Integrand size = 11, antiderivative size = 74 \[ \int \frac {1}{\left (a+\frac {b}{x^{2/3}}\right )^2} \, dx=-\frac {6 b \sqrt [3]{x}}{a^3}-\frac {3 b^2 \sqrt [3]{x}}{2 a^3 \left (b+a x^{2/3}\right )}+\frac {x}{a^2}+\frac {15 b^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt [3]{x}}{\sqrt {b}}\right )}{2 a^{7/2}} \] Output:
-6*b*x^(1/3)/a^3-3/2*b^2*x^(1/3)/a^3/(b+a*x^(2/3))+x/a^2+15/2*b^(3/2)*arct an(a^(1/2)*x^(1/3)/b^(1/2))/a^(7/2)
Time = 0.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\left (a+\frac {b}{x^{2/3}}\right )^2} \, dx=\frac {-15 b^2 \sqrt [3]{x}-10 a b x+2 a^2 x^{5/3}}{2 a^3 \left (b+a x^{2/3}\right )}+\frac {15 b^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt [3]{x}}{\sqrt {b}}\right )}{2 a^{7/2}} \] Input:
Integrate[(a + b/x^(2/3))^(-2),x]
Output:
(-15*b^2*x^(1/3) - 10*a*b*x + 2*a^2*x^(5/3))/(2*a^3*(b + a*x^(2/3))) + (15 *b^(3/2)*ArcTan[(Sqrt[a]*x^(1/3))/Sqrt[b]])/(2*a^(7/2))
Time = 0.33 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {774, 795, 252, 254, 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.
\(\displaystyle \int \frac {1}{\left (a+\frac {b}{x^{2/3}}\right )^2} \, dx\) |
\(\Big \downarrow \) 774 |
\(\displaystyle 3 \int \frac {x^{2/3}}{\left (a+\frac {b}{x^{2/3}}\right )^2}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 795 |
\(\displaystyle 3 \int \frac {x^2}{\left (x^{2/3} a+b\right )^2}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle 3 \left (\frac {5 \int \frac {x^{4/3}}{x^{2/3} a+b}d\sqrt [3]{x}}{2 a}-\frac {x^{5/3}}{2 a \left (a x^{2/3}+b\right )}\right )\) |
\(\Big \downarrow \) 254 |
\(\displaystyle 3 \left (\frac {5 \int \left (\frac {b^2}{a^2 \left (x^{2/3} a+b\right )}-\frac {b}{a^2}+\frac {x^{2/3}}{a}\right )d\sqrt [3]{x}}{2 a}-\frac {x^{5/3}}{2 a \left (a x^{2/3}+b\right )}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \left (\frac {5 \left (\frac {b^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt [3]{x}}{\sqrt {b}}\right )}{a^{5/2}}-\frac {b \sqrt [3]{x}}{a^2}+\frac {x}{3 a}\right )}{2 a}-\frac {x^{5/3}}{2 a \left (a x^{2/3}+b\right )}\right )\) |
Input:
Int[(a + b/x^(2/3))^(-2),x]
Output:
3*(-1/2*x^(5/3)/(a*(b + a*x^(2/3))) + (5*(-((b*x^(1/3))/a^2) + x/(3*a) + ( b^(3/2)*ArcTan[(Sqrt[a]*x^(1/3))/Sqrt[b]])/a^(5/2)))/(2*a))
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
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] :> Int[x^(m + n*p)* (b + a/x^n)^p, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && NegQ[n]
