Integrand size = 13, antiderivative size = 86 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x \, dx=-2 b^2 \sqrt {a+\frac {b}{x}}+\frac {9}{4} a b \sqrt {a+\frac {b}{x}} x+\frac {1}{2} a^2 \sqrt {a+\frac {b}{x}} x^2+\frac {15}{4} \sqrt {a} b^2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right ) \] Output:
-2*b^2*(a+b/x)^(1/2)+9/4*a*b*(a+b/x)^(1/2)*x+1/2*a^2*(a+b/x)^(1/2)*x^2+15/ 4*a^(1/2)*b^2*arctanh((a+b/x)^(1/2)/a^(1/2))
Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.74 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x \, dx=\frac {1}{4} \left (\sqrt {a+\frac {b}{x}} \left (-8 b^2+9 a b x+2 a^2 x^2\right )+15 \sqrt {a} b^2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )\right ) \] Input:
Integrate[(a + b/x)^(5/2)*x,x]
Output:
(Sqrt[a + b/x]*(-8*b^2 + 9*a*b*x + 2*a^2*x^2) + 15*Sqrt[a]*b^2*ArcTanh[Sqr t[a + b/x]/Sqrt[a]])/4
Time = 0.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {798, 51, 51, 60, 73, 221}
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.
\(\displaystyle \int x \left (a+\frac {b}{x}\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\int \left (a+\frac {b}{x}\right )^{5/2} x^3d\frac {1}{x}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} x^2 \left (a+\frac {b}{x}\right )^{5/2}-\frac {5}{4} b \int \left (a+\frac {b}{x}\right )^{3/2} x^2d\frac {1}{x}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} x^2 \left (a+\frac {b}{x}\right )^{5/2}-\frac {5}{4} b \left (\frac {3}{2} b \int \sqrt {a+\frac {b}{x}} xd\frac {1}{x}-x \left (a+\frac {b}{x}\right )^{3/2}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} x^2 \left (a+\frac {b}{x}\right )^{5/2}-\frac {5}{4} b \left (\frac {3}{2} b \left (a \int \frac {x}{\sqrt {a+\frac {b}{x}}}d\frac {1}{x}+2 \sqrt {a+\frac {b}{x}}\right )-x \left (a+\frac {b}{x}\right )^{3/2}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} x^2 \left (a+\frac {b}{x}\right )^{5/2}-\frac {5}{4} b \left (\frac {3}{2} b \left (\frac {2 a \int \frac {1}{\frac {1}{b x^2}-\frac {a}{b}}d\sqrt {a+\frac {b}{x}}}{b}+2 \sqrt {a+\frac {b}{x}}\right )-x \left (a+\frac {b}{x}\right )^{3/2}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} x^2 \left (a+\frac {b}{x}\right )^{5/2}-\frac {5}{4} b \left (\frac {3}{2} b \left (2 \sqrt {a+\frac {b}{x}}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )\right )-x \left (a+\frac {b}{x}\right )^{3/2}\right )\) |
Input:
Int[(a + b/x)^(5/2)*x,x]
Output:
((a + b/x)^(5/2)*x^2)/2 - (5*b*(-((a + b/x)^(3/2)*x) + (3*b*(2*Sqrt[a + b/ x] - 2*Sqrt[a]*ArcTanh[Sqrt[a + b/x]/Sqrt[a]]))/2))/4
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, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.16 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.08
method | result | size |
risch | \(\frac {\left (2 a^{2} x^{2}+9 a b x -8 b^{2}\right ) \sqrt {\frac {a x +b}{x}}}{4}+\frac {15 \sqrt {a}\, b^{2} \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right ) \sqrt {\frac {a x +b}{x}}\, \sqrt {x \left (a x +b \right )}}{8 \left (a x +b \right )}\) | \(93\) |
default | \(-\frac {\sqrt {\frac {a x +b}{x}}\, \left (-4 \sqrt {a \,x^{2}+b x}\, a^{\frac {7}{2}} x^{3}-34 \sqrt {a \,x^{2}+b x}\, a^{\frac {5}{2}} b \,x^{2}-15 a^{2} \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) b^{2} x^{2}+16 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {3}{2}} b \right )}{8 x \sqrt {x \left (a x +b \right )}\, a^{\frac {3}{2}}}\) | \(125\) |
Input:
int((a+b/x)^(5/2)*x,x,method=_RETURNVERBOSE)
Output:
1/4*(2*a^2*x^2+9*a*b*x-8*b^2)*((a*x+b)/x)^(1/2)+15/8*a^(1/2)*b^2*ln((1/2*b +a*x)/a^(1/2)+(a*x^2+b*x)^(1/2))*((a*x+b)/x)^(1/2)*(x*(a*x+b))^(1/2)/(a*x+ b)
