Integrand size = 15, antiderivative size = 86 \[ \int \frac {x^{5/2}}{(2+b x)^{3/2}} \, dx=-\frac {2 x^{5/2}}{b \sqrt {2+b x}}-\frac {15 \sqrt {x} \sqrt {2+b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2+b x}}{2 b^2}+\frac {15 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}} \]
15*arcsinh(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(7/2)-2*x^(5/2)/b/(b*x+2)^(1/2)+ 5/2*x^(3/2)*(b*x+2)^(1/2)/b^2-15/2*x^(1/2)*(b*x+2)^(1/2)/b^3
Time = 0.22 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.85 \[ \int \frac {x^{5/2}}{(2+b x)^{3/2}} \, dx=\frac {\sqrt {x} \left (-30-5 b x+b^2 x^2\right )}{2 b^3 \sqrt {2+b x}}-\frac {30 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2+b x}}\right )}{b^{7/2}} \]
(Sqrt[x]*(-30 - 5*b*x + b^2*x^2))/(2*b^3*Sqrt[2 + b*x]) - (30*ArcTanh[(Sqr t[b]*Sqrt[x])/(Sqrt[2] - Sqrt[2 + b*x])])/b^(7/2)
Time = 0.18 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {57, 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.
\(\displaystyle \int \frac {x^{5/2}}{(b x+2)^{3/2}} \, dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {5 \int \frac {x^{3/2}}{\sqrt {b x+2}}dx}{b}-\frac {2 x^{5/2}}{b \sqrt {b x+2}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5 \left (\frac {x^{3/2} \sqrt {b x+2}}{2 b}-\frac {3 \int \frac {\sqrt {x}}{\sqrt {b x+2}}dx}{2 b}\right )}{b}-\frac {2 x^{5/2}}{b \sqrt {b x+2}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5 \left (\frac {x^{3/2} \sqrt {b x+2}}{2 b}-\frac {3 \left (\frac {\sqrt {x} \sqrt {b x+2}}{b}-\frac {\int \frac {1}{\sqrt {x} \sqrt {b x+2}}dx}{b}\right )}{2 b}\right )}{b}-\frac {2 x^{5/2}}{b \sqrt {b x+2}}\) |
\(\Big \downarrow \) 63 |
\(\displaystyle \frac {5 \left (\frac {x^{3/2} \sqrt {b x+2}}{2 b}-\frac {3 \left (\frac {\sqrt {x} \sqrt {b x+2}}{b}-\frac {2 \int \frac {1}{\sqrt {b x+2}}d\sqrt {x}}{b}\right )}{2 b}\right )}{b}-\frac {2 x^{5/2}}{b \sqrt {b x+2}}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {5 \left (\frac {x^{3/2} \sqrt {b x+2}}{2 b}-\frac {3 \left (\frac {\sqrt {x} \sqrt {b x+2}}{b}-\frac {2 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}}\right )}{2 b}\right )}{b}-\frac {2 x^{5/2}}{b \sqrt {b x+2}}\) |
(-2*x^(5/2))/(b*Sqrt[2 + b*x]) + (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)))/b
3.7.16.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c , d, m, n, x]
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]
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]
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]
Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.73
method | result | size |
meijerg | \(\frac {-\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {b}\, \left (-\frac {7}{2} b^{2} x^{2}+\frac {35}{2} b x +105\right )}{14 \sqrt {\frac {b x}{2}+1}}+15 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{b^{\frac {7}{2}} \sqrt {\pi }}\) | \(63\) |
risch | \(\frac {\left (b x -7\right ) \sqrt {x}\, \sqrt {b x +2}}{2 b^{3}}+\frac {\left (\frac {15 \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right )}{2 b^{\frac {7}{2}}}-\frac {8 \sqrt {b \left (x +\frac {2}{b}\right )^{2}-2 x -\frac {4}{b}}}{b^{4} \left (x +\frac {2}{b}\right )}\right ) \sqrt {x \left (b x +2\right )}}{\sqrt {x}\, \sqrt {b x +2}}\) | \(106\) |
