3.2701 \(\int x^m (a+b x^{-\frac{3}{2} (1+m)})^{2/3} \, dx\)

Optimal. Leaf size=139 \[ \frac{b^{2/3} \log \left (\sqrt [3]{b} x^{\frac{1}{2} (-m-1)}-\sqrt [3]{a+b x^{-\frac{3}{2} (m+1)}}\right )}{m+1}-\frac{2 b^{2/3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x^{\frac{1}{2} (-m-1)}}{\sqrt [3]{a+b x^{-\frac{3}{2} (m+1)}}}+1}{\sqrt{3}}\right )}{\sqrt{3} (m+1)}+\frac{x^{m+1} \left (a+b x^{-\frac{3}{2} (m+1)}\right )^{2/3}}{m+1} \]

[Out]

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

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Rubi [A]  time = 0.11475, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {349, 345, 239} \[ \frac{b^{2/3} \log \left (\sqrt [3]{b} x^{\frac{1}{2} (-m-1)}-\sqrt [3]{a+b x^{-\frac{3}{2} (m+1)}}\right )}{m+1}-\frac{2 b^{2/3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x^{\frac{1}{2} (-m-1)}}{\sqrt [3]{a+b x^{-\frac{3}{2} (m+1)}}}+1}{\sqrt{3}}\right )}{\sqrt{3} (m+1)}+\frac{x^{m+1} \left (a+b x^{-\frac{3}{2} (m+1)}\right )^{2/3}}{m+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 349

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]

Rule 345

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 239

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]

Rubi steps

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

Mathematica [C]  time = 0.0670602, size = 73, normalized size = 0.53 \[ \frac{x^{m+1} \left (a+b x^{-\frac{3}{2} (m+1)}\right )^{2/3} \, _2F_1\left (-\frac{2}{3},-\frac{2}{3};\frac{1}{3};-\frac{b x^{-\frac{3}{2} (m+1)}}{a}\right )}{(m+1) \left (\frac{b x^{-\frac{3}{2} (m+1)}}{a}+1\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.082, size = 0, normalized size = 0. \begin{align*} \int{x}^{m} \left ( a+{b \left ({x}^{{\frac{3}{2}}+{\frac{3\,m}{2}}} \right ) ^{-1}} \right ) ^{{\frac{2}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{-\frac{3}{2} \, m - \frac{3}{2}} + a\right )}^{\frac{2}{3}} x^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*(a+b/(x**(3/2+3/2*m)))**(2/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a + \frac{b}{x^{\frac{3}{2} \, m + \frac{3}{2}}}\right )}^{\frac{2}{3}} x^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a + b/x^(3/2*m + 3/2))^(2/3)*x^m, x)