\(\int x^{3/2} (2+b x)^{5/2} \, dx\) [472]

Optimal result

Integrand size = 15, antiderivative size = 127 \[ \int x^{3/2} (2+b x)^{5/2} \, dx=-\frac {3 \sqrt {x} \sqrt {2+b x}}{8 b^2}+\frac {x^{3/2} \sqrt {2+b x}}{8 b}+\frac {31}{20} x^{5/2} \sqrt {2+b x}+\frac {21}{20} b x^{7/2} \sqrt {2+b x}+\frac {1}{5} b^2 x^{9/2} \sqrt {2+b x}+\frac {3 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{5/2}} \] Output:

-3/8*x^(1/2)*(b*x+2)^(1/2)/b^2+1/8*x^(3/2)*(b*x+2)^(1/2)/b+31/20*x^(5/2)*( 
b*x+2)^(1/2)+21/20*b*x^(7/2)*(b*x+2)^(1/2)+1/5*b^2*x^(9/2)*(b*x+2)^(1/2)+3 
/4*arcsinh(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.72 \[ \int x^{3/2} (2+b x)^{5/2} \, dx=\frac {\sqrt {x} \sqrt {2+b x} \left (-15+5 b x+62 b^2 x^2+42 b^3 x^3+8 b^4 x^4\right )}{40 b^2}-\frac {3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2+b x}}\right )}{2 b^{5/2}} \] Input:

Integrate[x^(3/2)*(2 + b*x)^(5/2),x]
 

Output:

(Sqrt[x]*Sqrt[2 + b*x]*(-15 + 5*b*x + 62*b^2*x^2 + 42*b^3*x^3 + 8*b^4*x^4) 
)/(40*b^2) - (3*ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt[2] - Sqrt[2 + b*x])])/(2*b 
^(5/2))
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {60, 60, 60, 60, 60, 63, 222}

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.

\(\displaystyle \int x^{3/2} (b x+2)^{5/2} \, dx\)

\(\Big \downarrow \) 60

\(\displaystyle \int x^{3/2} (b x+2)^{3/2}dx+\frac {1}{5} x^{5/2} (b x+2)^{5/2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {3}{4} \int x^{3/2} \sqrt {b x+2}dx+\frac {1}{5} x^{5/2} (b x+2)^{5/2}+\frac {1}{4} x^{5/2} (b x+2)^{3/2}\)

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 63

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

\(\Big \downarrow \) 222

\(\displaystyle \frac {3}{4} \left (\frac {1}{3} \left (\frac {x^{3/2} \sqrt {b x+2}}{2 b}-\frac {3 \left (\frac {\sqrt {x} \sqrt {b x+2}}{b}-\frac {2 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}}\right )}{2 b}\right )+\frac {1}{3} x^{5/2} \sqrt {b x+2}\right )+\frac {1}{5} x^{5/2} (b x+2)^{5/2}+\frac {1}{4} x^{5/2} (b x+2)^{3/2}\)

Input:

Int[x^(3/2)*(2 + b*x)^(5/2),x]
 

Output:

(x^(5/2)*(2 + b*x)^(3/2))/4 + (x^(5/2)*(2 + b*x)^(5/2))/5 + (3*((x^(5/2)*S 
qrt[2 + b*x])/3 + ((x^(3/2)*Sqrt[2 + b*x])/(2*b) - (3*((Sqrt[x]*Sqrt[2 + b 
*x])/b - (2*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(3/2)))/(2*b))/3))/4
 