Time = 0.36 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.80
method | result | size |
derivativedivides | \(\frac {a x -6 b \,x^{\frac {1}{3}}}{a^{3}}+\frac {3 b^{2} \left (-\frac {x^{\frac {1}{3}}}{2 \left (b +a \,x^{\frac {2}{3}}\right )}+\frac {5 \arctan \left (\frac {a \,x^{\frac {1}{3}}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}\) | \(59\) |
default | \(\frac {a x -6 b \,x^{\frac {1}{3}}}{a^{3}}+\frac {3 b^{2} \left (-\frac {x^{\frac {1}{3}}}{2 \left (b +a \,x^{\frac {2}{3}}\right )}+\frac {5 \arctan \left (\frac {a \,x^{\frac {1}{3}}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}\) | \(59\) |
Input:
int(1/(a+b/x^(2/3))^2,x,method=_RETURNVERBOSE)
Output:
3/a^3*(1/3*a*x-2*b*x^(1/3))+3*b^2/a^3*(-1/2*x^(1/3)/(b+a*x^(2/3))+5/2/(a*b )^(1/2)*arctan(a*x^(1/3)/(a*b)^(1/2)))
Time = 0.08 (sec) , antiderivative size = 285, normalized size of antiderivative = 3.85 \[ \int \frac {1}{\left (a+\frac {b}{x^{2/3}}\right )^2} \, dx=\left [\frac {4 \, a^{4} x^{3} - 6 \, a^{2} b^{2} x^{\frac {5}{3}} + 10 \, a b^{3} x + 15 \, {\left (a^{3} b x^{2} + b^{4}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {a^{3} x^{2} - 2 \, a^{2} b x \sqrt {-\frac {b}{a}} - b^{3} + 2 \, {\left (a^{3} x \sqrt {-\frac {b}{a}} + a b^{2}\right )} x^{\frac {2}{3}} - 2 \, {\left (a^{2} b x - a b^{2} \sqrt {-\frac {b}{a}}\right )} x^{\frac {1}{3}}}{a^{3} x^{2} + b^{3}}\right ) - 6 \, {\left (4 \, a^{3} b x^{2} + 5 \, b^{4}\right )} x^{\frac {1}{3}}}{4 \, {\left (a^{6} x^{2} + a^{3} b^{3}\right )}}, \frac {2 \, a^{4} x^{3} - 3 \, a^{2} b^{2} x^{\frac {5}{3}} + 5 \, a b^{3} x + 15 \, {\left (a^{3} b x^{2} + b^{4}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a x^{\frac {1}{3}} \sqrt {\frac {b}{a}}}{b}\right ) - 3 \, {\left (4 \, a^{3} b x^{2} + 5 \, b^{4}\right )} x^{\frac {1}{3}}}{2 \, {\left (a^{6} x^{2} + a^{3} b^{3}\right )}}\right ] \] Input:
integrate(1/(a+b/x^(2/3))^2,x, algorithm="fricas")
Output:
[1/4*(4*a^4*x^3 - 6*a^2*b^2*x^(5/3) + 10*a*b^3*x + 15*(a^3*b*x^2 + b^4)*sq rt(-b/a)*log((a^3*x^2 - 2*a^2*b*x*sqrt(-b/a) - b^3 + 2*(a^3*x*sqrt(-b/a) + a*b^2)*x^(2/3) - 2*(a^2*b*x - a*b^2*sqrt(-b/a))*x^(1/3))/(a^3*x^2 + b^3)) - 6*(4*a^3*b*x^2 + 5*b^4)*x^(1/3))/(a^6*x^2 + a^3*b^3), 1/2*(2*a^4*x^3 - 3*a^2*b^2*x^(5/3) + 5*a*b^3*x + 15*(a^3*b*x^2 + b^4)*sqrt(b/a)*arctan(a*x^ (1/3)*sqrt(b/a)/b) - 3*(4*a^3*b*x^2 + 5*b^4)*x^(1/3))/(a^6*x^2 + a^3*b^3)]
Leaf count of result is larger than twice the leaf count of optimal. 410 vs. \(2 (70) = 140\).
