Integrand size = 17, antiderivative size = 109 \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x^{9/2}} \, dx=-\frac {\sqrt {a+\frac {b}{x}}}{3 b x^{5/2}}+\frac {5 a \sqrt {a+\frac {b}{x}}}{12 b^2 x^{3/2}}-\frac {5 a^2 \sqrt {a+\frac {b}{x}}}{8 b^3 \sqrt {x}}+\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{8 b^{7/2}} \] Output:
-1/3*(a+b/x)^(1/2)/b/x^(5/2)+5/12*a*(a+b/x)^(1/2)/b^2/x^(3/2)-5/8*a^2*(a+b /x)^(1/2)/b^3/x^(1/2)+5/8*a^3*arctanh(b^(1/2)/(a+b/x)^(1/2)/x^(1/2))/b^(7/ 2)
Time = 10.08 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.97 \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x^{9/2}} \, dx=\frac {-\sqrt {b} \left (8 b^3-2 a b^2 x+5 a^2 b x^2+15 a^3 x^3\right )+15 a^{7/2} \sqrt {1+\frac {b}{a x}} x^{7/2} \text {arcsinh}\left (\frac {\sqrt {b}}{\sqrt {a} \sqrt {x}}\right )}{24 b^{7/2} \sqrt {a+\frac {b}{x}} x^{7/2}} \] Input:
Integrate[1/(Sqrt[a + b/x]*x^(9/2)),x]
Output:
(-(Sqrt[b]*(8*b^3 - 2*a*b^2*x + 5*a^2*b*x^2 + 15*a^3*x^3)) + 15*a^(7/2)*Sq rt[1 + b/(a*x)]*x^(7/2)*ArcSinh[Sqrt[b]/(Sqrt[a]*Sqrt[x])])/(24*b^(7/2)*Sq rt[a + b/x]*x^(7/2))
Time = 0.35 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {860, 262, 262, 262, 224, 219}
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.
\(\displaystyle \int \frac {1}{x^{9/2} \sqrt {a+\frac {b}{x}}} \, dx\) |
\(\Big \downarrow \) 860 |
\(\displaystyle -2 \int \frac {1}{\sqrt {a+\frac {b}{x}} x^3}d\frac {1}{\sqrt {x}}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -2 \left (\frac {\sqrt {a+\frac {b}{x}}}{6 b x^{5/2}}-\frac {5 a \int \frac {1}{\sqrt {a+\frac {b}{x}} x^2}d\frac {1}{\sqrt {x}}}{6 b}\right )\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -2 \left (\frac {\sqrt {a+\frac {b}{x}}}{6 b x^{5/2}}-\frac {5 a \left (\frac {\sqrt {a+\frac {b}{x}}}{4 b x^{3/2}}-\frac {3 a \int \frac {1}{\sqrt {a+\frac {b}{x}} x}d\frac {1}{\sqrt {x}}}{4 b}\right )}{6 b}\right )\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -2 \left (\frac {\sqrt {a+\frac {b}{x}}}{6 b x^{5/2}}-\frac {5 a \left (\frac {\sqrt {a+\frac {b}{x}}}{4 b x^{3/2}}-\frac {3 a \left (\frac {\sqrt {a+\frac {b}{x}}}{2 b \sqrt {x}}-\frac {a \int \frac {1}{\sqrt {a+\frac {b}{x}}}d\frac {1}{\sqrt {x}}}{2 b}\right )}{4 b}\right )}{6 b}\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -2 \left (\frac {\sqrt {a+\frac {b}{x}}}{6 b x^{5/2}}-\frac {5 a \left (\frac {\sqrt {a+\frac {b}{x}}}{4 b x^{3/2}}-\frac {3 a \left (\frac {\sqrt {a+\frac {b}{x}}}{2 b \sqrt {x}}-\frac {a \int \frac {1}{1-\frac {b}{x}}d\frac {1}{\sqrt {a+\frac {b}{x}} \sqrt {x}}}{2 b}\right )}{4 b}\right )}{6 b}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -2 \left (\frac {\sqrt {a+\frac {b}{x}}}{6 b x^{5/2}}-\frac {5 a \left (\frac {\sqrt {a+\frac {b}{x}}}{4 b x^{3/2}}-\frac {3 a \left (\frac {\sqrt {a+\frac {b}{x}}}{2 b \sqrt {x}}-\frac {a \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )}{2 b^{3/2}}\right )}{4 b}\right )}{6 b}\right )\) |
Input:
