\(\int \frac {(b x^2)^{5/2}}{x^7} \, dx\) [31]

Optimal result

Integrand size = 13, antiderivative size = 17 \[ \int \frac {\left (b x^2\right )^{5/2}}{x^7} \, dx=-\frac {b^2 \sqrt {b x^2}}{x^2} \]

[Out]

-b^2*(b*x^2)^(1/2)/x^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} \[ \int \frac {\left (b x^2\right )^{5/2}}{x^7} \, dx=-\frac {b^2 \sqrt {b x^2}}{x^2} \]

[In]

Int[(b*x^2)^(5/2)/x^7,x]

[Out]

-((b^2*Sqrt[b*x^2])/x^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 \frac {1}{x^2} \, dx}{x} \\ & = -\frac {b^2 \sqrt {b x^2}}{x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {\left (b x^2\right )^{5/2}}{x^7} \, dx=-\frac {\left (b x^2\right )^{5/2}}{x^6} \]

[In]

Integrate[(b*x^2)^(5/2)/x^7,x]

[Out]

-((b*x^2)^(5/2)/x^6)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76

method	result	size
gosper	\(-\frac {\left (b \,x^{2}\right )^{\frac {5}{2}}}{x^{6}}\)	\(13\)
default	\(-\frac {\left (b \,x^{2}\right )^{\frac {5}{2}}}{x^{6}}\)	\(13\)
risch	\(-\frac {b^{2} \sqrt {b \,x^{2}}}{x^{2}}\)	\(16\)
pseudoelliptic	\(-\frac {b^{2} \sqrt {b \,x^{2}}}{x^{2}}\)	\(16\)
trager	\(\frac {b^{2} \left (-1+x \right ) \sqrt {b \,x^{2}}}{x^{2}}\)	\(18\)

[In]

int((b*x^2)^(5/2)/x^7,x,method=_RETURNVERBOSE)

[Out]

-(b*x^2)^(5/2)/x^6

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {\left (b x^2\right )^{5/2}}{x^7} \, dx=-\frac {\sqrt {b x^{2}} b^{2}}{x^{2}} \]

[In]

integrate((b*x^2)^(5/2)/x^7,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {\left (b x^2\right )^{5/2}}{x^7} \, dx=- \frac {\left (b x^{2}\right )^{\frac {5}{2}}}{x^{6}} \]

[In]

integrate((b*x**2)**(5/2)/x**7,x)

[Out]

-(b*x**2)**(5/2)/x**6

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (b x^2\right )^{5/2}}{x^7} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((b*x^2)^(5/2)/x^7,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59 \[ \int \frac {\left (b x^2\right )^{5/2}}{x^7} \, dx=-\frac {b^{\frac {5}{2}} \mathrm {sgn}\left (x\right )}{x} \]

[In]

integrate((b*x^2)^(5/2)/x^7,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.51 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59 \[ \int \frac {\left (b x^2\right )^{5/2}}{x^7} \, dx=-\frac {b^{5/2}}{\sqrt {x^2}} \]

[In]

int((b*x^2)^(5/2)/x^7,x)

[Out]

-b^(5/2)/(x^2)^(1/2)