Integrand size = 13, antiderivative size = 61 \[ \int \frac {1}{\sqrt {a+b \sqrt [3]{x}}} \, dx=\frac {6 a^2 \sqrt {a+b \sqrt [3]{x}}}{b^3}-\frac {4 a \left (a+b \sqrt [3]{x}\right )^{3/2}}{b^3}+\frac {6 \left (a+b \sqrt [3]{x}\right )^{5/2}}{5 b^3} \] Output:
6*a^2*(a+b*x^(1/3))^(1/2)/b^3-4*a*(a+b*x^(1/3))^(3/2)/b^3+6/5*(a+b*x^(1/3) )^(5/2)/b^3
Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\sqrt {a+b \sqrt [3]{x}}} \, dx=\frac {2 \sqrt {a+b \sqrt [3]{x}} \left (8 a^2-4 a b \sqrt [3]{x}+3 b^2 x^{2/3}\right )}{5 b^3} \] Input:
Integrate[1/Sqrt[a + b*x^(1/3)],x]
Output:
(2*Sqrt[a + b*x^(1/3)]*(8*a^2 - 4*a*b*x^(1/3) + 3*b^2*x^(2/3)))/(5*b^3)
Time = 0.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {774, 53, 2009}
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.
\(\displaystyle \int \frac {1}{\sqrt {a+b \sqrt [3]{x}}} \, dx\) |
\(\Big \downarrow \) 774 |
\(\displaystyle 3 \int \frac {x^{2/3}}{\sqrt {a+b \sqrt [3]{x}}}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle 3 \int \left (\frac {a^2}{b^2 \sqrt {a+b \sqrt [3]{x}}}-\frac {2 \sqrt {a+b \sqrt [3]{x}} a}{b^2}+\frac {\left (a+b \sqrt [3]{x}\right )^{3/2}}{b^2}\right )d\sqrt [3]{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \left (\frac {2 a^2 \sqrt {a+b \sqrt [3]{x}}}{b^3}+\frac {2 \left (a+b \sqrt [3]{x}\right )^{5/2}}{5 b^3}-\frac {4 a \left (a+b \sqrt [3]{x}\right )^{3/2}}{3 b^3}\right )\) |
Input:
Int[1/Sqrt[a + b*x^(1/3)],x]
Output:
3*((2*a^2*Sqrt[a + b*x^(1/3)])/b^3 - (4*a*(a + b*x^(1/3))^(3/2))/(3*b^3) + (2*(a + b*x^(1/3))^(5/2))/(5*b^3))
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
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]
Time = 0.37 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.70
method | result | size |
derivativedivides | \(\frac {\frac {6 \left (a +b \,x^{\frac {1}{3}}\right )^{\frac {5}{2}}}{5}-4 a \left (a +b \,x^{\frac {1}{3}}\right )^{\frac {3}{2}}+6 a^{2} \sqrt {a +b \,x^{\frac {1}{3}}}}{b^{3}}\) | \(43\) |
default | \(\frac {\frac {6 \left (a +b \,x^{\frac {1}{3}}\right )^{\frac {5}{2}}}{5}-4 a \left (a +b \,x^{\frac {1}{3}}\right )^{\frac {3}{2}}+6 a^{2} \sqrt {a +b \,x^{\frac {1}{3}}}}{b^{3}}\) | \(43\) |
Input:
int(1/(a+b*x^(1/3))^(1/2),x,method=_RETURNVERBOSE)
Output:
6/b^3*(1/5*(a+b*x^(1/3))^(5/2)-2/3*a*(a+b*x^(1/3))^(3/2)+a^2*(a+b*x^(1/3)) ^(1/2))
Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.57 \[ \int \frac {1}{\sqrt {a+b \sqrt [3]{x}}} \, dx=\frac {2 \, {\left (3 \, b^{2} x^{\frac {2}{3}} - 4 \, a b x^{\frac {1}{3}} + 8 \, a^{2}\right )} \sqrt {b x^{\frac {1}{3}} + a}}{5 \, b^{3}} \] Input:
integrate(1/(a+b*x^(1/3))^(1/2),x, algorithm="fricas")
Output:
2/5*(3*b^2*x^(2/3) - 4*a*b*x^(1/3) + 8*a^2)*sqrt(b*x^(1/3) + a)/b^3
Leaf count of result is larger than twice the leaf count of optimal. 726 vs. \(2 (56) = 112\).
