3.7.25 \(\int \frac {\sqrt {x}}{(2+b x)^{5/2}} \, dx\) [625]

Optimal. Leaf size=18 \[ \frac {x^{3/2}}{3 (2+b x)^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {37} \begin {gather*} \frac {x^{3/2}}{3 (b x+2)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(2 + b*x)^(5/2),x]

[Out]

x^(3/2)/(3*(2 + b*x)^(3/2))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin {align*} \int \frac {\sqrt {x}}{(2+b x)^{5/2}} \, dx &=\frac {x^{3/2}}{3 (2+b x)^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 18, normalized size = 1.00 \begin {gather*} \frac {x^{3/2}}{3 (2+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(2 + b*x)^(5/2),x]

[Out]

x^(3/2)/(3*(2 + b*x)^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(45\) vs. \(2(12)=24\).
time = 0.10, size = 46, normalized size = 2.56

method	result	size
gosper	\(\frac {x^{\frac {3}{2}}}{3 \left (b x +2\right )^{\frac {3}{2}}}\)	\(13\)
meijerg	\(\frac {x^{\frac {3}{2}} \sqrt {2}}{12 \left (\frac {b x}{2}+1\right )^{\frac {3}{2}}}\)	\(17\)
default	\(-\frac {\sqrt {x}}{b \left (b x +2\right )^{\frac {3}{2}}}+\frac {\frac {\sqrt {x}}{3 \left (b x +2\right )^{\frac {3}{2}}}+\frac {\sqrt {x}}{3 \sqrt {b x +2}}}{b}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(b*x+2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 12, normalized size = 0.67 \begin {gather*} \frac {x^{\frac {3}{2}}}{3 \, {\left (b x + 2\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(b*x+2)^(5/2),x, algorithm="maxima")

[Out]

1/3*x^(3/2)/(b*x + 2)^(3/2)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (12) = 24\).
time = 0.68, size = 27, normalized size = 1.50 \begin {gather*} \frac {\sqrt {b x + 2} x^{\frac {3}{2}}}{3 \, {\left (b^{2} x^{2} + 4 \, b x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(b*x+2)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.74, size = 27, normalized size = 1.50 \begin {gather*} \frac {x^{\frac {3}{2}}}{3 b x \sqrt {b x + 2} + 6 \sqrt {b x + 2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(1/2)/(b*x+2)**(5/2),x)

[Out]

x**(3/2)/(3*b*x*sqrt(b*x + 2) + 6*sqrt(b*x + 2))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (12) = 24\).
time = 2.05, size = 82, normalized size = 4.56 \begin {gather*} \frac {4 \, {\left (3 \, {\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{4} \sqrt {b} + 4 \, b^{\frac {5}{2}}\right )} {\left | b \right |}}{3 \, {\left ({\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2} + 2 \, b\right )}^{3} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(b*x+2)^(5/2),x, algorithm="giac")

[Out]

4/3*(3*(sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^4*sqrt(b) + 4*b^(5/2))*abs(b)/(((sqrt(b*x + 2)*sqrt(b
) - sqrt((b*x + 2)*b - 2*b))^2 + 2*b)^3*b^2)

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Mupad [B]
time = 0.25, size = 12, normalized size = 0.67 \begin {gather*} \frac {x^{3/2}}{3\,{\left (b\,x+2\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(b*x + 2)^(5/2),x)

[Out]

x^(3/2)/(3*(b*x + 2)^(3/2))

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