Integrand size = 15, antiderivative size = 115 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^3 \, dx=-\frac {3 b^3 \sqrt {a+\frac {b}{x}} x}{64 a^2}+\frac {b^2 \sqrt {a+\frac {b}{x}} x^2}{32 a}+\frac {3}{8} b \sqrt {a+\frac {b}{x}} x^3+\frac {1}{4} a \sqrt {a+\frac {b}{x}} x^4+\frac {3 b^4 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{64 a^{5/2}} \] Output:
-3/64*b^3*(a+b/x)^(1/2)*x/a^2+1/32*b^2*(a+b/x)^(1/2)*x^2/a+3/8*b*(a+b/x)^( 1/2)*x^3+1/4*a*(a+b/x)^(1/2)*x^4+3/64*b^4*arctanh((a+b/x)^(1/2)/a^(1/2))/a ^(5/2)
Time = 0.11 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.70 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^3 \, dx=\frac {\sqrt {a} \sqrt {a+\frac {b}{x}} x \left (-3 b^3+2 a b^2 x+24 a^2 b x^2+16 a^3 x^3\right )+3 b^4 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{64 a^{5/2}} \] Input:
Integrate[(a + b/x)^(3/2)*x^3,x]
Output:
(Sqrt[a]*Sqrt[a + b/x]*x*(-3*b^3 + 2*a*b^2*x + 24*a^2*b*x^2 + 16*a^3*x^3) + 3*b^4*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(64*a^(5/2))
Time = 0.34 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {798, 51, 51, 52, 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.
\(\displaystyle \int x^3 \left (a+\frac {b}{x}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\int \left (a+\frac {b}{x}\right )^{3/2} x^5d\frac {1}{x}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{4} x^4 \left (a+\frac {b}{x}\right )^{3/2}-\frac {3}{8} b \int \sqrt {a+\frac {b}{x}} x^4d\frac {1}{x}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{4} x^4 \left (a+\frac {b}{x}\right )^{3/2}-\frac {3}{8} b \left (\frac {1}{6} b \int \frac {x^3}{\sqrt {a+\frac {b}{x}}}d\frac {1}{x}-\frac {1}{3} x^3 \sqrt {a+\frac {b}{x}}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{4} x^4 \left (a+\frac {b}{x}\right )^{3/2}-\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {3 b \int \frac {x^2}{\sqrt {a+\frac {b}{x}}}d\frac {1}{x}}{4 a}-\frac {x^2 \sqrt {a+\frac {b}{x}}}{2 a}\right )-\frac {1}{3} x^3 \sqrt {a+\frac {b}{x}}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{4} x^4 \left (a+\frac {b}{x}\right )^{3/2}-\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {3 b \left (-\frac {b \int \frac {x}{\sqrt {a+\frac {b}{x}}}d\frac {1}{x}}{2 a}-\frac {x \sqrt {a+\frac {b}{x}}}{a}\right )}{4 a}-\frac {x^2 \sqrt {a+\frac {b}{x}}}{2 a}\right )-\frac {1}{3} x^3 \sqrt {a+\frac {b}{x}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} x^4 \left (a+\frac {b}{x}\right )^{3/2}-\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {3 b \left (-\frac {\int \frac {1}{\frac {1}{b x^2}-\frac {a}{b}}d\sqrt {a+\frac {b}{x}}}{a}-\frac {x \sqrt {a+\frac {b}{x}}}{a}\right )}{4 a}-\frac {x^2 \sqrt {a+\frac {b}{x}}}{2 a}\right )-\frac {1}{3} x^3 \sqrt {a+\frac {b}{x}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{4} x^4 \left (a+\frac {b}{x}\right )^{3/2}-\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {x \sqrt {a+\frac {b}{x}}}{a}\right )}{4 a}-\frac {x^2 \sqrt {a+\frac {b}{x}}}{2 a}\right )-\frac {1}{3} x^3 \sqrt {a+\frac {b}{x}}\right )\) |
Input:
Int[(a + b/x)^(3/2)*x^3,x]
Output:
((a + b/x)^(3/2)*x^4)/4 - (3*b*(-1/3*(Sqrt[a + b/x]*x^3) + (b*(-1/2*(Sqrt[ a + b/x]*x^2)/a - (3*b*(-((Sqrt[a + b/x]*x)/a) + (b*ArcTanh[Sqrt[a + b/x]/ Sqrt[a]])/a^(3/2)))/(4*a)))/6))/8
