Integrand size = 17, antiderivative size = 136 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^3} \, dx=-\frac {\sqrt {a+b \sqrt {x}}}{2 a x^2}+\frac {7 b \sqrt {a+b \sqrt {x}}}{12 a^2 x^{3/2}}-\frac {35 b^2 \sqrt {a+b \sqrt {x}}}{48 a^3 x}+\frac {35 b^3 \sqrt {a+b \sqrt {x}}}{32 a^4 \sqrt {x}}-\frac {35 b^4 \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )}{32 a^{9/2}} \]
-35/32*b^4*arctanh((a+b*x^(1/2))^(1/2)/a^(1/2))/a^(9/2)-1/2*(a+b*x^(1/2))^ (1/2)/a/x^2+7/12*b*(a+b*x^(1/2))^(1/2)/a^2/x^(3/2)-35/48*b^2*(a+b*x^(1/2)) ^(1/2)/a^3/x+35/32*b^3*(a+b*x^(1/2))^(1/2)/a^4/x^(1/2)
Time = 0.16 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.66 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^3} \, dx=\frac {\sqrt {a+b \sqrt {x}} \left (-48 a^3+56 a^2 b \sqrt {x}-70 a b^2 x+105 b^3 x^{3/2}\right )}{96 a^4 x^2}-\frac {35 b^4 \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )}{32 a^{9/2}} \]
(Sqrt[a + b*Sqrt[x]]*(-48*a^3 + 56*a^2*b*Sqrt[x] - 70*a*b^2*x + 105*b^3*x^ (3/2)))/(96*a^4*x^2) - (35*b^4*ArcTanh[Sqrt[a + b*Sqrt[x]]/Sqrt[a]])/(32*a ^(9/2))
Time = 0.22 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {798, 52, 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.
\(\displaystyle \int \frac {1}{x^3 \sqrt {a+b \sqrt {x}}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle 2 \int \frac {1}{\sqrt {a+b \sqrt {x}} x^{5/2}}d\sqrt {x}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle 2 \left (-\frac {7 b \int \frac {1}{\sqrt {a+b \sqrt {x}} x^2}d\sqrt {x}}{8 a}-\frac {\sqrt {a+b \sqrt {x}}}{4 a x^2}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle 2 \left (-\frac {7 b \left (-\frac {5 b \int \frac {1}{\sqrt {a+b \sqrt {x}} x^{3/2}}d\sqrt {x}}{6 a}-\frac {\sqrt {a+b \sqrt {x}}}{3 a x^{3/2}}\right )}{8 a}-\frac {\sqrt {a+b \sqrt {x}}}{4 a x^2}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle 2 \left (-\frac {7 b \left (-\frac {5 b \left (-\frac {3 b \int \frac {1}{\sqrt {a+b \sqrt {x}} x}d\sqrt {x}}{4 a}-\frac {\sqrt {a+b \sqrt {x}}}{2 a x}\right )}{6 a}-\frac {\sqrt {a+b \sqrt {x}}}{3 a x^{3/2}}\right )}{8 a}-\frac {\sqrt {a+b \sqrt {x}}}{4 a x^2}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle 2 \left (-\frac {7 b \left (-\frac {5 b \left (-\frac {3 b \left (-\frac {b \int \frac {1}{\sqrt {a+b \sqrt {x}} \sqrt {x}}d\sqrt {x}}{2 a}-\frac {\sqrt {a+b \sqrt {x}}}{a \sqrt {x}}\right )}{4 a}-\frac {\sqrt {a+b \sqrt {x}}}{2 a x}\right )}{6 a}-\frac {\sqrt {a+b \sqrt {x}}}{3 a x^{3/2}}\right )}{8 a}-\frac {\sqrt {a+b \sqrt {x}}}{4 a x^2}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle 2 \left (-\frac {7 b \left (-\frac {5 b \left (-\frac {3 b \left (-\frac {\int \frac {1}{\frac {x}{b}-\frac {a}{b}}d\sqrt {a+b \sqrt {x}}}{a}-\frac {\sqrt {a+b \sqrt {x}}}{a \sqrt {x}}\right )}{4 a}-\frac {\sqrt {a+b \sqrt {x}}}{2 a x}\right )}{6 a}-\frac {\sqrt {a+b \sqrt {x}}}{3 a x^{3/2}}\right )}{8 a}-\frac {\sqrt {a+b \sqrt {x}}}{4 a x^2}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle 2 \left (-\frac {7 b \left (-\frac {5 b \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b \sqrt {x}}}{a \sqrt {x}}\right )}{4 a}-\frac {\sqrt {a+b \sqrt {x}}}{2 a x}\right )}{6 a}-\frac {\sqrt {a+b \sqrt {x}}}{3 a x^{3/2}}\right )}{8 a}-\frac {\sqrt {a+b \sqrt {x}}}{4 a x^2}\right )\) |
