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

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

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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 = {45, 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]

-Sqrt[2 + b*x]/(5*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 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} \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 = 0.01, size = 32, normalized size = 0.54 \begin {gather*} -\frac {\sqrt {b x+2} \left (2 b^2 x^2-2 b x+3\right )}{15 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.07, size = 32, normalized size = 0.54 \begin {gather*} \frac {\sqrt {b x+2} \left (-2 b^2 x^2+2 b x-3\right )}{15 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[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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fricas [A]  time = 0.87, 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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giac [A]  time = 1.06, size = 55, normalized size = 0.93 \begin {gather*} -\frac {{\left (15 \, b^{5} + 2 \, {\left ({\left (b x + 2\right )} b^{5} - 5 \, b^{5}\right )} {\left (b x + 2\right )}\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(1/x^(7/2)/(b*x+2)^(1/2),x, algorithm="giac")

[Out]

-1/15*(15*b^5 + 2*((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 = 0.01, size = 27, normalized size = 0.46 \begin {gather*} -\frac {\sqrt {b x +2}\, \left (2 b^{2} x^{2}-2 b x +3\right )}{15 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)^(1/2),x)

[Out]

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

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maxima [A]  time = 1.31, 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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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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sympy [B]  time = 6.10, 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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