3.5.80 \(\int x^{3/2} \sqrt {a+b x} (A+B x) \, dx\) [480]

3.5.80.1 Optimal result

Integrand size = 20, antiderivative size = 159 \[ \int x^{3/2} \sqrt {a+b x} (A+B x) \, dx=-\frac {a^2 (8 A b-5 a B) \sqrt {x} \sqrt {a+b x}}{64 b^3}+\frac {a (8 A b-5 a B) x^{3/2} \sqrt {a+b x}}{96 b^2}+\frac {(8 A b-5 a B) x^{5/2} \sqrt {a+b x}}{24 b}+\frac {B x^{5/2} (a+b x)^{3/2}}{4 b}+\frac {a^3 (8 A b-5 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{64 b^{7/2}} \]

output

1/4*B*x^(5/2)*(b*x+a)^(3/2)/b+1/64*a^3*(8*A*b-5*B*a)*arctanh(b^(1/2)*x^(1/ 
2)/(b*x+a)^(1/2))/b^(7/2)+1/96*a*(8*A*b-5*B*a)*x^(3/2)*(b*x+a)^(1/2)/b^2+1 
/24*(8*A*b-5*B*a)*x^(5/2)*(b*x+a)^(1/2)/b-1/64*a^2*(8*A*b-5*B*a)*x^(1/2)*( 
b*x+a)^(1/2)/b^3
 

3.5.80.2 Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.79 \[ \int x^{3/2} \sqrt {a+b x} (A+B x) \, dx=\frac {\sqrt {b} \sqrt {x} \sqrt {a+b x} \left (15 a^3 B+8 a b^2 x (2 A+B x)+16 b^3 x^2 (4 A+3 B x)-2 a^2 b (12 A+5 B x)\right )+6 a^3 (-8 A b+5 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )}{192 b^{7/2}} \]

input

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

output

(Sqrt[b]*Sqrt[x]*Sqrt[a + b*x]*(15*a^3*B + 8*a*b^2*x*(2*A + B*x) + 16*b^3* 
x^2*(4*A + 3*B*x) - 2*a^2*b*(12*A + 5*B*x)) + 6*a^3*(-8*A*b + 5*a*B)*ArcTa 
nh[(Sqrt[b]*Sqrt[x])/(Sqrt[a] - Sqrt[a + b*x])])/(192*b^(7/2))
 

3.5.80.3 Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {90, 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.

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 65

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

\(\Big \downarrow \) 219

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

input

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

output

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

3.5.80.3.1 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])
 

3.5.80.4 Maple [A] (verified)

Time = 1.42 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.85

method	result	size
risch	\(-\frac {\left (-48 b^{3} B \,x^{3}-64 A \,b^{3} x^{2}-8 B a \,b^{2} x^{2}-16 a \,b^{2} A x +10 a^{2} b B x +24 a^{2} b A -15 a^{3} B \right ) \sqrt {x}\, \sqrt {b x +a}}{192 b^{3}}+\frac {a^{3} \left (8 A b -5 B a \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{128 b^{\frac {7}{2}} \sqrt {x}\, \sqrt {b x +a}}\)	\(135\)
default	\(\frac {\sqrt {x}\, \sqrt {b x +a}\, \left (96 B \,b^{\frac {7}{2}} x^{3} \sqrt {x \left (b x +a \right )}+128 A \,b^{\frac {7}{2}} x^{2} \sqrt {x \left (b x +a \right )}+16 B a \,b^{\frac {5}{2}} x^{2} \sqrt {x \left (b x +a \right )}+32 A \sqrt {x \left (b x +a \right )}\, b^{\frac {5}{2}} a x -20 B \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a^{2} x +24 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{3} b -48 A \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} a^{2}-15 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{4}+30 B \sqrt {x \left (b x +a \right )}\, \sqrt {b}\, a^{3}\right )}{384 b^{\frac {7}{2}} \sqrt {x \left (b x +a \right )}}\)	\(218\)

input

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

output

-1/192*(-48*B*b^3*x^3-64*A*b^3*x^2-8*B*a*b^2*x^2-16*A*a*b^2*x+10*B*a^2*b*x 
+24*A*a^2*b-15*B*a^3)*x^(1/2)*(b*x+a)^(1/2)/b^3+1/128*a^3*(8*A*b-5*B*a)/b^ 
(7/2)*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)
 

3.5.80.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.55 \[ \int x^{3/2} \sqrt {a+b x} (A+B x) \, dx=\left [-\frac {3 \, {\left (5 \, B a^{4} - 8 \, A a^{3} b\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (48 \, B b^{4} x^{3} + 15 \, B a^{3} b - 24 \, A a^{2} b^{2} + 8 \, {\left (B a b^{3} + 8 \, A b^{4}\right )} x^{2} - 2 \, {\left (5 \, B a^{2} b^{2} - 8 \, A a b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{384 \, b^{4}}, \frac {3 \, {\left (5 \, B a^{4} - 8 \, A a^{3} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (48 \, B b^{4} x^{3} + 15 \, B a^{3} b - 24 \, A a^{2} b^{2} + 8 \, {\left (B a b^{3} + 8 \, A b^{4}\right )} x^{2} - 2 \, {\left (5 \, B a^{2} b^{2} - 8 \, A a b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{192 \, b^{4}}\right ] \]