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 = 0.09 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.94
method | result | size |
risch | \(\frac {\left (16 a^{3} x^{3}+24 a^{2} b \,x^{2}+2 a \,b^{2} x -3 b^{3}\right ) x \sqrt {\frac {a x +b}{x}}}{64 a^{2}}+\frac {3 b^{4} \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right ) \sqrt {\frac {a x +b}{x}}\, \sqrt {x \left (a x +b \right )}}{128 a^{\frac {5}{2}} \left (a x +b \right )}\) | \(108\) |
default | \(\frac {\sqrt {\frac {a x +b}{x}}\, x \left (32 x \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {7}{2}}+16 a^{\frac {5}{2}} \left (a \,x^{2}+b x \right )^{\frac {3}{2}} b -12 a^{\frac {5}{2}} \sqrt {a \,x^{2}+b x}\, b^{2} x -6 a^{\frac {3}{2}} \sqrt {a \,x^{2}+b x}\, b^{3}+3 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a \,b^{4}\right )}{128 a^{\frac {7}{2}} \sqrt {x \left (a x +b \right )}}\) | \(135\) |
Input:
int((a+b/x)^(3/2)*x^3,x,method=_RETURNVERBOSE)
Output:
1/64*(16*a^3*x^3+24*a^2*b*x^2+2*a*b^2*x-3*b^3)*x/a^2*((a*x+b)/x)^(1/2)+3/1 28*b^4/a^(5/2)*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 = 179, normalized size of antiderivative = 1.56 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^3 \, dx=\left [\frac {3 \, \sqrt {a} b^{4} \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (16 \, a^{4} x^{4} + 24 \, a^{3} b x^{3} + 2 \, a^{2} b^{2} x^{2} - 3 \, a b^{3} x\right )} \sqrt {\frac {a x + b}{x}}}{128 \, a^{3}}, -\frac {3 \, \sqrt {-a} b^{4} \arctan \left (\frac {\sqrt {-a} x \sqrt {\frac {a x + b}{x}}}{a x + b}\right ) - {\left (16 \, a^{4} x^{4} + 24 \, a^{3} b x^{3} + 2 \, a^{2} b^{2} x^{2} - 3 \, a b^{3} x\right )} \sqrt {\frac {a x + b}{x}}}{64 \, a^{3}}\right ] \] Input:
integrate((a+b/x)^(3/2)*x^3,x, algorithm="fricas")
Output:
[1/128*(3*sqrt(a)*b^4*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*( 16*a^4*x^4 + 24*a^3*b*x^3 + 2*a^2*b^2*x^2 - 3*a*b^3*x)*sqrt((a*x + b)/x))/ a^3, -1/64*(3*sqrt(-a)*b^4*arctan(sqrt(-a)*x*sqrt((a*x + b)/x)/(a*x + b)) - (16*a^4*x^4 + 24*a^3*b*x^3 + 2*a^2*b^2*x^2 - 3*a*b^3*x)*sqrt((a*x + b)/x ))/a^3]
Time = 16.21 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.33 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^3 \, dx=\frac {a^{2} x^{\frac {9}{2}}}{4 \sqrt {b} \sqrt {\frac {a x}{b} + 1}} + \frac {5 a \sqrt {b} x^{\frac {7}{2}}}{8 \sqrt {\frac {a x}{b} + 1}} + \frac {13 b^{\frac {3}{2}} x^{\frac {5}{2}}}{32 \sqrt {\frac {a x}{b} + 1}} - \frac {b^{\frac {5}{2}} x^{\frac {3}{2}}}{64 a \sqrt {\frac {a x}{b} + 1}} - \frac {3 b^{\frac {7}{2}} \sqrt {x}}{64 a^{2} \sqrt {\frac {a x}{b} + 1}} + \frac {3 b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{64 a^{\frac {5}{2}}} \] Input:
integrate((a+b/x)**(3/2)*x**3,x)
Output:
a**2*x**(9/2)/(4*sqrt(b)*sqrt(a*x/b + 1)) + 5*a*sqrt(b)*x**(7/2)/(8*sqrt(a *x/b + 1)) + 13*b**(3/2)*x**(5/2)/(32*sqrt(a*x/b + 1)) - b**(5/2)*x**(3/2) /(64*a*sqrt(a*x/b + 1)) - 3*b**(7/2)*sqrt(x)/(64*a**2*sqrt(a*x/b + 1)) + 3 *b**4*asinh(sqrt(a)*sqrt(x)/sqrt(b))/(64*a**(5/2))
