\(\int \frac {x^2}{(b x^n)^{3/2}} \, dx\) [149]

Optimal result

Integrand size = 13, antiderivative size = 30 \[ \int \frac {x^2}{\left (b x^n\right )^{3/2}} \, dx=\frac {2 x^{3-n}}{3 b (2-n) \sqrt {b x^n}} \]

[Out]

2/3*x^(3-n)/b/(2-n)/(b*x^n)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {15, 30} \[ \int \frac {x^2}{\left (b x^n\right )^{3/2}} \, dx=\frac {2 x^{3-n}}{3 b (2-n) \sqrt {b x^n}} \]

[In]

Int[x^2/(b*x^n)^(3/2),x]

[Out]

(2*x^(3 - n))/(3*b*(2 - n)*Sqrt[b*x^n])

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {x^{n/2} \int x^{2-\frac {3 n}{2}} \, dx}{b \sqrt {b x^n}} \\ & = \frac {2 x^{3-n}}{3 b (2-n) \sqrt {b x^n}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {x^2}{\left (b x^n\right )^{3/2}} \, dx=\frac {x^3}{\left (3-\frac {3 n}{2}\right ) \left (b x^n\right )^{3/2}} \]

[In]

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

[Out]

x^3/((3 - (3*n)/2)*(b*x^n)^(3/2))

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.60

method	result	size
gosper	\(-\frac {2 x^{3}}{3 \left (-2+n \right ) \left (b \,x^{n}\right )^{\frac {3}{2}}}\)	\(18\)

[In]

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

[Out]

-2/3*x^3/(-2+n)/(b*x^n)^(3/2)

Fricas [F(-2)]

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

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).

Time = 0.63 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {x^2}{\left (b x^n\right )^{3/2}} \, dx=\begin {cases} - \frac {2 x^{3}}{3 n \left (b x^{n}\right )^{\frac {3}{2}} - 6 \left (b x^{n}\right )^{\frac {3}{2}}} & \text {for}\: n \neq 2 \\\frac {x^{3} \log {\left (x \right )}}{\left (b x^{2}\right )^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**2/(b*x**n)**(3/2),x)

[Out]

Piecewise((-2*x**3/(3*n*(b*x**n)**(3/2) - 6*(b*x**n)**(3/2)), Ne(n, 2)), (x**3*log(x)/(b*x**2)**(3/2), True))

Maxima [F(-2)]

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

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(2-(3*n)/2>0)', see `assume?` f
or more deta

Giac [F]

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

[In]

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

[Out]

integrate(x^2/(b*x^n)^(3/2), x)

Mupad [F(-1)]

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

[In]

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

[Out]

int(x^2/(b*x^n)^(3/2), x)