Integrand size = 15, antiderivative size = 91 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^2 \, dx=\frac {b^2 \sqrt {a+\frac {b}{x}} x}{8 a}+\frac {7}{12} b \sqrt {a+\frac {b}{x}} x^2+\frac {1}{3} a \sqrt {a+\frac {b}{x}} x^3-\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{8 a^{3/2}} \] Output:
1/8*b^2*(a+b/x)^(1/2)*x/a+7/12*b*(a+b/x)^(1/2)*x^2+1/3*a*(a+b/x)^(1/2)*x^3 -1/8*b^3*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(3/2)
Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.77 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^2 \, dx=\frac {\sqrt {a} \sqrt {a+\frac {b}{x}} x \left (3 b^2+14 a b x+8 a^2 x^2\right )-3 b^3 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{24 a^{3/2}} \] Input:
Integrate[(a + b/x)^(3/2)*x^2,x]
Output:
(Sqrt[a]*Sqrt[a + b/x]*x*(3*b^2 + 14*a*b*x + 8*a^2*x^2) - 3*b^3*ArcTanh[Sq rt[a + b/x]/Sqrt[a]])/(24*a^(3/2))
Time = 0.31 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, 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, 52, 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.
\(\displaystyle \int x^2 \left (a+\frac {b}{x}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\int \left (a+\frac {b}{x}\right )^{3/2} x^4d\frac {1}{x}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{3} x^3 \left (a+\frac {b}{x}\right )^{3/2}-\frac {1}{2} b \int \sqrt {a+\frac {b}{x}} x^3d\frac {1}{x}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{3} x^3 \left (a+\frac {b}{x}\right )^{3/2}-\frac {1}{2} b \left (\frac {1}{4} b \int \frac {x^2}{\sqrt {a+\frac {b}{x}}}d\frac {1}{x}-\frac {1}{2} x^2 \sqrt {a+\frac {b}{x}}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{3} x^3 \left (a+\frac {b}{x}\right )^{3/2}-\frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {b \int \frac {x}{\sqrt {a+\frac {b}{x}}}d\frac {1}{x}}{2 a}-\frac {x \sqrt {a+\frac {b}{x}}}{a}\right )-\frac {1}{2} x^2 \sqrt {a+\frac {b}{x}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} x^3 \left (a+\frac {b}{x}\right )^{3/2}-\frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {\int \frac {1}{\frac {1}{b x^2}-\frac {a}{b}}d\sqrt {a+\frac {b}{x}}}{a}-\frac {x \sqrt {a+\frac {b}{x}}}{a}\right )-\frac {1}{2} x^2 \sqrt {a+\frac {b}{x}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{3} x^3 \left (a+\frac {b}{x}\right )^{3/2}-\frac {1}{2} b \left (\frac {1}{4} b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {x \sqrt {a+\frac {b}{x}}}{a}\right )-\frac {1}{2} x^2 \sqrt {a+\frac {b}{x}}\right )\) |
Input:
Int[(a + b/x)^(3/2)*x^2,x]
Output:
((a + b/x)^(3/2)*x^3)/3 - (b*(-1/2*(Sqrt[a + b/x]*x^2) + (b*(-((Sqrt[a + b /x]*x)/a) + (b*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/a^(3/2)))/4))/2
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 + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[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 = 0.14 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.07
method | result | size |
risch | \(\frac {\left (8 a^{2} x^{2}+14 a b x +3 b^{2}\right ) x \sqrt {\frac {a x +b}{x}}}{24 a}-\frac {b^{3} \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right ) \sqrt {\frac {a x +b}{x}}\, \sqrt {x \left (a x +b \right )}}{16 a^{\frac {3}{2}} \left (a x +b \right )}\) | \(97\) |
default | \(\frac {\sqrt {\frac {a x +b}{x}}\, x \left (16 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {5}{2}}+12 \sqrt {a \,x^{2}+b x}\, a^{\frac {5}{2}} b x +6 \sqrt {a \,x^{2}+b x}\, a^{\frac {3}{2}} b^{2}-3 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a \,b^{3}\right )}{48 a^{\frac {5}{2}} \sqrt {x \left (a x +b \right )}}\) | \(115\) |
Input:
int((a+b/x)^(3/2)*x^2,x,method=_RETURNVERBOSE)
Output:
1/24*(8*a^2*x^2+14*a*b*x+3*b^2)*x/a*((a*x+b)/x)^(1/2)-1/16*b^3/a^(3/2)*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 = 0.09 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.71 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^2 \, dx=\left [\frac {3 \, \sqrt {a} b^{3} \log \left (2 \, a x - 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (8 \, a^{3} x^{3} + 14 \, a^{2} b x^{2} + 3 \, a b^{2} x\right )} \sqrt {\frac {a x + b}{x}}}{48 \, a^{2}}, \frac {3 \, \sqrt {-a} b^{3} \arctan \left (\frac {\sqrt {-a} x \sqrt {\frac {a x + b}{x}}}{a x + b}\right ) + {\left (8 \, a^{3} x^{3} + 14 \, a^{2} b x^{2} + 3 \, a b^{2} x\right )} \sqrt {\frac {a x + b}{x}}}{24 \, a^{2}}\right ] \] Input:
integrate((a+b/x)^(3/2)*x^2,x, algorithm="fricas")
Output:
[1/48*(3*sqrt(a)*b^3*log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(8 *a^3*x^3 + 14*a^2*b*x^2 + 3*a*b^2*x)*sqrt((a*x + b)/x))/a^2, 1/24*(3*sqrt( -a)*b^3*arctan(sqrt(-a)*x*sqrt((a*x + b)/x)/(a*x + b)) + (8*a^3*x^3 + 14*a ^2*b*x^2 + 3*a*b^2*x)*sqrt((a*x + b)/x))/a^2]
Time = 4.97 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.36 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^2 \, dx=\frac {a^{2} x^{\frac {7}{2}}}{3 \sqrt {b} \sqrt {\frac {a x}{b} + 1}} + \frac {11 a \sqrt {b} x^{\frac {5}{2}}}{12 \sqrt {\frac {a x}{b} + 1}} + \frac {17 b^{\frac {3}{2}} x^{\frac {3}{2}}}{24 \sqrt {\frac {a x}{b} + 1}} + \frac {b^{\frac {5}{2}} \sqrt {x}}{8 a \sqrt {\frac {a x}{b} + 1}} - \frac {b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{8 a^{\frac {3}{2}}} \] Input:
integrate((a+b/x)**(3/2)*x**2,x)
Output:
a**2*x**(7/2)/(3*sqrt(b)*sqrt(a*x/b + 1)) + 11*a*sqrt(b)*x**(5/2)/(12*sqrt (a*x/b + 1)) + 17*b**(3/2)*x**(3/2)/(24*sqrt(a*x/b + 1)) + b**(5/2)*sqrt(x )/(8*a*sqrt(a*x/b + 1)) - b**3*asinh(sqrt(a)*sqrt(x)/sqrt(b))/(8*a**(3/2))
Time = 0.11 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.48 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^2 \, dx=\frac {b^{3} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{16 \, a^{\frac {3}{2}}} + \frac {3 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} b^{3} + 8 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a b^{3} - 3 \, \sqrt {a + \frac {b}{x}} a^{2} b^{3}}{24 \, {\left ({\left (a + \frac {b}{x}\right )}^{3} a - 3 \, {\left (a + \frac {b}{x}\right )}^{2} a^{2} + 3 \, {\left (a + \frac {b}{x}\right )} a^{3} - a^{4}\right )}} \] Input:
integrate((a+b/x)^(3/2)*x^2,x, algorithm="maxima")
Output:
1/16*b^3*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/a^(3/2) + 1/24*(3*(a + b/x)^(5/2)*b^3 + 8*(a + b/x)^(3/2)*a*b^3 - 3*sqrt(a + b/x)* a^2*b^3)/((a + b/x)^3*a - 3*(a + b/x)^2*a^2 + 3*(a + b/x)*a^3 - a^4)
Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^2 \, dx=\frac {b^{3} \log \left ({\left | 2 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} + b \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, a^{\frac {3}{2}}} - \frac {b^{3} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, a^{\frac {3}{2}}} + \frac {1}{24} \, \sqrt {a x^{2} + b x} {\left (2 \, {\left (4 \, a x \mathrm {sgn}\left (x\right ) + 7 \, b \mathrm {sgn}\left (x\right )\right )} x + \frac {3 \, b^{2} \mathrm {sgn}\left (x\right )}{a}\right )} \] Input:
integrate((a+b/x)^(3/2)*x^2,x, algorithm="giac")
Output:
1/16*b^3*log(abs(2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a) + b))*sgn(x)/a^ (3/2) - 1/16*b^3*log(abs(b))*sgn(x)/a^(3/2) + 1/24*sqrt(a*x^2 + b*x)*(2*(4 *a*x*sgn(x) + 7*b*sgn(x))*x + 3*b^2*sgn(x)/a)
Time = 0.56 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.79 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^2 \, dx=\frac {x^3\,{\left (a+\frac {b}{x}\right )}^{3/2}}{3}-\frac {a\,x^3\,\sqrt {a+\frac {b}{x}}}{8}+\frac {x^3\,{\left (a+\frac {b}{x}\right )}^{5/2}}{8\,a}+\frac {b^3\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{8\,a^{3/2}} \] Input:
int(x^2*(a + b/x)^(3/2),x)
Output:
(x^3*(a + b/x)^(3/2))/3 + (b^3*atan(((a + b/x)^(1/2)*1i)/a^(1/2))*1i)/(8*a ^(3/2)) - (a*x^3*(a + b/x)^(1/2))/8 + (x^3*(a + b/x)^(5/2))/(8*a)
Time = 0.22 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^2 \, dx=\frac {8 \sqrt {x}\, \sqrt {a x +b}\, a^{3} x^{2}+14 \sqrt {x}\, \sqrt {a x +b}\, a^{2} b x +3 \sqrt {x}\, \sqrt {a x +b}\, a \,b^{2}-3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {a x +b}+\sqrt {x}\, \sqrt {a}}{\sqrt {b}}\right ) b^{3}}{24 a^{2}} \] Input:
int((a+b/x)^(3/2)*x^2,x)
Output:
(8*sqrt(x)*sqrt(a*x + b)*a**3*x**2 + 14*sqrt(x)*sqrt(a*x + b)*a**2*b*x + 3 *sqrt(x)*sqrt(a*x + b)*a*b**2 - 3*sqrt(a)*log((sqrt(a*x + b) + sqrt(x)*sqr t(a))/sqrt(b))*b**3)/(24*a**2)