2*(-1/4*Sqrt[a + b*Sqrt[x]]/(a*x^2) - (7*b*(-1/3*Sqrt[a + b*Sqrt[x]]/(a*x^ (3/2)) - (5*b*(-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)) )/(6*a)))/(8*a))
3.23.44.3.1 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 + 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 = 3.62 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.91
method | result | size |
derivativedivides | \(4 b^{4} \left (-\frac {\sqrt {a +b \sqrt {x}}}{8 a \,b^{4} x^{2}}-\frac {7 \left (-\frac {\sqrt {a +b \sqrt {x}}}{6 a \,b^{3} x^{\frac {3}{2}}}+\frac {\frac {5 \sqrt {a +b \sqrt {x}}}{24 a \,b^{2} x}+\frac {5 \left (-\frac {3 \sqrt {a +b \sqrt {x}}}{8 a b \sqrt {x}}+\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \sqrt {x}}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )}{6 a}}{a}\right )}{8 a}\right )\) | \(124\) |
default | \(4 b^{4} \left (-\frac {\sqrt {a +b \sqrt {x}}}{8 a \,b^{4} x^{2}}-\frac {7 \left (-\frac {\sqrt {a +b \sqrt {x}}}{6 a \,b^{3} x^{\frac {3}{2}}}+\frac {\frac {5 \sqrt {a +b \sqrt {x}}}{24 a \,b^{2} x}+\frac {5 \left (-\frac {3 \sqrt {a +b \sqrt {x}}}{8 a b \sqrt {x}}+\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \sqrt {x}}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )}{6 a}}{a}\right )}{8 a}\right )\) | \(124\) |
4*b^4*(-1/8*(a+b*x^(1/2))^(1/2)/a/b^4/x^2-7/8/a*(-1/6*(a+b*x^(1/2))^(1/2)/ a/b^3/x^(3/2)+5/6/a*(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 = 0.27 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.35 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^3} \, dx=\left [\frac {105 \, \sqrt {a} b^{4} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b \sqrt {x} + a} \sqrt {a} \sqrt {x} + 2 \, a \sqrt {x}}{x}\right ) - 2 \, {\left (70 \, a^{2} b^{2} x + 48 \, a^{4} - 7 \, {\left (15 \, a b^{3} x + 8 \, a^{3} b\right )} \sqrt {x}\right )} \sqrt {b \sqrt {x} + a}}{192 \, a^{5} x^{2}}, \frac {105 \, \sqrt {-a} b^{4} x^{2} \arctan \left (\frac {\sqrt {b \sqrt {x} + a} \sqrt {-a}}{a}\right ) - {\left (70 \, a^{2} b^{2} x + 48 \, a^{4} - 7 \, {\left (15 \, a b^{3} x + 8 \, a^{3} b\right )} \sqrt {x}\right )} \sqrt {b \sqrt {x} + a}}{96 \, a^{5} x^{2}}\right ] \]
[1/192*(105*sqrt(a)*b^4*x^2*log((b*x - 2*sqrt(b*sqrt(x) + a)*sqrt(a)*sqrt( x) + 2*a*sqrt(x))/x) - 2*(70*a^2*b^2*x + 48*a^4 - 7*(15*a*b^3*x + 8*a^3*b) *sqrt(x))*sqrt(b*sqrt(x) + a))/(a^5*x^2), 1/96*(105*sqrt(-a)*b^4*x^2*arcta n(sqrt(b*sqrt(x) + a)*sqrt(-a)/a) - (70*a^2*b^2*x + 48*a^4 - 7*(15*a*b^3*x + 8*a^3*b)*sqrt(x))*sqrt(b*sqrt(x) + a))/(a^5*x^2)]
