3.1.26 \(\int \frac {(b x^2)^{5/2}}{x^2} \, dx\) [26]

Optimal. Leaf size=19 \[ \frac {1}{4} b^2 x^3 \sqrt {b x^2} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {15, 30} \begin {gather*} \frac {1}{4} b^2 x^3 \sqrt {b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*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*} \int \frac {\left (b x^2\right )^{5/2}}{x^2} \, dx &=\frac {\left (b^2 \sqrt {b x^2}\right ) \int x^3 \, dx}{x}\\ &=\frac {1}{4} b^2 x^3 \sqrt {b x^2}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 15, normalized size = 0.79 \begin {gather*} \frac {1}{4} b x \left (b x^2\right )^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.02, size = 13, normalized size = 0.68

method	result	size
gosper	\(\frac {\left (b \,x^{2}\right )^{\frac {5}{2}}}{4 x}\)	\(13\)
default	\(\frac {\left (b \,x^{2}\right )^{\frac {5}{2}}}{4 x}\)	\(13\)
risch	\(\frac {b^{2} x^{3} \sqrt {b \,x^{2}}}{4}\)	\(16\)
trager	\(\frac {b^{2} \left (x^{3}+x^{2}+x +1\right ) \left (x -1\right ) \sqrt {b \,x^{2}}}{4 x}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 12, normalized size = 0.63 \begin {gather*} \frac {\left (b x^{2}\right )^{\frac {5}{2}}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.33, size = 15, normalized size = 0.79 \begin {gather*} \frac {1}{4} \, \sqrt {b x^{2}} b^{2} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.22, size = 10, normalized size = 0.53 \begin {gather*} \frac {\left (b x^{2}\right )^{\frac {5}{2}}}{4 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 2.11, size = 10, normalized size = 0.53 \begin {gather*} \frac {1}{4} \, b^{\frac {5}{2}} x^{4} \mathrm {sgn}\left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.97, size = 13, normalized size = 0.68 \begin {gather*} \frac {b^{5/2}\,x^3\,\sqrt {x^2}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^(5/2)*x^3*(x^2)^(1/2))/4

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