Integrand size = 11, antiderivative size = 100 \[ \int \frac {1}{\left (a+\frac {b}{x^{2/3}}\right )^3} \, dx=-\frac {9 b \sqrt [3]{x}}{a^4}+\frac {3 b^3 \sqrt [3]{x}}{4 a^4 \left (b+a x^{2/3}\right )^2}-\frac {39 b^2 \sqrt [3]{x}}{8 a^4 \left (b+a x^{2/3}\right )}+\frac {x}{a^3}+\frac {105 b^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt [3]{x}}{\sqrt {b}}\right )}{8 a^{9/2}} \] Output:
-9*b*x^(1/3)/a^4+3/4*b^3*x^(1/3)/a^4/(b+a*x^(2/3))^2-39/8*b^2*x^(1/3)/a^4/ (b+a*x^(2/3))+x/a^3+105/8*b^(3/2)*arctan(a^(1/2)*x^(1/3)/b^(1/2))/a^(9/2)
Time = 0.14 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\left (a+\frac {b}{x^{2/3}}\right )^3} \, dx=\frac {-105 b^3 \sqrt [3]{x}-175 a b^2 x-56 a^2 b x^{5/3}+8 a^3 x^{7/3}}{8 a^4 \left (b+a x^{2/3}\right )^2}+\frac {105 b^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt [3]{x}}{\sqrt {b}}\right )}{8 a^{9/2}} \] Input:
Integrate[(a + b/x^(2/3))^(-3),x]
Output:
(-105*b^3*x^(1/3) - 175*a*b^2*x - 56*a^2*b*x^(5/3) + 8*a^3*x^(7/3))/(8*a^4 *(b + a*x^(2/3))^2) + (105*b^(3/2)*ArcTan[(Sqrt[a]*x^(1/3))/Sqrt[b]])/(8*a ^(9/2))
Time = 0.36 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {774, 795, 252, 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.
\(\displaystyle \int \frac {1}{\left (a+\frac {b}{x^{2/3}}\right )^3} \, dx\) |
\(\Big \downarrow \) 774 |
\(\displaystyle 3 \int \frac {x^{2/3}}{\left (a+\frac {b}{x^{2/3}}\right )^3}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 795 |
\(\displaystyle 3 \int \frac {x^{8/3}}{\left (x^{2/3} a+b\right )^3}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle 3 \left (\frac {7 \int \frac {x^2}{\left (x^{2/3} a+b\right )^2}d\sqrt [3]{x}}{4 a}-\frac {x^{7/3}}{4 a \left (a x^{2/3}+b\right )^2}\right )\) |
\(\Big \downarrow \) 252 |
\(\displaystyle 3 \left (\frac {7 \left (\frac {5 \int \frac {x^{4/3}}{x^{2/3} a+b}d\sqrt [3]{x}}{2 a}-\frac {x^{5/3}}{2 a \left (a x^{2/3}+b\right )}\right )}{4 a}-\frac {x^{7/3}}{4 a \left (a x^{2/3}+b\right )^2}\right )\) |
\(\Big \downarrow \) 254 |
\(\displaystyle 3 \left (\frac {7 \left (\frac {5 \int \left (\frac {b^2}{a^2 \left (x^{2/3} a+b\right )}-\frac {b}{a^2}+\frac {x^{2/3}}{a}\right )d\sqrt [3]{x}}{2 a}-\frac {x^{5/3}}{2 a \left (a x^{2/3}+b\right )}\right )}{4 a}-\frac {x^{7/3}}{4 a \left (a x^{2/3}+b\right )^2}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \left (\frac {7 \left (\frac {5 \left (\frac {b^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt [3]{x}}{\sqrt {b}}\right )}{a^{5/2}}-\frac {b \sqrt [3]{x}}{a^2}+\frac {x}{3 a}\right )}{2 a}-\frac {x^{5/3}}{2 a \left (a x^{2/3}+b\right )}\right )}{4 a}-\frac {x^{7/3}}{4 a \left (a x^{2/3}+b\right )^2}\right )\) |
Input:
Int[(a + b/x^(2/3))^(-3),x]
Output:
3*(-1/4*x^(7/3)/(a*(b + a*x^(2/3))^2) + (7*(-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)))/(4*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 = 0.37 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.66
method | result | size |
derivativedivides | \(\frac {a x -9 b \,x^{\frac {1}{3}}}{a^{4}}+\frac {3 b^{2} \left (\frac {-\frac {13 a x}{8}-\frac {11 b \,x^{\frac {1}{3}}}{8}}{\left (b +a \,x^{\frac {2}{3}}\right )^{2}}+\frac {35 \arctan \left (\frac {a \,x^{\frac {1}{3}}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{4}}\) | \(66\) |
default | \(\frac {a x -9 b \,x^{\frac {1}{3}}}{a^{4}}+\frac {3 b^{2} \left (\frac {-\frac {13 a x}{8}-\frac {11 b \,x^{\frac {1}{3}}}{8}}{\left (b +a \,x^{\frac {2}{3}}\right )^{2}}+\frac {35 \arctan \left (\frac {a \,x^{\frac {1}{3}}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{4}}\) | \(66\) |
Input:
int(1/(a+b/x^(2/3))^3,x,method=_RETURNVERBOSE)
Output:
3/a^4*(1/3*a*x-3*b*x^(1/3))+3/a^4*b^2*((-13/8*a*x-11/8*b*x^(1/3))/(b+a*x^( 2/3))^2+35/8/(a*b)^(1/2)*arctan(a*x^(1/3)/(a*b)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (74) = 148\).
