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

Optimal. Leaf size=38 \[ \frac{b (b x+2)^{3/2}}{15 x^{3/2}}-\frac{(b x+2)^{3/2}}{5 x^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.0039924, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {45, 37} \[ \frac{b (b x+2)^{3/2}}{15 x^{3/2}}-\frac{(b x+2)^{3/2}}{5 x^{5/2}} \]

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 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{\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*}

Mathematica [A]  time = 0.0074926, size = 23, normalized size = 0.61 \[ \frac{(b x-3) (b x+2)^{3/2}}{15 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[2 + b*x]/x^(7/2),x]

[Out]

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

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Maple [A]  time = 0.004, size = 18, normalized size = 0.5 \begin{align*}{\frac{bx-3}{15} \left ( bx+2 \right ) ^{{\frac{3}{2}}}{x}^{-{\frac{5}{2}}}} \end{align*}

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

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Maxima [A]  time = 1.05509, size = 35, normalized size = 0.92 \begin{align*} \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{align*}

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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Fricas [A]  time = 1.58403, size = 65, normalized size = 1.71 \begin{align*} \frac{{\left (b^{2} x^{2} - b x - 6\right )} \sqrt{b x + 2}}{15 \, x^{\frac{5}{2}}} \end{align*}

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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Sympy [A]  time = 17.2822, size = 56, normalized size = 1.47 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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Giac [A]  time = 1.21857, size = 57, normalized size = 1.5 \begin{align*} \frac{{\left ({\left (b x + 2\right )} b^{5} - 5 \, b^{5}\right )}{\left (b x + 2\right )}^{\frac{3}{2}} b}{15 \,{\left ({\left (b x + 2\right )} b - 2 \, b\right )}^{\frac{5}{2}}{\left | b \right |}} \end{align*}

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