\(\int x (b x^2)^{5/2} \, dx\) [23]

Optimal result

Integrand size = 11, antiderivative size = 19 \[ \int x \left (b x^2\right )^{5/2} \, dx=\frac {1}{7} b^2 x^6 \sqrt {b x^2} \]

[Out]

1/7*b^2*x^6*(b*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {15, 30} \[ \int x \left (b x^2\right )^{5/2} \, dx=\frac {1}{7} b^2 x^6 \sqrt {b x^2} \]

[In]

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

[Out]

(b^2*x^6*Sqrt[b*x^2])/7

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^6 \, dx}{x} \\ & = \frac {1}{7} b^2 x^6 \sqrt {b x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int x \left (b x^2\right )^{5/2} \, dx=\frac {1}{7} x^2 \left (b x^2\right )^{5/2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68

method	result	size
gosper	\(\frac {x^{2} \left (b \,x^{2}\right )^{\frac {5}{2}}}{7}\)	\(13\)
derivativedivides	\(\frac {\left (b \,x^{2}\right )^{\frac {7}{2}}}{7 b}\)	\(13\)
default	\(\frac {x^{2} \left (b \,x^{2}\right )^{\frac {5}{2}}}{7}\)	\(13\)
risch	\(\frac {b^{2} x^{6} \sqrt {b \,x^{2}}}{7}\)	\(16\)
pseudoelliptic	\(\frac {b^{2} x^{6} \sqrt {b \,x^{2}}}{7}\)	\(16\)
trager	\(\frac {b^{2} \left (x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x +1\right ) \left (-1+x \right ) \sqrt {b \,x^{2}}}{7 x}\)	\(37\)

[In]

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

[Out]

1/7*x^2*(b*x^2)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int x \left (b x^2\right )^{5/2} \, dx=\frac {1}{7} \, \sqrt {b x^{2}} b^{2} x^{6} \]

[In]

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

[Out]

1/7*sqrt(b*x^2)*b^2*x^6

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int x \left (b x^2\right )^{5/2} \, dx=\frac {x^{2} \left (b x^{2}\right )^{\frac {5}{2}}}{7} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int x \left (b x^2\right )^{5/2} \, dx=\frac {\left (b x^{2}\right )^{\frac {7}{2}}}{7 \, b} \]

[In]

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

[Out]

1/7*(b*x^2)^(7/2)/b

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53 \[ \int x \left (b x^2\right )^{5/2} \, dx=\frac {1}{7} \, b^{\frac {5}{2}} x^{7} \mathrm {sgn}\left (x\right ) \]

[In]

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

[Out]

1/7*b^(5/2)*x^7*sgn(x)

Mupad [B] (verification not implemented)

Time = 5.74 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53 \[ \int x \left (b x^2\right )^{5/2} \, dx=\frac {b^{5/2}\,\sqrt {x^{14}}}{7} \]

[In]

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

[Out]

(b^(5/2)*(x^14)^(1/2))/7