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 63

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2/b   S 
ubst[Int[1/Sqrt[c + d*(x^2/b)], x], x, Sqrt[b*x]], x] /; FreeQ[{b, c, d}, x 
] && GtQ[c, 0]
 

rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.62

method	result	size
meijerg	\(-\frac {60 \left (\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {b}\, \left (-8 b^{4} x^{4}-42 b^{3} x^{3}-62 b^{2} x^{2}-5 b x +15\right ) \sqrt {\frac {b x}{2}+1}}{2400}-\frac {\sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{80}\right )}{b^{\frac {5}{2}} \sqrt {\pi }}\)	\(79\)
risch	\(\frac {\left (8 b^{4} x^{4}+42 b^{3} x^{3}+62 b^{2} x^{2}+5 b x -15\right ) \sqrt {x}\, \sqrt {b x +2}}{40 b^{2}}+\frac {3 \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right ) \sqrt {x \left (b x +2\right )}}{8 b^{\frac {5}{2}} \sqrt {x}\, \sqrt {b x +2}}\)	\(93\)
default	\(\frac {x^{\frac {3}{2}} \left (b x +2\right )^{\frac {7}{2}}}{5 b}-\frac {3 \left (\frac {\sqrt {x}\, \left (b x +2\right )^{\frac {7}{2}}}{4 b}-\frac {\frac {\sqrt {x}\, \left (b x +2\right )^{\frac {5}{2}}}{3}+\frac {5 \sqrt {x}\, \left (b x +2\right )^{\frac {3}{2}}}{6}+\frac {5 \sqrt {x}\, \sqrt {b x +2}}{2}+\frac {5 \sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right )}{2 \sqrt {b x +2}\, \sqrt {x}\, \sqrt {b}}}{4 b}\right )}{5 b}\)	\(126\)

Input:

int(x^(3/2)*(b*x+2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-60/b^(5/2)/Pi^(1/2)*(1/2400*Pi^(1/2)*x^(1/2)*2^(1/2)*b^(1/2)*(-8*b^4*x^4- 
42*b^3*x^3-62*b^2*x^2-5*b*x+15)*(1/2*b*x+1)^(1/2)-1/80*Pi^(1/2)*arcsinh(1/ 
2*b^(1/2)*x^(1/2)*2^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.20 \[ \int x^{3/2} (2+b x)^{5/2} \, dx=\left [\frac {{\left (8 \, b^{5} x^{4} + 42 \, b^{4} x^{3} + 62 \, b^{3} x^{2} + 5 \, b^{2} x - 15 \, b\right )} \sqrt {b x + 2} \sqrt {x} + 15 \, \sqrt {b} \log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right )}{40 \, b^{3}}, \frac {{\left (8 \, b^{5} x^{4} + 42 \, b^{4} x^{3} + 62 \, b^{3} x^{2} + 5 \, b^{2} x - 15 \, b\right )} \sqrt {b x + 2} \sqrt {x} - 30 \, \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {b x + 2}}\right )}{40 \, b^{3}}\right ] \] Input:

integrate(x^(3/2)*(b*x+2)^(5/2),x, algorithm="fricas")
 

Output:

[1/40*((8*b^5*x^4 + 42*b^4*x^3 + 62*b^3*x^2 + 5*b^2*x - 15*b)*sqrt(b*x + 2 
)*sqrt(x) + 15*sqrt(b)*log(b*x + sqrt(b*x + 2)*sqrt(b)*sqrt(x) + 1))/b^3, 
1/40*((8*b^5*x^4 + 42*b^4*x^3 + 62*b^3*x^2 + 5*b^2*x - 15*b)*sqrt(b*x + 2) 
*sqrt(x) - 30*sqrt(-b)*arctan(sqrt(-b)*sqrt(x)/sqrt(b*x + 2)))/b^3]
 

Sympy [A] (verification not implemented)

Time = 27.76 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.09 \[ \int x^{3/2} (2+b x)^{5/2} \, dx=\frac {b^{3} x^{\frac {11}{2}}}{5 \sqrt {b x + 2}} + \frac {29 b^{2} x^{\frac {9}{2}}}{20 \sqrt {b x + 2}} + \frac {73 b x^{\frac {7}{2}}}{20 \sqrt {b x + 2}} + \frac {129 x^{\frac {5}{2}}}{40 \sqrt {b x + 2}} - \frac {x^{\frac {3}{2}}}{8 b \sqrt {b x + 2}} - \frac {3 \sqrt {x}}{4 b^{2} \sqrt {b x + 2}} + \frac {3 \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{4 b^{\frac {5}{2}}} \] Input:

integrate(x**(3/2)*(b*x+2)**(5/2),x)
 