Time = 1.88 (sec) , antiderivative size = 410, normalized size of antiderivative = 5.54 \[ \int \frac {1}{\left (a+\frac {b}{x^{2/3}}\right )^2} \, dx=\begin {cases} \tilde {\infty } x^{\frac {7}{3}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {3 x^{\frac {7}{3}}}{7 b^{2}} & \text {for}\: a = 0 \\\frac {x}{a^{2}} & \text {for}\: b = 0 \\\frac {4 a^{3} x^{\frac {5}{3}} \sqrt {- \frac {b}{a}}}{4 a^{5} x^{\frac {2}{3}} \sqrt {- \frac {b}{a}} + 4 a^{4} b \sqrt {- \frac {b}{a}}} - \frac {20 a^{2} b x \sqrt {- \frac {b}{a}}}{4 a^{5} x^{\frac {2}{3}} \sqrt {- \frac {b}{a}} + 4 a^{4} b \sqrt {- \frac {b}{a}}} + \frac {15 a b^{2} x^{\frac {2}{3}} \log {\left (\sqrt [3]{x} - \sqrt {- \frac {b}{a}} \right )}}{4 a^{5} x^{\frac {2}{3}} \sqrt {- \frac {b}{a}} + 4 a^{4} b \sqrt {- \frac {b}{a}}} - \frac {15 a b^{2} x^{\frac {2}{3}} \log {\left (\sqrt [3]{x} + \sqrt {- \frac {b}{a}} \right )}}{4 a^{5} x^{\frac {2}{3}} \sqrt {- \frac {b}{a}} + 4 a^{4} b \sqrt {- \frac {b}{a}}} - \frac {30 a b^{2} \sqrt [3]{x} \sqrt {- \frac {b}{a}}}{4 a^{5} x^{\frac {2}{3}} \sqrt {- \frac {b}{a}} + 4 a^{4} b \sqrt {- \frac {b}{a}}} + \frac {15 b^{3} \log {\left (\sqrt [3]{x} - \sqrt {- \frac {b}{a}} \right )}}{4 a^{5} x^{\frac {2}{3}} \sqrt {- \frac {b}{a}} + 4 a^{4} b \sqrt {- \frac {b}{a}}} - \frac {15 b^{3} \log {\left (\sqrt [3]{x} + \sqrt {- \frac {b}{a}} \right )}}{4 a^{5} x^{\frac {2}{3}} \sqrt {- \frac {b}{a}} + 4 a^{4} b \sqrt {- \frac {b}{a}}} & \text {otherwise} \end {cases} \] Input:
integrate(1/(a+b/x**(2/3))**2,x)
Output:
Piecewise((zoo*x**(7/3), Eq(a, 0) & Eq(b, 0)), (3*x**(7/3)/(7*b**2), Eq(a, 0)), (x/a**2, Eq(b, 0)), (4*a**3*x**(5/3)*sqrt(-b/a)/(4*a**5*x**(2/3)*sqr t(-b/a) + 4*a**4*b*sqrt(-b/a)) - 20*a**2*b*x*sqrt(-b/a)/(4*a**5*x**(2/3)*s qrt(-b/a) + 4*a**4*b*sqrt(-b/a)) + 15*a*b**2*x**(2/3)*log(x**(1/3) - sqrt( -b/a))/(4*a**5*x**(2/3)*sqrt(-b/a) + 4*a**4*b*sqrt(-b/a)) - 15*a*b**2*x**( 2/3)*log(x**(1/3) + sqrt(-b/a))/(4*a**5*x**(2/3)*sqrt(-b/a) + 4*a**4*b*sqr t(-b/a)) - 30*a*b**2*x**(1/3)*sqrt(-b/a)/(4*a**5*x**(2/3)*sqrt(-b/a) + 4*a **4*b*sqrt(-b/a)) + 15*b**3*log(x**(1/3) - sqrt(-b/a))/(4*a**5*x**(2/3)*sq rt(-b/a) + 4*a**4*b*sqrt(-b/a)) - 15*b**3*log(x**(1/3) + sqrt(-b/a))/(4*a* *5*x**(2/3)*sqrt(-b/a) + 4*a**4*b*sqrt(-b/a)), True))
Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\left (a+\frac {b}{x^{2/3}}\right )^2} \, dx=\frac {2 \, a^{2} - \frac {10 \, a b}{x^{\frac {2}{3}}} - \frac {15 \, b^{2}}{x^{\frac {4}{3}}}}{2 \, {\left (\frac {a^{4}}{x} + \frac {a^{3} b}{x^{\frac {5}{3}}}\right )}} - \frac {15 \, b^{2} \arctan \left (\frac {b}{\sqrt {a b} x^{\frac {1}{3}}}\right )}{2 \, \sqrt {a b} a^{3}} \] Input:
integrate(1/(a+b/x^(2/3))^2,x, algorithm="maxima")
Output:
1/2*(2*a^2 - 10*a*b/x^(2/3) - 15*b^2/x^(4/3))/(a^4/x + a^3*b/x^(5/3)) - 15 /2*b^2*arctan(b/(sqrt(a*b)*x^(1/3)))/(sqrt(a*b)*a^3)
Time = 0.12 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (a+\frac {b}{x^{2/3}}\right )^2} \, dx=\frac {15 \, b^{2} \arctan \left (\frac {a x^{\frac {1}{3}}}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3}} - \frac {3 \, b^{2} x^{\frac {1}{3}}}{2 \, {\left (a x^{\frac {2}{3}} + b\right )} a^{3}} + \frac {a^{4} x - 6 \, a^{3} b x^{\frac {1}{3}}}{a^{6}} \] Input:
integrate(1/(a+b/x^(2/3))^2,x, algorithm="giac")
Output:
15/2*b^2*arctan(a*x^(1/3)/sqrt(a*b))/(sqrt(a*b)*a^3) - 3/2*b^2*x^(1/3)/((a *x^(2/3) + b)*a^3) + (a^4*x - 6*a^3*b*x^(1/3))/a^6
Time = 0.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\left (a+\frac {b}{x^{2/3}}\right )^2} \, dx=\frac {x}{a^2}-\frac {3\,b^2\,x^{1/3}}{2\,\left (a^3\,b+a^4\,x^{2/3}\right )}-\frac {6\,b\,x^{1/3}}{a^3}+\frac {15\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,x^{1/3}}{\sqrt {b}}\right )}{2\,a^{7/2}} \] Input:
int(1/(a + b/x^(2/3))^2,x)
Output:
x/a^2 - (3*b^2*x^(1/3))/(2*(a^3*b + a^4*x^(2/3))) - (6*b*x^(1/3))/a^3 + (1 5*b^(3/2)*atan((a^(1/2)*x^(1/3))/b^(1/2)))/(2*a^(7/2))
Time = 0.24 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.18 \[ \int \frac {1}{\left (a+\frac {b}{x^{2/3}}\right )^2} \, dx=\frac {15 x^{\frac {2}{3}} \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {x^{\frac {1}{3}} a}{\sqrt {b}\, \sqrt {a}}\right ) a b +15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {x^{\frac {1}{3}} a}{\sqrt {b}\, \sqrt {a}}\right ) b^{2}+2 x^{\frac {5}{3}} a^{3}-15 x^{\frac {1}{3}} a \,b^{2}-10 a^{2} b x}{2 a^{4} \left (x^{\frac {2}{3}} a +b \right )} \] Input:
int(1/(a+b/x^(2/3))^2,x)
Output:
(15*x**(2/3)*sqrt(b)*sqrt(a)*atan((x**(1/3)*a)/(sqrt(b)*sqrt(a)))*a*b + 15 *sqrt(b)*sqrt(a)*atan((x**(1/3)*a)/(sqrt(b)*sqrt(a)))*b**2 + 2*x**(2/3)*a* *3*x - 15*x**(1/3)*a*b**2 - 10*a**2*b*x)/(2*a**4*(x**(2/3)*a + b))