Time = 0.08 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.57 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x \, dx=\left [\frac {15}{8} \, \sqrt {a} b^{2} \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + \frac {1}{4} \, {\left (2 \, a^{2} x^{2} + 9 \, a b x - 8 \, b^{2}\right )} \sqrt {\frac {a x + b}{x}}, -\frac {15}{4} \, \sqrt {-a} b^{2} \arctan \left (\frac {\sqrt {-a} x \sqrt {\frac {a x + b}{x}}}{a x + b}\right ) + \frac {1}{4} \, {\left (2 \, a^{2} x^{2} + 9 \, a b x - 8 \, b^{2}\right )} \sqrt {\frac {a x + b}{x}}\right ] \] Input:
integrate((a+b/x)^(5/2)*x,x, algorithm="fricas")
Output:
[15/8*sqrt(a)*b^2*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 1/4*(2* a^2*x^2 + 9*a*b*x - 8*b^2)*sqrt((a*x + b)/x), -15/4*sqrt(-a)*b^2*arctan(sq rt(-a)*x*sqrt((a*x + b)/x)/(a*x + b)) + 1/4*(2*a^2*x^2 + 9*a*b*x - 8*b^2)* sqrt((a*x + b)/x)]
Time = 3.33 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.47 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x \, dx=\frac {15 \sqrt {a} b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{4} + \frac {a^{3} x^{\frac {5}{2}}}{2 \sqrt {b} \sqrt {\frac {a x}{b} + 1}} + \frac {11 a^{2} \sqrt {b} x^{\frac {3}{2}}}{4 \sqrt {\frac {a x}{b} + 1}} + \frac {a b^{\frac {3}{2}} \sqrt {x}}{4 \sqrt {\frac {a x}{b} + 1}} - \frac {2 b^{\frac {5}{2}}}{\sqrt {x} \sqrt {\frac {a x}{b} + 1}} \] Input:
integrate((a+b/x)**(5/2)*x,x)
Output:
15*sqrt(a)*b**2*asinh(sqrt(a)*sqrt(x)/sqrt(b))/4 + a**3*x**(5/2)/(2*sqrt(b )*sqrt(a*x/b + 1)) + 11*a**2*sqrt(b)*x**(3/2)/(4*sqrt(a*x/b + 1)) + a*b**( 3/2)*sqrt(x)/(4*sqrt(a*x/b + 1)) - 2*b**(5/2)/(sqrt(x)*sqrt(a*x/b + 1))
Time = 0.11 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.34 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x \, dx=-\frac {15}{8} \, \sqrt {a} b^{2} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right ) - 2 \, \sqrt {a + \frac {b}{x}} b^{2} + \frac {9 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a b^{2} - 7 \, \sqrt {a + \frac {b}{x}} a^{2} b^{2}}{4 \, {\left ({\left (a + \frac {b}{x}\right )}^{2} - 2 \, {\left (a + \frac {b}{x}\right )} a + a^{2}\right )}} \] Input:
integrate((a+b/x)^(5/2)*x,x, algorithm="maxima")
Output:
-15/8*sqrt(a)*b^2*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a))) - 2*sqrt(a + b/x)*b^2 + 1/4*(9*(a + b/x)^(3/2)*a*b^2 - 7*sqrt(a + b/x)*a^ 2*b^2)/((a + b/x)^2 - 2*(a + b/x)*a + a^2)
Exception generated. \[ \int \left (a+\frac {b}{x}\right )^{5/2} x \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+b/x)^(5/2)*x,x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value
Time = 0.51 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.84 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x \, dx=\frac {9\,a\,x^2\,{\left (a+\frac {b}{x}\right )}^{3/2}}{4}-2\,b^2\,\sqrt {a+\frac {b}{x}}-\frac {7\,a^2\,x^2\,\sqrt {a+\frac {b}{x}}}{4}-\frac {\sqrt {a}\,b^2\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,15{}\mathrm {i}}{4} \] Input:
int(x*(a + b/x)^(5/2),x)
Output:
(9*a*x^2*(a + b/x)^(3/2))/4 - (a^(1/2)*b^2*atan(((a + b/x)^(1/2)*1i)/a^(1/ 2))*15i)/4 - 2*b^2*(a + b/x)^(1/2) - (7*a^2*x^2*(a + b/x)^(1/2))/4
Time = 0.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.95 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x \, dx=\frac {2 \sqrt {x}\, \sqrt {a x +b}\, a^{2} x^{2}+9 \sqrt {x}\, \sqrt {a x +b}\, a b x -8 \sqrt {x}\, \sqrt {a x +b}\, b^{2}+15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {a x +b}+\sqrt {x}\, \sqrt {a}}{\sqrt {b}}\right ) b^{2} x -10 \sqrt {a}\, b^{2} x}{4 x} \] Input:
int((a+b/x)^(5/2)*x,x)
Output:
(2*sqrt(x)*sqrt(a*x + b)*a**2*x**2 + 9*sqrt(x)*sqrt(a*x + b)*a*b*x - 8*sqr t(x)*sqrt(a*x + b)*b**2 + 15*sqrt(a)*log((sqrt(a*x + b) + sqrt(x)*sqrt(a)) /sqrt(b))*b**2*x - 10*sqrt(a)*b**2*x)/(4*x)