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

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

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Rubi [A]  time = 0.01, antiderivative size = 74, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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}} \end {gather*}

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 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^{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*}

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Mathematica [A]  time = 0.01, size = 39, normalized size = 0.53 \begin {gather*} \frac {-2 b^3 x^3-2 b^2 x^2+b x-1}{5 x^{5/2} \sqrt {b x+2}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.10, size = 39, normalized size = 0.53 \begin {gather*} \frac {-2 b^3 x^3-2 b^2 x^2+b x-1}{5 x^{5/2} \sqrt {b x+2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[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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fricas [A]  time = 1.22, size = 47, normalized size = 0.64 \begin {gather*} -\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 {gather*}

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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giac [B]  time = 1.11, size = 107, normalized size = 1.45 \begin {gather*} -\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 {gather*}

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)

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maple [A]  time = 0.00, size = 35, normalized size = 0.47 \begin {gather*} -\frac {2 b^{3} x^{3}+2 b^{2} x^{2}-b x +1}{5 \sqrt {b x +2}\, x^{\frac {5}{2}}} \end {gather*}

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

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maxima [A]  time = 1.30, size = 56, normalized size = 0.76 \begin {gather*} -\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 {gather*}

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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mupad [B]  time = 0.43, size = 46, normalized size = 0.62 \begin {gather*} -\frac {\sqrt {b\,x+2}\,\left (\frac {2\,b\,x^2}{5}-\frac {x}{5}+\frac {1}{5\,b}+\frac {2\,b^2\,x^3}{5}\right )}{x^{7/2}+\frac {2\,x^{5/2}}{b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(7/2)*(b*x + 2)^(3/2)),x)

[Out]

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

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sympy [B]  time = 10.97, size = 269, normalized size = 3.64 \begin {gather*} - \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 {gather*}

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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