Integrand size = 11, antiderivative size = 54 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}\right )^3} \, dx=-\frac {3 a^2}{2 b^3 \left (a+b \sqrt [3]{x}\right )^2}+\frac {6 a}{b^3 \left (a+b \sqrt [3]{x}\right )}+\frac {3 \log \left (a+b \sqrt [3]{x}\right )}{b^3} \] Output:
-3/2*a^2/b^3/(a+b*x^(1/3))^2+6*a/b^3/(a+b*x^(1/3))+3*ln(a+b*x^(1/3))/b^3
Time = 0.04 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}\right )^3} \, dx=\frac {3 a \left (3 a+4 b \sqrt [3]{x}\right )}{2 b^3 \left (a+b \sqrt [3]{x}\right )^2}+\frac {3 \log \left (a+b \sqrt [3]{x}\right )}{b^3} \] Input:
Integrate[(a + b*x^(1/3))^(-3),x]
Output:
(3*a*(3*a + 4*b*x^(1/3)))/(2*b^3*(a + b*x^(1/3))^2) + (3*Log[a + b*x^(1/3) ])/b^3
Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {774, 49, 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.
\(\displaystyle \int \frac {1}{\left (a+b \sqrt [3]{x}\right )^3} \, dx\) |
\(\Big \downarrow \) 774 |
\(\displaystyle 3 \int \frac {x^{2/3}}{\left (a+b \sqrt [3]{x}\right )^3}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle 3 \int \left (\frac {a^2}{b^2 \left (a+b \sqrt [3]{x}\right )^3}-\frac {2 a}{b^2 \left (a+b \sqrt [3]{x}\right )^2}+\frac {1}{b^2 \left (a+b \sqrt [3]{x}\right )}\right )d\sqrt [3]{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \left (-\frac {a^2}{2 b^3 \left (a+b \sqrt [3]{x}\right )^2}+\frac {2 a}{b^3 \left (a+b \sqrt [3]{x}\right )}+\frac {\log \left (a+b \sqrt [3]{x}\right )}{b^3}\right )\) |
Input:
Int[(a + b*x^(1/3))^(-3),x]
Output:
3*(-1/2*a^2/(b^3*(a + b*x^(1/3))^2) + (2*a)/(b^3*(a + b*x^(1/3))) + Log[a + b*x^(1/3)]/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}, x] && IGtQ[m, 0] && IGtQ[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 = 0.38 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87
method | result | size |
derivativedivides | \(-\frac {3 a^{2}}{2 b^{3} \left (a +b \,x^{\frac {1}{3}}\right )^{2}}+\frac {6 a}{b^{3} \left (a +b \,x^{\frac {1}{3}}\right )}+\frac {3 \ln \left (a +b \,x^{\frac {1}{3}}\right )}{b^{3}}\) | \(47\) |
default | \(-\frac {a^{6}}{\left (b^{3} x +a^{3}\right )^{2} b^{3}}+\frac {\ln \left (b^{3} x +a^{3}\right )}{b^{3}}+\frac {2 a^{3}}{b^{3} \left (b^{3} x +a^{3}\right )}-3 a \,b^{5} \left (\frac {\frac {-8 a \,b^{2} x +\frac {23 a^{2} b \,x^{\frac {2}{3}}}{2}-10 x^{\frac {1}{3}} a^{3}+\frac {7 a^{4}}{2 b}}{\left (b^{2} x^{\frac {2}{3}}-a b \,x^{\frac {1}{3}}+a^{2}\right )^{2}}+\frac {5 \ln \left (b^{2} x^{\frac {2}{3}}-a b \,x^{\frac {1}{3}}+a^{2}\right )}{2 b}+\frac {5 \sqrt {3}\, \arctan \left (\frac {\left (2 x^{\frac {1}{3}} b^{2}-a b \right ) \sqrt {3}}{3 a b}\right )}{b}}{9 a \,b^{7}}-\frac {4}{9 b^{8} \left (a +b \,x^{\frac {1}{3}}\right )}-\frac {5 \ln \left (a +b \,x^{\frac {1}{3}}\right )}{9 a \,b^{8}}+\frac {a}{18 b^{8} \left (a +b \,x^{\frac {1}{3}}\right )^{2}}\right )+6 a^{2} b^{4} \left (-\frac {\frac {3 a b x +x^{\frac {2}{3}} a^{2}-\frac {a^{3} x^{\frac {1}{3}}}{b}+\frac {5 a^{4}}{2 b^{2}}}{\left (b^{2} x^{\frac {2}{3}}-a b \,x^{\frac {1}{3}}+a^{2}\right )^{2}}+\frac {\frac {\ln \left (b^{2} x^{\frac {2}{3}}-a b \,x^{\frac {1}{3}}+a^{2}\right )}{b}-\frac {2 \sqrt {3}\, \arctan \left (\frac {\left (2 x^{\frac {1}{3}} b^{2}-a b \right ) \sqrt {3}}{3 a b}\right )}{b}}{b}}{9 a^{2} b^{5}}-\frac {1}{18 b^{7} \left (a +b \,x^{\frac {1}{3}}\right )^{2}}+\frac {2 \ln \left (a +b \,x^{\frac {1}{3}}\right )}{9 a^{2} b^{7}}+\frac {1}{3 a \,b^{7} \left (a +b \,x^{\frac {1}{3}}\right )}\right )-7 a^{3} b^{3} \left (\frac {a^{3}}{2 b^{6} \left (b^{3} x +a^{3}\right )^{2}}-\frac {1}{\left (b^{3} x +a^{3}\right ) b^{6}}\right )+6 a^{4} b^{2} \left (\frac {\frac {2 a \,b^{2} x -\frac {5 a^{2} b \,x^{\frac {2}{3}}}{2}+x^{\frac {1}{3}} a^{3}-\frac {a^{4}}{2 b}}{\left (b^{2} x^{\frac {2}{3}}-a b \,x^{\frac {1}{3}}+a^{2}\right )^{2}}+\frac {\ln \left (b^{2} x^{\frac {2}{3}}-a b \,x^{\frac {1}{3}}+a^{2}\right )}{2 b}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{\frac {1}{3}} b^{2}-a b \right ) \sqrt {3}}{3 a b}\right )}{b}}{9 a^{4} b^{4}}-\frac {\ln \left (a +b \,x^{\frac {1}{3}}\right )}{9 a^{4} b^{5}}+\frac {1}{9 a^{3} b^{5} \left (a +b \,x^{\frac {1}{3}}\right )}-\frac {1}{18 a^{2} b^{5} \left (a +b \,x^{\frac {1}{3}}\right )^{2}}\right )-3 a^{5} b \left (-\frac {\frac {-x^{\frac {2}{3}} a^{2}+\frac {a^{3} x^{\frac {1}{3}}}{b}+\frac {a^{4}}{2 b^{2}}}{\left (b^{2} x^{\frac {2}{3}}-a b \,x^{\frac {1}{3}}+a^{2}\right )^{2}}+\frac {\frac {\ln \left (b^{2} x^{\frac {2}{3}}-a b \,x^{\frac {1}{3}}+a^{2}\right )}{2 b}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{\frac {1}{3}} b^{2}-a b \right ) \sqrt {3}}{3 a b}\right )}{b}}{b}}{9 a^{5} b^{2}}+\frac {\ln \left (a +b \,x^{\frac {1}{3}}\right )}{9 a^{5} b^{4}}+\frac {1}{18 a^{3} b^{4} \left (a +b \,x^{\frac {1}{3}}\right )^{2}}\right )\) | \(787\) |
Input:
int(1/(a+b*x^(1/3))^3,x,method=_RETURNVERBOSE)
Output:
-3/2*a^2/b^3/(a+b*x^(1/3))^2+6*a/b^3/(a+b*x^(1/3))+3*ln(a+b*x^(1/3))/b^3
Leaf count of result is larger than twice the leaf count of optimal. 113 vs. \(2 (46) = 92\).
Time = 0.09 (sec) , antiderivative size = 113, normalized size of antiderivative = 2.09 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}\right )^3} \, dx=\frac {3 \, {\left (6 \, a^{3} b^{3} x + 3 \, a^{6} + 2 \, {\left (b^{6} x^{2} + 2 \, a^{3} b^{3} x + a^{6}\right )} \log \left (b x^{\frac {1}{3}} + a\right ) + {\left (4 \, a b^{5} x + a^{4} b^{2}\right )} x^{\frac {2}{3}} - {\left (5 \, a^{2} b^{4} x + 2 \, a^{5} b\right )} x^{\frac {1}{3}}\right )}}{2 \, {\left (b^{9} x^{2} + 2 \, a^{3} b^{6} x + a^{6} b^{3}\right )}} \] Input:
integrate(1/(a+b*x^(1/3))^3,x, algorithm="fricas")
Output:
3/2*(6*a^3*b^3*x + 3*a^6 + 2*(b^6*x^2 + 2*a^3*b^3*x + a^6)*log(b*x^(1/3) + a) + (4*a*b^5*x + a^4*b^2)*x^(2/3) - (5*a^2*b^4*x + 2*a^5*b)*x^(1/3))/(b^ 9*x^2 + 2*a^3*b^6*x + a^6*b^3)
Leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (49) = 98\).
