\(\int (b x^2)^{3/2} \, dx\) [14]

Optimal result

Integrand size = 9, antiderivative size = 17 \[ \int \left (b x^2\right )^{3/2} \, dx=\frac {1}{4} b x^3 \sqrt {b x^2} \]

[Out]

1/4*b*x^3*(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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {15, 30} \[ \int \left (b x^2\right )^{3/2} \, dx=\frac {1}{4} b x^3 \sqrt {b x^2} \]

[In]

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

[Out]

(b*x^3*Sqrt[b*x^2])/4

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 \sqrt {b x^2}\right ) \int x^3 \, dx}{x} \\ & = \frac {1}{4} b x^3 \sqrt {b x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \left (b x^2\right )^{3/2} \, dx=\frac {1}{4} x \left (b x^2\right )^{3/2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.65

method	result	size
gosper	\(\frac {x \left (b \,x^{2}\right )^{\frac {3}{2}}}{4}\)	\(11\)
default	\(\frac {x \left (b \,x^{2}\right )^{\frac {3}{2}}}{4}\)	\(11\)
risch	\(\frac {b \,x^{3} \sqrt {b \,x^{2}}}{4}\)	\(14\)
trager	\(\frac {b \left (x^{3}+x^{2}+x +1\right ) \left (-1+x \right ) \sqrt {b \,x^{2}}}{4 x}\)	\(26\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \left (b x^2\right )^{3/2} \, dx=\frac {1}{4} \, \sqrt {b x^{2}} b x^{3} \]

[In]

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

[Out]

1/4*sqrt(b*x^2)*b*x^3

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59 \[ \int \left (b x^2\right )^{3/2} \, dx=\frac {x \left (b x^{2}\right )^{\frac {3}{2}}}{4} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59 \[ \int \left (b x^2\right )^{3/2} \, dx=\frac {1}{4} \, \left (b x^{2}\right )^{\frac {3}{2}} x \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59 \[ \int \left (b x^2\right )^{3/2} \, dx=\frac {1}{4} \, b^{\frac {3}{2}} x^{4} \mathrm {sgn}\left (x\right ) \]

[In]

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

[Out]

1/4*b^(3/2)*x^4*sgn(x)

Mupad [F(-1)]

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

[In]

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

[Out]

int((b*x^2)^(3/2), x)