Integrand size = 11, antiderivative size = 46 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^p \, dx=-\frac {3 b^3 \left (a+\frac {b}{\sqrt [3]{x}}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (4,1+p,2+p,1+\frac {b}{a \sqrt [3]{x}}\right )}{a^4 (1+p)} \] Output:
-3*b^3*(a+b/x^(1/3))^(p+1)*hypergeom([4, p+1],[2+p],1+b/a/x^(1/3))/a^4/(p+ 1)
Time = 0.12 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^p \, dx=-\frac {3 b^3 \left (a+\frac {b}{\sqrt [3]{x}}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (4,1+p,2+p,1+\frac {b}{a \sqrt [3]{x}}\right )}{a^4 (1+p)} \] Input:
Integrate[(a + b/x^(1/3))^p,x]
Output:
(-3*b^3*(a + b/x^(1/3))^(1 + p)*Hypergeometric2F1[4, 1 + p, 2 + p, 1 + b/( a*x^(1/3))])/(a^4*(1 + p))
Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {774, 798, 75}
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.
\(\displaystyle \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^p \, dx\) |
\(\Big \downarrow \) 774 |
\(\displaystyle 3 \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^p x^{2/3}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -3 \int \frac {\left (a+\frac {b}{\sqrt [3]{x}}\right )^p}{x^{4/3}}d\frac {1}{\sqrt [3]{x}}\) |
\(\Big \downarrow \) 75 |
\(\displaystyle -\frac {3 b^3 \left (a+\frac {b}{\sqrt [3]{x}}\right )^{p+1} \operatorname {Hypergeometric2F1}\left (4,p+1,p+2,\frac {b}{a \sqrt [3]{x}}+1\right )}{a^4 (p+1)}\) |
Input:
Int[(a + b/x^(1/3))^p,x]
Output:
(-3*b^3*(a + b/x^(1/3))^(1 + p)*Hypergeometric2F1[4, 1 + p, 2 + p, 1 + b/( a*x^(1/3))])/(a^4*(1 + p))
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x )^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[n] && (IntegerQ[m] || GtQ[-d/(b*c), 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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
\[\int \left (a +\frac {b}{x^{\frac {1}{3}}}\right )^{p}d x\]
Input:
int((a+b/x^(1/3))^p,x)
Output:
int((a+b/x^(1/3))^p,x)
\[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^p \, dx=\int { {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{p} \,d x } \] Input:
integrate((a+b/x^(1/3))^p,x, algorithm="fricas")
Output:
integral(((a*x + b*x^(2/3))/x)^p, x)
Result contains complex when optimal does not.
Time = 10.76 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.89 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^p \, dx=\frac {3 b^{p} x^{1 - \frac {p}{3}} \Gamma \left (3 - p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 3 - p \\ 4 - p \end {matrix}\middle | {\frac {a \sqrt [3]{x} e^{i \pi }}{b}} \right )}}{\Gamma \left (4 - p\right )} \] Input:
integrate((a+b/x**(1/3))**p,x)
Output:
3*b**p*x**(1 - p/3)*gamma(3 - p)*hyper((-p, 3 - p), (4 - p,), a*x**(1/3)*e xp_polar(I*pi)/b)/gamma(4 - p)
\[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^p \, dx=\int { {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{p} \,d x } \] Input:
integrate((a+b/x^(1/3))^p,x, algorithm="maxima")
Output:
integrate((a + b/x^(1/3))^p, x)
\[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^p \, dx=\int { {\left (a + \frac {b}{x^{\frac {1}{3}}}\right )}^{p} \,d x } \] Input:
integrate((a+b/x^(1/3))^p,x, algorithm="giac")
Output:
integrate((a + b/x^(1/3))^p, x)
Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.24 \[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^p \, dx=-\frac {x\,{\left (a+\frac {b}{x^{1/3}}\right )}^p\,{{}}_2{\mathrm {F}}_1\left (3-p,-p;\ 4-p;\ -\frac {a\,x^{1/3}}{b}\right )}{\left (\frac {p}{3}-1\right )\,{\left (\frac {a\,x^{1/3}}{b}+1\right )}^p} \] Input:
int((a + b/x^(1/3))^p,x)
Output:
-(x*(a + b/x^(1/3))^p*hypergeom([3 - p, -p], 4 - p, -(a*x^(1/3))/b))/((p/3 - 1)*((a*x^(1/3))/b + 1)^p)
\[ \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^p \, dx=\frac {3 x^{\frac {2}{3}} \left (x^{\frac {1}{3}} a +b \right )^{p} a b p +3 x^{\frac {1}{3}} \left (x^{\frac {1}{3}} a +b \right )^{p} b^{2} p^{2}-6 x^{\frac {1}{3}} \left (x^{\frac {1}{3}} a +b \right )^{p} b^{2} p +6 \left (x^{\frac {1}{3}} a +b \right )^{p} a^{2} x +x^{\frac {p}{3}} \left (\int \frac {\left (x^{\frac {1}{3}} a +b \right )^{p}}{x^{\frac {p}{3}+\frac {2}{3}} b +x^{\frac {p}{3}} a x}d x \right ) b^{3} p^{3}-3 x^{\frac {p}{3}} \left (\int \frac {\left (x^{\frac {1}{3}} a +b \right )^{p}}{x^{\frac {p}{3}+\frac {2}{3}} b +x^{\frac {p}{3}} a x}d x \right ) b^{3} p^{2}+2 x^{\frac {p}{3}} \left (\int \frac {\left (x^{\frac {1}{3}} a +b \right )^{p}}{x^{\frac {p}{3}+\frac {2}{3}} b +x^{\frac {p}{3}} a x}d x \right ) b^{3} p}{6 x^{\frac {p}{3}} a^{2}} \] Input:
int((a+b/x^(1/3))^p,x)
Output:
(3*x**(2/3)*(x**(1/3)*a + b)**p*a*b*p + 3*x**(1/3)*(x**(1/3)*a + b)**p*b** 2*p**2 - 6*x**(1/3)*(x**(1/3)*a + b)**p*b**2*p + 6*(x**(1/3)*a + b)**p*a** 2*x + x**(p/3)*int((x**(1/3)*a + b)**p/(x**((p + 2)/3)*b + x**(p/3)*a*x),x )*b**3*p**3 - 3*x**(p/3)*int((x**(1/3)*a + b)**p/(x**((p + 2)/3)*b + x**(p /3)*a*x),x)*b**3*p**2 + 2*x**(p/3)*int((x**(1/3)*a + b)**p/(x**((p + 2)/3) *b + x**(p/3)*a*x),x)*b**3*p)/(6*x**(p/3)*a**2)