Integrand size = 13, antiderivative size = 102 \[ \int \sqrt {a+\frac {b}{\sqrt [3]{x}}} \, dx=-\frac {3 b^2 \sqrt {a+\frac {b}{\sqrt [3]{x}}} \sqrt [3]{x}}{8 a^2}+\frac {b \sqrt {a+\frac {b}{\sqrt [3]{x}}} x^{2/3}}{4 a}+\sqrt {a+\frac {b}{\sqrt [3]{x}}} x+\frac {3 b^3 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{\sqrt {a}}\right )}{8 a^{5/2}} \] Output:
-3/8*b^2*(a+b/x^(1/3))^(1/2)*x^(1/3)/a^2+1/4*b*(a+b/x^(1/3))^(1/2)*x^(2/3) /a+(a+b/x^(1/3))^(1/2)*x+3/8*b^3*arctanh((a+b/x^(1/3))^(1/2)/a^(1/2))/a^(5 /2)
Time = 0.10 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.82 \[ \int \sqrt {a+\frac {b}{\sqrt [3]{x}}} \, dx=\frac {\sqrt {a} \sqrt {a+\frac {b}{\sqrt [3]{x}}} \left (-3 b^2+2 a b \sqrt [3]{x}+8 a^2 x^{2/3}\right ) \sqrt [3]{x}+3 b^3 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{\sqrt {a}}\right )}{8 a^{5/2}} \] Input:
Integrate[Sqrt[a + b/x^(1/3)],x]
Output:
(Sqrt[a]*Sqrt[a + b/x^(1/3)]*(-3*b^2 + 2*a*b*x^(1/3) + 8*a^2*x^(2/3))*x^(1 /3) + 3*b^3*ArcTanh[Sqrt[a + b/x^(1/3)]/Sqrt[a]])/(8*a^(5/2))
Time = 0.34 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {774, 798, 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.
\(\displaystyle \int \sqrt {a+\frac {b}{\sqrt [3]{x}}} \, dx\) |
\(\Big \downarrow \) 774 |
\(\displaystyle 3 \int \sqrt {a+\frac {b}{\sqrt [3]{x}}} x^{2/3}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -3 \int \frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{x^{4/3}}d\frac {1}{\sqrt [3]{x}}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -3 \left (\frac {1}{6} b \int \frac {1}{\sqrt {a+\frac {b}{\sqrt [3]{x}}} x}d\frac {1}{\sqrt [3]{x}}-\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{3 x}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -3 \left (\frac {1}{6} b \left (-\frac {3 b \int \frac {1}{\sqrt {a+\frac {b}{\sqrt [3]{x}}} x^{2/3}}d\frac {1}{\sqrt [3]{x}}}{4 a}-\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{2 a x^{2/3}}\right )-\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{3 x}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -3 \left (\frac {1}{6} b \left (-\frac {3 b \left (-\frac {b \int \frac {1}{\sqrt {a+\frac {b}{\sqrt [3]{x}}} \sqrt [3]{x}}d\frac {1}{\sqrt [3]{x}}}{2 a}-\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{a \sqrt [3]{x}}\right )}{4 a}-\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{2 a x^{2/3}}\right )-\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{3 x}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -3 \left (\frac {1}{6} b \left (-\frac {3 b \left (-\frac {\int \frac {1}{\frac {x^{2/3}}{b}-\frac {a}{b}}d\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{a}-\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{a \sqrt [3]{x}}\right )}{4 a}-\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{2 a x^{2/3}}\right )-\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{3 x}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -3 \left (\frac {1}{6} b \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{a \sqrt [3]{x}}\right )}{4 a}-\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{2 a x^{2/3}}\right )-\frac {\sqrt {a+\frac {b}{\sqrt [3]{x}}}}{3 x}\right )\) |
Input:
Int[Sqrt[a + b/x^(1/3)],x]
Output:
-3*(-1/3*Sqrt[a + b/x^(1/3)]/x + (b*(-1/2*Sqrt[a + b/x^(1/3)]/(a*x^(2/3)) - (3*b*(-(Sqrt[a + b/x^(1/3)]/(a*x^(1/3))) + (b*ArcTanh[Sqrt[a + b/x^(1/3) ]/Sqrt[a]])/a^(3/2)))/(4*a)))/6)
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[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Simp[k Subst[Int[x^(k - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; Fre eQ[{a, b, p}, x] && FractionQ[n]
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 = 0.37 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.32
method | result | size |
