\(\int \frac {x^3}{(b x^2)^{3/2}} \, dx\) [43]

Optimal result

Integrand size = 13, antiderivative size = 16 \[ \int \frac {x^3}{\left (b x^2\right )^{3/2}} \, dx=\frac {x^2}{b \sqrt {b x^2}} \]

[Out]

x^2/b/(b*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, 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, 8} \[ \int \frac {x^3}{\left (b x^2\right )^{3/2}} \, dx=\frac {x^2}{b \sqrt {b x^2}} \]

[In]

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

[Out]

x^2/(b*Sqrt[b*x^2])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {x \int 1 \, dx}{b \sqrt {b x^2}} \\ & = \frac {x^2}{b \sqrt {b x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int \frac {x^3}{\left (b x^2\right )^{3/2}} \, dx=\frac {\sqrt {b x^2}}{b^2} \]

[In]

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

[Out]

Sqrt[b*x^2]/b^2

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75

method	result	size
default	\(\frac {x^{4}}{\left (b \,x^{2}\right )^{\frac {3}{2}}}\)	\(12\)
risch	\(\frac {x^{2}}{b \sqrt {b \,x^{2}}}\)	\(15\)
pseudoelliptic	\(\frac {x^{2}}{b \sqrt {b \,x^{2}}}\)	\(15\)
trager	\(\frac {\left (-1+x \right ) \sqrt {b \,x^{2}}}{b^{2} x}\)	\(18\)

[In]

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

[Out]

x^4/(b*x^2)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69 \[ \int \frac {x^3}{\left (b x^2\right )^{3/2}} \, dx=\frac {\sqrt {b x^{2}}}{b^{2}} \]

[In]

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

[Out]

sqrt(b*x^2)/b^2

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \frac {x^3}{\left (b x^2\right )^{3/2}} \, dx=\frac {x^{4}}{\left (b x^{2}\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

x**4/(b*x**2)**(3/2)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {x^3}{\left (b x^2\right )^{3/2}} \, dx=\frac {x^{2}}{\sqrt {b x^{2}} b} \]

[In]

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

[Out]

x^2/(sqrt(b*x^2)*b)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.56 \[ \int \frac {x^3}{\left (b x^2\right )^{3/2}} \, dx=\frac {x}{b^{\frac {3}{2}} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

x/(b^(3/2)*sgn(x))

Mupad [B] (verification not implemented)

Time = 6.13 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.38 \[ \int \frac {x^3}{\left (b x^2\right )^{3/2}} \, dx=\frac {\left |x\right |}{b^{3/2}} \]

[In]

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

[Out]

abs(x)/b^(3/2)