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

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Rubi [A]  time = 0.11, antiderivative size = 139, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [C]  time = 0.08, 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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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a + \frac {b}{x^{\frac {3}{2} \, m + \frac {3}{2}}}\right )}^{\frac {2}{3}} x^{m}\,{d x} \]

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)

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maple [F]  time = 0.24, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{-\frac {3 m}{2}-\frac {3}{2}}+a \right )^{\frac {2}{3}} x^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{-\frac {3}{2} \, m - \frac {3}{2}} + a\right )}^{\frac {2}{3}} x^{m}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^m\,{\left (a+\frac {b}{x^{\frac {3\,m}{2}+\frac {3}{2}}}\right )}^{2/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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