Time = 0.12 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.44 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^3 \, dx=-\frac {3 \, b^{4} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{128 \, a^{\frac {5}{2}}} - \frac {3 \, {\left (a + \frac {b}{x}\right )}^{\frac {7}{2}} b^{4} - 11 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} a b^{4} - 11 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{2} b^{4} + 3 \, \sqrt {a + \frac {b}{x}} a^{3} b^{4}}{64 \, {\left ({\left (a + \frac {b}{x}\right )}^{4} a^{2} - 4 \, {\left (a + \frac {b}{x}\right )}^{3} a^{3} + 6 \, {\left (a + \frac {b}{x}\right )}^{2} a^{4} - 4 \, {\left (a + \frac {b}{x}\right )} a^{5} + a^{6}\right )}} \] Input:
integrate((a+b/x)^(3/2)*x^3,x, algorithm="maxima")
Output:
-3/128*b^4*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/a^(5/2 ) - 1/64*(3*(a + b/x)^(7/2)*b^4 - 11*(a + b/x)^(5/2)*a*b^4 - 11*(a + b/x)^ (3/2)*a^2*b^4 + 3*sqrt(a + b/x)*a^3*b^4)/((a + b/x)^4*a^2 - 4*(a + b/x)^3* a^3 + 6*(a + b/x)^2*a^4 - 4*(a + b/x)*a^5 + a^6)
Time = 0.14 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.90 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^3 \, dx=-\frac {3 \, b^{4} \log \left ({\left | 2 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} + b \right |}\right ) \mathrm {sgn}\left (x\right )}{128 \, a^{\frac {5}{2}}} + \frac {3 \, b^{4} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{128 \, a^{\frac {5}{2}}} + \frac {1}{64} \, \sqrt {a x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, a x \mathrm {sgn}\left (x\right ) + 3 \, b \mathrm {sgn}\left (x\right )\right )} x + \frac {b^{2} \mathrm {sgn}\left (x\right )}{a}\right )} x - \frac {3 \, b^{3} \mathrm {sgn}\left (x\right )}{a^{2}}\right )} \] Input:
integrate((a+b/x)^(3/2)*x^3,x, algorithm="giac")
Output:
-3/128*b^4*log(abs(2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a) + b))*sgn(x)/ a^(5/2) + 3/128*b^4*log(abs(b))*sgn(x)/a^(5/2) + 1/64*sqrt(a*x^2 + b*x)*(2 *(4*(2*a*x*sgn(x) + 3*b*sgn(x))*x + b^2*sgn(x)/a)*x - 3*b^3*sgn(x)/a^2)
Time = 0.60 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.77 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^3 \, dx=\frac {11\,x^4\,{\left (a+\frac {b}{x}\right )}^{3/2}}{64}-\frac {3\,a\,x^4\,\sqrt {a+\frac {b}{x}}}{64}+\frac {11\,x^4\,{\left (a+\frac {b}{x}\right )}^{5/2}}{64\,a}-\frac {3\,x^4\,{\left (a+\frac {b}{x}\right )}^{7/2}}{64\,a^2}-\frac {b^4\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,3{}\mathrm {i}}{64\,a^{5/2}} \] Input:
int(x^3*(a + b/x)^(3/2),x)
Output:
(11*x^4*(a + b/x)^(3/2))/64 - (b^4*atan(((a + b/x)^(1/2)*1i)/a^(1/2))*3i)/ (64*a^(5/2)) - (3*a*x^4*(a + b/x)^(1/2))/64 + (11*x^4*(a + b/x)^(5/2))/(64 *a) - (3*x^4*(a + b/x)^(7/2))/(64*a^2)
Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.83 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^3 \, dx=\frac {16 \sqrt {x}\, \sqrt {a x +b}\, a^{4} x^{3}+24 \sqrt {x}\, \sqrt {a x +b}\, a^{3} b \,x^{2}+2 \sqrt {x}\, \sqrt {a x +b}\, a^{2} b^{2} x -3 \sqrt {x}\, \sqrt {a x +b}\, a \,b^{3}+3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {a x +b}+\sqrt {x}\, \sqrt {a}}{\sqrt {b}}\right ) b^{4}}{64 a^{3}} \] Input:
int((a+b/x)^(3/2)*x^3,x)
Output:
(16*sqrt(x)*sqrt(a*x + b)*a**4*x**3 + 24*sqrt(x)*sqrt(a*x + b)*a**3*b*x**2 + 2*sqrt(x)*sqrt(a*x + b)*a**2*b**2*x - 3*sqrt(x)*sqrt(a*x + b)*a*b**3 + 3*sqrt(a)*log((sqrt(a*x + b) + sqrt(x)*sqrt(a))/sqrt(b))*b**4)/(64*a**3)