\(\int \frac {(b x^n)^{3/2}}{x^3} \, dx\) [141]

Optimal result

Integrand size = 13, antiderivative size = 24 \[ \int \frac {\left (b x^n\right )^{3/2}}{x^3} \, dx=-\frac {2 b x^{-2+n} \sqrt {b x^n}}{4-3 n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, 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} \[ \int \frac {\left (b x^n\right )^{3/2}}{x^3} \, dx=-\frac {2 b x^{n-2} \sqrt {b x^n}}{4-3 n} \]

[In]

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

[Out]

(-2*b*x^(-2 + n)*Sqrt[b*x^n])/(4 - 3*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}& = \left (b x^{-n/2} \sqrt {b x^n}\right ) \int x^{-3+\frac {3 n}{2}} \, dx \\ & = -\frac {2 b x^{-2+n} \sqrt {b x^n}}{4-3 n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {\left (b x^n\right )^{3/2}}{x^3} \, dx=\frac {\left (b x^n\right )^{3/2}}{\left (-2+\frac {3 n}{2}\right ) x^2} \]

[In]

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

[Out]

(b*x^n)^(3/2)/((-2 + (3*n)/2)*x^2)

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83

method	result	size
gosper	\(\frac {2 \left (b \,x^{n}\right )^{\frac {3}{2}}}{x^{2} \left (-4+3 n \right )}\)	\(20\)
risch	\(\frac {2 b^{2} x^{2 n}}{\left (-4+3 n \right ) x^{2} \sqrt {b \,x^{n}}}\)	\(28\)

[In]

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

[Out]

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

Fricas [F(-2)]

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

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 10.77 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {\left (b x^n\right )^{3/2}}{x^3} \, dx=\begin {cases} \frac {2 \left (b x^{n}\right )^{\frac {3}{2}}}{3 n x^{2} - 4 x^{2}} & \text {for}\: n \neq \frac {4}{3} \\\frac {3 \left (b x^{\frac {4}{3}}\right )^{\frac {3}{2}} \log {\left (\sqrt [3]{x} \right )}}{x^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(b*x**n)**(3/2)/(3*n*x**2 - 4*x**2), Ne(n, 4/3)), (3*(b*x**(4/3))**(3/2)*log(x**(1/3))/x**2, True
))

Maxima [F(-2)]

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

[In]

integrate((b*x^n)^(3/2)/x^3,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((3*n)/2-3>0)', see `assume?` f
or more deta

Giac [F]

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

[In]

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

[Out]

integrate((b*x^n)^(3/2)/x^3, x)

Mupad [B] (verification not implemented)

Time = 5.43 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {\left (b x^n\right )^{3/2}}{x^3} \, dx=\frac {2\,b\,x^{n-2}\,\sqrt {b\,x^n}}{3\,n-4} \]

[In]

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

[Out]

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