Integrand size = 15, antiderivative size = 81 \[ \int \frac {(2+b x)^{5/2}}{x^{5/2}} \, dx=5 b^2 \sqrt {x} \sqrt {2+b x}-\frac {10 b (2+b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (2+b x)^{5/2}}{3 x^{3/2}}+10 b^{3/2} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right ) \]
-2/3*(b*x+2)^(5/2)/x^(3/2)+10*b^(3/2)*arcsinh(1/2*b^(1/2)*x^(1/2)*2^(1/2)) -10/3*b*(b*x+2)^(3/2)/x^(1/2)+5*b^2*x^(1/2)*(b*x+2)^(1/2)
Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.78 \[ \int \frac {(2+b x)^{5/2}}{x^{5/2}} \, dx=\frac {\sqrt {2+b x} \left (-8-28 b x+3 b^2 x^2\right )}{3 x^{3/2}}-10 b^{3/2} \log \left (-\sqrt {b} \sqrt {x}+\sqrt {2+b x}\right ) \]
(Sqrt[2 + b*x]*(-8 - 28*b*x + 3*b^2*x^2))/(3*x^(3/2)) - 10*b^(3/2)*Log[-(S qrt[b]*Sqrt[x]) + Sqrt[2 + b*x]]
Time = 0.17 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {57, 57, 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.
\(\displaystyle \int \frac {(b x+2)^{5/2}}{x^{5/2}} \, dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {5}{3} b \int \frac {(b x+2)^{3/2}}{x^{3/2}}dx-\frac {2 (b x+2)^{5/2}}{3 x^{3/2}}\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {5}{3} b \left (3 b \int \frac {\sqrt {b x+2}}{\sqrt {x}}dx-\frac {2 (b x+2)^{3/2}}{\sqrt {x}}\right )-\frac {2 (b x+2)^{5/2}}{3 x^{3/2}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5}{3} b \left (3 b \left (\int \frac {1}{\sqrt {x} \sqrt {b x+2}}dx+\sqrt {x} \sqrt {b x+2}\right )-\frac {2 (b x+2)^{3/2}}{\sqrt {x}}\right )-\frac {2 (b x+2)^{5/2}}{3 x^{3/2}}\) |
\(\Big \downarrow \) 63 |
\(\displaystyle \frac {5}{3} b \left (3 b \left (2 \int \frac {1}{\sqrt {b x+2}}d\sqrt {x}+\sqrt {x} \sqrt {b x+2}\right )-\frac {2 (b x+2)^{3/2}}{\sqrt {x}}\right )-\frac {2 (b x+2)^{5/2}}{3 x^{3/2}}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {5}{3} b \left (3 b \left (\frac {2 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}}+\sqrt {x} \sqrt {b x+2}\right )-\frac {2 (b x+2)^{3/2}}{\sqrt {x}}\right )-\frac {2 (b x+2)^{5/2}}{3 x^{3/2}}\) |
(-2*(2 + b*x)^(5/2))/(3*x^(3/2)) + (5*b*((-2*(2 + b*x)^(3/2))/Sqrt[x] + 3* b*(Sqrt[x]*Sqrt[2 + b*x] + (2*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/Sqrt[b]) ))/3
3.6.62.3.1 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 = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.78
method | result | size |
meijerg | \(-\frac {15 b^{\frac {3}{2}} \left (\frac {32 \sqrt {\pi }\, \sqrt {2}\, \left (-\frac {3}{8} b^{2} x^{2}+\frac {7}{2} b x +1\right ) \sqrt {\frac {b x}{2}+1}}{45 x^{\frac {3}{2}} b^{\frac {3}{2}}}-\frac {8 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{3}\right )}{4 \sqrt {\pi }}\) | \(63\) |
risch | \(\frac {3 b^{3} x^{3}-22 b^{2} x^{2}-64 b x -16}{3 x^{\frac {3}{2}} \sqrt {b x +2}}+\frac {5 b^{\frac {3}{2}} \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right ) \sqrt {x \left (b x +2\right )}}{\sqrt {x}\, \sqrt {b x +2}}\) | \(82\) |
