Integrand size = 15, antiderivative size = 85 \[ \int \frac {x^{5/2}}{(2+b x)^{5/2}} \, dx=-\frac {8 \sqrt {x}}{3 b^3 (2+b x)^{3/2}}+\frac {28 \sqrt {x}}{3 b^3 \sqrt {2+b x}}+\frac {\sqrt {x} \sqrt {2+b x}}{b^3}-\frac {10 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}} \] Output:
-8/3*x^(1/2)/b^3/(b*x+2)^(3/2)+28/3*x^(1/2)/b^3/(b*x+2)^(1/2)+x^(1/2)*(b*x +2)^(1/2)/b^3-10*arcsinh(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(7/2)
Time = 0.20 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.87 \[ \int \frac {x^{5/2}}{(2+b x)^{5/2}} \, dx=\frac {\sqrt {x} \left (60+40 b x+3 b^2 x^2\right )}{3 b^3 (2+b x)^{3/2}}+\frac {20 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2+b x}}\right )}{b^{7/2}} \] Input:
Integrate[x^(5/2)/(2 + b*x)^(5/2),x]
Output:
(Sqrt[x]*(60 + 40*b*x + 3*b^2*x^2))/(3*b^3*(2 + b*x)^(3/2)) + (20*ArcTanh[ (Sqrt[b]*Sqrt[x])/(Sqrt[2] - Sqrt[2 + b*x])])/b^(7/2)
Time = 0.17 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.14, 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.
\(\displaystyle \int \frac {x^{5/2}}{(b x+2)^{5/2}} \, dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {5 \int \frac {x^{3/2}}{(b x+2)^{3/2}}dx}{3 b}-\frac {2 x^{5/2}}{3 b (b x+2)^{3/2}}\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {5 \left (\frac {3 \int \frac {\sqrt {x}}{\sqrt {b x+2}}dx}{b}-\frac {2 x^{3/2}}{b \sqrt {b x+2}}\right )}{3 b}-\frac {2 x^{5/2}}{3 b (b x+2)^{3/2}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\sqrt {x} \sqrt {b x+2}}{b}-\frac {\int \frac {1}{\sqrt {x} \sqrt {b x+2}}dx}{b}\right )}{b}-\frac {2 x^{3/2}}{b \sqrt {b x+2}}\right )}{3 b}-\frac {2 x^{5/2}}{3 b (b x+2)^{3/2}}\) |
\(\Big \downarrow \) 63 |
\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\sqrt {x} \sqrt {b x+2}}{b}-\frac {2 \int \frac {1}{\sqrt {b x+2}}d\sqrt {x}}{b}\right )}{b}-\frac {2 x^{3/2}}{b \sqrt {b x+2}}\right )}{3 b}-\frac {2 x^{5/2}}{3 b (b x+2)^{3/2}}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\sqrt {x} \sqrt {b x+2}}{b}-\frac {2 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}}\right )}{b}-\frac {2 x^{3/2}}{b \sqrt {b x+2}}\right )}{3 b}-\frac {2 x^{5/2}}{3 b (b x+2)^{3/2}}\) |
Input:
Int[x^(5/2)/(2 + b*x)^(5/2),x]
Output:
(-2*x^(5/2))/(3*b*(2 + b*x)^(3/2)) + (5*((-2*x^(3/2))/(b*Sqrt[2 + b*x]) + (3*((Sqrt[x]*Sqrt[2 + b*x])/b - (2*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^( 3/2)))/b))/(3*b)
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 = 0.74
method | result | size |
meijerg | \(\frac {\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {b}\, \left (\frac {21}{4} b^{2} x^{2}+70 b x +105\right )}{21 \left (\frac {b x}{2}+1\right )^{\frac {3}{2}}}-10 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{b^{\frac {7}{2}} \sqrt {\pi }}\) | \(63\) |
risch | \(\frac {\sqrt {x}\, \sqrt {b x +2}}{b^{3}}+\frac {\left (-\frac {5 \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right )}{b^{\frac {7}{2}}}-\frac {8 \sqrt {b \left (x +\frac {2}{b}\right )^{2}-2 x -\frac {4}{b}}}{3 b^{5} \left (x +\frac {2}{b}\right )^{2}}+\frac {28 \sqrt {b \left (x +\frac {2}{b}\right )^{2}-2 x -\frac {4}{b}}}{3 b^{4} \left (x +\frac {2}{b}\right )}\right ) \sqrt {x \left (b x +2\right )}}{\sqrt {x}\, \sqrt {b x +2}}\) | \(136\) |
Input:
int(x^(5/2)/(b*x+2)^(5/2),x,method=_RETURNVERBOSE)
Output:
8/3/b^(7/2)/Pi^(1/2)*(1/56*Pi^(1/2)*x^(1/2)*2^(1/2)*b^(1/2)*(21/4*b^2*x^2+ 70*b*x+105)/(1/2*b*x+1)^(3/2)-15/4*Pi^(1/2)*arcsinh(1/2*b^(1/2)*x^(1/2)*2^ (1/2)))
