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

Optimal. Leaf size=55 \[ -\frac {2 \sqrt {b x+2}}{3 \sqrt {x}}+\frac {2}{3 \sqrt {x} \sqrt {b x+2}}+\frac {1}{3 \sqrt {x} (b x+2)^{3/2}} \]

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Rubi [A]  time = 0.01, antiderivative size = 55, 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 \sqrt {x}}+\frac {2}{3 \sqrt {x} \sqrt {b x+2}}+\frac {1}{3 \sqrt {x} (b x+2)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*(2 + b*x)^(5/2)),x]

[Out]

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

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Mathematica [A]  time = 0.01, size = 32, normalized size = 0.58 \begin {gather*} \frac {-2 b^2 x^2-6 b x-3}{3 \sqrt {x} (b x+2)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3 - 6*b*x - 2*b^2*x^2)/(3*Sqrt[x]*(2 + b*x)^(3/2))

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IntegrateAlgebraic [A]  time = 0.09, size = 32, normalized size = 0.58 \begin {gather*} \frac {-2 b^2 x^2-6 b x-3}{3 \sqrt {x} (b x+2)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3 - 6*b*x - 2*b^2*x^2)/(3*Sqrt[x]*(2 + b*x)^(3/2))

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fricas [A]  time = 1.47, size = 45, normalized size = 0.82 \begin {gather*} -\frac {{\left (2 \, b^{2} x^{2} + 6 \, b x + 3\right )} \sqrt {b x + 2} \sqrt {x}}{3 \, {\left (b^{2} x^{3} + 4 \, b x^{2} + 4 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(2*b^2*x^2 + 6*b*x + 3)*sqrt(b*x + 2)*sqrt(x)/(b^2*x^3 + 4*b*x^2 + 4*x)

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giac [B]  time = 1.36, size = 145, normalized size = 2.64 \begin {gather*} -\frac {\sqrt {b x + 2} b^{2}}{4 \, \sqrt {{\left (b x + 2\right )} b - 2 \, b} {\left | b \right |}} - \frac {3 \, {\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{4} b^{\frac {5}{2}} + 24 \, {\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2} b^{\frac {7}{2}} + 20 \, b^{\frac {9}{2}}}{3 \, {\left ({\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2} + 2 \, b\right )}^{3} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(b*x + 2)*b^2/(sqrt((b*x + 2)*b - 2*b)*abs(b)) - 1/3*(3*(sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2
*b))^4*b^(5/2) + 24*(sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^2*b^(7/2) + 20*b^(9/2))/(((sqrt(b*x + 2)
*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^2 + 2*b)^3*abs(b))

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maple [A]  time = 0.00, size = 27, normalized size = 0.49 \begin {gather*} -\frac {2 b^{2} x^{2}+6 b x +3}{3 \left (b x +2\right )^{\frac {3}{2}} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(2*b^2*x^2+6*b*x+3)/(b*x+2)^(3/2)/x^(1/2)

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maxima [A]  time = 1.40, size = 40, normalized size = 0.73 \begin {gather*} \frac {{\left (b^{2} - \frac {6 \, {\left (b x + 2\right )} b}{x}\right )} x^{\frac {3}{2}}}{12 \, {\left (b x + 2\right )}^{\frac {3}{2}}} - \frac {\sqrt {b x + 2}}{4 \, \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(b^2 - 6*(b*x + 2)*b/x)*x^(3/2)/(b*x + 2)^(3/2) - 1/4*sqrt(b*x + 2)/sqrt(x)

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mupad [B]  time = 0.38, size = 57, normalized size = 1.04 \begin {gather*} -\frac {3\,\sqrt {b\,x+2}+6\,b\,x\,\sqrt {b\,x+2}+2\,b^2\,x^2\,\sqrt {b\,x+2}}{\sqrt {x}\,\left (x\,\left (3\,x\,b^2+12\,b\right )+12\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(3/2)*(b*x + 2)^(5/2)),x)

[Out]

-(3*(b*x + 2)^(1/2) + 6*b*x*(b*x + 2)^(1/2) + 2*b^2*x^2*(b*x + 2)^(1/2))/(x^(1/2)*(x*(12*b + 3*b^2*x) + 12))

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sympy [B]  time = 3.89, size = 117, normalized size = 2.13 \begin {gather*} - \frac {2 b^{\frac {13}{2}} x^{2} \sqrt {1 + \frac {2}{b x}}}{3 b^{6} x^{2} + 12 b^{5} x + 12 b^{4}} - \frac {6 b^{\frac {11}{2}} x \sqrt {1 + \frac {2}{b x}}}{3 b^{6} x^{2} + 12 b^{5} x + 12 b^{4}} - \frac {3 b^{\frac {9}{2}} \sqrt {1 + \frac {2}{b x}}}{3 b^{6} x^{2} + 12 b^{5} x + 12 b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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