\(\int x^{-1-\frac {2 n}{3}} (a+b x^n)^{2/3} \, dx\) [2703]

Optimal result

Integrand size = 21, antiderivative size = 114 \[ \int x^{-1-\frac {2 n}{3}} \left (a+b x^n\right )^{2/3} \, dx=-\frac {3 x^{-2 n/3} \left (a+b x^n\right )^{2/3}}{2 n}+\frac {\sqrt {3} b^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x^{n/3}}{\sqrt [3]{a+b x^n}}}{\sqrt {3}}\right )}{n}-\frac {3 b^{2/3} \log \left (\sqrt [3]{b} x^{n/3}-\sqrt [3]{a+b x^n}\right )}{2 n} \]

[Out]

-3/2*(a+b*x^n)^(2/3)/n/(x^(2/3*n))-3/2*b^(2/3)*ln(b^(1/3)*x^(1/3*n)-(a+b*x^n)^(1/3))/n+b^(2/3)*arctan(1/3*(1+2
*b^(1/3)*x^(1/3*n)/(a+b*x^n)^(1/3))*3^(1/2))*3^(1/2)/n

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {356, 352, 245} \[ \int x^{-1-\frac {2 n}{3}} \left (a+b x^n\right )^{2/3} \, dx=\frac {\sqrt {3} b^{2/3} \arctan \left (\frac {\frac {2 \sqrt [3]{b} x^{n/3}}{\sqrt [3]{a+b x^n}}+1}{\sqrt {3}}\right )}{n}-\frac {3 b^{2/3} \log \left (\sqrt [3]{b} x^{n/3}-\sqrt [3]{a+b x^n}\right )}{2 n}-\frac {3 x^{-2 n/3} \left (a+b x^n\right )^{2/3}}{2 n} \]

[In]

Int[x^(-1 - (2*n)/3)*(a + b*x^n)^(2/3),x]

[Out]

(-3*(a + b*x^n)^(2/3))/(2*n*x^((2*n)/3)) + (Sqrt[3]*b^(2/3)*ArcTan[(1 + (2*b^(1/3)*x^(n/3))/(a + b*x^n)^(1/3))
/Sqrt[3]])/n - (3*b^(2/3)*Log[b^(1/3)*x^(n/3) - (a + b*x^n)^(1/3)])/(2*n)

Rule 245

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rule 352

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/(m + 1), Subst[Int[(a + b*x^Simplify[n/(m +
1)])^p, x], x, x^(m + 1)], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[n/(m + 1)]] &&  !IntegerQ[n]

Rule 356

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^p/(m + 1)), x] - Dist[b
*n*(p/(m + 1)), Int[x^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, m, n}, x] && EqQ[(m + 1)/n + p, 0] &
& GtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {3 x^{-2 n/3} \left (a+b x^n\right )^{2/3}}{2 n}+b \int \frac {x^{-1+\frac {n}{3}}}{\sqrt [3]{a+b x^n}} \, dx \\ & = -\frac {3 x^{-2 n/3} \left (a+b x^n\right )^{2/3}}{2 n}+\frac {(3 b) \text {Subst}\left (\int \frac {1}{\sqrt [3]{a+b x^3}} \, dx,x,x^{n/3}\right )}{n} \\ & = -\frac {3 x^{-2 n/3} \left (a+b x^n\right )^{2/3}}{2 n}+\frac {\sqrt {3} b^{2/3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x^{n/3}}{\sqrt [3]{a+b x^n}}}{\sqrt {3}}\right )}{n}-\frac {3 b^{2/3} \log \left (\sqrt [3]{b} x^{n/3}-\sqrt [3]{a+b x^n}\right )}{2 n} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.06 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.51 \[ \int x^{-1-\frac {2 n}{3}} \left (a+b x^n\right )^{2/3} \, dx=-\frac {3 x^{-2 n/3} \left (a+b x^n\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},-\frac {2}{3},\frac {1}{3},-\frac {b x^n}{a}\right )}{2 n \left (1+\frac {b x^n}{a}\right )^{2/3}} \]

[In]

Integrate[x^(-1 - (2*n)/3)*(a + b*x^n)^(2/3),x]

[Out]

(-3*(a + b*x^n)^(2/3)*Hypergeometric2F1[-2/3, -2/3, 1/3, -((b*x^n)/a)])/(2*n*x^((2*n)/3)*(1 + (b*x^n)/a)^(2/3)
)

Maple [F]

\[\int x^{-1-\frac {2 n}{3}} \left (a +b \,x^{n}\right )^{\frac {2}{3}}d x\]

[In]

int(x^(-1-2/3*n)*(a+b*x^n)^(2/3),x)

[Out]

int(x^(-1-2/3*n)*(a+b*x^n)^(2/3),x)

Fricas [F(-2)]

Exception generated. \[ \int x^{-1-\frac {2 n}{3}} \left (a+b x^n\right )^{2/3} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x^(-1-2/3*n)*(a+b*x^n)^(2/3),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.13 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.40 \[ \int x^{-1-\frac {2 n}{3}} \left (a+b x^n\right )^{2/3} \, dx=\frac {a^{\frac {2}{3}} x^{- \frac {2 n}{3}} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {2}{3} \\ \frac {1}{3} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (\frac {1}{3}\right )} \]

[In]

integrate(x**(-1-2/3*n)*(a+b*x**n)**(2/3),x)

[Out]

a**(2/3)*gamma(-2/3)*hyper((-2/3, -2/3), (1/3,), b*x**n*exp_polar(I*pi)/a)/(n*x**(2*n/3)*gamma(1/3))

Maxima [F]

\[ \int x^{-1-\frac {2 n}{3}} \left (a+b x^n\right )^{2/3} \, dx=\int { {\left (b x^{n} + a\right )}^{\frac {2}{3}} x^{-\frac {2}{3} \, n - 1} \,d x } \]

[In]

integrate(x^(-1-2/3*n)*(a+b*x^n)^(2/3),x, algorithm="maxima")

[Out]

integrate((b*x^n + a)^(2/3)*x^(-2/3*n - 1), x)

Giac [F]

\[ \int x^{-1-\frac {2 n}{3}} \left (a+b x^n\right )^{2/3} \, dx=\int { {\left (b x^{n} + a\right )}^{\frac {2}{3}} x^{-\frac {2}{3} \, n - 1} \,d x } \]

[In]

integrate(x^(-1-2/3*n)*(a+b*x^n)^(2/3),x, algorithm="giac")

[Out]

integrate((b*x^n + a)^(2/3)*x^(-2/3*n - 1), x)

Mupad [F(-1)]

Timed out. \[ \int x^{-1-\frac {2 n}{3}} \left (a+b x^n\right )^{2/3} \, dx=\int \frac {{\left (a+b\,x^n\right )}^{2/3}}{x^{\frac {2\,n}{3}+1}} \,d x \]

[In]

int((a + b*x^n)^(2/3)/x^((2*n)/3 + 1),x)

[Out]

int((a + b*x^n)^(2/3)/x^((2*n)/3 + 1), x)