3.7.14 \(\int \frac {1}{x^{7/2} \sqrt {2+b x}} \, dx\) [614]

Optimal. Leaf size=59 \[ -\frac {\sqrt {2+b x}}{5 x^{5/2}}+\frac {2 b \sqrt {2+b x}}{15 x^{3/2}}-\frac {2 b^2 \sqrt {2+b x}}{15 \sqrt {x}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, 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 {2 b^2 \sqrt {b x+2}}{15 \sqrt {x}}+\frac {2 b \sqrt {b x+2}}{15 x^{3/2}}-\frac {\sqrt {b x+2}}{5 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(7/2)*Sqrt[2 + b*x]),x]

[Out]

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

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Mathematica [A]
time = 0.05, size = 32, normalized size = 0.54 \begin {gather*} \frac {\sqrt {2+b x} \left (-3+2 b x-2 b^2 x^2\right )}{15 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathics [A]
time = 6.51, size = 67, normalized size = 1.14 \begin {gather*} \frac {\sqrt {b} \left (-12-4 b x-3 b^2 x^2-6 b^3 x^3-2 b^4 x^4\right ) \sqrt {\frac {2+b x}{b x}}}{15 x^2 \left (4+4 b x+b^2 x^2\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[1/(x^(7/2)*Sqrt[2 + b*x]),x]')

[Out]

Sqrt[b] (-12 - 4 b x - 3 b ^ 2 x ^ 2 - 6 b ^ 3 x ^ 3 - 2 b ^ 4 x ^ 4) Sqrt[(2 + b x) / (b x)] / (15 x ^ 2 (4 +
4 b x + b ^ 2 x ^ 2))

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Maple [A]
time = 0.12, size = 43, normalized size = 0.73

method	result	size
gosper	\(-\frac {\sqrt {b x +2}\, \left (2 x^{2} b^{2}-2 b x +3\right )}{15 x^{\frac {5}{2}}}\)	\(27\)
meijerg	\(-\frac {\sqrt {2}\, \left (\frac {2}{3} x^{2} b^{2}-\frac {2}{3} b x +1\right ) \sqrt {\frac {b x}{2}+1}}{5 x^{\frac {5}{2}}}\)	\(31\)
risch	\(-\frac {2 b^{3} x^{3}+2 x^{2} b^{2}-b x +6}{15 x^{\frac {5}{2}} \sqrt {b x +2}}\)	\(35\)
default	\(-\frac {\sqrt {b x +2}}{5 x^{\frac {5}{2}}}-\frac {2 b \left (-\frac {\sqrt {b x +2}}{3 x^{\frac {3}{2}}}+\frac {b \sqrt {b x +2}}{3 \sqrt {x}}\right )}{5}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(b*x+2)^(1/2)/x^(5/2)-2/5*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 = 0.27, size = 41, normalized size = 0.69 \begin {gather*} -\frac {\sqrt {b x + 2} b^{2}}{4 \, \sqrt {x}} + \frac {{\left (b x + 2\right )}^{\frac {3}{2}} b}{6 \, x^{\frac {3}{2}}} - \frac {{\left (b x + 2\right )}^{\frac {5}{2}}}{20 \, 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)^(1/2),x, algorithm="maxima")

[Out]

-1/4*sqrt(b*x + 2)*b^2/sqrt(x) + 1/6*(b*x + 2)^(3/2)*b/x^(3/2) - 1/20*(b*x + 2)^(5/2)/x^(5/2)

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Fricas [A]
time = 0.32, size = 26, normalized size = 0.44 \begin {gather*} -\frac {{\left (2 \, b^{2} x^{2} - 2 \, b x + 3\right )} \sqrt {b x + 2}}{15 \, 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)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (53) = 106\).
time = 3.83, size = 224, normalized size = 3.80 \begin {gather*} - \frac {2 b^{\frac {17}{2}} x^{4} \sqrt {1 + \frac {2}{b x}}}{15 b^{6} x^{4} + 60 b^{5} x^{3} + 60 b^{4} x^{2}} - \frac {6 b^{\frac {15}{2}} x^{3} \sqrt {1 + \frac {2}{b x}}}{15 b^{6} x^{4} + 60 b^{5} x^{3} + 60 b^{4} x^{2}} - \frac {3 b^{\frac {13}{2}} x^{2} \sqrt {1 + \frac {2}{b x}}}{15 b^{6} x^{4} + 60 b^{5} x^{3} + 60 b^{4} x^{2}} - \frac {4 b^{\frac {11}{2}} x \sqrt {1 + \frac {2}{b x}}}{15 b^{6} x^{4} + 60 b^{5} x^{3} + 60 b^{4} x^{2}} - \frac {12 b^{\frac {9}{2}} \sqrt {1 + \frac {2}{b x}}}{15 b^{6} x^{4} + 60 b^{5} x^{3} + 60 b^{4} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b**(17/2)*x**4*sqrt(1 + 2/(b*x))/(15*b**6*x**4 + 60*b**5*x**3 + 60*b**4*x**2) - 6*b**(15/2)*x**3*sqrt(1 + 2
/(b*x))/(15*b**6*x**4 + 60*b**5*x**3 + 60*b**4*x**2) - 3*b**(13/2)*x**2*sqrt(1 + 2/(b*x))/(15*b**6*x**4 + 60*b
**5*x**3 + 60*b**4*x**2) - 4*b**(11/2)*x*sqrt(1 + 2/(b*x))/(15*b**6*x**4 + 60*b**5*x**3 + 60*b**4*x**2) - 12*b
**(9/2)*sqrt(1 + 2/(b*x))/(15*b**6*x**4 + 60*b**5*x**3 + 60*b**4*x**2)

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Giac [A]
time = 0.01, size = 97, normalized size = 1.64 \begin {gather*} -\frac {128 \sqrt {b} b^{2} \left (-5 \left (\sqrt {b x+2}-\sqrt {b} \sqrt {x}\right )^{4}+5 \left (\sqrt {b x+2}-\sqrt {b} \sqrt {x}\right )^{2}-2\right )}{30 \left (\left (\sqrt {b x+2}-\sqrt {b} \sqrt {x}\right )^{2}-2\right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

64/15*(5*(sqrt(b)*sqrt(x) - sqrt(b*x + 2))^4 - 5*(sqrt(b)*sqrt(x) - sqrt(b*x + 2))^2 + 2)*b^(5/2)/((sqrt(b)*sq
rt(x) - sqrt(b*x + 2))^2 - 2)^5

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Mupad [B]
time = 0.32, size = 26, normalized size = 0.44 \begin {gather*} -\frac {\sqrt {b\,x+2}\,\left (\frac {2\,b^2\,x^2}{15}-\frac {2\,b\,x}{15}+\frac {1}{5}\right )}{x^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(7/2)*(b*x + 2)^(1/2)),x)

[Out]

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

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