-15/4*b^(3/2)/Pi^(1/2)*(32/45*Pi^(1/2)/x^(3/2)*2^(1/2)/b^(3/2)*(-3/8*b^2*x ^2+7/2*b*x+1)*(1/2*b*x+1)^(1/2)-8/3*Pi^(1/2)*arcsinh(1/2*b^(1/2)*x^(1/2)*2 ^(1/2)))
Time = 0.23 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.52 \[ \int \frac {(2+b x)^{5/2}}{x^{5/2}} \, dx=\left [\frac {15 \, b^{\frac {3}{2}} x^{2} \log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right ) + {\left (3 \, b^{2} x^{2} - 28 \, b x - 8\right )} \sqrt {b x + 2} \sqrt {x}}{3 \, x^{2}}, -\frac {30 \, \sqrt {-b} b x^{2} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (3 \, b^{2} x^{2} - 28 \, b x - 8\right )} \sqrt {b x + 2} \sqrt {x}}{3 \, x^{2}}\right ] \]
[1/3*(15*b^(3/2)*x^2*log(b*x + sqrt(b*x + 2)*sqrt(b)*sqrt(x) + 1) + (3*b^2 *x^2 - 28*b*x - 8)*sqrt(b*x + 2)*sqrt(x))/x^2, -1/3*(30*sqrt(-b)*b*x^2*arc tan(sqrt(b*x + 2)*sqrt(-b)/(b*sqrt(x))) - (3*b^2*x^2 - 28*b*x - 8)*sqrt(b* x + 2)*sqrt(x))/x^2]
Time = 4.57 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.09 \[ \int \frac {(2+b x)^{5/2}}{x^{5/2}} \, dx=b^{\frac {5}{2}} x \sqrt {1 + \frac {2}{b x}} - \frac {28 b^{\frac {3}{2}} \sqrt {1 + \frac {2}{b x}}}{3} - 5 b^{\frac {3}{2}} \log {\left (\frac {1}{b x} \right )} + 10 b^{\frac {3}{2}} \log {\left (\sqrt {1 + \frac {2}{b x}} + 1 \right )} - \frac {8 \sqrt {b} \sqrt {1 + \frac {2}{b x}}}{3 x} \]
b**(5/2)*x*sqrt(1 + 2/(b*x)) - 28*b**(3/2)*sqrt(1 + 2/(b*x))/3 - 5*b**(3/2 )*log(1/(b*x)) + 10*b**(3/2)*log(sqrt(1 + 2/(b*x)) + 1) - 8*sqrt(b)*sqrt(1 + 2/(b*x))/(3*x)
Time = 0.32 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.19 \[ \int \frac {(2+b x)^{5/2}}{x^{5/2}} \, dx=-5 \, b^{\frac {3}{2}} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right ) - \frac {8 \, \sqrt {b x + 2} b}{\sqrt {x}} - \frac {2 \, \sqrt {b x + 2} b^{2}}{{\left (b - \frac {b x + 2}{x}\right )} \sqrt {x}} - \frac {4 \, {\left (b x + 2\right )}^{\frac {3}{2}}}{3 \, x^{\frac {3}{2}}} \]
-5*b^(3/2)*log(-(sqrt(b) - sqrt(b*x + 2)/sqrt(x))/(sqrt(b) + sqrt(b*x + 2) /sqrt(x))) - 8*sqrt(b*x + 2)*b/sqrt(x) - 2*sqrt(b*x + 2)*b^2/((b - (b*x + 2)/x)*sqrt(x)) - 4/3*(b*x + 2)^(3/2)/x^(3/2)
Time = 6.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.12 \[ \int \frac {(2+b x)^{5/2}}{x^{5/2}} \, dx=-\frac {{\left (30 \, b^{\frac {3}{2}} \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right ) - \frac {{\left (60 \, b^{3} + {\left (3 \, {\left (b x + 2\right )} b^{3} - 40 \, b^{3}\right )} {\left (b x + 2\right )}\right )} \sqrt {b x + 2}}{{\left ({\left (b x + 2\right )} b - 2 \, b\right )}^{\frac {3}{2}}}\right )} b}{3 \, {\left | b \right |}} \]
-1/3*(30*b^(3/2)*log(abs(-sqrt(b*x + 2)*sqrt(b) + sqrt((b*x + 2)*b - 2*b)) ) - (60*b^3 + (3*(b*x + 2)*b^3 - 40*b^3)*(b*x + 2))*sqrt(b*x + 2)/((b*x + 2)*b - 2*b)^(3/2))*b/abs(b)
Timed out. \[ \int \frac {(2+b x)^{5/2}}{x^{5/2}} \, dx=\int \frac {{\left (b\,x+2\right )}^{5/2}}{x^{5/2}} \,d x \]