Integrand size = 15, antiderivative size = 117 \[ \int \sqrt {a+\frac {b}{x}} x^3 \, dx=\frac {5 b^3 \sqrt {a+\frac {b}{x}} x}{64 a^3}-\frac {5 b^2 \sqrt {a+\frac {b}{x}} x^2}{96 a^2}+\frac {b \sqrt {a+\frac {b}{x}} x^3}{24 a}+\frac {1}{4} \sqrt {a+\frac {b}{x}} x^4-\frac {5 b^4 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{64 a^{7/2}} \]
-5/64*b^4*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(7/2)+5/64*b^3*x*(a+b/x)^(1/2)/ a^3-5/96*b^2*x^2*(a+b/x)^(1/2)/a^2+1/24*b*x^3*(a+b/x)^(1/2)/a+1/4*x^4*(a+b /x)^(1/2)
Time = 0.17 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.69 \[ \int \sqrt {a+\frac {b}{x}} x^3 \, dx=\frac {\sqrt {a} \sqrt {a+\frac {b}{x}} x \left (15 b^3-10 a b^2 x+8 a^2 b x^2+48 a^3 x^3\right )-15 b^4 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{192 a^{7/2}} \]
(Sqrt[a]*Sqrt[a + b/x]*x*(15*b^3 - 10*a*b^2*x + 8*a^2*b*x^2 + 48*a^3*x^3) - 15*b^4*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(192*a^(7/2))
Time = 0.22 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {798, 51, 52, 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.
\(\displaystyle \int x^3 \sqrt {a+\frac {b}{x}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\int \sqrt {a+\frac {b}{x}} x^5d\frac {1}{x}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{4} x^4 \sqrt {a+\frac {b}{x}}-\frac {1}{8} b \int \frac {x^4}{\sqrt {a+\frac {b}{x}}}d\frac {1}{x}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{4} x^4 \sqrt {a+\frac {b}{x}}-\frac {1}{8} b \left (-\frac {5 b \int \frac {x^3}{\sqrt {a+\frac {b}{x}}}d\frac {1}{x}}{6 a}-\frac {x^3 \sqrt {a+\frac {b}{x}}}{3 a}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{4} x^4 \sqrt {a+\frac {b}{x}}-\frac {1}{8} b \left (-\frac {5 b \left (-\frac {3 b \int \frac {x^2}{\sqrt {a+\frac {b}{x}}}d\frac {1}{x}}{4 a}-\frac {x^2 \sqrt {a+\frac {b}{x}}}{2 a}\right )}{6 a}-\frac {x^3 \sqrt {a+\frac {b}{x}}}{3 a}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{4} x^4 \sqrt {a+\frac {b}{x}}-\frac {1}{8} b \left (-\frac {5 b \left (-\frac {3 b \left (-\frac {b \int \frac {x}{\sqrt {a+\frac {b}{x}}}d\frac {1}{x}}{2 a}-\frac {x \sqrt {a+\frac {b}{x}}}{a}\right )}{4 a}-\frac {x^2 \sqrt {a+\frac {b}{x}}}{2 a}\right )}{6 a}-\frac {x^3 \sqrt {a+\frac {b}{x}}}{3 a}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} x^4 \sqrt {a+\frac {b}{x}}-\frac {1}{8} b \left (-\frac {5 b \left (-\frac {3 b \left (-\frac {\int \frac {1}{\frac {1}{b x^2}-\frac {a}{b}}d\sqrt {a+\frac {b}{x}}}{a}-\frac {x \sqrt {a+\frac {b}{x}}}{a}\right )}{4 a}-\frac {x^2 \sqrt {a+\frac {b}{x}}}{2 a}\right )}{6 a}-\frac {x^3 \sqrt {a+\frac {b}{x}}}{3 a}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{4} x^4 \sqrt {a+\frac {b}{x}}-\frac {1}{8} b \left (-\frac {5 b \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {x \sqrt {a+\frac {b}{x}}}{a}\right )}{4 a}-\frac {x^2 \sqrt {a+\frac {b}{x}}}{2 a}\right )}{6 a}-\frac {x^3 \sqrt {a+\frac {b}{x}}}{3 a}\right )\) |