Time = 0.95 (sec) , antiderivative size = 726, normalized size of antiderivative = 11.90 \[ \int \frac {1}{\sqrt {a+b \sqrt [3]{x}}} \, dx=\frac {16 a^{\frac {21}{2}} x^{3} \sqrt {1 + \frac {b \sqrt [3]{x}}{a}}}{5 a^{8} b^{3} x^{3} + 15 a^{7} b^{4} x^{\frac {10}{3}} + 15 a^{6} b^{5} x^{\frac {11}{3}} + 5 a^{5} b^{6} x^{4}} - \frac {16 a^{\frac {21}{2}} x^{3}}{5 a^{8} b^{3} x^{3} + 15 a^{7} b^{4} x^{\frac {10}{3}} + 15 a^{6} b^{5} x^{\frac {11}{3}} + 5 a^{5} b^{6} x^{4}} + \frac {40 a^{\frac {19}{2}} b x^{\frac {10}{3}} \sqrt {1 + \frac {b \sqrt [3]{x}}{a}}}{5 a^{8} b^{3} x^{3} + 15 a^{7} b^{4} x^{\frac {10}{3}} + 15 a^{6} b^{5} x^{\frac {11}{3}} + 5 a^{5} b^{6} x^{4}} - \frac {48 a^{\frac {19}{2}} b x^{\frac {10}{3}}}{5 a^{8} b^{3} x^{3} + 15 a^{7} b^{4} x^{\frac {10}{3}} + 15 a^{6} b^{5} x^{\frac {11}{3}} + 5 a^{5} b^{6} x^{4}} + \frac {30 a^{\frac {17}{2}} b^{2} x^{\frac {11}{3}} \sqrt {1 + \frac {b \sqrt [3]{x}}{a}}}{5 a^{8} b^{3} x^{3} + 15 a^{7} b^{4} x^{\frac {10}{3}} + 15 a^{6} b^{5} x^{\frac {11}{3}} + 5 a^{5} b^{6} x^{4}} - \frac {48 a^{\frac {17}{2}} b^{2} x^{\frac {11}{3}}}{5 a^{8} b^{3} x^{3} + 15 a^{7} b^{4} x^{\frac {10}{3}} + 15 a^{6} b^{5} x^{\frac {11}{3}} + 5 a^{5} b^{6} x^{4}} + \frac {10 a^{\frac {15}{2}} b^{3} x^{4} \sqrt {1 + \frac {b \sqrt [3]{x}}{a}}}{5 a^{8} b^{3} x^{3} + 15 a^{7} b^{4} x^{\frac {10}{3}} + 15 a^{6} b^{5} x^{\frac {11}{3}} + 5 a^{5} b^{6} x^{4}} - \frac {16 a^{\frac {15}{2}} b^{3} x^{4}}{5 a^{8} b^{3} x^{3} + 15 a^{7} b^{4} x^{\frac {10}{3}} + 15 a^{6} b^{5} x^{\frac {11}{3}} + 5 a^{5} b^{6} x^{4}} + \frac {10 a^{\frac {13}{2}} b^{4} x^{\frac {13}{3}} \sqrt {1 + \frac {b \sqrt [3]{x}}{a}}}{5 a^{8} b^{3} x^{3} + 15 a^{7} b^{4} x^{\frac {10}{3}} + 15 a^{6} b^{5} x^{\frac {11}{3}} + 5 a^{5} b^{6} x^{4}} + \frac {6 a^{\frac {11}{2}} b^{5} x^{\frac {14}{3}} \sqrt {1 + \frac {b \sqrt [3]{x}}{a}}}{5 a^{8} b^{3} x^{3} + 15 a^{7} b^{4} x^{\frac {10}{3}} + 15 a^{6} b^{5} x^{\frac {11}{3}} + 5 a^{5} b^{6} x^{4}} \] Input:
integrate(1/(a+b*x**(1/3))**(1/2),x)
Output:
16*a**(21/2)*x**3*sqrt(1 + b*x**(1/3)/a)/(5*a**8*b**3*x**3 + 15*a**7*b**4* x**(10/3) + 15*a**6*b**5*x**(11/3) + 5*a**5*b**6*x**4) - 16*a**(21/2)*x**3 /(5*a**8*b**3*x**3 + 15*a**7*b**4*x**(10/3) + 15*a**6*b**5*x**(11/3) + 5*a **5*b**6*x**4) + 40*a**(19/2)*b*x**(10/3)*sqrt(1 + b*x**(1/3)/a)/(5*a**8*b **3*x**3 + 15*a**7*b**4*x**(10/3) + 15*a**6*b**5*x**(11/3) + 5*a**5*b**6*x **4) - 48*a**(19/2)*b*x**(10/3)/(5*a**8*b**3*x**3 + 15*a**7*b**4*x**(10/3) + 15*a**6*b**5*x**(11/3) + 5*a**5*b**6*x**4) + 