derivativedivides | \(\frac {\sqrt {\frac {b +a \,x^{\frac {1}{3}}}{x^{\frac {1}{3}}}}\, x^{\frac {1}{3}} \left (16 \left (a \,x^{\frac {2}{3}}+b \,x^{\frac {1}{3}}\right )^{\frac {3}{2}} a^{\frac {5}{2}}-12 a^{\frac {5}{2}} \sqrt {a \,x^{\frac {2}{3}}+b \,x^{\frac {1}{3}}}\, b \,x^{\frac {1}{3}}-6 a^{\frac {3}{2}} \sqrt {a \,x^{\frac {2}{3}}+b \,x^{\frac {1}{3}}}\, b^{2}+3 a \ln \left (\frac {2 \sqrt {a \,x^{\frac {2}{3}}+b \,x^{\frac {1}{3}}}\, \sqrt {a}+2 a \,x^{\frac {1}{3}}+b}{2 \sqrt {a}}\right ) b^{3}\right )}{16 \sqrt {\left (b +a \,x^{\frac {1}{3}}\right ) x^{\frac {1}{3}}}\, a^{\frac {7}{2}}}\) | \(135\) |
default | \(\frac {\sqrt {\frac {b +a \,x^{\frac {1}{3}}}{x^{\frac {1}{3}}}}\, x^{\frac {1}{3}} \left (16 \left (a \,x^{\frac {2}{3}}+b \,x^{\frac {1}{3}}\right )^{\frac {3}{2}} a^{\frac {5}{2}}-12 a^{\frac {5}{2}} \sqrt {a \,x^{\frac {2}{3}}+b \,x^{\frac {1}{3}}}\, b \,x^{\frac {1}{3}}-6 a^{\frac {3}{2}} \sqrt {a \,x^{\frac {2}{3}}+b \,x^{\frac {1}{3}}}\, b^{2}+3 a \ln \left (\frac {2 \sqrt {a \,x^{\frac {2}{3}}+b \,x^{\frac {1}{3}}}\, \sqrt {a}+2 a \,x^{\frac {1}{3}}+b}{2 \sqrt {a}}\right ) b^{3}\right )}{16 \sqrt {\left (b +a \,x^{\frac {1}{3}}\right ) x^{\frac {1}{3}}}\, a^{\frac {7}{2}}}\) | \(135\) |
Input:
int((a+b/x^(1/3))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/16*((b+a*x^(1/3))/x^(1/3))^(1/2)*x^(1/3)*(16*(a*x^(2/3)+b*x^(1/3))^(3/2) *a^(5/2)-12*a^(5/2)*(a*x^(2/3)+b*x^(1/3))^(1/2)*b*x^(1/3)-6*a^(3/2)*(a*x^( 2/3)+b*x^(1/3))^(1/2)*b^2+3*a*ln(1/2*(2*(a*x^(2/3)+b*x^(1/3))^(1/2)*a^(1/2 )+2*a*x^(1/3)+b)/a^(1/2))*b^3)/((b+a*x^(1/3))*x^(1/3))^(1/2)/a^(7/2)
Timed out. \[ \int \sqrt {a+\frac {b}{\sqrt [3]{x}}} \, dx=\text {Timed out} \] Input:
integrate((a+b/x^(1/3))^(1/2),x, algorithm="fricas")
Output:
Timed out
Time = 4.99 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.35 \[ \int \sqrt {a+\frac {b}{\sqrt [3]{x}}} \, dx=\frac {a x^{\frac {7}{6}}}{\sqrt {b} \sqrt {\frac {a \sqrt [3]{x}}{b} + 1}} + \frac {5 \sqrt {b} x^{\frac {5}{6}}}{4 \sqrt {\frac {a \sqrt [3]{x}}{b} + 1}} - \frac {b^{\frac {3}{2}} \sqrt {x}}{8 a \sqrt {\frac {a \sqrt [3]{x}}{b} + 1}} - \frac {3 b^{\frac {5}{2}} \sqrt [6]{x}}{8 a^{2} \sqrt {\frac {a \sqrt [3]{x}}{b} + 1}} + \frac {3 b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt [6]{x}}{\sqrt {b}} \right )}}{8 a^{\frac {5}{2}}} \] Input:
integrate((a+b/x**(1/3))**(1/2),x)
Output:
a*x**(7/6)/(sqrt(b)*sqrt(a*x**(1/3)/b + 1)) + 5*sqrt(b)*x**(5/6)/(4*sqrt(a *x**(1/3)/b + 1)) - b**(3/2)*sqrt(x)/(8*a*sqrt(a*x**(1/3)/b + 1)) - 3*b**( 5/2)*x**(1/6)/(8*a**2*sqrt(a*x**(1/3)/b + 1)) + 3*b**3*asinh(sqrt(a)*x**(1 /6)/sqrt(b))/(8*a**(5/2))
Time = 0.11 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.34 \[ \int \sqrt {a+\frac {b}{\sqrt [3]{x}}} \, dx=-\frac {3 \, b^{3} \log \left (\frac {\sqrt {a + \frac {b}{x^{\frac {1}{3}}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{\frac {1}{3}}}} + \sqrt {a}}\right )}{16 \, a^{\frac {5}{2}}} - \frac {3 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{\frac {5}{2}} b^{3} - 8 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{\frac {3}{2}} a b^{3} - 3 \, \sqrt {a + \frac {b}{x^{\frac {1}{3}}}} a^{2} b^{3}}{8 \, {\left ({\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{3} a^{2} - 3 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{2} a^{3} + 3 \, {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )} a^{4} - a^{5}\right )}} \] Input:
integrate((a+b/x^(1/3))^(1/2),x, algorithm="maxima")
Output:
-3/16*b^3*log((sqrt(a + b/x^(1/3)) - sqrt(a))/(sqrt(a + b/x^(1/3)) + sqrt( a)))/a^(5/2) - 1/8*(3*(a + b/x^(1/3))^(5/2)*b^3 - 8*(a + b/x^(1/3))^(3/2)* a*b^3 - 3*sqrt(a + b/x^(1/3))*a^2*b^3)/((a + b/x^(1/3))^3*a^2 - 3*(a + b/x ^(1/3))^2*a^3 + 3*(a + b/x^(1/3))*a^4 - a^5)
Result contains complex when optimal does not.
