\(\int \frac {(a+b x)^{3/2} (A+B x)}{\sqrt {x}} \, dx\) [298]

Optimal result

Integrand size = 20, antiderivative size = 123 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{\sqrt {x}} \, dx=\frac {5 a (6 A b-a B) \sqrt {x} \sqrt {a+b x}}{24 b}+\frac {1}{12} (6 A b-a B) x^{3/2} \sqrt {a+b x}+\frac {B \sqrt {x} (a+b x)^{5/2}}{3 b}+\frac {a^2 (6 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{8 b^{3/2}} \] Output:

5/24*a*(6*A*b-B*a)*x^(1/2)*(b*x+a)^(1/2)/b+1/12*(6*A*b-B*a)*x^(3/2)*(b*x+a 
)^(1/2)+1/3*B*x^(1/2)*(b*x+a)^(5/2)/b+1/8*a^2*(6*A*b-B*a)*arctanh(b^(1/2)* 
x^(1/2)/(b*x+a)^(1/2))/b^(3/2)
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{\sqrt {x}} \, dx=\frac {\sqrt {x} \sqrt {a+b x} \left (30 a A b+3 a^2 B+12 A b^2 x+14 a b B x+8 b^2 B x^2\right )}{24 b}+\frac {a^2 (-6 A b+a B) \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )}{8 b^{3/2}} \] Input:

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

Output:

(Sqrt[x]*Sqrt[a + b*x]*(30*a*A*b + 3*a^2*B + 12*A*b^2*x + 14*a*b*B*x + 8*b 
^2*B*x^2))/(24*b) + (a^2*(-6*A*b + a*B)*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + 
b*x]])/(8*b^(3/2))
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.87, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {90, 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.

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 65

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(B*Sqrt[x]*(a + b*x)^(5/2))/(3*b) + ((6*A*b - a*B)*((Sqrt[x]*(a + b*x)^(3/ 
2))/2 + (3*a*(Sqrt[x]*Sqrt[a + b*x] + (a*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a 
+ b*x]])/Sqrt[b]))/4))/(6*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 = 111, normalized size of antiderivative = 0.90

method	result	size
risch	\(\frac {\left (8 b^{2} B \,x^{2}+12 A \,b^{2} x +14 B a b x +30 a b A +3 a^{2} B \right ) \sqrt {x}\, \sqrt {b x +a}}{24 b}+\frac {a^{2} \left (6 A b -B a \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{16 b^{\frac {3}{2}} \sqrt {x}\, \sqrt {b x +a}}\)	\(111\)
default	\(\frac {\sqrt {b x +a}\, \sqrt {x}\, \left (16 B \,b^{\frac {5}{2}} x^{2} \sqrt {x \left (b x +a \right )}+24 A \sqrt {x \left (b x +a \right )}\, b^{\frac {5}{2}} x +28 B \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a x +18 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2} b +60 A \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a -3 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{3}+6 B \sqrt {x \left (b x +a \right )}\, \sqrt {b}\, a^{2}\right )}{48 b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}}\)	\(176\)

Input:

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

Output:

1/24/b*(8*B*b^2*x^2+12*A*b^2*x+14*B*a*b*x+30*A*a*b+3*B*a^2)*x^(1/2)*(b*x+a 
)^(1/2)+1/16*a^2/b^(3/2)*(6*A*b-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 = 0.11 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.59 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{\sqrt {x}} \, dx=\left [-\frac {3 \, {\left (B a^{3} - 6 \, A a^{2} b\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (8 \, B b^{3} x^{2} + 3 \, B a^{2} b + 30 \, A a b^{2} + 2 \, {\left (7 \, B a b^{2} + 6 \, A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{48 \, b^{2}}, \frac {3 \, {\left (B a^{3} - 6 \, A a^{2} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {b x + a}}\right ) + {\left (8 \, B b^{3} x^{2} + 3 \, B a^{2} b + 30 \, A a b^{2} + 2 \, {\left (7 \, B a b^{2} + 6 \, A b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{24 \, b^{2}}\right ] \] Input:

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

Output:

[-1/48*(3*(B*a^3 - 6*A*a^2*b)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)* 
sqrt(x) + a) - 2*(8*B*b^3*x^2 + 3*B*a^2*b + 30*A*a*b^2 + 2*(7*B*a*b^2 + 6* 
A*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^2, 1/24*(3*(B*a^3 - 6*A*a^2*b)*sqrt(-b) 
*arctan(sqrt(-b)*sqrt(x)/sqrt(b*x + a)) + (8*B*b^3*x^2 + 3*B*a^2*b + 30*A* 
a*b^2 + 2*(7*B*a*b^2 + 6*A*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^2]
 

Sympy [A] (verification not implemented)

Time = 8.55 (sec) , antiderivative size = 394, normalized size of antiderivative = 3.20 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{\sqrt {x}} \, dx=A a^{\frac {3}{2}} \sqrt {x} \sqrt {1 + \frac {b x}{a}} + \frac {A a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{\sqrt {b}} + 2 A b \left (\begin {cases} - \frac {a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x} + 2 b \sqrt {x} \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {\sqrt {x} \log {\left (\sqrt {x} \right )}}{\sqrt {b x}} & \text {otherwise} \end {cases}\right )}{8 b} + \frac {a \sqrt {x} \sqrt {a + b x}}{8 b} + \frac {x^{\frac {3}{2}} \sqrt {a + b x}}{4} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{\frac {3}{2}}}{3} & \text {otherwise} \end {cases}\right ) - \frac {B a^{\frac {5}{2}} \sqrt {x}}{8 b \sqrt {1 + \frac {b x}{a}}} - \frac {B a^{\frac {3}{2}} x^{\frac {3}{2}}}{24 \sqrt {1 + \frac {b x}{a}}} + \frac {5 B \sqrt {a} b x^{\frac {5}{2}}}{12 \sqrt {1 + \frac {b x}{a}}} + \frac {B a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{8 b^{\frac {3}{2}}} + 2 B a \left (\begin {cases} - \frac {a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x} + 2 b \sqrt {x} \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {\sqrt {x} \log {\left (\sqrt {x} \right )}}{\sqrt {b x}} & \text {otherwise} \end {cases}\right )}{8 b} + \frac {a \sqrt {x} \sqrt {a + b x}}{8 b} + \frac {x^{\frac {3}{2}} \sqrt {a + b x}}{4} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{\frac {3}{2}}}{3} & \text {otherwise} \end {cases}\right ) + \frac {B b^{2} x^{\frac {7}{2}}}{3 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} \] Input:

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

Output:

A*a**(3/2)*sqrt(x)*sqrt(1 + b*x/a) + A*a**2*asinh(sqrt(b)*sqrt(x)/sqrt(a)) 
/sqrt(b) + 2*A*b*Piecewise((-a**2*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))/ 
(8*b) + a*sqrt(x)*sqrt(a + b*x)/(8*b) + x**(3/2)*sqrt(a + b*x)/4, Ne(b, 0) 
), (sqrt(a)*x**(3/2)/3, True)) - B*a**(5/2)*sqrt(x)/(8*b*sqrt(1 + b*x/a)) 
- B*a**(3/2)*x**(3/2)/(24*sqrt(1 + b*x/a)) + 5*B*sqrt(a)*b*x**(5/2)/(12*sq 
rt(1 + b*x/a)) + B*a**3*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(8*b**(3/2)) + 2*B* 
a*Piecewise((-a**2*Piecewise((log(2*sqrt(b)*sqrt(a + b*x) + 2*b*sqrt(x))/s 
qrt(b), Ne(a, 0)), (sqrt(x)*log(sqrt(x))/sqrt(b*x), True))/(8*b) + a*sqrt( 
x)*sqrt(a + b*x)/(8*b) + x**(3/2)*sqrt(a + b*x)/4, Ne(b, 0)), (sqrt(a)*x** 
(3/2)/3, True)) + B*b**2*x**(7/2)/(3*sqrt(a)*sqrt(1 + b*x/a))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 282 vs. \(2 (92) = 184\).

