\(\int x^{5/2} (2+b x)^{5/2} \, dx\) [471]

Optimal result

Integrand size = 15, antiderivative size = 148 \[ \int x^{5/2} (2+b x)^{5/2} \, dx=\frac {5 \sqrt {x} \sqrt {2+b x}}{16 b^3}-\frac {5 x^{3/2} \sqrt {2+b x}}{48 b^2}+\frac {x^{5/2} \sqrt {2+b x}}{24 b}+\frac {9}{8} x^{7/2} \sqrt {2+b x}+\frac {5}{6} b x^{9/2} \sqrt {2+b x}+\frac {1}{6} b^2 x^{11/2} \sqrt {2+b x}-\frac {5 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{8 b^{7/2}} \] Output:

5/16*x^(1/2)*(b*x+2)^(1/2)/b^3-5/48*x^(3/2)*(b*x+2)^(1/2)/b^2+1/24*x^(5/2) 
*(b*x+2)^(1/2)/b+9/8*x^(7/2)*(b*x+2)^(1/2)+5/6*b*x^(9/2)*(b*x+2)^(1/2)+1/6 
*b^2*x^(11/2)*(b*x+2)^(1/2)-5/8*arcsinh(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(7/ 
2)
 

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.68 \[ \int x^{5/2} (2+b x)^{5/2} \, dx=\frac {\sqrt {x} \sqrt {2+b x} \left (15-5 b x+2 b^2 x^2+54 b^3 x^3+40 b^4 x^4+8 b^5 x^5\right )}{48 b^3}+\frac {5 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2+b x}}\right )}{4 b^{7/2}} \] Input:

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

Output:

(Sqrt[x]*Sqrt[2 + b*x]*(15 - 5*b*x + 2*b^2*x^2 + 54*b^3*x^3 + 40*b^4*x^4 + 
 8*b^5*x^5))/(48*b^3) + (5*ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt[2] - Sqrt[2 + b 
*x])])/(4*b^(7/2))
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.15, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {60, 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.

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

\(\Big \downarrow \) 60

\(\displaystyle \frac {5}{6} \int x^{5/2} (b x+2)^{3/2}dx+\frac {1}{6} x^{7/2} (b x+2)^{5/2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {5}{6} \left (\frac {3}{5} \int x^{5/2} \sqrt {b x+2}dx+\frac {1}{5} x^{7/2} (b x+2)^{3/2}\right )+\frac {1}{6} x^{7/2} (b x+2)^{5/2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {5}{6} \left (\frac {3}{5} \left (\frac {1}{4} \int \frac {x^{5/2}}{\sqrt {b x+2}}dx+\frac {1}{4} x^{7/2} \sqrt {b x+2}\right )+\frac {1}{5} x^{7/2} (b x+2)^{3/2}\right )+\frac {1}{6} x^{7/2} (b x+2)^{5/2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {5}{6} \left (\frac {3}{5} \left (\frac {1}{4} \left (\frac {x^{5/2} \sqrt {b x+2}}{3 b}-\frac {5 \int \frac {x^{3/2}}{\sqrt {b x+2}}dx}{3 b}\right )+\frac {1}{4} x^{7/2} \sqrt {b x+2}\right )+\frac {1}{5} x^{7/2} (b x+2)^{3/2}\right )+\frac {1}{6} x^{7/2} (b x+2)^{5/2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {5}{6} \left (\frac {3}{5} \left (\frac {1}{4} \left (\frac {x^{5/2} \sqrt {b x+2}}{3 b}-\frac {5 \left (\frac {x^{3/2} \sqrt {b x+2}}{2 b}-\frac {3 \int \frac {\sqrt {x}}{\sqrt {b x+2}}dx}{2 b}\right )}{3 b}\right )+\frac {1}{4} x^{7/2} \sqrt {b x+2}\right )+\frac {1}{5} x^{7/2} (b x+2)^{3/2}\right )+\frac {1}{6} x^{7/2} (b x+2)^{5/2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {5}{6} \left (\frac {3}{5} \left (\frac {1}{4} \left (\frac {x^{5/2} \sqrt {b x+2}}{3 b}-\frac {5 \left (\frac {x^{3/2} \sqrt {b x+2}}{2 b}-\frac {3 \left (\frac {\sqrt {x} \sqrt {b x+2}}{b}-\frac {\int \frac {1}{\sqrt {x} \sqrt {b x+2}}dx}{b}\right )}{2 b}\right )}{3 b}\right )+\frac {1}{4} x^{7/2} \sqrt {b x+2}\right )+\frac {1}{5} x^{7/2} (b x+2)^{3/2}\right )+\frac {1}{6} x^{7/2} (b x+2)^{5/2}\)