input

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

output

[-1/384*(3*(5*B*a^4 - 8*A*a^3*b)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt( 
b)*sqrt(x) + a) - 2*(48*B*b^4*x^3 + 15*B*a^3*b - 24*A*a^2*b^2 + 8*(B*a*b^3 
 + 8*A*b^4)*x^2 - 2*(5*B*a^2*b^2 - 8*A*a*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^ 
4, 1/192*(3*(5*B*a^4 - 8*A*a^3*b)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/( 
b*sqrt(x))) + (48*B*b^4*x^3 + 15*B*a^3*b - 24*A*a^2*b^2 + 8*(B*a*b^3 + 8*A 
*b^4)*x^2 - 2*(5*B*a^2*b^2 - 8*A*a*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^4]
 

3.5.80.6 Sympy [A] (verification not implemented)

Time = 24.29 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.86 \[ \int x^{3/2} \sqrt {a+b x} (A+B x) \, dx=- \frac {A a^{\frac {5}{2}} \sqrt {x}}{8 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {A a^{\frac {3}{2}} x^{\frac {3}{2}}}{24 b \sqrt {1 + \frac {b x}{a}}} + \frac {5 A \sqrt {a} x^{\frac {5}{2}}}{12 \sqrt {1 + \frac {b x}{a}}} + \frac {A a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{8 b^{\frac {5}{2}}} + \frac {A b x^{\frac {7}{2}}}{3 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} + \frac {5 B a^{\frac {7}{2}} \sqrt {x}}{64 b^{3} \sqrt {1 + \frac {b x}{a}}} + \frac {5 B a^{\frac {5}{2}} x^{\frac {3}{2}}}{192 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {B a^{\frac {3}{2}} x^{\frac {5}{2}}}{96 b \sqrt {1 + \frac {b x}{a}}} + \frac {7 B \sqrt {a} x^{\frac {7}{2}}}{24 \sqrt {1 + \frac {b x}{a}}} - \frac {5 B a^{4} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{64 b^{\frac {7}{2}}} + \frac {B b x^{\frac {9}{2}}}{4 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} \]

input

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

output

-A*a**(5/2)*sqrt(x)/(8*b**2*sqrt(1 + b*x/a)) - A*a**(3/2)*x**(3/2)/(24*b*s 
qrt(1 + b*x/a)) + 5*A*sqrt(a)*x**(5/2)/(12*sqrt(1 + b*x/a)) + A*a**3*asinh 
(sqrt(b)*sqrt(x)/sqrt(a))/(8*b**(5/2)) + A*b*x**(7/2)/(3*sqrt(a)*sqrt(1 + 
b*x/a)) + 5*B*a**(7/2)*sqrt(x)/(64*b**3*sqrt(1 + b*x/a)) + 5*B*a**(5/2)*x* 
*(3/2)/(192*b**2*sqrt(1 + b*x/a)) - B*a**(3/2)*x**(5/2)/(96*b*sqrt(1 + b*x 
/a)) + 7*B*sqrt(a)*x**(7/2)/(24*sqrt(1 + b*x/a)) - 5*B*a**4*asinh(sqrt(b)* 
sqrt(x)/sqrt(a))/(64*b**(7/2)) + B*b*x**(9/2)/(4*sqrt(a)*sqrt(1 + b*x/a))
 

3.5.80.7 Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.25 \[ \int x^{3/2} \sqrt {a+b x} (A+B x) \, dx=\frac {5 \, \sqrt {b x^{2} + a x} B a^{2} x}{32 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B x}{4 \, b} - \frac {\sqrt {b x^{2} + a x} A a x}{4 \, b} - \frac {5 \, B a^{4} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{128 \, b^{\frac {7}{2}}} + \frac {A a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {5}{2}}} + \frac {5 \, \sqrt {b x^{2} + a x} B a^{3}}{64 \, b^{3}} - \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a}{24 \, b^{2}} - \frac {\sqrt {b x^{2} + a x} A a^{2}}{8 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A}{3 \, b} \]

input

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

output

5/32*sqrt(b*x^2 + a*x)*B*a^2*x/b^2 + 1/4*(b*x^2 + a*x)^(3/2)*B*x/b - 1/4*s 
qrt(b*x^2 + a*x)*A*a*x/b - 5/128*B*a^4*log(2*b*x + a + 2*sqrt(b*x^2 + a*x) 
*sqrt(b))/b^(7/2) + 1/16*A*a^3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b) 
)/b^(5/2) + 5/64*sqrt(b*x^2 + a*x)*B*a^3/b^3 - 5/24*(b*x^2 + a*x)^(3/2)*B* 
a/b^2 - 1/8*sqrt(b*x^2 + a*x)*A*a^2/b^2 + 1/3*(b*x^2 + a*x)^(3/2)*A/b
 

3.5.80.8 Giac [F(-1)]

Timed out. \[ \int x^{3/2} \sqrt {a+b x} (A+B x) \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.5.80.9 Mupad [F(-1)]

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

input

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

output

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