\(\int \frac {(b x^2)^{5/2}}{x^4} \, dx\) [28]

Optimal result

Integrand size = 13, antiderivative size = 17 \[ \int \frac {\left (b x^2\right )^{5/2}}{x^4} \, dx=\frac {1}{2} b^2 x \sqrt {b x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, 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} \[ \int \frac {\left (b x^2\right )^{5/2}}{x^4} \, dx=\frac {1}{2} b^2 x \sqrt {b x^2} \]

[In]

Int[(b*x^2)^(5/2)/x^4,x]

[Out]

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

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 {\left (b^2 \sqrt {b x^2}\right ) \int x \, dx}{x} \\ & = \frac {1}{2} b^2 x \sqrt {b x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {\left (b x^2\right )^{5/2}}{x^4} \, dx=\frac {1}{2} b^2 x \sqrt {b x^2} \]

[In]

Integrate[(b*x^2)^(5/2)/x^4,x]

[Out]

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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76

method	result	size
gosper	\(\frac {\left (b \,x^{2}\right )^{\frac {5}{2}}}{2 x^{3}}\)	\(13\)
default	\(\frac {\left (b \,x^{2}\right )^{\frac {5}{2}}}{2 x^{3}}\)	\(13\)
risch	\(\frac {b^{2} x \sqrt {b \,x^{2}}}{2}\)	\(14\)
trager	\(\frac {b^{2} \left (-1+x \right ) \left (1+x \right ) \sqrt {b \,x^{2}}}{2 x}\)	\(22\)

[In]

int((b*x^2)^(5/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

1/2*(b*x^2)^(5/2)/x^3

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {\left (b x^2\right )^{5/2}}{x^4} \, dx=\frac {1}{2} \, \sqrt {b x^{2}} b^{2} x \]

[In]

integrate((b*x^2)^(5/2)/x^4,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {\left (b x^2\right )^{5/2}}{x^4} \, dx=\frac {\left (b x^{2}\right )^{\frac {5}{2}}}{2 x^{3}} \]

[In]

integrate((b*x**2)**(5/2)/x**4,x)

[Out]

(b*x**2)**(5/2)/(2*x**3)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (b x^2\right )^{5/2}}{x^4} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((b*x^2)^(5/2)/x^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59 \[ \int \frac {\left (b x^2\right )^{5/2}}{x^4} \, dx=\frac {1}{2} \, b^{\frac {5}{2}} x^{2} \mathrm {sgn}\left (x\right ) \]

[In]

integrate((b*x^2)^(5/2)/x^4,x, algorithm="giac")

[Out]

1/2*b^(5/2)*x^2*sgn(x)

Mupad [B] (verification not implemented)

Time = 5.58 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.47 \[ \int \frac {\left (b x^2\right )^{5/2}}{x^4} \, dx=\frac {b^{5/2}\,x\,\left |x\right |}{2} \]

[In]

int((b*x^2)^(5/2)/x^4,x)

[Out]

(b^(5/2)*x*abs(x))/2