {b}\, a^{4}\right )}{3840 b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}}\)	\(260\)

Input:

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

Output:

1/1920/b^2*(384*B*b^4*x^4+480*A*b^4*x^3+1008*B*a*b^3*x^3+1360*A*a*b^3*x^2+ 
744*B*a^2*b^2*x^2+1180*A*a^2*b^2*x+30*B*a^3*b*x+150*A*a^3*b-45*B*a^4)*x^(1 
/2)*(b*x+a)^(1/2)-1/256*a^4/b^(5/2)*(10*A*b-3*B*a)*ln((1/2*a+b*x)/b^(1/2)+ 
(b*x^2+a*x)^(1/2))*(x*(b*x+a))^(1/2)/x^(1/2)/(b*x+a)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.57 \[ \int \sqrt {x} (a+b x)^{5/2} (A+B x) \, dx=\left [-\frac {15 \, {\left (3 \, B a^{5} - 10 \, A a^{4} b\right )} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (384 \, B b^{5} x^{4} - 45 \, B a^{4} b + 150 \, A a^{3} b^{2} + 48 \, {\left (21 \, B a b^{4} + 10 \, A b^{5}\right )} x^{3} + 8 \, {\left (93 \, B a^{2} b^{3} + 170 \, A a b^{4}\right )} x^{2} + 10 \, {\left (3 \, B a^{3} b^{2} + 118 \, A a^{2} b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{3840 \, b^{3}}, -\frac {15 \, {\left (3 \, B a^{5} - 10 \, A a^{4} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {b x + a}}\right ) - {\left (384 \, B b^{5} x^{4} - 45 \, B a^{4} b + 150 \, A a^{3} b^{2} + 48 \, {\left (21 \, B a b^{4} + 10 \, A b^{5}\right )} x^{3} + 8 \, {\left (93 \, B a^{2} b^{3} + 170 \, A a b^{4}\right )} x^{2} + 10 \, {\left (3 \, B a^{3} b^{2} + 118 \, A a^{2} b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{1920 \, b^{3}}\right ] \] Input:

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

Output:

[-1/3840*(15*(3*B*a^5 - 10*A*a^4*b)*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sq 
rt(b)*sqrt(x) + a) - 2*(384*B*b^5*x^4 - 45*B*a^4*b + 150*A*a^3*b^2 + 48*(2 
1*B*a*b^4 + 10*A*b^5)*x^3 + 8*(93*B*a^2*b^3 + 170*A*a*b^4)*x^2 + 10*(3*B*a 
^3*b^2 + 118*A*a^2*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^3, -1/1920*(15*(3*B*a^ 
5 - 10*A*a^4*b)*sqrt(-b)*arctan(sqrt(-b)*sqrt(x)/sqrt(b*x + a)) - (384*B*b 
^5*x^4 - 45*B*a^4*b + 150*A*a^3*b^2 + 48*(21*B*a*b^4 + 10*A*b^5)*x^3 + 8*( 
93*B*a^2*b^3 + 170*A*a*b^4)*x^2 + 10*(3*B*a^3*b^2 + 118*A*a^2*b^3)*x)*sqrt 
(b*x + a)*sqrt(x))/b^3]
 

Sympy [A] (verification not implemented)

Time = 2.80 (sec) , antiderivative size = 748, normalized size of antiderivative = 4.00 \[ \int \sqrt {x} (a+b x)^{5/2} (A+B x) \, dx=\text {Too large to display} \] Input:

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

Output:

2*A*a**2*Piecewise((-a**2*Piecewise((log(2*sqrt(b)*sqrt(a + b*x) + 2*b*sqr 
t(x))/sqrt(b), Ne(a, 0)), (sqrt(x)*log(sqrt(x))/sqrt(b*x), True))/(8*b) + 
sqrt(a + b*x)*(a*sqrt(x)/(8*b) + x**(3/2)/4), Ne(b, 0)), (sqrt(a)*x**(3/2) 
/3, True)) + 4*A*a*b*Piecewise((a**3*Piecewise((log(2*sqrt(b)*sqrt(a + b*x 
) + 2*b*sqrt(x))/sqrt(b), Ne(a, 0)), (sqrt(x)*log(sqrt(x))/sqrt(b*x), True 
))/(16*b**2) + sqrt(a + b*x)*(-a**2*sqrt(x)/(16*b**2) + a*x**(3/2)/(24*b) 
+ x**(5/2)/6), Ne(b, 0)), (sqrt(a)*x**(5/2)/5, True)) + 2*A*b**2*Piecewise 
((-5*a**4*Piecewise((log(2*sqrt(b)*sqrt(a + b*x) + 2*b*sqrt(x))/sqrt(b), N 
e(a, 0)), (sqrt(x)*log(sqrt(x))/sqrt(b*x), True))/(128*b**3) + sqrt(a + b* 
x)*(5*a**3*sqrt(x)/(128*b**3) - 5*a**2*x**(3/2)/(192*b**2) + a*x**(5/2)/(4 
8*b) + x**(7/2)/8), Ne(b, 0)), (sqrt(a)*x**(7/2)/7, True)) + 2*B*a**2*Piec 
ewise((a**3*Piecewise((log(2*sqrt(b)*sqrt(a + b*x) + 2*b*sqrt(x))/sqrt(b), 
 Ne(a, 0)), (sqrt(x)*log(sqrt(x))/sqrt(b*x), True))/(16*b**2) + sqrt(a + b 
*x)*(-a**2*sqrt(x)/(16*b**2) + a*x**(3/2)/(24*b) + x**(5/2)/6), Ne(b, 0)), 
 (sqrt(a)*x**(5/2)/5, True)) + 4*B*a*b*Piecewise((-5*a**4*Piecewise((log(2 
*sqrt(b)*sqrt(a + b*x) + 2*b*sqrt(x))/sqrt(b), Ne(a, 0)), (sqrt(x)*log(sqr 
t(x))/sqrt(b*x), True))/(128*b**3) + sqrt(a + b*x)*(5*a**3*sqrt(x)/(128*b* 
*3) - 5*a**2*x**(3/2)/(192*b**2) + a*x**(5/2)/(48*b) + x**(7/2)/8), Ne(b, 
0)), (sqrt(a)*x**(7/2)/7, True)) + 2*B*b**2*Piecewise((7*a**5*Piecewise((l 
og(2*sqrt(b)*sqrt(a + b*x) + 2*b*sqrt(x))/sqrt(b), Ne(a, 0)), (sqrt(x)*...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 485 vs. \(2 (147) = 294\).