\(\Big \downarrow \) 63

\(\displaystyle \frac {5}{6} \left (\frac {3}{5} \left (\frac {1}{4} \left (\frac {x^{5/2} \sqrt {b x+2}}{3 b}-\frac {5 \left (\frac {x^{3/2} \sqrt {b x+2}}{2 b}-\frac {3 \left (\frac {\sqrt {x} \sqrt {b x+2}}{b}-\frac {2 \int \frac {1}{\sqrt {b x+2}}d\sqrt {x}}{b}\right )}{2 b}\right )}{3 b}\right )+\frac {1}{4} x^{7/2} \sqrt {b x+2}\right )+\frac {1}{5} x^{7/2} (b x+2)^{3/2}\right )+\frac {1}{6} x^{7/2} (b x+2)^{5/2}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {5}{6} \left (\frac {3}{5} \left (\frac {1}{4} \left (\frac {x^{5/2} \sqrt {b x+2}}{3 b}-\frac {5 \left (\frac {x^{3/2} \sqrt {b x+2}}{2 b}-\frac {3 \left (\frac {\sqrt {x} \sqrt {b x+2}}{b}-\frac {2 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}}\right )}{2 b}\right )}{3 b}\right )+\frac {1}{4} x^{7/2} \sqrt {b x+2}\right )+\frac {1}{5} x^{7/2} (b x+2)^{3/2}\right )+\frac {1}{6} x^{7/2} (b x+2)^{5/2}\)

Input:

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

Output:

(x^(7/2)*(2 + b*x)^(5/2))/6 + (5*((x^(7/2)*(2 + b*x)^(3/2))/5 + (3*((x^(7/ 
2)*Sqrt[2 + b*x])/4 + ((x^(5/2)*Sqrt[2 + b*x])/(3*b) - (5*((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*b))/4))/5))/6
 

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 = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.59

method	result	size
meijerg	\(-\frac {120 \left (-\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {b}\, \left (56 b^{5} x^{5}+280 b^{4} x^{4}+378 b^{3} x^{3}+14 b^{2} x^{2}-35 b x +105\right ) \sqrt {\frac {b x}{2}+1}}{40320}+\frac {\sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{192}\right )}{b^{\frac {7}{2}} \sqrt {\pi }}\)	\(87\)
risch	\(\frac {\left (8 b^{5} x^{5}+40 b^{4} x^{4}+54 b^{3} x^{3}+2 b^{2} x^{2}-5 b x +15\right ) \sqrt {x}\, \sqrt {b x +2}}{48 b^{3}}-\frac {5 \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right ) \sqrt {x \left (b x +2\right )}}{16 b^{\frac {7}{2}} \sqrt {x}\, \sqrt {b x +2}}\)	\(101\)
default	\(\frac {x^{\frac {5}{2}} \left (b x +2\right )^{\frac {7}{2}}}{6 b}-\frac {5 \left (\frac {x^{\frac {3}{2}} \left (b x +2\right )^{\frac {7}{2}}}{5 b}-\frac {3 \left (\frac {\sqrt {x}\, \left (b x +2\right )^{\frac {7}{2}}}{4 b}-\frac {\frac {\sqrt {x}\, \left (b x +2\right )^{\frac {5}{2}}}{3}+\frac {5 \sqrt {x}\, \left (b x +2\right )^{\frac {3}{2}}}{6}+\frac {5 \sqrt {x}\, \sqrt {b x +2}}{2}+\frac {5 \sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right )}{2 \sqrt {b x +2}\, \sqrt {x}\, \sqrt {b}}}{4 b}\right )}{5 b}\right )}{6 b}\)	\(147\)

Input:

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

Output:

-120/b^(7/2)/Pi^(1/2)*(-1/40320*Pi^(1/2)*x^(1/2)*2^(1/2)*b^(1/2)*(56*b^5*x 
^5+280*b^4*x^4+378*b^3*x^3+14*b^2*x^2-35*b*x+105)*(1/2*b*x+1)^(1/2)+1/192* 
Pi^(1/2)*arcsinh(1/2*b^(1/2)*x^(1/2)*2^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.14 \[ \int x^{5/2} (2+b x)^{5/2} \, dx=\left [\frac {{\left (8 \, b^{6} x^{5} + 40 \, b^{5} x^{4} + 54 \, b^{4} x^{3} + 2 \, b^{3} x^{2} - 5 \, b^{2} x + 15 \, b\right )} \sqrt {b x + 2} \sqrt {x} + 15 \, \sqrt {b} \log \left (b x - \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right )}{48 \, b^{4}}, \frac {{\left (8 \, b^{6} x^{5} + 40 \, b^{5} x^{4} + 54 \, b^{4} x^{3} + 2 \, b^{3} x^{2} - 5 \, b^{2} x + 15 \, b\right )} \sqrt {b x + 2} \sqrt {x} + 30 \, \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {b x + 2}}\right )}{48 \, b^{4}}\right ] \] Input:

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

Output:

[1/48*((8*b^6*x^5 + 40*b^5*x^4 + 54*b^4*x^3 + 2*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^4, 1/48*((8*b^6*x^5 + 40*b^5*x^4 + 54*b^4*x^3 + 2*b^3*x^2 - 5*b^2 
*x + 15*b)*sqrt(b*x + 2)*sqrt(x) + 30*sqrt(-b)*arctan(sqrt(-b)*sqrt(x)/sqr 
t(b*x + 2)))/b^4]
 

Sympy [A] (verification not implemented)

Time = 153.94 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.07 \[ \int x^{5/2} (2+b x)^{5/2} \, dx=\frac {b^{3} x^{\frac {13}{2}}}{6 \sqrt {b x + 2}} + \frac {7 b^{2} x^{\frac {11}{2}}}{6 \sqrt {b x + 2}} + \frac {67 b x^{\frac {9}{2}}}{24 \sqrt {b x + 2}} + \frac {55 x^{\frac {7}{2}}}{24 \sqrt {b x + 2}} - \frac {x^{\frac {5}{2}}}{48 b \sqrt {b x + 2}} + \frac {5 x^{\frac {3}{2}}}{48 b^{2} \sqrt {b x + 2}} + \frac {5 \sqrt {x}}{8 b^{3} \sqrt {b x + 2}} - \frac {5 \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{8 b^{\frac {7}{2}}} \] Input:

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

Output:

b**3*x**(13/2)/(6*sqrt(b*x + 2)) + 7*b**2*x**(11/2)/(6*sqrt(b*x + 2)) + 67 
*b*x**(9/2)/(24*sqrt(b*x + 2)) + 55*x**(7/2)/(24*sqrt(b*x + 2)) - x**(5/2) 
/(48*b*sqrt(b*x + 2)) + 5*x**(3/2)/(48*b**2*sqrt(b*x + 2)) + 5*sqrt(x)/(8* 
b**3*sqrt(b*x + 2)) - 5*asinh(sqrt(2)*sqrt(b)*sqrt(x)/2)/(8*b**(7/2))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (103) = 206\).

Time = 0.13 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.51 \[ \int x^{5/2} (2+b x)^{5/2} \, dx=\frac {\frac {15 \, \sqrt {b x + 2} b^{5}}{\sqrt {x}} - \frac {85 \, {\left (b x + 2\right )}^{\frac {3}{2}} b^{4}}{x^{\frac {3}{2}}} + \frac {198 \, {\left (b x + 2\right )}^{\frac {5}{2}} b^{3}}{x^{\frac {5}{2}}} + \frac {198 \, {\left (b x + 2\right )}^{\frac {7}{2}} b^{2}}{x^{\frac {7}{2}}} - \frac {85 \, {\left (b x + 2\right )}^{\frac {9}{2}} b}{x^{\frac {9}{2}}} + \frac {15 \, {\left (b x + 2\right )}^{\frac {11}{2}}}{x^{\frac {11}{2}}}}{24 \, {\left (b^{9} - \frac {6 \, {\left (b x + 2\right )} b^{8}}{x} + \frac {15 \, {\left (b x + 2\right )}^{2} b^{7}}{x^{2}} - \frac {20 \, {\left (b x + 2\right )}^{3} b^{6}}{x^{3}} + \frac {15 \, {\left (b x + 2\right )}^{4} b^{5}}{x^{4}} - \frac {6 \, {\left (b x + 2\right )}^{5} b^{4}}{x^{5}} + \frac {{\left (b x + 2\right )}^{6} b^{3}}{x^{6}}\right )}} + \frac {5 \, \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{16 \, b^{\frac {7}{2}}} \] Input:

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

Output:

1/24*(15*sqrt(b*x + 2)*b^5/sqrt(x) - 85*(b*x + 2)^(3/2)*b^4/x^(3/2) + 198* 
(b*x + 2)^(5/2)*b^3/x^(5/2) + 198*(b*x + 2)^(7/2)*b^2/x^(7/2) - 85*(b*x + 
2)^(9/2)*b/x^(9/2) + 15*(b*x + 2)^(11/2)/x^(11/2))/(b^9 - 6*(b*x + 2)*b^8/ 
x + 15*(b*x + 2)^2*b^7/x^2 - 20*(b*x + 2)^3*b^6/x^3 + 15*(b*x + 2)^4*b^5/x 
^4 - 6*(b*x + 2)^5*b^4/x^5 + (b*x + 2)^6*b^3/x^6) + 5/16*log(-(sqrt(b) - s 
qrt(b*x + 2)/sqrt(x))/(sqrt(b) + sqrt(b*x + 2)/sqrt(x)))/b^(7/2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 403 vs. \(2 (103) = 206\).

Time = 22.49 (sec) , antiderivative size = 403, normalized size of antiderivative = 2.72 \[ \int x^{5/2} (2+b x)^{5/2} \, dx=\frac {{\left ({\left ({\left (2 \, {\left ({\left (b x + 2\right )} {\left (4 \, {\left (b x + 2\right )} {\left (\frac {5 \, {\left (b x + 2\right )}}{b^{5}} - \frac {61}{b^{5}}\right )} + \frac {1251}{b^{5}}\right )} - \frac {3481}{b^{5}}\right )} {\left (b x + 2\right )} + \frac {11395}{b^{5}}\right )} {\left (b x + 2\right )} - \frac {11895}{b^{5}}\right )} \sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} - \frac {6930 \, \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )}{b^{\frac {9}{2}}}\right )} b {\left | b \right |} + \frac {120 \, {\left ({\left ({\left (b x + 2\right )} {\left (2 \, {\left (b x + 2\right )} {\left (\frac {3 \, {\left (b x + 2\right )}}{b^{3}} - \frac {25}{b^{3}}\right )} + \frac {163}{b^{3}}\right )} - \frac {279}{b^{3}}\right )} \sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} - \frac {210 \, \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )}{b^{\frac {5}{2}}}\right )} {\left | b \right |}}{b} + \frac {36 \, {\left ({\left ({\left (2 \, {\left ({\left (4 \, b x - 33\right )} {\left (b x + 2\right )} + 171\right )} {\left (b x + 2\right )} - 745\right )} {\left (b x + 2\right )} + 965\right )} \sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} + 630 \, \sqrt {b} \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )\right )} {\left | b \right |}}{b^{4}} + \frac {320 \, {\left ({\left ({\left (2 \, b x - 9\right )} {\left (b x + 2\right )} + 33\right )} \sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} + 30 \, \sqrt {b} \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )\right )} {\left | b \right |}}{b^{4}}}{240 \, b} \] Input:

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

Output:

1/240*((((2*((b*x + 2)*(4*(b*x + 2)*(5*(b*x + 2)/b^5 - 61/b^5) + 1251/b^5) 
 - 3481/b^5)*(b*x + 2) + 11395/b^5)*(b*x + 2) - 11895/b^5)*sqrt((b*x + 2)* 
b - 2*b)*sqrt(b*x + 2) - 6930*log(abs(-sqrt(b*x + 2)*sqrt(b) + sqrt((b*x + 
 2)*b - 2*b)))/b^(9/2))*b*abs(b) + 120*(((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)/b + 36*(((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^4 + 320*(((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) + sqrt((b*x + 2)*b - 2*b))))*abs(b)/b^4)/b
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.79 \[ \int x^{5/2} (2+b x)^{5/2} \, dx=\frac {8 \sqrt {x}\, \sqrt {b x +2}\, b^{6} x^{5}+40 \sqrt {x}\, \sqrt {b x +2}\, b^{5} x^{4}+54 \sqrt {x}\, \sqrt {b x +2}\, b^{4} x^{3}+2 \sqrt {x}\, \sqrt {b x +2}\, b^{3} x^{2}-5 \sqrt {x}\, \sqrt {b x +2}\, b^{2} x +15 \sqrt {x}\, \sqrt {b x +2}\, b -30 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +2}+\sqrt {x}\, \sqrt {b}}{\sqrt {2}}\right )}{48 b^{4}} \] Input:

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

Output:

(8*sqrt(x)*sqrt(b*x + 2)*b**6*x**5 + 40*sqrt(x)*sqrt(b*x + 2)*b**5*x**4 + 
54*sqrt(x)*sqrt(b*x + 2)*b**4*x**3 + 2*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)*lo 
g((sqrt(b*x + 2) + sqrt(x)*sqrt(b))/sqrt(2)))/(48*b**4)