Integrand size = 19, antiderivative size = 61 \[ \int \frac {1}{\left (a \sqrt [3]{x}+b x^{2/3}\right )^{2/3}} \, dx=\frac {9 \left (1+\frac {b \sqrt [3]{x}}{a}\right )^{2/3} x \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {7}{3},\frac {10}{3},-\frac {b \sqrt [3]{x}}{a}\right )}{7 \left (a \sqrt [3]{x}+b x^{2/3}\right )^{2/3}} \] Output:
9/7*(1+b*x^(1/3)/a)^(2/3)*x*hypergeom([2/3, 7/3],[10/3],-b*x^(1/3)/a)/(a*x ^(1/3)+b*x^(2/3))^(2/3)
Time = 10.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a \sqrt [3]{x}+b x^{2/3}\right )^{2/3}} \, dx=\frac {9 \left (1+\frac {b \sqrt [3]{x}}{a}\right )^{2/3} x \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {7}{3},\frac {10}{3},-\frac {b \sqrt [3]{x}}{a}\right )}{7 \left (\left (a+b \sqrt [3]{x}\right ) \sqrt [3]{x}\right )^{2/3}} \] Input:
Integrate[(a*x^(1/3) + b*x^(2/3))^(-2/3),x]
Output:
(9*(1 + (b*x^(1/3))/a)^(2/3)*x*Hypergeometric2F1[2/3, 7/3, 10/3, -((b*x^(1 /3))/a)])/(7*((a + b*x^(1/3))*x^(1/3))^(2/3))
Time = 0.20 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1917, 864, 76, 74}
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.
\(\displaystyle \int \frac {1}{\left (a \sqrt [3]{x}+b x^{2/3}\right )^{2/3}} \, dx\) |
\(\Big \downarrow \) 1917 |
\(\displaystyle \frac {x^{2/9} \left (a+b \sqrt [3]{x}\right )^{2/3} \int \frac {1}{\left (a+b \sqrt [3]{x}\right )^{2/3} x^{2/9}}dx}{\left (a \sqrt [3]{x}+b x^{2/3}\right )^{2/3}}\) |
\(\Big \downarrow \) 864 |
\(\displaystyle \frac {3 x^{2/9} \left (a+b \sqrt [3]{x}\right )^{2/3} \int \frac {x^{4/9}}{\left (a+b \sqrt [3]{x}\right )^{2/3}}d\sqrt [3]{x}}{\left (a \sqrt [3]{x}+b x^{2/3}\right )^{2/3}}\) |
\(\Big \downarrow \) 76 |
\(\displaystyle \frac {3 x^{2/9} \left (\frac {b \sqrt [3]{x}}{a}+1\right )^{2/3} \int \frac {x^{4/9}}{\left (\frac {\sqrt [3]{x} b}{a}+1\right )^{2/3}}d\sqrt [3]{x}}{\left (a \sqrt [3]{x}+b x^{2/3}\right )^{2/3}}\) |
\(\Big \downarrow \) 74 |
\(\displaystyle \frac {9 x \left (\frac {b \sqrt [3]{x}}{a}+1\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {7}{3},\frac {10}{3},-\frac {b \sqrt [3]{x}}{a}\right )}{7 \left (a \sqrt [3]{x}+b x^{2/3}\right )^{2/3}}\) |
Input:
Int[(a*x^(1/3) + b*x^(2/3))^(-2/3),x]
Output:
(9*(1 + (b*x^(1/3))/a)^(2/3)*x*Hypergeometric2F1[2/3, 7/3, 10/3, -((b*x^(1 /3))/a)])/(7*(a*x^(1/3) + b*x^(2/3))^(2/3))
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x )^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] && !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^IntPart [n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) Int[(b*x)^m*(1 + d* (x/c))^n, x], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && !Integer Q[n] && !GtQ[c, 0] && !GtQ[-d/(b*c), 0] && ((RationalQ[m] && !(EqQ[n, -2 ^(-1)] && EqQ[c^2 - d^2, 0])) || !RationalQ[n])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denomi nator[n]}, Simp[k Subst[Int[x^(k*(m + 1) - 1)*(a + b*x^(k*n))^p, x], x, x ^(1/k)], x]] /; FreeQ[{a, b, m, p}, x] && FractionQ[n]
Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]) Int[ x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] && !Integ erQ[p] && NeQ[n, j] && PosQ[n - j]
\[\int \frac {1}{\left (a \,x^{\frac {1}{3}}+b \,x^{\frac {2}{3}}\right )^{\frac {2}{3}}}d x\]
Input:
int(1/(a*x^(1/3)+b*x^(2/3))^(2/3),x)
Output:
int(1/(a*x^(1/3)+b*x^(2/3))^(2/3),x)
Timed out. \[ \int \frac {1}{\left (a \sqrt [3]{x}+b x^{2/3}\right )^{2/3}} \, dx=\text {Timed out} \] Input:
integrate(1/(a*x^(1/3)+b*x^(2/3))^(2/3),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{\left (a \sqrt [3]{x}+b x^{2/3}\right )^{2/3}} \, dx=\int \frac {1}{\left (a \sqrt [3]{x} + b x^{\frac {2}{3}}\right )^{\frac {2}{3}}}\, dx \] Input:
integrate(1/(a*x**(1/3)+b*x**(2/3))**(2/3),x)
Output:
Integral((a*x**(1/3) + b*x**(2/3))**(-2/3), x)
\[ \int \frac {1}{\left (a \sqrt [3]{x}+b x^{2/3}\right )^{2/3}} \, dx=\int { \frac {1}{{\left (b x^{\frac {2}{3}} + a x^{\frac {1}{3}}\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate(1/(a*x^(1/3)+b*x^(2/3))^(2/3),x, algorithm="maxima")
Output:
integrate((b*x^(2/3) + a*x^(1/3))^(-2/3), x)
\[ \int \frac {1}{\left (a \sqrt [3]{x}+b x^{2/3}\right )^{2/3}} \, dx=\int { \frac {1}{{\left (b x^{\frac {2}{3}} + a x^{\frac {1}{3}}\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate(1/(a*x^(1/3)+b*x^(2/3))^(2/3),x, algorithm="giac")
Output:
integrate((b*x^(2/3) + a*x^(1/3))^(-2/3), x)
Time = 11.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.69 \[ \int \frac {1}{\left (a \sqrt [3]{x}+b x^{2/3}\right )^{2/3}} \, dx=\frac {9\,x\,{\left (\frac {b\,x^{1/3}}{a}+1\right )}^{2/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {2}{3},\frac {7}{3};\ \frac {10}{3};\ -\frac {b\,x^{1/3}}{a}\right )}{7\,{\left (a\,x^{1/3}+b\,x^{2/3}\right )}^{2/3}} \] Input:
int(1/(a*x^(1/3) + b*x^(2/3))^(2/3),x)
Output:
(9*x*((b*x^(1/3))/a + 1)^(2/3)*hypergeom([2/3, 7/3], 10/3, -(b*x^(1/3))/a) )/(7*(a*x^(1/3) + b*x^(2/3))^(2/3))
\[ \int \frac {1}{\left (a \sqrt [3]{x}+b x^{2/3}\right )^{2/3}} \, dx=\int \frac {1}{x^{\frac {2}{9}} \left (x^{\frac {1}{3}} b +a \right )^{\frac {2}{3}}}d x \] Input:
int(1/(a*x^(1/3)+b*x^(2/3))^(2/3),x)
Output:
int(1/(x**(2/9)*(x**(1/3)*b + a)**(2/3)),x)