Integrand size = 11, antiderivative size = 50 \[ \int \left (a+b x^{2/3}\right )^p \, dx=\left (a+b x^{2/3}\right )^p \left (1+\frac {b x^{2/3}}{a}\right )^{-p} x \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^{2/3}}{a}\right ) \] Output:
(a+b*x^(2/3))^p*x*hypergeom([3/2, -p],[5/2],-b*x^(2/3)/a)/((1+b*x^(2/3)/a) ^p)
Time = 0.06 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int \left (a+b x^{2/3}\right )^p \, dx=\left (a+b x^{2/3}\right )^p \left (1+\frac {b x^{2/3}}{a}\right )^{-p} x \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^{2/3}}{a}\right ) \] Input:
Integrate[(a + b*x^(2/3))^p,x]
Output:
((a + b*x^(2/3))^p*x*Hypergeometric2F1[3/2, -p, 5/2, -((b*x^(2/3))/a)])/(1 + (b*x^(2/3))/a)^p
Time = 0.28 (sec) , antiderivative size = 50, 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, 279, 278}
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.
\(\displaystyle \int \left (a+b x^{2/3}\right )^p \, dx\) |
\(\Big \downarrow \) 774 |
\(\displaystyle 3 \int \left (a+b x^{2/3}\right )^p x^{2/3}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 279 |
\(\displaystyle 3 \left (a+b x^{2/3}\right )^p \left (\frac {b x^{2/3}}{a}+1\right )^{-p} \int \left (\frac {x^{2/3} b}{a}+1\right )^p x^{2/3}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle x \left (a+b x^{2/3}\right )^p \left (\frac {b x^{2/3}}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^{2/3}}{a}\right )\) |
Input:
Int[(a + b*x^(2/3))^p,x]
Output:
((a + b*x^(2/3))^p*x*Hypergeometric2F1[3/2, -p, 5/2, -((b*x^(2/3))/a)])/(1 + (b*x^(2/3))/a)^p
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(c*x)^m* (1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && !(ILtQ[p, 0] || GtQ[a, 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]
\[\int \left (a +b \,x^{\frac {2}{3}}\right )^{p}d x\]
Input:
int((a+b*x^(2/3))^p,x)
Output:
int((a+b*x^(2/3))^p,x)
\[ \int \left (a+b x^{2/3}\right )^p \, dx=\int { {\left (b x^{\frac {2}{3}} + a\right )}^{p} \,d x } \] Input:
integrate((a+b*x^(2/3))^p,x, algorithm="fricas")
Output:
integral((b*x^(2/3) + a)^p, x)
Result contains complex when optimal does not.
Time = 2.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.48 \[ \int \left (a+b x^{2/3}\right )^p \, dx=a^{p} x {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{\frac {2}{3}} e^{i \pi }}{a}} \right )} \] Input:
integrate((a+b*x**(2/3))**p,x)
Output:
a**p*x*hyper((3/2, -p), (5/2,), b*x**(2/3)*exp_polar(I*pi)/a)
\[ \int \left (a+b x^{2/3}\right )^p \, dx=\int { {\left (b x^{\frac {2}{3}} + a\right )}^{p} \,d x } \] Input:
integrate((a+b*x^(2/3))^p,x, algorithm="maxima")
Output:
integrate((b*x^(2/3) + a)^p, x)
\[ \int \left (a+b x^{2/3}\right )^p \, dx=\int { {\left (b x^{\frac {2}{3}} + a\right )}^{p} \,d x } \] Input:
integrate((a+b*x^(2/3))^p,x, algorithm="giac")
Output:
integrate((b*x^(2/3) + a)^p, x)
Time = 0.23 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.82 \[ \int \left (a+b x^{2/3}\right )^p \, dx=\frac {x\,{\left (a+b\,x^{2/3}\right )}^p\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},-p;\ \frac {5}{2};\ -\frac {b\,x^{2/3}}{a}\right )}{{\left (\frac {b\,x^{2/3}}{a}+1\right )}^p} \] Input:
int((a + b*x^(2/3))^p,x)
Output:
(x*(a + b*x^(2/3))^p*hypergeom([3/2, -p], 5/2, -(b*x^(2/3))/a))/((b*x^(2/3 ))/a + 1)^p
\[ \int \left (a+b x^{2/3}\right )^p \, dx=\frac {6 x^{\frac {1}{3}} \left (x^{\frac {2}{3}} b +a \right )^{p} a p +6 \left (x^{\frac {2}{3}} b +a \right )^{p} b p x +3 \left (x^{\frac {2}{3}} b +a \right )^{p} b x -8 \left (\int \frac {\left (x^{\frac {2}{3}} b +a \right )^{p}}{4 x^{\frac {2}{3}} a \,p^{2}+8 x^{\frac {2}{3}} a p +3 x^{\frac {2}{3}} a +4 x^{\frac {4}{3}} b \,p^{2}+8 x^{\frac {4}{3}} b p +3 x^{\frac {4}{3}} b}d x \right ) a^{2} p^{3}-16 \left (\int \frac {\left (x^{\frac {2}{3}} b +a \right )^{p}}{4 x^{\frac {2}{3}} a \,p^{2}+8 x^{\frac {2}{3}} a p +3 x^{\frac {2}{3}} a +4 x^{\frac {4}{3}} b \,p^{2}+8 x^{\frac {4}{3}} b p +3 x^{\frac {4}{3}} b}d x \right ) a^{2} p^{2}-6 \left (\int \frac {\left (x^{\frac {2}{3}} b +a \right )^{p}}{4 x^{\frac {2}{3}} a \,p^{2}+8 x^{\frac {2}{3}} a p +3 x^{\frac {2}{3}} a +4 x^{\frac {4}{3}} b \,p^{2}+8 x^{\frac {4}{3}} b p +3 x^{\frac {4}{3}} b}d x \right ) a^{2} p}{b \left (4 p^{2}+8 p +3\right )} \] Input:
int((a+b*x^(2/3))^p,x)
Output:
(6*x**(1/3)*(x**(2/3)*b + a)**p*a*p + 6*(x**(2/3)*b + a)**p*b*p*x + 3*(x** (2/3)*b + a)**p*b*x - 8*int((x**(2/3)*b + a)**p/(4*x**(2/3)*a*p**2 + 8*x** (2/3)*a*p + 3*x**(2/3)*a + 4*x**(1/3)*b*p**2*x + 8*x**(1/3)*b*p*x + 3*x**( 1/3)*b*x),x)*a**2*p**3 - 16*int((x**(2/3)*b + a)**p/(4*x**(2/3)*a*p**2 + 8 *x**(2/3)*a*p + 3*x**(2/3)*a + 4*x**(1/3)*b*p**2*x + 8*x**(1/3)*b*p*x + 3* x**(1/3)*b*x),x)*a**2*p**2 - 6*int((x**(2/3)*b + a)**p/(4*x**(2/3)*a*p**2 + 8*x**(2/3)*a*p + 3*x**(2/3)*a + 4*x**(1/3)*b*p**2*x + 8*x**(1/3)*b*p*x + 3*x**(1/3)*b*x),x)*a**2*p)/(b*(4*p**2 + 8*p + 3))