3.622 \(\int \frac{1}{x^{7/2} (2+b x)^{3/2}} \, dx\)

Optimal. Leaf size=74 \[ -\frac{2 b^2 \sqrt{b x+2}}{5 \sqrt{x}}+\frac{2 b \sqrt{b x+2}}{5 x^{3/2}}-\frac{3 \sqrt{b x+2}}{5 x^{5/2}}+\frac{1}{x^{5/2} \sqrt{b x+2}} \]

[Out]

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

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Rubi [A]  time = 0.013534, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {45, 37} \[ -\frac{2 b^2 \sqrt{b x+2}}{5 \sqrt{x}}+\frac{2 b \sqrt{b x+2}}{5 x^{3/2}}-\frac{3 \sqrt{b x+2}}{5 x^{5/2}}+\frac{1}{x^{5/2} \sqrt{b x+2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(7/2)*(2 + b*x)^(3/2)),x]

[Out]

1/(x^(5/2)*Sqrt[2 + b*x]) - (3*Sqrt[2 + b*x])/(5*x^(5/2)) + (2*b*Sqrt[2 + b*x])/(5*x^(3/2)) - (2*b^2*Sqrt[2 +
b*x])/(5*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^{7/2} (2+b x)^{3/2}} \, dx &=\frac{1}{x^{5/2} \sqrt{2+b x}}+3 \int \frac{1}{x^{7/2} \sqrt{2+b x}} \, dx\\ &=\frac{1}{x^{5/2} \sqrt{2+b x}}-\frac{3 \sqrt{2+b x}}{5 x^{5/2}}-\frac{1}{5} (6 b) \int \frac{1}{x^{5/2} \sqrt{2+b x}} \, dx\\ &=\frac{1}{x^{5/2} \sqrt{2+b x}}-\frac{3 \sqrt{2+b x}}{5 x^{5/2}}+\frac{2 b \sqrt{2+b x}}{5 x^{3/2}}+\frac{1}{5} \left (2 b^2\right ) \int \frac{1}{x^{3/2} \sqrt{2+b x}} \, dx\\ &=\frac{1}{x^{5/2} \sqrt{2+b x}}-\frac{3 \sqrt{2+b x}}{5 x^{5/2}}+\frac{2 b \sqrt{2+b x}}{5 x^{3/2}}-\frac{2 b^2 \sqrt{2+b x}}{5 \sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.009784, size = 39, normalized size = 0.53 \[ \frac{-2 b^3 x^3-2 b^2 x^2+b x-1}{5 x^{5/2} \sqrt{b x+2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(7/2)*(2 + b*x)^(3/2)),x]

[Out]

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

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Maple [A]  time = 0.004, size = 35, normalized size = 0.5 \begin{align*} -{\frac{2\,{b}^{3}{x}^{3}+2\,{b}^{2}{x}^{2}-bx+1}{5}{x}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{bx+2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(7/2)/(b*x+2)^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.02739, size = 76, normalized size = 1.03 \begin{align*} -\frac{b^{3} \sqrt{x}}{8 \, \sqrt{b x + 2}} - \frac{3 \, \sqrt{b x + 2} b^{2}}{8 \, \sqrt{x}} + \frac{{\left (b x + 2\right )}^{\frac{3}{2}} b}{8 \, x^{\frac{3}{2}}} - \frac{{\left (b x + 2\right )}^{\frac{5}{2}}}{40 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/(b*x+2)^(3/2),x, algorithm="maxima")

[Out]

-1/8*b^3*sqrt(x)/sqrt(b*x + 2) - 3/8*sqrt(b*x + 2)*b^2/sqrt(x) + 1/8*(b*x + 2)^(3/2)*b/x^(3/2) - 1/40*(b*x + 2
)^(5/2)/x^(5/2)

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Fricas [A]  time = 1.58721, size = 105, normalized size = 1.42 \begin{align*} -\frac{{\left (2 \, b^{3} x^{3} + 2 \, b^{2} x^{2} - b x + 1\right )} \sqrt{b x + 2} \sqrt{x}}{5 \,{\left (b x^{4} + 2 \, x^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/(b*x+2)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 55.9985, size = 269, normalized size = 3.64 \begin{align*} - \frac{2 b^{\frac{29}{2}} x^{5} \sqrt{1 + \frac{2}{b x}}}{5 b^{12} x^{5} + 30 b^{11} x^{4} + 60 b^{10} x^{3} + 40 b^{9} x^{2}} - \frac{10 b^{\frac{27}{2}} x^{4} \sqrt{1 + \frac{2}{b x}}}{5 b^{12} x^{5} + 30 b^{11} x^{4} + 60 b^{10} x^{3} + 40 b^{9} x^{2}} - \frac{15 b^{\frac{25}{2}} x^{3} \sqrt{1 + \frac{2}{b x}}}{5 b^{12} x^{5} + 30 b^{11} x^{4} + 60 b^{10} x^{3} + 40 b^{9} x^{2}} - \frac{5 b^{\frac{23}{2}} x^{2} \sqrt{1 + \frac{2}{b x}}}{5 b^{12} x^{5} + 30 b^{11} x^{4} + 60 b^{10} x^{3} + 40 b^{9} x^{2}} - \frac{4 b^{\frac{19}{2}} \sqrt{1 + \frac{2}{b x}}}{5 b^{12} x^{5} + 30 b^{11} x^{4} + 60 b^{10} x^{3} + 40 b^{9} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(7/2)/(b*x+2)**(3/2),x)

[Out]

-2*b**(29/2)*x**5*sqrt(1 + 2/(b*x))/(5*b**12*x**5 + 30*b**11*x**4 + 60*b**10*x**3 + 40*b**9*x**2) - 10*b**(27/
2)*x**4*sqrt(1 + 2/(b*x))/(5*b**12*x**5 + 30*b**11*x**4 + 60*b**10*x**3 + 40*b**9*x**2) - 15*b**(25/2)*x**3*sq
rt(1 + 2/(b*x))/(5*b**12*x**5 + 30*b**11*x**4 + 60*b**10*x**3 + 40*b**9*x**2) - 5*b**(23/2)*x**2*sqrt(1 + 2/(b
*x))/(5*b**12*x**5 + 30*b**11*x**4 + 60*b**10*x**3 + 40*b**9*x**2) - 4*b**(19/2)*sqrt(1 + 2/(b*x))/(5*b**12*x*
*5 + 30*b**11*x**4 + 60*b**10*x**3 + 40*b**9*x**2)

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Giac [B]  time = 1.09133, size = 144, normalized size = 1.95 \begin{align*} -\frac{b^{\frac{9}{2}}}{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 (\frac{60 \, b^{6}}{{\left | b \right |}} +{\left (\frac{11 \,{\left (b x + 2\right )} b^{6}}{{\left | b \right |}} - \frac{50 \, b^{6}}{{\left | b \right |}}\right )}{\left (b x + 2\right )}\right )} \sqrt{b x + 2}}{40 \,{\left ({\left (b x + 2\right )} b - 2 \, b\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/(b*x+2)^(3/2),x, algorithm="giac")

[Out]

-1/2*b^(9/2)/(((sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^2 + 2*b)*abs(b)) - 1/40*(60*b^6/abs(b) + (11*
(b*x + 2)*b^6/abs(b) - 50*b^6/abs(b))*(b*x + 2))*sqrt(b*x + 2)/((b*x + 2)*b - 2*b)^(5/2)