Time = 0.34 (sec) , antiderivative size = 228, normalized size of antiderivative = 4.22 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}\right )^3} \, dx=\begin {cases} \frac {6 a^{2} \log {\left (\frac {a}{b} + \sqrt [3]{x} \right )}}{2 a^{2} b^{3} + 4 a b^{4} \sqrt [3]{x} + 2 b^{5} x^{\frac {2}{3}}} + \frac {9 a^{2}}{2 a^{2} b^{3} + 4 a b^{4} \sqrt [3]{x} + 2 b^{5} x^{\frac {2}{3}}} + \frac {12 a b \sqrt [3]{x} \log {\left (\frac {a}{b} + \sqrt [3]{x} \right )}}{2 a^{2} b^{3} + 4 a b^{4} \sqrt [3]{x} + 2 b^{5} x^{\frac {2}{3}}} + \frac {12 a b \sqrt [3]{x}}{2 a^{2} b^{3} + 4 a b^{4} \sqrt [3]{x} + 2 b^{5} x^{\frac {2}{3}}} + \frac {6 b^{2} x^{\frac {2}{3}} \log {\left (\frac {a}{b} + \sqrt [3]{x} \right )}}{2 a^{2} b^{3} + 4 a b^{4} \sqrt [3]{x} + 2 b^{5} x^{\frac {2}{3}}} & \text {for}\: b \neq 0 \\\frac {x}{a^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(1/(a+b*x**(1/3))**3,x)
Output:
Piecewise((6*a**2*log(a/b + x**(1/3))/(2*a**2*b**3 + 4*a*b**4*x**(1/3) + 2 *b**5*x**(2/3)) + 9*a**2/(2*a**2*b**3 + 4*a*b**4*x**(1/3) + 2*b**5*x**(2/3 )) + 12*a*b*x**(1/3)*log(a/b + x**(1/3))/(2*a**2*b**3 + 4*a*b**4*x**(1/3) + 2*b**5*x**(2/3)) + 12*a*b*x**(1/3)/(2*a**2*b**3 + 4*a*b**4*x**(1/3) + 2* b**5*x**(2/3)) + 6*b**2*x**(2/3)*log(a/b + x**(1/3))/(2*a**2*b**3 + 4*a*b* *4*x**(1/3) + 2*b**5*x**(2/3)), Ne(b, 0)), (x/a**3, True))
Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}\right )^3} \, dx=\frac {3 \, \log \left (b x^{\frac {1}{3}} + a\right )}{b^{3}} + \frac {6 \, a}{{\left (b x^{\frac {1}{3}} + a\right )} b^{3}} - \frac {3 \, a^{2}}{2 \, {\left (b x^{\frac {1}{3}} + a\right )}^{2} b^{3}} \] Input:
integrate(1/(a+b*x^(1/3))^3,x, algorithm="maxima")
Output:
3*log(b*x^(1/3) + a)/b^3 + 6*a/((b*x^(1/3) + a)*b^3) - 3/2*a^2/((b*x^(1/3) + a)^2*b^3)
Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}\right )^3} \, dx=\frac {3 \, \log \left ({\left | b x^{\frac {1}{3}} + a \right |}\right )}{b^{3}} + \frac {3 \, {\left (4 \, a x^{\frac {1}{3}} + \frac {3 \, a^{2}}{b}\right )}}{2 \, {\left (b x^{\frac {1}{3}} + a\right )}^{2} b^{2}} \] Input:
integrate(1/(a+b*x^(1/3))^3,x, algorithm="giac")
Output:
3*log(abs(b*x^(1/3) + a))/b^3 + 3/2*(4*a*x^(1/3) + 3*a^2/b)/((b*x^(1/3) + a)^2*b^2)
Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}\right )^3} \, dx=\frac {\frac {9\,a^2}{2\,b^3}+\frac {6\,a\,x^{1/3}}{b^2}}{a^2+b^2\,x^{2/3}+2\,a\,b\,x^{1/3}}+\frac {3\,\ln \left (a+b\,x^{1/3}\right )}{b^3} \] Input:
int(1/(a + b*x^(1/3))^3,x)
Output:
((9*a^2)/(2*b^3) + (6*a*x^(1/3))/b^2)/(a^2 + b^2*x^(2/3) + 2*a*b*x^(1/3)) + (3*log(a + b*x^(1/3)))/b^3
Time = 0.22 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.50 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}\right )^3} \, dx=\frac {3 x^{\frac {2}{3}} \mathrm {log}\left (x^{\frac {1}{3}} b +a \right ) b^{2}-3 x^{\frac {2}{3}} b^{2}+6 x^{\frac {1}{3}} \mathrm {log}\left (x^{\frac {1}{3}} b +a \right ) a b +3 \,\mathrm {log}\left (x^{\frac {1}{3}} b +a \right ) a^{2}+\frac {3 a^{2}}{2}}{b^{3} \left (x^{\frac {2}{3}} b^{2}+2 x^{\frac {1}{3}} a b +a^{2}\right )} \] Input:
int(1/(a+b*x^(1/3))^3,x)
Output:
(3*(2*x**(2/3)*log(x**(1/3)*b + a)*b**2 - 2*x**(2/3)*b**2 + 4*x**(1/3)*log (x**(1/3)*b + a)*a*b + 2*log(x**(1/3)*b + a)*a**2 + a**2))/(2*b**3*(x**(2/ 3)*b**2 + 2*x**(1/3)*a*b + a**2))