Int[1/(Sqrt[a + b/x]*x^(9/2)),x]
Output:
-2*(Sqrt[a + b/x]/(6*b*x^(5/2)) - (5*a*(Sqrt[a + b/x]/(4*b*x^(3/2)) - (3*a *(Sqrt[a + b/x]/(2*b*Sqrt[x]) - (a*ArcTanh[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x]) ])/(2*b^(3/2))))/(4*b)))/(6*b))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[-k/c Subst[Int[(a + b/(c^n*x^(k*n)))^p/x^(k*(m + 1 ) + 1), x], x, 1/(c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && ILtQ[n, 0] && FractionQ[m]
Time = 0.25 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.79
method | result | size |
risch | \(-\frac {\left (a x +b \right ) \left (15 a^{2} x^{2}-10 a b x +8 b^{2}\right )}{24 b^{3} x^{\frac {7}{2}} \sqrt {\frac {a x +b}{x}}}+\frac {5 a^{3} \operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right ) \sqrt {a x +b}}{8 b^{\frac {7}{2}} \sqrt {\frac {a x +b}{x}}\, \sqrt {x}}\) | \(86\) |
default | \(-\frac {\sqrt {\frac {a x +b}{x}}\, \left (-15 \,\operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right ) a^{3} x^{3}+8 \sqrt {a x +b}\, b^{\frac {5}{2}}-10 a \,b^{\frac {3}{2}} x \sqrt {a x +b}+15 a^{2} x^{2} \sqrt {a x +b}\, \sqrt {b}\right )}{24 x^{\frac {5}{2}} b^{\frac {7}{2}} \sqrt {a x +b}}\) | \(92\) |
Input:
int(1/(a+b/x)^(1/2)/x^(9/2),x,method=_RETURNVERBOSE)
Output:
-1/24*(a*x+b)*(15*a^2*x^2-10*a*b*x+8*b^2)/b^3/x^(7/2)/((a*x+b)/x)^(1/2)+5/ 8*a^3/b^(7/2)*arctanh((a*x+b)^(1/2)/b^(1/2))/((a*x+b)/x)^(1/2)/x^(1/2)*(a* x+b)^(1/2)
Time = 0.10 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.62 \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x^{9/2}} \, dx=\left [\frac {15 \, a^{3} \sqrt {b} x^{3} \log \left (\frac {a x + 2 \, \sqrt {b} \sqrt {x} \sqrt {\frac {a x + b}{x}} + 2 \, b}{x}\right ) - 2 \, {\left (15 \, a^{2} b x^{2} - 10 \, a b^{2} x + 8 \, b^{3}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{48 \, b^{4} x^{3}}, -\frac {15 \, a^{3} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {-b} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{a x + b}\right ) + {\left (15 \, a^{2} b x^{2} - 10 \, a b^{2} x + 8 \, b^{3}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{24 \, b^{4} x^{3}}\right ] \] Input:
integrate(1/(a+b/x)^(1/2)/x^(9/2),x, algorithm="fricas")
Output:
[1/48*(15*a^3*sqrt(b)*x^3*log((a*x + 2*sqrt(b)*sqrt(x)*sqrt((a*x + b)/x) + 2*b)/x) - 2*(15*a^2*b*x^2 - 10*a*b^2*x + 8*b^3)*sqrt(x)*sqrt((a*x + b)/x) )/(b^4*x^3), -1/24*(15*a^3*sqrt(-b)*x^3*arctan(sqrt(-b)*sqrt(x)*sqrt((a*x + b)/x)/(a*x + b)) + (15*a^2*b*x^2 - 10*a*b^2*x + 8*b^3)*sqrt(x)*sqrt((a*x + b)/x))/(b^4*x^3)]
Time = 34.00 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.18 \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x^{9/2}} \, dx=- \frac {5 a^{\frac {5}{2}}}{8 b^{3} \sqrt {x} \sqrt {1 + \frac {b}{a x}}} - \frac {5 a^{\frac {3}{2}}}{24 b^{2} x^{\frac {3}{2}} \sqrt {1 + \frac {b}{a x}}} + \frac {\sqrt {a}}{12 b x^{\frac {5}{2}} \sqrt {1 + \frac {b}{a x}}} + \frac {5 a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} \sqrt {x}} \right )}}{8 b^{\frac {7}{2}}} - \frac {1}{3 \sqrt {a} x^{\frac {7}{2}} \sqrt {1 + \frac {b}{a x}}} \] Input:
