Integrand size = 15, antiderivative size = 91 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x^2 \, dx=\frac {11}{8} b^2 \sqrt {a+\frac {b}{x}} x+\frac {13}{12} a b \sqrt {a+\frac {b}{x}} x^2+\frac {1}{3} a^2 \sqrt {a+\frac {b}{x}} x^3+\frac {5 b^3 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{8 \sqrt {a}} \] Output:
11/8*b^2*(a+b/x)^(1/2)*x+13/12*a*b*(a+b/x)^(1/2)*x^2+1/3*a^2*(a+b/x)^(1/2) *x^3+5/8*b^3*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(1/2)
Time = 0.10 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x^2 \, dx=\frac {1}{24} \sqrt {a+\frac {b}{x}} x \left (33 b^2+26 a b x+8 a^2 x^2\right )+\frac {5 b^3 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{8 \sqrt {a}} \] Input:
Integrate[(a + b/x)^(5/2)*x^2,x]
Output:
(Sqrt[a + b/x]*x*(33*b^2 + 26*a*b*x + 8*a^2*x^2))/24 + (5*b^3*ArcTanh[Sqrt [a + b/x]/Sqrt[a]])/(8*Sqrt[a])
Time = 0.31 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {798, 51, 51, 51, 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.
\(\displaystyle \int x^2 \left (a+\frac {b}{x}\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\int \left (a+\frac {b}{x}\right )^{5/2} x^4d\frac {1}{x}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{3} x^3 \left (a+\frac {b}{x}\right )^{5/2}-\frac {5}{6} b \int \left (a+\frac {b}{x}\right )^{3/2} x^3d\frac {1}{x}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{3} x^3 \left (a+\frac {b}{x}\right )^{5/2}-\frac {5}{6} b \left (\frac {3}{4} b \int \sqrt {a+\frac {b}{x}} x^2d\frac {1}{x}-\frac {1}{2} x^2 \left (a+\frac {b}{x}\right )^{3/2}\right )\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{3} x^3 \left (a+\frac {b}{x}\right )^{5/2}-\frac {5}{6} b \left (\frac {3}{4} b \left (\frac {1}{2} b \int \frac {x}{\sqrt {a+\frac {b}{x}}}d\frac {1}{x}-x \sqrt {a+\frac {b}{x}}\right )-\frac {1}{2} x^2 \left (a+\frac {b}{x}\right )^{3/2}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} x^3 \left (a+\frac {b}{x}\right )^{5/2}-\frac {5}{6} b \left (\frac {3}{4} b \left (\int \frac {1}{\frac {1}{b x^2}-\frac {a}{b}}d\sqrt {a+\frac {b}{x}}-x \sqrt {a+\frac {b}{x}}\right )-\frac {1}{2} x^2 \left (a+\frac {b}{x}\right )^{3/2}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{3} x^3 \left (a+\frac {b}{x}\right )^{5/2}-\frac {5}{6} b \left (\frac {3}{4} b \left (x \left (-\sqrt {a+\frac {b}{x}}\right )-\frac {b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{\sqrt {a}}\right )-\frac {1}{2} x^2 \left (a+\frac {b}{x}\right )^{3/2}\right )\) |
Input:
Int[(a + b/x)^(5/2)*x^2,x]
Output:
((a + b/x)^(5/2)*x^3)/3 - (5*b*(-1/2*((a + b/x)^(3/2)*x^2) + (3*b*(-(Sqrt[ a + b/x]*x) - (b*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/Sqrt[a]))/4))/6
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] :> 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 = 0.17 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.03
method | result | size |
risch | \(\frac {\left (8 a^{2} x^{2}+26 a b x +33 b^{2}\right ) x \sqrt {\frac {a x +b}{x}}}{24}+\frac {5 b^{3} \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right ) \sqrt {\frac {a x +b}{x}}\, \sqrt {x \left (a x +b \right )}}{16 \sqrt {a}\, \left (a x +b \right )}\) | \(94\) |
default | \(\frac {\sqrt {\frac {a x +b}{x}}\, x \left (16 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {5}{2}}+36 \sqrt {a \,x^{2}+b x}\, a^{\frac {5}{2}} b x +66 \sqrt {a \,x^{2}+b x}\, a^{\frac {3}{2}} b^{2}+15 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a \,b^{3}\right )}{48 \sqrt {x \left (a x +b \right )}\, a^{\frac {3}{2}}}\) | \(115\) |
Input:
int((a+b/x)^(5/2)*x^2,x,method=_RETURNVERBOSE)
Output:
1/24*(8*a^2*x^2+26*a*b*x+33*b^2)*x*((a*x+b)/x)^(1/2)+5/16*b^3*ln((1/2*b+a* x)/a^(1/2)+(a*x^2+b*x)^(1/2))/a^(1/2)*((a*x+b)/x)^(1/2)*(x*(a*x+b))^(1/2)/ (a*x+b)
Time = 0.09 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.73 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x^2 \, dx=\left [\frac {15 \, \sqrt {a} b^{3} \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (8 \, a^{3} x^{3} + 26 \, a^{2} b x^{2} + 33 \, a b^{2} x\right )} \sqrt {\frac {a x + b}{x}}}{48 \, a}, -\frac {15 \, \sqrt {-a} b^{3} \arctan \left (\frac {\sqrt {-a} x \sqrt {\frac {a x + b}{x}}}{a x + b}\right ) - {\left (8 \, a^{3} x^{3} + 26 \, a^{2} b x^{2} + 33 \, a b^{2} x\right )} \sqrt {\frac {a x + b}{x}}}{24 \, a}\right ] \] Input:
integrate((a+b/x)^(5/2)*x^2,x, algorithm="fricas")
Output:
[1/48*(15*sqrt(a)*b^3*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*( 8*a^3*x^3 + 26*a^2*b*x^2 + 33*a*b^2*x)*sqrt((a*x + b)/x))/a, -1/24*(15*sqr t(-a)*b^3*arctan(sqrt(-a)*x*sqrt((a*x + b)/x)/(a*x + b)) - (8*a^3*x^3 + 26 *a^2*b*x^2 + 33*a*b^2*x)*sqrt((a*x + b)/x))/a]
Time = 3.72 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.12 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x^2 \, dx=\frac {a^{2} \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a x}{b} + 1}}{3} + \frac {13 a b^{\frac {3}{2}} x^{\frac {3}{2}} \sqrt {\frac {a x}{b} + 1}}{12} + \frac {11 b^{\frac {5}{2}} \sqrt {x} \sqrt {\frac {a x}{b} + 1}}{8} + \frac {5 b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{8 \sqrt {a}} \] Input:
integrate((a+b/x)**(5/2)*x**2,x)
Output:
a**2*sqrt(b)*x**(5/2)*sqrt(a*x/b + 1)/3 + 13*a*b**(3/2)*x**(3/2)*sqrt(a*x/ b + 1)/12 + 11*b**(5/2)*sqrt(x)*sqrt(a*x/b + 1)/8 + 5*b**3*asinh(sqrt(a)*s qrt(x)/sqrt(b))/(8*sqrt(a))
Time = 0.12 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.44 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x^2 \, dx=-\frac {5 \, b^{3} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{16 \, \sqrt {a}} + \frac {33 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} b^{3} - 40 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a b^{3} + 15 \, \sqrt {a + \frac {b}{x}} a^{2} b^{3}}{24 \, {\left ({\left (a + \frac {b}{x}\right )}^{3} - 3 \, {\left (a + \frac {b}{x}\right )}^{2} a + 3 \, {\left (a + \frac {b}{x}\right )} a^{2} - a^{3}\right )}} \] Input:
integrate((a+b/x)^(5/2)*x^2,x, algorithm="maxima")
Output:
-5/16*b^3*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/sqrt(a) + 1/24*(33*(a + b/x)^(5/2)*b^3 - 40*(a + b/x)^(3/2)*a*b^3 + 15*sqrt(a + b /x)*a^2*b^3)/((a + b/x)^3 - 3*(a + b/x)^2*a + 3*(a + b/x)*a^2 - a^3)
Time = 0.14 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x^2 \, dx=-\frac {5 \, b^{3} \log \left ({\left | 2 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} + b \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, \sqrt {a}} + \frac {5 \, b^{3} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, \sqrt {a}} + \frac {1}{24} \, \sqrt {a x^{2} + b x} {\left (33 \, b^{2} \mathrm {sgn}\left (x\right ) + 2 \, {\left (4 \, a^{2} x \mathrm {sgn}\left (x\right ) + 13 \, a b \mathrm {sgn}\left (x\right )\right )} x\right )} \] Input:
integrate((a+b/x)^(5/2)*x^2,x, algorithm="giac")
Output:
-5/16*b^3*log(abs(2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a) + b))*sgn(x)/s qrt(a) + 5/16*b^3*log(abs(b))*sgn(x)/sqrt(a) + 1/24*sqrt(a*x^2 + b*x)*(33* b^2*sgn(x) + 2*(4*a^2*x*sgn(x) + 13*a*b*sgn(x))*x)
Time = 0.57 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.79 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x^2 \, dx=\frac {11\,x^3\,{\left (a+\frac {b}{x}\right )}^{5/2}}{8}-\frac {5\,a\,x^3\,{\left (a+\frac {b}{x}\right )}^{3/2}}{3}+\frac {5\,a^2\,x^3\,\sqrt {a+\frac {b}{x}}}{8}-\frac {b^3\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{8\,\sqrt {a}} \] Input:
int(x^2*(a + b/x)^(5/2),x)
Output:
(11*x^3*(a + b/x)^(5/2))/8 - (b^3*atan(((a + b/x)^(1/2)*1i)/a^(1/2))*5i)/( 8*a^(1/2)) - (5*a*x^3*(a + b/x)^(3/2))/3 + (5*a^2*x^3*(a + b/x)^(1/2))/8
Time = 0.23 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84 \[ \int \left (a+\frac {b}{x}\right )^{5/2} x^2 \, dx=\frac {8 \sqrt {x}\, \sqrt {a x +b}\, a^{3} x^{2}+26 \sqrt {x}\, \sqrt {a x +b}\, a^{2} b x +33 \sqrt {x}\, \sqrt {a x +b}\, a \,b^{2}+15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {a x +b}+\sqrt {x}\, \sqrt {a}}{\sqrt {b}}\right ) b^{3}}{24 a} \] Input:
int((a+b/x)^(5/2)*x^2,x)
Output:
(8*sqrt(x)*sqrt(a*x + b)*a**3*x**2 + 26*sqrt(x)*sqrt(a*x + b)*a**2*b*x + 3 3*sqrt(x)*sqrt(a*x + b)*a*b**2 + 15*sqrt(a)*log((sqrt(a*x + b) + sqrt(x)*s qrt(a))/sqrt(b))*b**3)/(24*a)