Time = 0.19 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.21 \[ \int \sqrt {a+\frac {b}{\sqrt [3]{x}}} \, dx=\frac {1}{8} \, \sqrt {a x^{\frac {1}{3}} + b} {\left (2 \, x^{\frac {1}{3}} {\left (\frac {4 \, x^{\frac {1}{3}}}{\mathrm {sgn}\left (x\right )^{\frac {1}{3}}} + \frac {b}{a \mathrm {sgn}\left (x\right )^{\frac {1}{3}}}\right )} - \frac {3 \, b^{2}}{a^{2} \mathrm {sgn}\left (x\right )^{\frac {1}{3}}}\right )} x^{\frac {1}{6}} - \frac {3 \, b^{3} \log \left ({\left | -\sqrt {a} x^{\frac {1}{6}} + \sqrt {a x^{\frac {1}{3}} + b} \right |}\right )}{8 \, a^{\frac {5}{2}} \mathrm {sgn}\left (x\right )^{\frac {1}{3}}} - \frac {3 \, {\left (-i \, \sqrt {3} b^{3} \log \left ({\left | b \right |}\right ) + b^{3} \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{32 i \, \sqrt {3} a^{\frac {5}{2}} + 32 \, a^{\frac {5}{2}}} \] Input:
integrate((a+b/x^(1/3))^(1/2),x, algorithm="giac")
Output:
1/8*sqrt(a*x^(1/3) + b)*(2*x^(1/3)*(4*x^(1/3)/sgn(x)^(1/3) + b/(a*sgn(x)^( 1/3))) - 3*b^2/(a^2*sgn(x)^(1/3)))*x^(1/6) - 3/8*b^3*log(abs(-sqrt(a)*x^(1 /6) + sqrt(a*x^(1/3) + b)))/(a^(5/2)*sgn(x)^(1/3)) - 3*(-I*sqrt(3)*b^3*log (abs(b)) + b^3*log(abs(b)))*sgn(x)/(32*I*sqrt(3)*a^(5/2) + 32*a^(5/2))
Time = 0.16 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.37 \[ \int \sqrt {a+\frac {b}{\sqrt [3]{x}}} \, dx=\frac {6\,x\,\sqrt {a+\frac {b}{x^{1/3}}}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},\frac {5}{2};\ \frac {7}{2};\ -\frac {a\,x^{1/3}}{b}\right )}{5\,\sqrt {\frac {a\,x^{1/3}}{b}+1}} \] Input:
int((a + b/x^(1/3))^(1/2),x)
Output:
(6*x*(a + b/x^(1/3))^(1/2)*hypergeom([-1/2, 5/2], 7/2, -(a*x^(1/3))/b))/(5 *((a*x^(1/3))/b + 1)^(1/2))
Time = 0.21 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81 \[ \int \sqrt {a+\frac {b}{\sqrt [3]{x}}} \, dx=\frac {8 x^{\frac {5}{6}} \sqrt {x^{\frac {1}{3}} a +b}\, a^{3}-3 x^{\frac {1}{6}} \sqrt {x^{\frac {1}{3}} a +b}\, a \,b^{2}+2 \sqrt {x}\, \sqrt {x^{\frac {1}{3}} a +b}\, a^{2} b +3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {x^{\frac {1}{3}} a +b}+x^{\frac {1}{6}} \sqrt {a}}{\sqrt {b}}\right ) b^{3}}{8 a^{3}} \] Input:
int((a+b/x^(1/3))^(1/2),x)
Output:
(8*x**(5/6)*sqrt(x**(1/3)*a + b)*a**3 - 3*x**(1/6)*sqrt(x**(1/3)*a + b)*a* b**2 + 2*sqrt(x)*sqrt(x**(1/3)*a + b)*a**2*b + 3*sqrt(a)*log((sqrt(x**(1/3 )*a + b) + x**(1/6)*sqrt(a))/sqrt(b))*b**3)/(8*a**3)