Time = 0.08 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.15 \[ \int \frac {x^{5/2}}{(2+b x)^{5/2}} \, dx=\left [\frac {15 \, {\left (b^{2} x^{2} + 4 \, b x + 4\right )} \sqrt {b} \log \left (b x - \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right ) + {\left (3 \, b^{3} x^{2} + 40 \, b^{2} x + 60 \, b\right )} \sqrt {b x + 2} \sqrt {x}}{3 \, {\left (b^{6} x^{2} + 4 \, b^{5} x + 4 \, b^{4}\right )}}, \frac {30 \, {\left (b^{2} x^{2} + 4 \, b x + 4\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {b x + 2}}\right ) + {\left (3 \, b^{3} x^{2} + 40 \, b^{2} x + 60 \, b\right )} \sqrt {b x + 2} \sqrt {x}}{3 \, {\left (b^{6} x^{2} + 4 \, b^{5} x + 4 \, b^{4}\right )}}\right ] \] Input:
integrate(x^(5/2)/(b*x+2)^(5/2),x, algorithm="fricas")
Output:
[1/3*(15*(b^2*x^2 + 4*b*x + 4)*sqrt(b)*log(b*x - sqrt(b*x + 2)*sqrt(b)*sqr t(x) + 1) + (3*b^3*x^2 + 40*b^2*x + 60*b)*sqrt(b*x + 2)*sqrt(x))/(b^6*x^2 + 4*b^5*x + 4*b^4), 1/3*(30*(b^2*x^2 + 4*b*x + 4)*sqrt(-b)*arctan(sqrt(-b) *sqrt(x)/sqrt(b*x + 2)) + (3*b^3*x^2 + 40*b^2*x + 60*b)*sqrt(b*x + 2)*sqrt (x))/(b^6*x^2 + 4*b^5*x + 4*b^4)]
Leaf count of result is larger than twice the leaf count of optimal. 308 vs. \(2 (82) = 164\).
Time = 3.06 (sec) , antiderivative size = 308, normalized size of antiderivative = 3.62 \[ \int \frac {x^{5/2}}{(2+b x)^{5/2}} \, dx=\frac {3 b^{\frac {23}{2}} x^{15}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {b x + 2} + 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {b x + 2}} + \frac {40 b^{\frac {21}{2}} x^{14}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {b x + 2} + 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {b x + 2}} + \frac {60 b^{\frac {19}{2}} x^{13}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {b x + 2} + 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {b x + 2}} - \frac {30 b^{10} x^{\frac {27}{2}} \sqrt {b x + 2} \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {b x + 2} + 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {b x + 2}} - \frac {60 b^{9} x^{\frac {25}{2}} \sqrt {b x + 2} \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {b x + 2} + 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {b x + 2}} \] Input:
integrate(x**(5/2)/(b*x+2)**(5/2),x)
Output:
3*b**(23/2)*x**15/(3*b**(27/2)*x**(27/2)*sqrt(b*x + 2) + 6*b**(25/2)*x**(2 5/2)*sqrt(b*x + 2)) + 40*b**(21/2)*x**14/(3*b**(27/2)*x**(27/2)*sqrt(b*x + 2) + 6*b**(25/2)*x**(25/2)*sqrt(b*x + 2)) + 60*b**(19/2)*x**13/(3*b**(27/ 2)*x**(27/2)*sqrt(b*x + 2) + 6*b**(25/2)*x**(25/2)*sqrt(b*x + 2)) - 30*b** 10*x**(27/2)*sqrt(b*x + 2)*asinh(sqrt(2)*sqrt(b)*sqrt(x)/2)/(3*b**(27/2)*x **(27/2)*sqrt(b*x + 2) + 6*b**(25/2)*x**(25/2)*sqrt(b*x + 2)) - 60*b**9*x* *(25/2)*sqrt(b*x + 2)*asinh(sqrt(2)*sqrt(b)*sqrt(x)/2)/(3*b**(27/2)*x**(27 /2)*sqrt(b*x + 2) + 6*b**(25/2)*x**(25/2)*sqrt(b*x + 2))
Time = 0.11 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.24 \[ \int \frac {x^{5/2}}{(2+b x)^{5/2}} \, dx=\frac {2 \, {\left (2 \, b^{2} + \frac {10 \, {\left (b x + 2\right )} b}{x} - \frac {15 \, {\left (b x + 2\right )}^{2}}{x^{2}}\right )}}{3 \, {\left (\frac {{\left (b x + 2\right )}^{\frac {3}{2}} b^{4}}{x^{\frac {3}{2}}} - \frac {{\left (b x + 2\right )}^{\frac {5}{2}} b^{3}}{x^{\frac {5}{2}}}\right )}} + \frac {5 \, \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{b^{\frac {7}{2}}} \] Input:
integrate(x^(5/2)/(b*x+2)^(5/2),x, algorithm="maxima")
Output:
2/3*(2*b^2 + 10*(b*x + 2)*b/x - 15*(b*x + 2)^2/x^2)/((b*x + 2)^(3/2)*b^4/x ^(3/2) - (b*x + 2)^(5/2)*b^3/x^(5/2)) + 5*log(-(sqrt(b) - sqrt(b*x + 2)/sq rt(x))/(sqrt(b) + sqrt(b*x + 2)/sqrt(x)))/b^(7/2)
Leaf count of result is larger than twice the leaf count of optimal. 179 vs. \(2 (62) = 124\).
