3.6.19 \(\int \frac {\sqrt {2-b x}}{x^{7/2}} \, dx\)

Optimal. Leaf size=40 \[ -\frac {b (2-b x)^{3/2}}{15 x^{3/2}}-\frac {(2-b x)^{3/2}}{5 x^{5/2}} \]

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Rubi [A]  time = 0.00, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {45, 37} \begin {gather*} -\frac {b (2-b x)^{3/2}}{15 x^{3/2}}-\frac {(2-b x)^{3/2}}{5 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[2 - b*x]/x^(7/2),x]

[Out]

-(2 - b*x)^(3/2)/(5*x^(5/2)) - (b*(2 - b*x)^(3/2))/(15*x^(3/2))

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 {\sqrt {2-b x}}{x^{7/2}} \, dx &=-\frac {(2-b x)^{3/2}}{5 x^{5/2}}+\frac {1}{5} b \int \frac {\sqrt {2-b x}}{x^{5/2}} \, dx\\ &=-\frac {(2-b x)^{3/2}}{5 x^{5/2}}-\frac {b (2-b x)^{3/2}}{15 x^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 24, normalized size = 0.60 \begin {gather*} -\frac {(2-b x)^{3/2} (b x+3)}{15 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[2 - b*x]/x^(7/2),x]

[Out]

-1/15*((2 - b*x)^(3/2)*(3 + b*x))/x^(5/2)

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IntegrateAlgebraic [A]  time = 0.08, size = 31, normalized size = 0.78 \begin {gather*} \frac {\sqrt {2-b x} \left (b^2 x^2+b x-6\right )}{15 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[2 - b*x]/x^(7/2),x]

[Out]

(Sqrt[2 - b*x]*(-6 + b*x + b^2*x^2))/(15*x^(5/2))

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fricas [A]  time = 0.67, size = 25, normalized size = 0.62 \begin {gather*} \frac {{\left (b^{2} x^{2} + b x - 6\right )} \sqrt {-b x + 2}}{15 \, x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(b^2*x^2 + b*x - 6)*sqrt(-b*x + 2)/x^(5/2)

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giac [A]  time = 0.85, size = 48, normalized size = 1.20 \begin {gather*} \frac {{\left ({\left (b x - 2\right )} b^{5} + 5 \, b^{5}\right )} {\left (b x - 2\right )} \sqrt {-b x + 2} b}{15 \, {\left ({\left (b x - 2\right )} b + 2 \, b\right )}^{\frac {5}{2}} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*((b*x - 2)*b^5 + 5*b^5)*(b*x - 2)*sqrt(-b*x + 2)*b/(((b*x - 2)*b + 2*b)^(5/2)*abs(b))

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maple [A]  time = 0.00, size = 19, normalized size = 0.48 \begin {gather*} -\frac {\left (b x +3\right ) \left (-b x +2\right )^{\frac {3}{2}}}{15 x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(b*x+3)*(-b*x+2)^(3/2)/x^(5/2)

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maxima [A]  time = 1.33, size = 28, normalized size = 0.70 \begin {gather*} -\frac {{\left (-b x + 2\right )}^{\frac {3}{2}} b}{6 \, x^{\frac {3}{2}}} - \frac {{\left (-b x + 2\right )}^{\frac {5}{2}}}{10 \, x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(-b*x + 2)^(3/2)*b/x^(3/2) - 1/10*(-b*x + 2)^(5/2)/x^(5/2)

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mupad [B]  time = 0.22, size = 26, normalized size = 0.65 \begin {gather*} \frac {\sqrt {2-b\,x}\,\left (\frac {b^2\,x^2}{15}+\frac {b\,x}{15}-\frac {2}{5}\right )}{x^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2 - b*x)^(1/2)/x^(7/2),x)

[Out]

((2 - b*x)^(1/2)*((b*x)/15 + (b^2*x^2)/15 - 2/5))/x^(5/2)

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sympy [A]  time = 4.91, size = 194, normalized size = 4.85 \begin {gather*} \begin {cases} \frac {b^{\frac {5}{2}} \sqrt {-1 + \frac {2}{b x}}}{15} + \frac {b^{\frac {3}{2}} \sqrt {-1 + \frac {2}{b x}}}{15 x} - \frac {2 \sqrt {b} \sqrt {-1 + \frac {2}{b x}}}{5 x^{2}} & \text {for}\: \frac {2}{\left |{b x}\right |} > 1 \\- \frac {i b^{\frac {9}{2}} x^{2} \sqrt {1 - \frac {2}{b x}}}{- 15 b^{2} x^{2} + 30 b x} + \frac {i b^{\frac {7}{2}} x \sqrt {1 - \frac {2}{b x}}}{- 15 b^{2} x^{2} + 30 b x} + \frac {8 i b^{\frac {5}{2}} \sqrt {1 - \frac {2}{b x}}}{- 15 b^{2} x^{2} + 30 b x} - \frac {12 i b^{\frac {3}{2}} \sqrt {1 - \frac {2}{b x}}}{x \left (- 15 b^{2} x^{2} + 30 b x\right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((b**(5/2)*sqrt(-1 + 2/(b*x))/15 + b**(3/2)*sqrt(-1 + 2/(b*x))/(15*x) - 2*sqrt(b)*sqrt(-1 + 2/(b*x))/
(5*x**2), 2/Abs(b*x) > 1), (-I*b**(9/2)*x**2*sqrt(1 - 2/(b*x))/(-15*b**2*x**2 + 30*b*x) + I*b**(7/2)*x*sqrt(1
- 2/(b*x))/(-15*b**2*x**2 + 30*b*x) + 8*I*b**(5/2)*sqrt(1 - 2/(b*x))/(-15*b**2*x**2 + 30*b*x) - 12*I*b**(3/2)*
sqrt(1 - 2/(b*x))/(x*(-15*b**2*x**2 + 30*b*x)), True))

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