\(\int \frac {x^2}{(a+b x^{3/2})^{2/3}} \, dx\) [2278]

Optimal result

Integrand size = 17, antiderivative size = 40 \[ \int \frac {x^2}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {2 a \sqrt [3]{a+b x^{3/2}}}{b^2}+\frac {\left (a+b x^{3/2}\right )^{4/3}}{2 b^2} \]

[Out]

-2*a*(a+b*x^(3/2))^(1/3)/b^2+1/2*(a+b*x^(3/2))^(4/3)/b^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {272, 45} \[ \int \frac {x^2}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {\left (a+b x^{3/2}\right )^{4/3}}{2 b^2}-\frac {2 a \sqrt [3]{a+b x^{3/2}}}{b^2} \]

[In]

Int[x^2/(a + b*x^(3/2))^(2/3),x]

[Out]

(-2*a*(a + b*x^(3/2))^(1/3))/b^2 + (a + b*x^(3/2))^(4/3)/(2*b^2)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

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

Rubi steps \begin{align*} \text {integral}& = \frac {2}{3} \text {Subst}\left (\int \frac {x}{(a+b x)^{2/3}} \, dx,x,x^{3/2}\right ) \\ & = \frac {2}{3} \text {Subst}\left (\int \left (-\frac {a}{b (a+b x)^{2/3}}+\frac {\sqrt [3]{a+b x}}{b}\right ) \, dx,x,x^{3/2}\right ) \\ & = -\frac {2 a \sqrt [3]{a+b x^{3/2}}}{b^2}+\frac {\left (a+b x^{3/2}\right )^{4/3}}{2 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.78 \[ \int \frac {x^2}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {\left (-3 a+b x^{3/2}\right ) \sqrt [3]{a+b x^{3/2}}}{2 b^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.41 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {\frac {\left (a +b \,x^{\frac {3}{2}}\right )^{\frac {4}{3}}}{2}-2 a \left (a +b \,x^{\frac {3}{2}}\right )^{\frac {1}{3}}}{b^{2}}\)	\(30\)
default	\(\frac {\frac {\left (a +b \,x^{\frac {3}{2}}\right )^{\frac {4}{3}}}{2}-2 a \left (a +b \,x^{\frac {3}{2}}\right )^{\frac {1}{3}}}{b^{2}}\)	\(30\)

[In]

int(x^2/(a+b*x^(3/2))^(2/3),x,method=_RETURNVERBOSE)

[Out]

2/b^2*(1/4*(a+b*x^(3/2))^(4/3)-a*(a+b*x^(3/2))^(1/3))

Fricas [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.58 \[ \int \frac {x^2}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {{\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} {\left (b x^{\frac {3}{2}} - 3 \, a\right )}}{2 \, b^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.22 \[ \int \frac {x^2}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\begin {cases} - \frac {3 a \sqrt [3]{a + b x^{\frac {3}{2}}}}{2 b^{2}} + \frac {x^{\frac {3}{2}} \sqrt [3]{a + b x^{\frac {3}{2}}}}{2 b} & \text {for}\: b \neq 0 \\\frac {x^{3}}{3 a^{\frac {2}{3}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-3*a*(a + b*x**(3/2))**(1/3)/(2*b**2) + x**(3/2)*(a + b*x**(3/2))**(1/3)/(2*b), Ne(b, 0)), (x**3/(3
*a**(2/3)), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.75 \[ \int \frac {x^2}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {{\left (b x^{\frac {3}{2}} + a\right )}^{\frac {4}{3}}}{2 \, b^{2}} - \frac {2 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} a}{b^{2}} \]

[In]

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

[Out]

1/2*(b*x^(3/2) + a)^(4/3)/b^2 - 2*(b*x^(3/2) + a)^(1/3)*a/b^2

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.75 \[ \int \frac {x^2}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {{\left (b x^{\frac {3}{2}} + a\right )}^{\frac {4}{3}}}{2 \, b^{2}} - \frac {2 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} a}{b^{2}} \]

[In]

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

[Out]

1/2*(b*x^(3/2) + a)^(4/3)/b^2 - 2*(b*x^(3/2) + a)^(1/3)*a/b^2

Mupad [B] (verification not implemented)

Time = 5.87 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.72 \[ \int \frac {x^2}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {4\,a\,{\left (a+b\,x^{3/2}\right )}^{1/3}-{\left (a+b\,x^{3/2}\right )}^{4/3}}{2\,b^2} \]

[In]

int(x^2/(a + b*x^(3/2))^(2/3),x)

[Out]

-(4*a*(a + b*x^(3/2))^(1/3) - (a + b*x^(3/2))^(4/3))/(2*b^2)