3.7.21 \(\int \frac {1}{x^{5/2} (2+b x)^{3/2}} \, dx\)

Optimal. Leaf size=53 \[ -\frac {2 \sqrt {b x+2}}{3 x^{3/2}}+\frac {1}{x^{3/2} \sqrt {b x+2}}+\frac {2 b \sqrt {b x+2}}{3 \sqrt {x}} \]

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Rubi [A]  time = 0.01, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {45, 37} \begin {gather*} -\frac {2 \sqrt {b x+2}}{3 x^{3/2}}+\frac {1}{x^{3/2} \sqrt {b x+2}}+\frac {2 b \sqrt {b x+2}}{3 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*(2 + b*x)^(3/2)),x]

[Out]

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

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]

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])

Rubi steps

\begin {align*} \int \frac {1}{x^{5/2} (2+b x)^{3/2}} \, dx &=\frac {1}{x^{3/2} \sqrt {2+b x}}+2 \int \frac {1}{x^{5/2} \sqrt {2+b x}} \, dx\\ &=\frac {1}{x^{3/2} \sqrt {2+b x}}-\frac {2 \sqrt {2+b x}}{3 x^{3/2}}-\frac {1}{3} (2 b) \int \frac {1}{x^{3/2} \sqrt {2+b x}} \, dx\\ &=\frac {1}{x^{3/2} \sqrt {2+b x}}-\frac {2 \sqrt {2+b x}}{3 x^{3/2}}+\frac {2 b \sqrt {2+b x}}{3 \sqrt {x}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 32, normalized size = 0.60 \begin {gather*} \frac {2 b^2 x^2+2 b x-1}{3 x^{3/2} \sqrt {b x+2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*(2 + b*x)^(3/2)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.09, size = 32, normalized size = 0.60 \begin {gather*} \frac {2 b^2 x^2+2 b x-1}{3 x^{3/2} \sqrt {b x+2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^(5/2)*(2 + b*x)^(3/2)),x]

[Out]

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

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fricas [A]  time = 0.64, size = 39, normalized size = 0.74 \begin {gather*} \frac {{\left (2 \, b^{2} x^{2} + 2 \, b x - 1\right )} \sqrt {b x + 2} \sqrt {x}}{3 \, {\left (b x^{3} + 2 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x+2)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.23, size = 86, normalized size = 1.62 \begin {gather*} \frac {b^{\frac {7}{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 |}} + \frac {{\left (5 \, {\left (b x + 2\right )} b^{2} {\left | b \right |} - 12 \, b^{2} {\left | b \right |}\right )} \sqrt {b x + 2}}{12 \, {\left ({\left (b x + 2\right )} b - 2 \, b\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x+2)^(3/2),x, algorithm="giac")

[Out]

b^(7/2)/(((sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^2 + 2*b)*abs(b)) + 1/12*(5*(b*x + 2)*b^2*abs(b) -
12*b^2*abs(b))*sqrt(b*x + 2)/((b*x + 2)*b - 2*b)^(3/2)

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maple [A]  time = 0.00, size = 27, normalized size = 0.51 \begin {gather*} \frac {2 b^{2} x^{2}+2 b x -1}{3 \sqrt {b x +2}\, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/2)/(b*x+2)^(3/2),x)

[Out]

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

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maxima [A]  time = 1.37, size = 41, normalized size = 0.77 \begin {gather*} \frac {b^{2} \sqrt {x}}{4 \, \sqrt {b x + 2}} + \frac {\sqrt {b x + 2} b}{2 \, \sqrt {x}} - \frac {{\left (b x + 2\right )}^{\frac {3}{2}}}{12 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x+2)^(3/2),x, algorithm="maxima")

[Out]

1/4*b^2*sqrt(x)/sqrt(b*x + 2) + 1/2*sqrt(b*x + 2)*b/sqrt(x) - 1/12*(b*x + 2)^(3/2)/x^(3/2)

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mupad [B]  time = 0.38, size = 37, normalized size = 0.70 \begin {gather*} \frac {\sqrt {b\,x+2}\,\left (\frac {2\,x}{3}+\frac {2\,b\,x^2}{3}-\frac {1}{3\,b}\right )}{x^{5/2}+\frac {2\,x^{3/2}}{b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/2)*(b*x + 2)^(3/2)),x)

[Out]

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

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sympy [B]  time = 3.86, size = 170, normalized size = 3.21 \begin {gather*} \frac {2 b^{\frac {15}{2}} x^{3} \sqrt {1 + \frac {2}{b x}}}{3 b^{6} x^{3} + 12 b^{5} x^{2} + 12 b^{4} x} + \frac {6 b^{\frac {13}{2}} x^{2} \sqrt {1 + \frac {2}{b x}}}{3 b^{6} x^{3} + 12 b^{5} x^{2} + 12 b^{4} x} + \frac {3 b^{\frac {11}{2}} x \sqrt {1 + \frac {2}{b x}}}{3 b^{6} x^{3} + 12 b^{5} x^{2} + 12 b^{4} x} - \frac {2 b^{\frac {9}{2}} \sqrt {1 + \frac {2}{b x}}}{3 b^{6} x^{3} + 12 b^{5} x^{2} + 12 b^{4} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(5/2)/(b*x+2)**(3/2),x)

[Out]

2*b**(15/2)*x**3*sqrt(1 + 2/(b*x))/(3*b**6*x**3 + 12*b**5*x**2 + 12*b**4*x) + 6*b**(13/2)*x**2*sqrt(1 + 2/(b*x
))/(3*b**6*x**3 + 12*b**5*x**2 + 12*b**4*x) + 3*b**(11/2)*x*sqrt(1 + 2/(b*x))/(3*b**6*x**3 + 12*b**5*x**2 + 12
*b**4*x) - 2*b**(9/2)*sqrt(1 + 2/(b*x))/(3*b**6*x**3 + 12*b**5*x**2 + 12*b**4*x)

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