integrate(1/(a+b/x)**(1/2)/x**(9/2),x)
Output:
-5*a**(5/2)/(8*b**3*sqrt(x)*sqrt(1 + b/(a*x))) - 5*a**(3/2)/(24*b**2*x**(3 /2)*sqrt(1 + b/(a*x))) + sqrt(a)/(12*b*x**(5/2)*sqrt(1 + b/(a*x))) + 5*a** 3*asinh(sqrt(b)/(sqrt(a)*sqrt(x)))/(8*b**(7/2)) - 1/(3*sqrt(a)*x**(7/2)*sq rt(1 + b/(a*x)))
Time = 0.11 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.48 \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x^{9/2}} \, dx=-\frac {5 \, a^{3} \log \left (\frac {\sqrt {a + \frac {b}{x}} \sqrt {x} - \sqrt {b}}{\sqrt {a + \frac {b}{x}} \sqrt {x} + \sqrt {b}}\right )}{16 \, b^{\frac {7}{2}}} - \frac {15 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} a^{3} x^{\frac {5}{2}} - 40 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{3} b x^{\frac {3}{2}} + 33 \, \sqrt {a + \frac {b}{x}} a^{3} b^{2} \sqrt {x}}{24 \, {\left ({\left (a + \frac {b}{x}\right )}^{3} b^{3} x^{3} - 3 \, {\left (a + \frac {b}{x}\right )}^{2} b^{4} x^{2} + 3 \, {\left (a + \frac {b}{x}\right )} b^{5} x - b^{6}\right )}} \] Input:
integrate(1/(a+b/x)^(1/2)/x^(9/2),x, algorithm="maxima")
Output:
-5/16*a^3*log((sqrt(a + b/x)*sqrt(x) - sqrt(b))/(sqrt(a + b/x)*sqrt(x) + s qrt(b)))/b^(7/2) - 1/24*(15*(a + b/x)^(5/2)*a^3*x^(5/2) - 40*(a + b/x)^(3/ 2)*a^3*b*x^(3/2) + 33*sqrt(a + b/x)*a^3*b^2*sqrt(x))/((a + b/x)^3*b^3*x^3 - 3*(a + b/x)^2*b^4*x^2 + 3*(a + b/x)*b^5*x - b^6)
Time = 0.14 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x^{9/2}} \, dx=-\frac {a^{3} {\left (\frac {15 \, \arctan \left (\frac {\sqrt {a x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{3}} + \frac {15 \, {\left (a x + b\right )}^{\frac {5}{2}} - 40 \, {\left (a x + b\right )}^{\frac {3}{2}} b + 33 \, \sqrt {a x + b} b^{2}}{a^{3} b^{3} x^{3}}\right )}}{24 \, \mathrm {sgn}\left (x\right )} \] Input:
integrate(1/(a+b/x)^(1/2)/x^(9/2),x, algorithm="giac")
Output:
-1/24*a^3*(15*arctan(sqrt(a*x + b)/sqrt(-b))/(sqrt(-b)*b^3) + (15*(a*x + b )^(5/2) - 40*(a*x + b)^(3/2)*b + 33*sqrt(a*x + b)*b^2)/(a^3*b^3*x^3))/sgn( x)
Timed out. \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x^{9/2}} \, dx=\int \frac {1}{x^{9/2}\,\sqrt {a+\frac {b}{x}}} \,d x \] Input:
int(1/(x^(9/2)*(a + b/x)^(1/2)),x)
Output:
int(1/(x^(9/2)*(a + b/x)^(1/2)), x)
Time = 0.21 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x^{9/2}} \, dx=\frac {-30 \sqrt {a x +b}\, a^{2} b \,x^{2}+20 \sqrt {a x +b}\, a \,b^{2} x -16 \sqrt {a x +b}\, b^{3}-15 \sqrt {b}\, \mathrm {log}\left (\sqrt {a x +b}-\sqrt {b}\right ) a^{3} x^{3}+15 \sqrt {b}\, \mathrm {log}\left (\sqrt {a x +b}+\sqrt {b}\right ) a^{3} x^{3}}{48 b^{4} x^{3}} \] Input:
int(1/(a+b/x)^(1/2)/x^(9/2),x)
Output:
( - 30*sqrt(a*x + b)*a**2*b*x**2 + 20*sqrt(a*x + b)*a*b**2*x - 16*sqrt(a*x + b)*b**3 - 15*sqrt(b)*log(sqrt(a*x + b) - sqrt(b))*a**3*x**3 + 15*sqrt(b )*log(sqrt(a*x + b) + sqrt(b))*a**3*x**3)/(48*b**4*x**3)