Integrand size = 17, antiderivative size = 80 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^2} \, dx=-\frac {\sqrt {a+b \sqrt {x}}}{a x}+\frac {3 b \sqrt {a+b \sqrt {x}}}{2 a^2 \sqrt {x}}-\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )}{2 a^{5/2}} \] Output:
-(a+b*x^(1/2))^(1/2)/a/x+3/2*b*(a+b*x^(1/2))^(1/2)/a^2/x^(1/2)-3/2*b^2*arc tanh((a+b*x^(1/2))^(1/2)/a^(1/2))/a^(5/2)
Time = 0.09 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^2} \, dx=\frac {\sqrt {a+b \sqrt {x}} \left (-2 a+3 b \sqrt {x}\right )}{2 a^2 x}-\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )}{2 a^{5/2}} \] Input:
Integrate[1/(Sqrt[a + b*Sqrt[x]]*x^2),x]
Output:
(Sqrt[a + b*Sqrt[x]]*(-2*a + 3*b*Sqrt[x]))/(2*a^2*x) - (3*b^2*ArcTanh[Sqrt [a + b*Sqrt[x]]/Sqrt[a]])/(2*a^(5/2))
Time = 0.31 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {798, 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.
\(\displaystyle \int \frac {1}{x^2 \sqrt {a+b \sqrt {x}}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle 2 \int \frac {1}{\sqrt {a+b \sqrt {x}} x^{3/2}}d\sqrt {x}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle 2 \left (-\frac {3 b \int \frac {1}{\sqrt {a+b \sqrt {x}} x}d\sqrt {x}}{4 a}-\frac {\sqrt {a+b \sqrt {x}}}{2 a x}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle 2 \left (-\frac {3 b \left (-\frac {b \int \frac {1}{\sqrt {a+b \sqrt {x}} \sqrt {x}}d\sqrt {x}}{2 a}-\frac {\sqrt {a+b \sqrt {x}}}{a \sqrt {x}}\right )}{4 a}-\frac {\sqrt {a+b \sqrt {x}}}{2 a x}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle 2 \left (-\frac {3 b \left (-\frac {\int \frac {1}{\frac {x}{b}-\frac {a}{b}}d\sqrt {a+b \sqrt {x}}}{a}-\frac {\sqrt {a+b \sqrt {x}}}{a \sqrt {x}}\right )}{4 a}-\frac {\sqrt {a+b \sqrt {x}}}{2 a x}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle 2 \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b \sqrt {x}}}{a \sqrt {x}}\right )}{4 a}-\frac {\sqrt {a+b \sqrt {x}}}{2 a x}\right )\) |
Input:
Int[1/(Sqrt[a + b*Sqrt[x]]*x^2),x]
Output:
2*(-1/2*Sqrt[a + b*Sqrt[x]]/(a*x) - (3*b*(-(Sqrt[a + b*Sqrt[x]]/(a*Sqrt[x] )) + (b*ArcTanh[Sqrt[a + b*Sqrt[x]]/Sqrt[a]])/a^(3/2)))/(4*a))
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 = 0.44 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(4 b^{2} \left (-\frac {\sqrt {a +b \sqrt {x}}}{4 a \,b^{2} x}-\frac {3 \left (-\frac {\sqrt {a +b \sqrt {x}}}{2 a b \sqrt {x}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {a +b \sqrt {x}}}{\sqrt {a}}\right )}{2 a^{\frac {3}{2}}}\right )}{4 a}\right )\) | \(72\) |
default | \(4 b^{2} \left (-\frac {\sqrt {a +b \sqrt {x}}}{4 a \,b^{2} x}-\frac {3 \left (-\frac {\sqrt {a +b \sqrt {x}}}{2 a b \sqrt {x}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {a +b \sqrt {x}}}{\sqrt {a}}\right )}{2 a^{\frac {3}{2}}}\right )}{4 a}\right )\) | \(72\) |
Input:
int(1/(a+b*x^(1/2))^(1/2)/x^2,x,method=_RETURNVERBOSE)
Output:
4*b^2*(-1/4*(a+b*x^(1/2))^(1/2)/a/b^2/x-3/4/a*(-1/2*(a+b*x^(1/2))^(1/2)/a/ b/x^(1/2)+1/2/a^(3/2)*arctanh((a+b*x^(1/2))^(1/2)/a^(1/2))))
Time = 0.10 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.01 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^2} \, dx=\left [\frac {3 \, \sqrt {a} b^{2} x \log \left (\frac {b x - 2 \, \sqrt {b \sqrt {x} + a} \sqrt {a} \sqrt {x} + 2 \, a \sqrt {x}}{x}\right ) + 2 \, {\left (3 \, a b \sqrt {x} - 2 \, a^{2}\right )} \sqrt {b \sqrt {x} + a}}{4 \, a^{3} x}, \frac {3 \, \sqrt {-a} b^{2} x \arctan \left (\frac {{\left (\sqrt {-a} b \sqrt {x} - \sqrt {-a} a\right )} \sqrt {b \sqrt {x} + a}}{b^{2} x - a^{2}}\right ) + {\left (3 \, a b \sqrt {x} - 2 \, a^{2}\right )} \sqrt {b \sqrt {x} + a}}{2 \, a^{3} x}\right ] \] Input:
integrate(1/(a+b*x^(1/2))^(1/2)/x^2,x, algorithm="fricas")
Output:
[1/4*(3*sqrt(a)*b^2*x*log((b*x - 2*sqrt(b*sqrt(x) + a)*sqrt(a)*sqrt(x) + 2 *a*sqrt(x))/x) + 2*(3*a*b*sqrt(x) - 2*a^2)*sqrt(b*sqrt(x) + a))/(a^3*x), 1 /2*(3*sqrt(-a)*b^2*x*arctan((sqrt(-a)*b*sqrt(x) - sqrt(-a)*a)*sqrt(b*sqrt( x) + a)/(b^2*x - a^2)) + (3*a*b*sqrt(x) - 2*a^2)*sqrt(b*sqrt(x) + a))/(a^3 *x)]
Time = 3.07 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.38 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^2} \, dx=- \frac {1}{\sqrt {b} x^{\frac {5}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} + \frac {\sqrt {b}}{2 a x^{\frac {3}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} + \frac {3 b^{\frac {3}{2}}}{2 a^{2} \sqrt [4]{x} \sqrt {\frac {a}{b \sqrt {x}} + 1}} - \frac {3 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt [4]{x}} \right )}}{2 a^{\frac {5}{2}}} \] Input:
integrate(1/(a+b*x**(1/2))**(1/2)/x**2,x)
Output:
-1/(sqrt(b)*x**(5/4)*sqrt(a/(b*sqrt(x)) + 1)) + sqrt(b)/(2*a*x**(3/4)*sqrt (a/(b*sqrt(x)) + 1)) + 3*b**(3/2)/(2*a**2*x**(1/4)*sqrt(a/(b*sqrt(x)) + 1) ) - 3*b**2*asinh(sqrt(a)/(sqrt(b)*x**(1/4)))/(2*a**(5/2))
Time = 0.11 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.30 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^2} \, dx=\frac {3 \, b^{2} \log \left (\frac {\sqrt {b \sqrt {x} + a} - \sqrt {a}}{\sqrt {b \sqrt {x} + a} + \sqrt {a}}\right )}{4 \, a^{\frac {5}{2}}} + \frac {3 \, {\left (b \sqrt {x} + a\right )}^{\frac {3}{2}} b^{2} - 5 \, \sqrt {b \sqrt {x} + a} a b^{2}}{2 \, {\left ({\left (b \sqrt {x} + a\right )}^{2} a^{2} - 2 \, {\left (b \sqrt {x} + a\right )} a^{3} + a^{4}\right )}} \] Input:
integrate(1/(a+b*x^(1/2))^(1/2)/x^2,x, algorithm="maxima")
Output:
3/4*b^2*log((sqrt(b*sqrt(x) + a) - sqrt(a))/(sqrt(b*sqrt(x) + a) + sqrt(a) ))/a^(5/2) + 1/2*(3*(b*sqrt(x) + a)^(3/2)*b^2 - 5*sqrt(b*sqrt(x) + a)*a*b^ 2)/((b*sqrt(x) + a)^2*a^2 - 2*(b*sqrt(x) + a)*a^3 + a^4)
Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^2} \, dx=\frac {\frac {3 \, b^{3} \arctan \left (\frac {\sqrt {b \sqrt {x} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (b \sqrt {x} + a\right )}^{\frac {3}{2}} b^{3} - 5 \, \sqrt {b \sqrt {x} + a} a b^{3}}{a^{2} b^{2} x}}{2 \, b} \] Input:
integrate(1/(a+b*x^(1/2))^(1/2)/x^2,x, algorithm="giac")
Output:
1/2*(3*b^3*arctan(sqrt(b*sqrt(x) + a)/sqrt(-a))/(sqrt(-a)*a^2) + (3*(b*sqr t(x) + a)^(3/2)*b^3 - 5*sqrt(b*sqrt(x) + a)*a*b^3)/(a^2*b^2*x))/b
Time = 0.78 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.71 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^2} \, dx=\frac {3\,{\left (a+b\,\sqrt {x}\right )}^{3/2}}{2\,a^2\,x}-\frac {5\,\sqrt {a+b\,\sqrt {x}}}{2\,a\,x}-\frac {3\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,\sqrt {x}}}{\sqrt {a}}\right )}{2\,a^{5/2}} \] Input:
int(1/(x^2*(a + b*x^(1/2))^(1/2)),x)
Output:
(3*(a + b*x^(1/2))^(3/2))/(2*a^2*x) - (5*(a + b*x^(1/2))^(1/2))/(2*a*x) - (3*b^2*atanh((a + b*x^(1/2))^(1/2)/a^(1/2)))/(2*a^(5/2))
Time = 0.21 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^2} \, dx=\frac {6 \sqrt {x}\, \sqrt {\sqrt {x}\, b +a}\, a b -4 \sqrt {\sqrt {x}\, b +a}\, a^{2}+3 \sqrt {a}\, \mathrm {log}\left (\sqrt {\sqrt {x}\, b +a}-\sqrt {a}\right ) b^{2} x -3 \sqrt {a}\, \mathrm {log}\left (\sqrt {\sqrt {x}\, b +a}+\sqrt {a}\right ) b^{2} x}{4 a^{3} x} \] Input:
int(1/(a+b*x^(1/2))^(1/2)/x^2,x)
Output:
(6*sqrt(x)*sqrt(sqrt(x)*b + a)*a*b - 4*sqrt(sqrt(x)*b + a)*a**2 + 3*sqrt(a )*log(sqrt(sqrt(x)*b + a) - sqrt(a))*b**2*x - 3*sqrt(a)*log(sqrt(sqrt(x)*b + a) + sqrt(a))*b**2*x)/(4*a**3*x)