30*a**(17/2)*b**2*x**(11/3 )*sqrt(1 + b*x**(1/3)/a)/(5*a**8*b**3*x**3 + 15*a**7*b**4*x**(10/3) + 15*a **6*b**5*x**(11/3) + 5*a**5*b**6*x**4) - 48*a**(17/2)*b**2*x**(11/3)/(5*a* *8*b**3*x**3 + 15*a**7*b**4*x**(10/3) + 15*a**6*b**5*x**(11/3) + 5*a**5*b* *6*x**4) + 10*a**(15/2)*b**3*x**4*sqrt(1 + b*x**(1/3)/a)/(5*a**8*b**3*x**3 + 15*a**7*b**4*x**(10/3) + 15*a**6*b**5*x**(11/3) + 5*a**5*b**6*x**4) - 1 6*a**(15/2)*b**3*x**4/(5*a**8*b**3*x**3 + 15*a**7*b**4*x**(10/3) + 15*a**6 *b**5*x**(11/3) + 5*a**5*b**6*x**4) + 10*a**(13/2)*b**4*x**(13/3)*sqrt(1 + b*x**(1/3)/a)/(5*a**8*b**3*x**3 + 15*a**7*b**4*x**(10/3) + 15*a**6*b**5*x **(11/3) + 5*a**5*b**6*x**4) + 6*a**(11/2)*b**5*x**(14/3)*sqrt(1 + b*x**(1 /3)/a)/(5*a**8*b**3*x**3 + 15*a**7*b**4*x**(10/3) + 15*a**6*b**5*x**(11/3) + 5*a**5*b**6*x**4)
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.77 \[ \int \frac {1}{\sqrt {a+b \sqrt [3]{x}}} \, dx=\frac {6 \, {\left (b x^{\frac {1}{3}} + a\right )}^{\frac {5}{2}}}{5 \, b^{3}} - \frac {4 \, {\left (b x^{\frac {1}{3}} + a\right )}^{\frac {3}{2}} a}{b^{3}} + \frac {6 \, \sqrt {b x^{\frac {1}{3}} + a} a^{2}}{b^{3}} \] Input:
integrate(1/(a+b*x^(1/3))^(1/2),x, algorithm="maxima")
Output:
6/5*(b*x^(1/3) + a)^(5/2)/b^3 - 4*(b*x^(1/3) + a)^(3/2)*a/b^3 + 6*sqrt(b*x ^(1/3) + a)*a^2/b^3
Time = 0.12 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\sqrt {a+b \sqrt [3]{x}}} \, dx=\frac {2 \, {\left (3 \, {\left (b x^{\frac {1}{3}} + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x^{\frac {1}{3}} + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x^{\frac {1}{3}} + a} a^{2}\right )}}{5 \, b^{3}} \] Input:
integrate(1/(a+b*x^(1/3))^(1/2),x, algorithm="giac")
Output:
2/5*(3*(b*x^(1/3) + a)^(5/2) - 10*(b*x^(1/3) + a)^(3/2)*a + 15*sqrt(b*x^(1 /3) + a)*a^2)/b^3
Time = 0.17 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.61 \[ \int \frac {1}{\sqrt {a+b \sqrt [3]{x}}} \, dx=\frac {x\,\sqrt {\frac {b\,x^{1/3}}{a}+1}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},3;\ 4;\ -\frac {b\,x^{1/3}}{a}\right )}{\sqrt {a+b\,x^{1/3}}} \] Input:
int(1/(a + b*x^(1/3))^(1/2),x)
Output:
(x*((b*x^(1/3))/a + 1)^(1/2)*hypergeom([1/2, 3], 4, -(b*x^(1/3))/a))/(a + b*x^(1/3))^(1/2)
Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.56 \[ \int \frac {1}{\sqrt {a+b \sqrt [3]{x}}} \, dx=\frac {2 \sqrt {x^{\frac {1}{3}} b +a}\, \left (3 x^{\frac {2}{3}} b^{2}-4 x^{\frac {1}{3}} a b +8 a^{2}\right )}{5 b^{3}} \] Input:
int(1/(a+b*x^(1/3))^(1/2),x)
Output:
(2*sqrt(x**(1/3)*b + a)*(3*x**(2/3)*b**2 - 4*x**(1/3)*a*b + 8*a**2))/(5*b* *3)