Time = 0.09 (sec) , antiderivative size = 403, normalized size of antiderivative = 4.03 \[ \int \frac {1}{\left (a+\frac {b}{x^{2/3}}\right )^3} \, dx=\left [\frac {16 \, a^{7} x^{5} + 122 \, a^{4} b^{3} x^{3} + 70 \, a b^{6} x + 105 \, {\left (a^{6} b x^{4} + 2 \, a^{3} b^{4} x^{2} + b^{7}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {a^{3} x^{2} - 2 \, a^{2} b x \sqrt {-\frac {b}{a}} - b^{3} + 2 \, {\left (a^{3} x \sqrt {-\frac {b}{a}} + a b^{2}\right )} x^{\frac {2}{3}} - 2 \, {\left (a^{2} b x - a b^{2} \sqrt {-\frac {b}{a}}\right )} x^{\frac {1}{3}}}{a^{3} x^{2} + b^{3}}\right ) - 6 \, {\left (13 \, a^{5} b^{2} x^{3} + 7 \, a^{2} b^{5} x\right )} x^{\frac {2}{3}} - 6 \, {\left (24 \, a^{6} b x^{4} + 65 \, a^{3} b^{4} x^{2} + 35 \, b^{7}\right )} x^{\frac {1}{3}}}{16 \, {\left (a^{10} x^{4} + 2 \, a^{7} b^{3} x^{2} + a^{4} b^{6}\right )}}, \frac {8 \, a^{7} x^{5} + 61 \, a^{4} b^{3} x^{3} + 35 \, a b^{6} x + 105 \, {\left (a^{6} b x^{4} + 2 \, a^{3} b^{4} x^{2} + b^{7}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a x^{\frac {1}{3}} \sqrt {\frac {b}{a}}}{b}\right ) - 3 \, {\left (13 \, a^{5} b^{2} x^{3} + 7 \, a^{2} b^{5} x\right )} x^{\frac {2}{3}} - 3 \, {\left (24 \, a^{6} b x^{4} + 65 \, a^{3} b^{4} x^{2} + 35 \, b^{7}\right )} x^{\frac {1}{3}}}{8 \, {\left (a^{10} x^{4} + 2 \, a^{7} b^{3} x^{2} + a^{4} b^{6}\right )}}\right ] \] Input:
integrate(1/(a+b/x^(2/3))^3,x, algorithm="fricas")
Output:
[1/16*(16*a^7*x^5 + 122*a^4*b^3*x^3 + 70*a*b^6*x + 105*(a^6*b*x^4 + 2*a^3* b^4*x^2 + b^7)*sqrt(-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*(13*a^5*b^2*x^3 + 7*a^2*b^5*x)*x^(2/3) - 6*(24*a^6*b *x^4 + 65*a^3*b^4*x^2 + 35*b^7)*x^(1/3))/(a^10*x^4 + 2*a^7*b^3*x^2 + a^4*b ^6), 1/8*(8*a^7*x^5 + 61*a^4*b^3*x^3 + 35*a*b^6*x + 105*(a^6*b*x^4 + 2*a^3 *b^4*x^2 + b^7)*sqrt(b/a)*arctan(a*x^(1/3)*sqrt(b/a)/b) - 3*(13*a^5*b^2*x^ 3 + 7*a^2*b^5*x)*x^(2/3) - 3*(24*a^6*b*x^4 + 65*a^3*b^4*x^2 + 35*b^7)*x^(1 /3))/(a^10*x^4 + 2*a^7*b^3*x^2 + a^4*b^6)]
Leaf count of result is larger than twice the leaf count of optimal. 808 vs. \(2 (95) = 190\).
