3.620 \(\int \frac{1}{x^{3/2} (2+b x)^{3/2}} \, dx\)

Optimal. Leaf size=32 \[ \frac{1}{\sqrt{x} \sqrt{b x+2}}-\frac{\sqrt{b x+2}}{\sqrt{x}} \]

[Out]

1/(Sqrt[x]*Sqrt[2 + b*x]) - Sqrt[2 + b*x]/Sqrt[x]

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Rubi [A]  time = 0.0030264, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {45, 37} \[ \frac{1}{\sqrt{x} \sqrt{b x+2}}-\frac{\sqrt{b x+2}}{\sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*(2 + b*x)^(3/2)),x]

[Out]

1/(Sqrt[x]*Sqrt[2 + b*x]) - Sqrt[2 + b*x]/Sqrt[x]

Rule 45

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] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

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{1}{x^{3/2} (2+b x)^{3/2}} \, dx &=\frac{1}{\sqrt{x} \sqrt{2+b x}}+\int \frac{1}{x^{3/2} \sqrt{2+b x}} \, dx\\ &=\frac{1}{\sqrt{x} \sqrt{2+b x}}-\frac{\sqrt{2+b x}}{\sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.0061833, size = 21, normalized size = 0.66 \[ \frac{-b x-1}{\sqrt{x} \sqrt{b x+2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(3/2)*(2 + b*x)^(3/2)),x]

[Out]

(-1 - b*x)/(Sqrt[x]*Sqrt[2 + b*x])

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Maple [A]  time = 0.002, size = 18, normalized size = 0.6 \begin{align*} -{(bx+1){\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{bx+2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.04309, size = 35, normalized size = 1.09 \begin{align*} -\frac{b \sqrt{x}}{2 \, \sqrt{b x + 2}} - \frac{\sqrt{b x + 2}}{2 \, \sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*sqrt(x)/sqrt(b*x + 2) - 1/2*sqrt(b*x + 2)/sqrt(x)

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Fricas [A]  time = 1.56168, size = 65, normalized size = 2.03 \begin{align*} -\frac{\sqrt{b x + 2}{\left (b x + 1\right )} \sqrt{x}}{b x^{2} + 2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.52736, size = 34, normalized size = 1.06 \begin{align*} - \frac{\sqrt{b}}{\sqrt{1 + \frac{2}{b x}}} - \frac{1}{\sqrt{b} x \sqrt{1 + \frac{2}{b x}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(b)/sqrt(1 + 2/(b*x)) - 1/(sqrt(b)*x*sqrt(1 + 2/(b*x)))

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Giac [B]  time = 1.08926, size = 100, normalized size = 3.12 \begin{align*} -\frac{\sqrt{b x + 2} b^{2}}{2 \, \sqrt{{\left (b x + 2\right )} b - 2 \, b}{\left | b \right |}} - \frac{2 \, b^{\frac{5}{2}}}{{\left ({\left (\sqrt{b x + 2} \sqrt{b} - \sqrt{{\left (b x + 2\right )} b - 2 \, b}\right )}^{2} + 2 \, b\right )}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(b*x + 2)*b^2/(sqrt((b*x + 2)*b - 2*b)*abs(b)) - 2*b^(5/2)/(((sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*
b - 2*b))^2 + 2*b)*abs(b))