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

Optimal result

Integrand size = 13, antiderivative size = 16 \[ \int x^3 \sqrt {b x^2} \, dx=\frac {1}{5} x^4 \sqrt {b x^2} \]

[Out]

1/5*x^4*(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, 30} \[ \int x^3 \sqrt {b x^2} \, dx=\frac {1}{5} x^4 \sqrt {b x^2} \]

[In]

Int[x^3*Sqrt[b*x^2],x]

[Out]

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

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 {\sqrt {b x^2} \int x^4 \, dx}{x} \\ & = \frac {1}{5} x^4 \sqrt {b x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int x^3 \sqrt {b x^2} \, dx=\frac {1}{5} x^4 \sqrt {b x^2} \]

[In]

Integrate[x^3*Sqrt[b*x^2],x]

[Out]

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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81

method	result	size
gosper	\(\frac {x^{4} \sqrt {b \,x^{2}}}{5}\)	\(13\)
default	\(\frac {x^{4} \sqrt {b \,x^{2}}}{5}\)	\(13\)
risch	\(\frac {x^{4} \sqrt {b \,x^{2}}}{5}\)	\(13\)
pseudoelliptic	\(\frac {x^{4} \sqrt {b \,x^{2}}}{5}\)	\(13\)
trager	\(\frac {\left (x^{4}+x^{3}+x^{2}+x +1\right ) \left (-1+x \right ) \sqrt {b \,x^{2}}}{5 x}\)	\(28\)

[In]

int(x^3*(b*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/5*x^4*(b*x^2)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int x^3 \sqrt {b x^2} \, dx=\frac {1}{5} \, \sqrt {b x^{2}} x^{4} \]

[In]

integrate(x^3*(b*x^2)^(1/2),x, algorithm="fricas")

[Out]

1/5*sqrt(b*x^2)*x^4

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int x^3 \sqrt {b x^2} \, dx=\frac {x^{4} \sqrt {b x^{2}}}{5} \]

[In]

integrate(x**3*(b*x**2)**(1/2),x)

[Out]

x**4*sqrt(b*x**2)/5

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int x^3 \sqrt {b x^2} \, dx=\frac {\left (b x^{2}\right )^{\frac {3}{2}} x^{2}}{5 \, b} \]

[In]

integrate(x^3*(b*x^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int x^3 \sqrt {b x^2} \, dx=\frac {1}{5} \, \sqrt {b} x^{5} \mathrm {sgn}\left (x\right ) \]

[In]

integrate(x^3*(b*x^2)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.87 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int x^3 \sqrt {b x^2} \, dx=\frac {\sqrt {b}\,\sqrt {x^{10}}}{5} \]

[In]

int(x^3*(b*x^2)^(1/2),x)

[Out]

(b^(1/2)*(x^10)^(1/2))/5