Time = 7.09 (sec) , antiderivative size = 808, normalized size of antiderivative = 8.08 \[ \int \frac {1}{\left (a+\frac {b}{x^{2/3}}\right )^3} \, dx =\text {Too large to display} \] Input:
integrate(1/(a+b/x**(2/3))**3,x)
Output:
Piecewise((zoo*x**3, Eq(a, 0) & Eq(b, 0)), (x**3/(3*b**3), Eq(a, 0)), (x/a **3, Eq(b, 0)), (16*a**4*x**(7/3)*sqrt(-b/a)/(16*a**7*x**(4/3)*sqrt(-b/a) + 32*a**6*b*x**(2/3)*sqrt(-b/a) + 16*a**5*b**2*sqrt(-b/a)) - 112*a**3*b*x* *(5/3)*sqrt(-b/a)/(16*a**7*x**(4/3)*sqrt(-b/a) + 32*a**6*b*x**(2/3)*sqrt(- b/a) + 16*a**5*b**2*sqrt(-b/a)) + 105*a**2*b**2*x**(4/3)*log(x**(1/3) - sq rt(-b/a))/(16*a**7*x**(4/3)*sqrt(-b/a) + 32*a**6*b*x**(2/3)*sqrt(-b/a) + 1 6*a**5*b**2*sqrt(-b/a)) - 105*a**2*b**2*x**(4/3)*log(x**(1/3) + sqrt(-b/a) )/(16*a**7*x**(4/3)*sqrt(-b/a) + 32*a**6*b*x**(2/3)*sqrt(-b/a) + 16*a**5*b **2*sqrt(-b/a)) - 350*a**2*b**2*x*sqrt(-b/a)/(16*a**7*x**(4/3)*sqrt(-b/a) + 32*a**6*b*x**(2/3)*sqrt(-b/a) + 16*a**5*b**2*sqrt(-b/a)) + 210*a*b**3*x* *(2/3)*log(x**(1/3) - sqrt(-b/a))/(16*a**7*x**(4/3)*sqrt(-b/a) + 32*a**6*b *x**(2/3)*sqrt(-b/a) + 16*a**5*b**2*sqrt(-b/a)) - 210*a*b**3*x**(2/3)*log( x**(1/3) + sqrt(-b/a))/(16*a**7*x**(4/3)*sqrt(-b/a) + 32*a**6*b*x**(2/3)*s qrt(-b/a) + 16*a**5*b**2*sqrt(-b/a)) - 210*a*b**3*x**(1/3)*sqrt(-b/a)/(16* a**7*x**(4/3)*sqrt(-b/a) + 32*a**6*b*x**(2/3)*sqrt(-b/a) + 16*a**5*b**2*sq rt(-b/a)) + 105*b**4*log(x**(1/3) - sqrt(-b/a))/(16*a**7*x**(4/3)*sqrt(-b/ a) + 32*a**6*b*x**(2/3)*sqrt(-b/a) + 16*a**5*b**2*sqrt(-b/a)) - 105*b**4*l og(x**(1/3) + sqrt(-b/a))/(16*a**7*x**(4/3)*sqrt(-b/a) + 32*a**6*b*x**(2/3 )*sqrt(-b/a) + 16*a**5*b**2*sqrt(-b/a)), True))
Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (a+\frac {b}{x^{2/3}}\right )^3} \, dx=\frac {8 \, a^{3} - \frac {56 \, a^{2} b}{x^{\frac {2}{3}}} - \frac {175 \, a b^{2}}{x^{\frac {4}{3}}} - \frac {105 \, b^{3}}{x^{2}}}{8 \, {\left (\frac {a^{6}}{x} + \frac {2 \, a^{5} b}{x^{\frac {5}{3}}} + \frac {a^{4} b^{2}}{x^{\frac {7}{3}}}\right )}} - \frac {105 \, b^{2} \arctan \left (\frac {b}{\sqrt {a b} x^{\frac {1}{3}}}\right )}{8 \, \sqrt {a b} a^{4}} \] Input:
integrate(1/(a+b/x^(2/3))^3,x, algorithm="maxima")
Output:
1/8*(8*a^3 - 56*a^2*b/x^(2/3) - 175*a*b^2/x^(4/3) - 105*b^3/x^2)/(a^6/x + 2*a^5*b/x^(5/3) + a^4*b^2/x^(7/3)) - 105/8*b^2*arctan(b/(sqrt(a*b)*x^(1/3) ))/(sqrt(a*b)*a^4)