Time = 0.05 (sec) , antiderivative size = 485, normalized size of antiderivative = 2.59 \[ \int \sqrt {x} (a+b x)^{5/2} (A+B x) \, dx=\frac {1}{5} \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B b x^{2} - \frac {7}{40} \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a x + \frac {1}{2} \, \sqrt {b x^{2} + a x} A a^{2} x - \frac {7 \, \sqrt {b x^{2} + a x} B a^{3} x}{64 \, b} + \frac {7 \, B a^{5} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{256 \, b^{\frac {5}{2}}} - \frac {A a^{4} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {3}{2}}} - \frac {7 \, \sqrt {b x^{2} + a x} B a^{4}}{128 \, b^{2}} + \frac {7 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a^{2}}{48 \, b} + \frac {\sqrt {b x^{2} + a x} A a^{3}}{4 \, b} + \frac {5 \, {\left (2 \, B a b + A b^{2}\right )} \sqrt {b x^{2} + a x} a^{2} x}{32 \, b^{2}} + \frac {{\left (2 \, B a b + A b^{2}\right )} {\left (b x^{2} + a x\right )}^{\frac {3}{2}} x}{4 \, b} - \frac {{\left (B a^{2} + 2 \, A a b\right )} \sqrt {b x^{2} + a x} a x}{4 \, b} - \frac {5 \, {\left (2 \, B a b + A b^{2}\right )} a^{4} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{128 \, b^{\frac {7}{2}}} + \frac {{\left (B a^{2} + 2 \, A a b\right )} a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {5}{2}}} + \frac {5 \, {\left (2 \, B a b + A b^{2}\right )} \sqrt {b x^{2} + a x} a^{3}}{64 \, b^{3}} - \frac {5 \, {\left (2 \, B a b + A b^{2}\right )} {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a}{24 \, b^{2}} - \frac {{\left (B a^{2} + 2 \, A a b\right )} \sqrt {b x^{2} + a x} a^{2}}{8 \, b^{2}} + \frac {{\left (B a^{2} + 2 \, A a b\right )} {\left (b x^{2} + a x\right )}^{\frac {3}{2}}}{3 \, b} \] Input:

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

Output:

1/5*(b*x^2 + a*x)^(3/2)*B*b*x^2 - 7/40*(b*x^2 + a*x)^(3/2)*B*a*x + 1/2*sqr 
t(b*x^2 + a*x)*A*a^2*x - 7/64*sqrt(b*x^2 + a*x)*B*a^3*x/b + 7/256*B*a^5*lo 
g(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(5/2) - 1/8*A*a^4*log(2*b*x + 
 a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(3/2) - 7/128*sqrt(b*x^2 + a*x)*B*a^4/ 
b^2 + 7/48*(b*x^2 + a*x)^(3/2)*B*a^2/b + 1/4*sqrt(b*x^2 + a*x)*A*a^3/b + 5 
/32*(2*B*a*b + A*b^2)*sqrt(b*x^2 + a*x)*a^2*x/b^2 + 1/4*(2*B*a*b + A*b^2)* 
(b*x^2 + a*x)^(3/2)*x/b - 1/4*(B*a^2 + 2*A*a*b)*sqrt(b*x^2 + a*x)*a*x/b - 
5/128*(2*B*a*b + A*b^2)*a^4*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b 
^(7/2) + 1/16*(B*a^2 + 2*A*a*b)*a^3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sq 
rt(b))/b^(5/2) + 5/64*(2*B*a*b + A*b^2)*sqrt(b*x^2 + a*x)*a^3/b^3 - 5/24*( 
2*B*a*b + A*b^2)*(b*x^2 + a*x)^(3/2)*a/b^2 - 1/8*(B*a^2 + 2*A*a*b)*sqrt(b* 
x^2 + a*x)*a^2/b^2 + 1/3*(B*a^2 + 2*A*a*b)*(b*x^2 + a*x)^(3/2)/b
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.61 \[ \int \sqrt {x} (a+b x)^{5/2} (A+B x) \, dx=\frac {105 \sqrt {x}\, \sqrt {b x +a}\, a^{4} b +1210 \sqrt {x}\, \sqrt {b x +a}\, a^{3} b^{2} x +2104 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b^{3} x^{2}+1488 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{4} x^{3}+384 \sqrt {x}\, \sqrt {b x +a}\, b^{5} x^{4}-105 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{5}}{1920 b^{2}} \] Input:

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

Output:

(105*sqrt(x)*sqrt(a + b*x)*a**4*b + 1210*sqrt(x)*sqrt(a + b*x)*a**3*b**2*x 
 + 2104*sqrt(x)*sqrt(a + b*x)*a**2*b**3*x**2 + 1488*sqrt(x)*sqrt(a + b*x)* 
a*b**4*x**3 + 384*sqrt(x)*sqrt(a + b*x)*b**5*x**4 - 105*sqrt(b)*log((sqrt( 
a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a**5)/(1920*b**2)