Output:

b**3*x**(11/2)/(5*sqrt(b*x + 2)) + 29*b**2*x**(9/2)/(20*sqrt(b*x + 2)) + 7 
3*b*x**(7/2)/(20*sqrt(b*x + 2)) + 129*x**(5/2)/(40*sqrt(b*x + 2)) - x**(3/ 
2)/(8*b*sqrt(b*x + 2)) - 3*sqrt(x)/(4*b**2*sqrt(b*x + 2)) + 3*asinh(sqrt(2 
)*sqrt(b)*sqrt(x)/2)/(4*b**(5/2))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (88) = 176\).

Time = 0.11 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.53 \[ \int x^{3/2} (2+b x)^{5/2} \, dx=-\frac {\frac {15 \, \sqrt {b x + 2} b^{4}}{\sqrt {x}} - \frac {70 \, {\left (b x + 2\right )}^{\frac {3}{2}} b^{3}}{x^{\frac {3}{2}}} + \frac {128 \, {\left (b x + 2\right )}^{\frac {5}{2}} b^{2}}{x^{\frac {5}{2}}} + \frac {70 \, {\left (b x + 2\right )}^{\frac {7}{2}} b}{x^{\frac {7}{2}}} - \frac {15 \, {\left (b x + 2\right )}^{\frac {9}{2}}}{x^{\frac {9}{2}}}}{20 \, {\left (b^{7} - \frac {5 \, {\left (b x + 2\right )} b^{6}}{x} + \frac {10 \, {\left (b x + 2\right )}^{2} b^{5}}{x^{2}} - \frac {10 \, {\left (b x + 2\right )}^{3} b^{4}}{x^{3}} + \frac {5 \, {\left (b x + 2\right )}^{4} b^{3}}{x^{4}} - \frac {{\left (b x + 2\right )}^{5} b^{2}}{x^{5}}\right )}} - \frac {3 \, \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{8 \, b^{\frac {5}{2}}} \] Input:

integrate(x^(3/2)*(b*x+2)^(5/2),x, algorithm="maxima")
 

Output:

-1/20*(15*sqrt(b*x + 2)*b^4/sqrt(x) - 70*(b*x + 2)^(3/2)*b^3/x^(3/2) + 128 
*(b*x + 2)^(5/2)*b^2/x^(5/2) + 70*(b*x + 2)^(7/2)*b/x^(7/2) - 15*(b*x + 2) 
^(9/2)/x^(9/2))/(b^7 - 5*(b*x + 2)*b^6/x + 10*(b*x + 2)^2*b^5/x^2 - 10*(b* 
x + 2)^3*b^4/x^3 + 5*(b*x + 2)^4*b^3/x^4 - (b*x + 2)^5*b^2/x^5) - 3/8*log( 
-(sqrt(b) - sqrt(b*x + 2)/sqrt(x))/(sqrt(b) + sqrt(b*x + 2)/sqrt(x)))/b^(5 
/2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 341 vs. \(2 (88) = 176\).

Time = 22.35 (sec) , antiderivative size = 341, normalized size of antiderivative = 2.69 \[ \int x^{3/2} (2+b x)^{5/2} \, dx=\frac {10 \, {\left ({\left ({\left (b x + 2\right )} {\left (2 \, {\left (b x + 2\right )} {\left (\frac {3 \, {\left (b x + 2\right )}}{b^{3}} - \frac {25}{b^{3}}\right )} + \frac {163}{b^{3}}\right )} - \frac {279}{b^{3}}\right )} \sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} - \frac {210 \, \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )}{b^{\frac {5}{2}}}\right )} {\left | b \right |} + \frac {{\left ({\left ({\left (2 \, {\left ({\left (4 \, b x - 33\right )} {\left (b x + 2\right )} + 171\right )} {\left (b x + 2\right )} - 745\right )} {\left (b x + 2\right )} + 965\right )} \sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} + 630 \, \sqrt {b} \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )\right )} {\left | b \right |}}{b^{3}} + \frac {80 \, {\left ({\left ({\left (2 \, b x - 9\right )} {\left (b x + 2\right )} + 33\right )} \sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} + 30 \, \sqrt {b} \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )\right )} {\left | b \right |}}{b^{3}} + \frac {160 \, {\left (\sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} {\left (b x - 3\right )} - 6 \, \sqrt {b} \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )\right )} {\left | b \right |}}{b^{3}}}{40 \, b} \] Input:

integrate(x^(3/2)*(b*x+2)^(5/2),x, algorithm="giac")
 

Output:

1/40*(10*(((b*x + 2)*(2*(b*x + 2)*(3*(b*x + 2)/b^3 - 25/b^3) + 163/b^3) - 
279/b^3)*sqrt((b*x + 2)*b - 2*b)*sqrt(b*x + 2) - 210*log(abs(-sqrt(b*x + 2 
)*sqrt(b) + sqrt((b*x + 2)*b - 2*b)))/b^(5/2))*abs(b) + (((2*((4*b*x - 33) 
*(b*x + 2) + 171)*(b*x + 2) - 745)*(b*x + 2) + 965)*sqrt((b*x + 2)*b - 2*b 
)*sqrt(b*x + 2) + 630*sqrt(b)*log(abs(-sqrt(b*x + 2)*sqrt(b) + sqrt((b*x + 
 2)*b - 2*b))))*abs(b)/b^3 + 80*(((2*b*x - 9)*(b*x + 2) + 33)*sqrt((b*x + 
2)*b - 2*b)*sqrt(b*x + 2) + 30*sqrt(b)*log(abs(-sqrt(b*x + 2)*sqrt(b) + sq 
rt((b*x + 2)*b - 2*b))))*abs(b)/b^3 + 160*(sqrt((b*x + 2)*b - 2*b)*sqrt(b* 
x + 2)*(b*x - 3) - 6*sqrt(b)*log(abs(-sqrt(b*x + 2)*sqrt(b) + sqrt((b*x + 
2)*b - 2*b))))*abs(b)/b^3)/b
 

Mupad [F(-1)]

Timed out. \[ \int x^{3/2} (2+b x)^{5/2} \, dx=\int x^{3/2}\,{\left (b\,x+2\right )}^{5/2} \,d x \] Input:

int(x^(3/2)*(b*x + 2)^(5/2),x)
 

Output:

int(x^(3/2)*(b*x + 2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.80 \[ \int x^{3/2} (2+b x)^{5/2} \, dx=\frac {8 \sqrt {x}\, \sqrt {b x +2}\, b^{5} x^{4}+42 \sqrt {x}\, \sqrt {b x +2}\, b^{4} x^{3}+62 \sqrt {x}\, \sqrt {b x +2}\, b^{3} x^{2}+5 \sqrt {x}\, \sqrt {b x +2}\, b^{2} x -15 \sqrt {x}\, \sqrt {b x +2}\, b +30 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +2}+\sqrt {x}\, \sqrt {b}}{\sqrt {2}}\right )}{40 b^{3}} \] Input:

int(x^(3/2)*(b*x+2)^(5/2),x)
 

Output:

(8*sqrt(x)*sqrt(b*x + 2)*b**5*x**4 + 42*sqrt(x)*sqrt(b*x + 2)*b**4*x**3 + 
62*sqrt(x)*sqrt(b*x + 2)*b**3*x**2 + 5*sqrt(x)*sqrt(b*x + 2)*b**2*x - 15*s 
qrt(x)*sqrt(b*x + 2)*b + 30*sqrt(b)*log((sqrt(b*x + 2) + sqrt(x)*sqrt(b))/ 
sqrt(2)))/(40*b**3)