Time = 0.13 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\left (a+\frac {b}{x^{2/3}}\right )^3} \, dx=\frac {105 \, b^{2} \arctan \left (\frac {a x^{\frac {1}{3}}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4}} - \frac {3 \, {\left (13 \, a b^{2} x + 11 \, b^{3} x^{\frac {1}{3}}\right )}}{8 \, {\left (a x^{\frac {2}{3}} + b\right )}^{2} a^{4}} + \frac {a^{6} x - 9 \, a^{5} b x^{\frac {1}{3}}}{a^{9}} \] Input:
integrate(1/(a+b/x^(2/3))^3,x, algorithm="giac")
Output:
105/8*b^2*arctan(a*x^(1/3)/sqrt(a*b))/(sqrt(a*b)*a^4) - 3/8*(13*a*b^2*x + 11*b^3*x^(1/3))/((a*x^(2/3) + b)^2*a^4) + (a^6*x - 9*a^5*b*x^(1/3))/a^9
Time = 0.07 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\left (a+\frac {b}{x^{2/3}}\right )^3} \, dx=\frac {x}{a^3}-\frac {\frac {33\,b^3\,x^{1/3}}{8}+\frac {39\,a\,b^2\,x}{8}}{a^4\,b^2+a^6\,x^{4/3}+2\,a^5\,b\,x^{2/3}}-\frac {9\,b\,x^{1/3}}{a^4}+\frac {105\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,x^{1/3}}{\sqrt {b}}\right )}{8\,a^{9/2}} \] Input:
int(1/(a + b/x^(2/3))^3,x)
Output:
x/a^3 - ((33*b^3*x^(1/3))/8 + (39*a*b^2*x)/8)/(a^4*b^2 + a^6*x^(4/3) + 2*a ^5*b*x^(2/3)) - (9*b*x^(1/3))/a^4 + (105*b^(3/2)*atan((a^(1/2)*x^(1/3))/b^ (1/2)))/(8*a^(9/2))
Time = 0.23 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.38 \[ \int \frac {1}{\left (a+\frac {b}{x^{2/3}}\right )^3} \, dx=\frac {210 x^{\frac {2}{3}} \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {x^{\frac {1}{3}} a}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2}+105 x^{\frac {4}{3}} \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {x^{\frac {1}{3}} a}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b +105 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {x^{\frac {1}{3}} a}{\sqrt {b}\, \sqrt {a}}\right ) b^{3}-56 x^{\frac {5}{3}} a^{3} b +8 x^{\frac {7}{3}} a^{4}-105 x^{\frac {1}{3}} a \,b^{3}-175 a^{2} b^{2} x}{8 a^{5} \left (2 x^{\frac {2}{3}} a b +x^{\frac {4}{3}} a^{2}+b^{2}\right )} \] Input:
int(1/(a+b/x^(2/3))^3,x)
Output:
(210*x**(2/3)*sqrt(b)*sqrt(a)*atan((x**(1/3)*a)/(sqrt(b)*sqrt(a)))*a*b**2 + 105*x**(1/3)*sqrt(b)*sqrt(a)*atan((x**(1/3)*a)/(sqrt(b)*sqrt(a)))*a**2*b *x + 105*sqrt(b)*sqrt(a)*atan((x**(1/3)*a)/(sqrt(b)*sqrt(a)))*b**3 - 56*x* *(2/3)*a**3*b*x + 8*x**(1/3)*a**4*x**2 - 105*x**(1/3)*a*b**3 - 175*a**2*b* *2*x)/(8*a**5*(2*x**(2/3)*a*b + x**(1/3)*a**2*x + b**2))