3.6.11 \(\int \frac {\sqrt {2+b x}}{x^{7/2}} \, dx\) [511]

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

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 38, normalized size of antiderivative = 1.00, 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 = {47, 37} \begin {gather*} \frac {b (b x+2)^{3/2}}{15 x^{3/2}}-\frac {(b x+2)^{3/2}}{5 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[2 + b*x]/x^(7/2),x]

[Out]

-1/5*(2 + b*x)^(3/2)/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 47

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 = 0.06, size = 31, normalized size = 0.82 \begin {gather*} \frac {\sqrt {2+b x} \left (-6-b x+b^2 x^2\right )}{15 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[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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Maple [A]
time = 0.12, size = 43, normalized size = 1.13

method	result	size
gosper	\(\frac {\left (b x +2\right )^{\frac {3}{2}} \left (b x -3\right )}{15 x^{\frac {5}{2}}}\)	\(18\)
meijerg	\(-\frac {2 \sqrt {2}\, \left (-\frac {1}{6} x^{2} b^{2}+\frac {1}{6} b x +1\right ) \sqrt {\frac {b x}{2}+1}}{5 x^{\frac {5}{2}}}\)	\(31\)
risch	\(\frac {b^{3} x^{3}+x^{2} b^{2}-8 b x -12}{15 x^{\frac {5}{2}} \sqrt {b x +2}}\)	\(33\)
default	\(-\frac {2 \sqrt {b x +2}}{5 x^{\frac {5}{2}}}+\frac {b \left (-\frac {\sqrt {b x +2}}{3 x^{\frac {3}{2}}}+\frac {b \sqrt {b x +2}}{3 \sqrt {x}}\right )}{5}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+2)^(1/2)/x^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 26, normalized size = 0.68 \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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Fricas [A]
time = 0.54, size = 25, normalized size = 0.66 \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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Sympy [A]
time = 3.22, size = 56, normalized size = 1.47 \begin {gather*} \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 {gather*}

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 = 0.75, size = 42, normalized size = 1.11 \begin {gather*} \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 {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)^(3/2)*b/(((b*x + 2)*b - 2*b)^(5/2)*abs(b))

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Mupad [B]
time = 0.22, size = 26, normalized size = 0.68 \begin {gather*} -\frac {\sqrt {b\,x+2}\,\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((b*x + 2)^(1/2)/x^(7/2),x)

[Out]

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

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