(Sqrt[a + b/x]*x^4)/4 - (b*(-1/3*(Sqrt[a + b/x]*x^3)/a - (5*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*a)))/8
3.17.91.3.1 Defintions of rubi rules used
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 = 0.06 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.92
method | result | size |
risch | \(\frac {\left (48 a^{3} x^{3}+8 a^{2} b \,x^{2}-10 a \,b^{2} x +15 b^{3}\right ) x \sqrt {\frac {a x +b}{x}}}{192 a^{3}}-\frac {5 b^{4} \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right ) \sqrt {\frac {a x +b}{x}}\, \sqrt {x \left (a x +b \right )}}{128 a^{\frac {7}{2}} \left (a x +b \right )}\) | \(108\) |
default | \(\frac {\sqrt {\frac {a x +b}{x}}\, x \left (96 x \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {7}{2}}-80 a^{\frac {5}{2}} \left (a \,x^{2}+b x \right )^{\frac {3}{2}} b +60 a^{\frac {5}{2}} \sqrt {a \,x^{2}+b x}\, b^{2} x +30 a^{\frac {3}{2}} \sqrt {a \,x^{2}+b x}\, b^{3}-15 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a \,b^{4}\right )}{384 \sqrt {x \left (a x +b \right )}\, a^{\frac {9}{2}}}\) | \(135\) |
1/192*(48*a^3*x^3+8*a^2*b*x^2-10*a*b^2*x+15*b^3)*x/a^3*((a*x+b)/x)^(1/2)-5 /128*b^4/a^(7/2)*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 = 0.29 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.48 \[ \int \sqrt {a+\frac {b}{x}} x^3 \, dx=\left [\frac {15 \, \sqrt {a} b^{4} \log \left (2 \, a x - 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (48 \, a^{4} x^{4} + 8 \, a^{3} b x^{3} - 10 \, a^{2} b^{2} x^{2} + 15 \, a b^{3} x\right )} \sqrt {\frac {a x + b}{x}}}{384 \, a^{4}}, \frac {15 \, \sqrt {-a} b^{4} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + {\left (48 \, a^{4} x^{4} + 8 \, a^{3} b x^{3} - 10 \, a^{2} b^{2} x^{2} + 15 \, a b^{3} x\right )} \sqrt {\frac {a x + b}{x}}}{192 \, a^{4}}\right ] \]
[1/384*(15*sqrt(a)*b^4*log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2* (48*a^4*x^4 + 8*a^3*b*x^3 - 10*a^2*b^2*x^2 + 15*a*b^3*x)*sqrt((a*x + b)/x) )/a^4, 1/192*(15*sqrt(-a)*b^4*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (48*a ^4*x^4 + 8*a^3*b*x^3 - 10*a^2*b^2*x^2 + 15*a*b^3*x)*sqrt((a*x + b)/x))/a^4 ]
Time = 18.16 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.31 \[ \int \sqrt {a+\frac {b}{x}} x^3 \, dx=\frac {a x^{\frac {9}{2}}}{4 \sqrt {b} \sqrt {\frac {a x}{b} + 1}} + \frac {7 \sqrt {b} x^{\frac {7}{2}}}{24 \sqrt {\frac {a x}{b} + 1}} - \frac {b^{\frac {3}{2}} x^{\frac {5}{2}}}{96 a \sqrt {\frac {a x}{b} + 1}} + \frac {5 b^{\frac {5}{2}} x^{\frac {3}{2}}}{192 a^{2} \sqrt {\frac {a x}{b} + 1}} + \frac {5 b^{\frac {7}{2}} \sqrt {x}}{64 a^{3} \sqrt {\frac {a x}{b} + 1}} - \frac {5 b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{64 a^{\frac {7}{2}}} \]
