\(\int x^m (a+b x^{-\frac {3}{2} (1+m)})^{2/3} \, dx\) [2701]

Optimal result

Integrand size = 21, antiderivative size = 139 \[ \int x^m \left (a+b x^{-\frac {3}{2} (1+m)}\right )^{2/3} \, dx=\frac {x^{1+m} \left (a+b x^{-\frac {3}{2} (1+m)}\right )^{2/3}}{1+m}-\frac {2 b^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x^{\frac {1}{2} (-1-m)}}{\sqrt [3]{a+b x^{-\frac {3}{2} (1+m)}}}}{\sqrt {3}}\right )}{\sqrt {3} (1+m)}+\frac {b^{2/3} \log \left (\sqrt [3]{b} x^{\frac {1}{2} (-1-m)}-\sqrt [3]{a+b x^{-\frac {3}{2} (1+m)}}\right )}{1+m} \]

[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)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.07, 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} \[ \int x^m \left (a+b x^{-\frac {3}{2} (1+m)}\right )^{2/3} \, dx=-\frac {2 b^{2/3} \arctan \left (\frac {\frac {2 \sqrt [3]{b} x^{\frac {1}{2} (-m-1)}}{\sqrt [3]{a+b x^{-\frac {3}{2} (m+1)}}}+1}{\sqrt {3}}\right )}{\sqrt {3} (m+1)}+\frac {b^{2/3} \log \left (-x^{\frac {1}{2} (-m-1)} \left (\sqrt [3]{b}-x^{\frac {m+1}{2}} \sqrt [3]{a+b x^{-\frac {3}{2} (m+1)}}\right )\right )}{m+1}+\frac {x^{m+1} \left (a+b x^{-\frac {3}{2} (m+1)}\right )^{2/3}}{m+1} \]

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

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 {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) \text {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 (-x^{\frac {1}{2} (-1-m)} \left (\sqrt [3]{b}-x^{\frac {1+m}{2}} \sqrt [3]{a+b x^{-\frac {3}{2} (1+m)}}\right )\right )}{1+m} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(347\) vs. \(2(139)=278\).

Time = 0.94 (sec) , antiderivative size = 347, normalized size of antiderivative = 2.50 \[ \int x^m \left (a+b x^{-\frac {3}{2} (1+m)}\right )^{2/3} \, dx=-\frac {x^{\frac {1}{2} (-1-m)} \left (a+b x^{-\frac {3}{2} (1+m)}\right )^{2/3} \left (\sqrt [3]{b}-\sqrt [3]{b+a x^{\frac {3 (1+m)}{2}}}\right ) \left (b^{2/3}+\sqrt [3]{b} \sqrt [3]{b+a x^{\frac {3 (1+m)}{2}}}+\left (b+a x^{\frac {3 (1+m)}{2}}\right )^{2/3}\right )^2 \left (3 \left (b+a x^{\frac {3 (1+m)}{2}}\right )^{2/3}+2 \sqrt {3} b^{2/3} \arctan \left (\frac {\sqrt [3]{b}+2 \sqrt [3]{b+a x^{\frac {3 (1+m)}{2}}}}{\sqrt {3} \sqrt [3]{b}}\right )+2 b^{2/3} \log \left (\sqrt [3]{b}-\sqrt [3]{b+a x^{\frac {3 (1+m)}{2}}}\right )-b^{2/3} \log \left (b^{2/3}+\sqrt [3]{b} \sqrt [3]{b+a x^{\frac {3 (1+m)}{2}}}+\left (b+a x^{\frac {3 (1+m)}{2}}\right )^{2/3}\right )\right )}{3 a (1+m) \sqrt [3]{b+a x^{\frac {3 (1+m)}{2}}} \left (b+a x^{\frac {3 (1+m)}{2}}+b^{2/3} \sqrt [3]{b+a x^{\frac {3 (1+m)}{2}}}+\sqrt [3]{b} \left (b+a x^{\frac {3 (1+m)}{2}}\right )^{2/3}\right )} \]

[In]

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

[Out]

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

Maple [F]

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

[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)

Fricas [F(-2)]

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

[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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.38 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.94 \[ \int x^m \left (a+b x^{-\frac {3}{2} (1+m)}\right )^{2/3} \, dx=- \frac {2 a^{\frac {2 m}{3 m + 3} + \frac {2}{3} + \frac {2}{3 m + 3}} x^{m + 1} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {2}{3} \\ - \frac {2 m}{3 m + 3} + 1 - \frac {2}{3 m + 3} \end {matrix}\middle | {\frac {b x^{- \frac {3 m}{2} - \frac {3}{2}} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} m \Gamma \left (- \frac {2 m}{3 m + 3} + 1 - \frac {2}{3 m + 3}\right ) + 3 a^{\frac {2}{3}} \Gamma \left (- \frac {2 m}{3 m + 3} + 1 - \frac {2}{3 m + 3}\right )} \]

[In]

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

[Out]

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

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

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

[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)