Time = 0.04 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.29 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{\sqrt {x}} \, dx=\frac {1}{3} \, \sqrt {b x^{2} + a x} B b x^{2} - \frac {5}{12} \, \sqrt {b x^{2} + a x} B a x - \frac {5 \, B a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {3}{2}}} + \frac {A a^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{\sqrt {b}} + \frac {5 \, \sqrt {b x^{2} + a x} B a^{2}}{8 \, b} + \frac {{\left (2 \, B a b + A b^{2}\right )} \sqrt {b x^{2} + a x} x}{2 \, b} + \frac {3 \, {\left (2 \, B a b + A b^{2}\right )} a^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {5}{2}}} - \frac {{\left (B a^{2} + 2 \, A a b\right )} a \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{2 \, b^{\frac {3}{2}}} - \frac {3 \, {\left (2 \, B a b + A b^{2}\right )} \sqrt {b x^{2} + a x} a}{4 \, b^{2}} + \frac {{\left (B a^{2} + 2 \, A a b\right )} \sqrt {b x^{2} + a x}}{b} \] Input:

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

Output:

1/3*sqrt(b*x^2 + a*x)*B*b*x^2 - 5/12*sqrt(b*x^2 + a*x)*B*a*x - 5/16*B*a^3* 
log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(3/2) + A*a^2*log(2*b*x + a 
 + 2*sqrt(b*x^2 + a*x)*sqrt(b))/sqrt(b) + 5/8*sqrt(b*x^2 + a*x)*B*a^2/b + 
1/2*(2*B*a*b + A*b^2)*sqrt(b*x^2 + a*x)*x/b + 3/8*(2*B*a*b + A*b^2)*a^2*lo 
g(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(5/2) - 1/2*(B*a^2 + 2*A*a*b) 
*a*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(3/2) - 3/4*(2*B*a*b + A 
*b^2)*sqrt(b*x^2 + a*x)*a/b^2 + (B*a^2 + 2*A*a*b)*sqrt(b*x^2 + a*x)/b
 

Giac [A] (verification not implemented)

Time = 75.38 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.11 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{\sqrt {x}} \, dx=\frac {{\left (\sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )} B}{b^{2}} - \frac {B a b^{2} - 6 \, A b^{3}}{b^{4}}\right )} - \frac {3 \, {\left (B a^{2} b^{2} - 6 \, A a b^{3}\right )}}{b^{4}}\right )} + \frac {3 \, {\left (B a^{3} - 6 \, A a^{2} b\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{b^{\frac {3}{2}}}\right )} b}{24 \, {\left | b \right |}} \] Input:

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

Output:

1/24*(sqrt((b*x + a)*b - a*b)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)*B/b^ 
2 - (B*a*b^2 - 6*A*b^3)/b^4) - 3*(B*a^2*b^2 - 6*A*a*b^3)/b^4) + 3*(B*a^3 - 
 6*A*a^2*b)*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a*b)))/b^( 
3/2))*b/abs(b)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.62 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{\sqrt {x}} \, dx=\frac {33 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b +26 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{2} x +8 \sqrt {x}\, \sqrt {b x +a}\, b^{3} x^{2}+15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{3}}{24 b} \] Input:

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

Output:

(33*sqrt(x)*sqrt(a + b*x)*a**2*b + 26*sqrt(x)*sqrt(a + b*x)*a*b**2*x + 8*s 
qrt(x)*sqrt(a + b*x)*b**3*x**2 + 15*sqrt(b)*log((sqrt(a + b*x) + sqrt(x)*s 
qrt(b))/sqrt(a))*a**3)/(24*b)