a*x**(9/2)/(4*sqrt(b)*sqrt(a*x/b + 1)) + 7*sqrt(b)*x**(7/2)/(24*sqrt(a*x/b + 1)) - b**(3/2)*x**(5/2)/(96*a*sqrt(a*x/b + 1)) + 5*b**(5/2)*x**(3/2)/(1 92*a**2*sqrt(a*x/b + 1)) + 5*b**(7/2)*sqrt(x)/(64*a**3*sqrt(a*x/b + 1)) - 5*b**4*asinh(sqrt(a)*sqrt(x)/sqrt(b))/(64*a**(7/2))
Time = 0.28 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.42 \[ \int \sqrt {a+\frac {b}{x}} x^3 \, dx=\frac {5 \, b^{4} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{128 \, a^{\frac {7}{2}}} + \frac {15 \, {\left (a + \frac {b}{x}\right )}^{\frac {7}{2}} b^{4} - 55 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} a b^{4} + 73 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{2} b^{4} + 15 \, \sqrt {a + \frac {b}{x}} a^{3} b^{4}}{192 \, {\left ({\left (a + \frac {b}{x}\right )}^{4} a^{3} - 4 \, {\left (a + \frac {b}{x}\right )}^{3} a^{4} + 6 \, {\left (a + \frac {b}{x}\right )}^{2} a^{5} - 4 \, {\left (a + \frac {b}{x}\right )} a^{6} + a^{7}\right )}} \]
5/128*b^4*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/a^(7/2) + 1/192*(15*(a + b/x)^(7/2)*b^4 - 55*(a + b/x)^(5/2)*a*b^4 + 73*(a + b/x) ^(3/2)*a^2*b^4 + 15*sqrt(a + b/x)*a^3*b^4)/((a + b/x)^4*a^3 - 4*(a + b/x)^ 3*a^4 + 6*(a + b/x)^2*a^5 - 4*(a + b/x)*a^6 + a^7)
Time = 0.30 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.91 \[ \int \sqrt {a+\frac {b}{x}} x^3 \, dx=\frac {5 \, b^{4} \log \left ({\left | 2 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} + b \right |}\right ) \mathrm {sgn}\left (x\right )}{128 \, a^{\frac {7}{2}}} - \frac {5 \, b^{4} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{128 \, a^{\frac {7}{2}}} + \frac {1}{192} \, \sqrt {a x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, x \mathrm {sgn}\left (x\right ) + \frac {b \mathrm {sgn}\left (x\right )}{a}\right )} x - \frac {5 \, b^{2} \mathrm {sgn}\left (x\right )}{a^{2}}\right )} x + \frac {15 \, b^{3} \mathrm {sgn}\left (x\right )}{a^{3}}\right )} \]
5/128*b^4*log(abs(2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a) + b))*sgn(x)/a ^(7/2) - 5/128*b^4*log(abs(b))*sgn(x)/a^(7/2) + 1/192*sqrt(a*x^2 + b*x)*(2 *(4*(6*x*sgn(x) + b*sgn(x)/a)*x - 5*b^2*sgn(x)/a^2)*x + 15*b^3*sgn(x)/a^3)
Time = 6.50 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.78 \[ \int \sqrt {a+\frac {b}{x}} x^3 \, dx=\frac {5\,x^4\,\sqrt {a+\frac {b}{x}}}{64}+\frac {73\,x^4\,{\left (a+\frac {b}{x}\right )}^{3/2}}{192\,a}-\frac {55\,x^4\,{\left (a+\frac {b}{x}\right )}^{5/2}}{192\,a^2}+\frac {5\,x^4\,{\left (a+\frac {b}{x}\right )}^{7/2}}{64\,a^3}+\frac {b^4\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{64\,a^{7/2}} \]