Time = 1.43 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.11 \[ \int \frac {x^{5/2}}{(2+b x)^{5/2}} \, dx=\frac {{\left (\frac {15 \, \log \left ({\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2}\right )}{\sqrt {b}} + \frac {3 \, \sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2}}{b} + \frac {16 \, {\left (9 \, {\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{4} \sqrt {b} + 24 \, {\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2} b^{\frac {3}{2}} + 28 \, b^{\frac {5}{2}}\right )}}{{\left ({\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2} + 2 \, b\right )}^{3}}\right )} {\left | b \right |}}{3 \, b^{4}} \] Input:
integrate(x^(5/2)/(b*x+2)^(5/2),x, algorithm="giac")
Output:
1/3*(15*log((sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^2)/sqrt(b) + 3*sqrt((b*x + 2)*b - 2*b)*sqrt(b*x + 2)/b + 16*(9*(sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^4*sqrt(b) + 24*(sqrt(b*x + 2)*sqrt(b) - sqrt((b *x + 2)*b - 2*b))^2*b^(3/2) + 28*b^(5/2))/((sqrt(b*x + 2)*sqrt(b) - sqrt(( b*x + 2)*b - 2*b))^2 + 2*b)^3)*abs(b)/b^4
Timed out. \[ \int \frac {x^{5/2}}{(2+b x)^{5/2}} \, dx=\int \frac {x^{5/2}}{{\left (b\,x+2\right )}^{5/2}} \,d x \] Input:
int(x^(5/2)/(b*x + 2)^(5/2),x)
Output:
int(x^(5/2)/(b*x + 2)^(5/2), x)
Time = 0.15 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.46 \[ \int \frac {x^{5/2}}{(2+b x)^{5/2}} \, dx=\frac {-30 \sqrt {b}\, \sqrt {b x +2}\, \mathrm {log}\left (\frac {\sqrt {b x +2}+\sqrt {x}\, \sqrt {b}}{\sqrt {2}}\right ) b x -60 \sqrt {b}\, \sqrt {b x +2}\, \mathrm {log}\left (\frac {\sqrt {b x +2}+\sqrt {x}\, \sqrt {b}}{\sqrt {2}}\right )-5 \sqrt {b}\, \sqrt {b x +2}\, b x -10 \sqrt {b}\, \sqrt {b x +2}+3 \sqrt {x}\, b^{3} x^{2}+40 \sqrt {x}\, b^{2} x +60 \sqrt {x}\, b}{3 \sqrt {b x +2}\, b^{4} \left (b x +2\right )} \] Input:
int(x^(5/2)/(b*x+2)^(5/2),x)
Output:
( - 30*sqrt(b)*sqrt(b*x + 2)*log((sqrt(b*x + 2) + sqrt(x)*sqrt(b))/sqrt(2) )*b*x - 60*sqrt(b)*sqrt(b*x + 2)*log((sqrt(b*x + 2) + sqrt(x)*sqrt(b))/sqr t(2)) - 5*sqrt(b)*sqrt(b*x + 2)*b*x - 10*sqrt(b)*sqrt(b*x + 2) + 3*sqrt(x) *b**3*x**2 + 40*sqrt(x)*b**2*x + 60*sqrt(x)*b)/(3*sqrt(b*x + 2)*b**4*(b*x + 2))