Time = 34.21 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.27 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^3} \, dx=- \frac {1}{2 \sqrt {b} x^{\frac {9}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} + \frac {\sqrt {b}}{12 a x^{\frac {7}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} - \frac {7 b^{\frac {3}{2}}}{48 a^{2} x^{\frac {5}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} + \frac {35 b^{\frac {5}{2}}}{96 a^{3} x^{\frac {3}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} + \frac {35 b^{\frac {7}{2}}}{32 a^{4} \sqrt [4]{x} \sqrt {\frac {a}{b \sqrt {x}} + 1}} - \frac {35 b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt [4]{x}} \right )}}{32 a^{\frac {9}{2}}} \]
-1/(2*sqrt(b)*x**(9/4)*sqrt(a/(b*sqrt(x)) + 1)) + sqrt(b)/(12*a*x**(7/4)*s qrt(a/(b*sqrt(x)) + 1)) - 7*b**(3/2)/(48*a**2*x**(5/4)*sqrt(a/(b*sqrt(x)) + 1)) + 35*b**(5/2)/(96*a**3*x**(3/4)*sqrt(a/(b*sqrt(x)) + 1)) + 35*b**(7/ 2)/(32*a**4*x**(1/4)*sqrt(a/(b*sqrt(x)) + 1)) - 35*b**4*asinh(sqrt(a)/(sqr t(b)*x**(1/4)))/(32*a**(9/2))
Time = 0.28 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.22 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^3} \, dx=\frac {35 \, b^{4} \log \left (\frac {\sqrt {b \sqrt {x} + a} - \sqrt {a}}{\sqrt {b \sqrt {x} + a} + \sqrt {a}}\right )}{64 \, a^{\frac {9}{2}}} + \frac {105 \, {\left (b \sqrt {x} + a\right )}^{\frac {7}{2}} b^{4} - 385 \, {\left (b \sqrt {x} + a\right )}^{\frac {5}{2}} a b^{4} + 511 \, {\left (b \sqrt {x} + a\right )}^{\frac {3}{2}} a^{2} b^{4} - 279 \, \sqrt {b \sqrt {x} + a} a^{3} b^{4}}{96 \, {\left ({\left (b \sqrt {x} + a\right )}^{4} a^{4} - 4 \, {\left (b \sqrt {x} + a\right )}^{3} a^{5} + 6 \, {\left (b \sqrt {x} + a\right )}^{2} a^{6} - 4 \, {\left (b \sqrt {x} + a\right )} a^{7} + a^{8}\right )}} \]
35/64*b^4*log((sqrt(b*sqrt(x) + a) - sqrt(a))/(sqrt(b*sqrt(x) + a) + sqrt( a)))/a^(9/2) + 1/96*(105*(b*sqrt(x) + a)^(7/2)*b^4 - 385*(b*sqrt(x) + a)^( 5/2)*a*b^4 + 511*(b*sqrt(x) + a)^(3/2)*a^2*b^4 - 279*sqrt(b*sqrt(x) + a)*a ^3*b^4)/((b*sqrt(x) + a)^4*a^4 - 4*(b*sqrt(x) + a)^3*a^5 + 6*(b*sqrt(x) + a)^2*a^6 - 4*(b*sqrt(x) + a)*a^7 + a^8)
Time = 0.28 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^3} \, dx=\frac {\frac {105 \, b^{5} \arctan \left (\frac {\sqrt {b \sqrt {x} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{4}} + \frac {105 \, {\left (b \sqrt {x} + a\right )}^{\frac {7}{2}} b^{5} - 385 \, {\left (b \sqrt {x} + a\right )}^{\frac {5}{2}} a b^{5} + 511 \, {\left (b \sqrt {x} + a\right )}^{\frac {3}{2}} a^{2} b^{5} - 279 \, \sqrt {b \sqrt {x} + a} a^{3} b^{5}}{a^{4} b^{4} x^{2}}}{96 \, b} \]
1/96*(105*b^5*arctan(sqrt(b*sqrt(x) + a)/sqrt(-a))/(sqrt(-a)*a^4) + (105*( b*sqrt(x) + a)^(7/2)*b^5 - 385*(b*sqrt(x) + a)^(5/2)*a*b^5 + 511*(b*sqrt(x ) + a)^(3/2)*a^2*b^5 - 279*sqrt(b*sqrt(x) + a)*a^3*b^5)/(a^4*b^4*x^2))/b
Time = 5.86 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.69 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^3} \, dx=\frac {511\,{\left (a+b\,\sqrt {x}\right )}^{3/2}}{96\,a^2\,x^2}-\frac {93\,\sqrt {a+b\,\sqrt {x}}}{32\,a\,x^2}-\frac {385\,{\left (a+b\,\sqrt {x}\right )}^{5/2}}{96\,a^3\,x^2}+\frac {35\,{\left (a+b\,\sqrt {x}\right )}^{7/2}}{32\,a^4\,x^2}+\frac {b^4\,\mathrm {atan}\left (\frac {\sqrt {a+b\,\sqrt {x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,35{}\mathrm {i}}{32\,a^{9/2}} \]