8/b^(7/2)/Pi^(1/2)*(-1/112*Pi^(1/2)*x^(1/2)*2^(1/2)*b^(1/2)*(-7/2*b^2*x^2+ 35/2*b*x+105)/(1/2*b*x+1)^(1/2)+15/8*Pi^(1/2)*arcsinh(1/2*b^(1/2)*x^(1/2)* 2^(1/2)))
Time = 0.24 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.77 \[ \int \frac {x^{5/2}}{(2+b x)^{3/2}} \, dx=\left [\frac {15 \, {\left (b x + 2\right )} \sqrt {b} \log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right ) + {\left (b^{3} x^{2} - 5 \, b^{2} x - 30 \, b\right )} \sqrt {b x + 2} \sqrt {x}}{2 \, {\left (b^{5} x + 2 \, b^{4}\right )}}, -\frac {30 \, {\left (b x + 2\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (b^{3} x^{2} - 5 \, b^{2} x - 30 \, b\right )} \sqrt {b x + 2} \sqrt {x}}{2 \, {\left (b^{5} x + 2 \, b^{4}\right )}}\right ] \]
[1/2*(15*(b*x + 2)*sqrt(b)*log(b*x + sqrt(b*x + 2)*sqrt(b)*sqrt(x) + 1) + (b^3*x^2 - 5*b^2*x - 30*b)*sqrt(b*x + 2)*sqrt(x))/(b^5*x + 2*b^4), -1/2*(3 0*(b*x + 2)*sqrt(-b)*arctan(sqrt(b*x + 2)*sqrt(-b)/(b*sqrt(x))) - (b^3*x^2 - 5*b^2*x - 30*b)*sqrt(b*x + 2)*sqrt(x))/(b^5*x + 2*b^4)]
Time = 7.17 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.93 \[ \int \frac {x^{5/2}}{(2+b x)^{3/2}} \, dx=\frac {x^{\frac {5}{2}}}{2 b \sqrt {b x + 2}} - \frac {5 x^{\frac {3}{2}}}{2 b^{2} \sqrt {b x + 2}} - \frac {15 \sqrt {x}}{b^{3} \sqrt {b x + 2}} + \frac {15 \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {7}{2}}} \]
x**(5/2)/(2*b*sqrt(b*x + 2)) - 5*x**(3/2)/(2*b**2*sqrt(b*x + 2)) - 15*sqrt (x)/(b**3*sqrt(b*x + 2)) + 15*asinh(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(7/2)
Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.38 \[ \int \frac {x^{5/2}}{(2+b x)^{3/2}} \, dx=-\frac {8 \, b^{2} - \frac {25 \, {\left (b x + 2\right )} b}{x} + \frac {15 \, {\left (b x + 2\right )}^{2}}{x^{2}}}{\frac {\sqrt {b x + 2} b^{5}}{\sqrt {x}} - \frac {2 \, {\left (b x + 2\right )}^{\frac {3}{2}} b^{4}}{x^{\frac {3}{2}}} + \frac {{\left (b x + 2\right )}^{\frac {5}{2}} b^{3}}{x^{\frac {5}{2}}}} - \frac {15 \, \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{2 \, b^{\frac {7}{2}}} \]
-(8*b^2 - 25*(b*x + 2)*b/x + 15*(b*x + 2)^2/x^2)/(sqrt(b*x + 2)*b^5/sqrt(x ) - 2*(b*x + 2)^(3/2)*b^4/x^(3/2) + (b*x + 2)^(5/2)*b^3/x^(5/2)) - 15/2*lo g(-(sqrt(b) - sqrt(b*x + 2)/sqrt(x))/(sqrt(b) + sqrt(b*x + 2)/sqrt(x)))/b^ (7/2)
Time = 1.56 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.38 \[ \int \frac {x^{5/2}}{(2+b x)^{3/2}} \, dx=\frac {{\left (\sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} {\left (\frac {b x + 2}{b^{3}} - \frac {9}{b^{3}}\right )} - \frac {15 \, \log \left ({\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2}\right )}{b^{\frac {5}{2}}} - \frac {64}{{\left ({\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2} + 2 \, b\right )} b^{\frac {3}{2}}}\right )} {\left | b \right |}}{2 \, b^{2}} \]
1/2*(sqrt((b*x + 2)*b - 2*b)*sqrt(b*x + 2)*((b*x + 2)/b^3 - 9/b^3) - 15*lo g((sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^2)/b^(5/2) - 64/(((sqr t(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^2 + 2*b)*b^(3/2)))*abs(b)/b^ 2
Timed out. \[ \int \frac {x^{5/2}}{(2+b x)^{3/2}} \, dx=\int \frac {x^{5/2}}{{\left (b\,x+2\right )}^{3/2}} \,d x \]