3.7.28 \(\int \frac {1}{x^{5/2} (2+b x)^{5/2}} \, dx\)

Optimal. Leaf size=71 \[ -\frac {2 \sqrt {b x+2}}{3 x^{3/2}}+\frac {1}{x^{3/2} \sqrt {b x+2}}+\frac {1}{3 x^{3/2} (b x+2)^{3/2}}+\frac {2 b \sqrt {b x+2}}{3 \sqrt {x}} \]

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Rubi [A]  time = 0.01, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, 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 \sqrt {b x+2}}{3 x^{3/2}}+\frac {1}{x^{3/2} \sqrt {b x+2}}+\frac {1}{3 x^{3/2} (b x+2)^{3/2}}+\frac {2 b \sqrt {b x+2}}{3 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*(2 + b*x)^(5/2)),x]

[Out]

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

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Mathematica [A]  time = 0.01, size = 40, normalized size = 0.56 \begin {gather*} \frac {2 b^3 x^3+6 b^2 x^2+3 b x-1}{3 x^{3/2} (b x+2)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*(2 + b*x)^(5/2)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.10, size = 40, normalized size = 0.56 \begin {gather*} \frac {2 b^3 x^3+6 b^2 x^2+3 b x-1}{3 x^{3/2} (b x+2)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^(5/2)*(2 + b*x)^(5/2)),x]

[Out]

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

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fricas [A]  time = 1.05, size = 55, normalized size = 0.77 \begin {gather*} \frac {{\left (2 \, b^{3} x^{3} + 6 \, b^{2} x^{2} + 3 \, b x - 1\right )} \sqrt {b x + 2} \sqrt {x}}{3 \, {\left (b^{2} x^{4} + 4 \, b x^{3} + 4 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x+2)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.27, size = 158, normalized size = 2.23 \begin {gather*} \frac {{\left (4 \, {\left (b x + 2\right )} b^{2} {\left | b \right |} - 9 \, b^{2} {\left | b \right |}\right )} \sqrt {b x + 2}}{12 \, {\left ({\left (b x + 2\right )} b - 2 \, b\right )}^{\frac {3}{2}}} + \frac {3 \, {\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{4} b^{\frac {7}{2}} + 18 \, {\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2} b^{\frac {9}{2}} + 16 \, b^{\frac {11}{2}}}{3 \, {\left ({\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2} + 2 \, b\right )}^{3} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x+2)^(5/2),x, algorithm="giac")

[Out]

1/12*(4*(b*x + 2)*b^2*abs(b) - 9*b^2*abs(b))*sqrt(b*x + 2)/((b*x + 2)*b - 2*b)^(3/2) + 1/3*(3*(sqrt(b*x + 2)*s
qrt(b) - sqrt((b*x + 2)*b - 2*b))^4*b^(7/2) + 18*(sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^2*b^(9/2) +
16*b^(11/2))/(((sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^2 + 2*b)^3*abs(b))

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maple [A]  time = 0.00, size = 35, normalized size = 0.49 \begin {gather*} \frac {2 b^{3} x^{3}+6 b^{2} x^{2}+3 b x -1}{3 \left (b x +2\right )^{\frac {3}{2}} x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/2)/(b*x+2)^(5/2),x)

[Out]

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

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maxima [A]  time = 1.28, size = 55, normalized size = 0.77 \begin {gather*} \frac {3 \, \sqrt {b x + 2} b}{8 \, \sqrt {x}} - \frac {{\left (b^{3} - \frac {9 \, {\left (b x + 2\right )} b^{2}}{x}\right )} x^{\frac {3}{2}}}{24 \, {\left (b x + 2\right )}^{\frac {3}{2}}} - \frac {{\left (b x + 2\right )}^{\frac {3}{2}}}{24 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x+2)^(5/2),x, algorithm="maxima")

[Out]

3/8*sqrt(b*x + 2)*b/sqrt(x) - 1/24*(b^3 - 9*(b*x + 2)*b^2/x)*x^(3/2)/(b*x + 2)^(3/2) - 1/24*(b*x + 2)^(3/2)/x^
(3/2)

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mupad [B]  time = 0.42, size = 71, normalized size = 1.00 \begin {gather*} \frac {3\,b\,x\,\sqrt {b\,x+2}-\sqrt {b\,x+2}+6\,b^2\,x^2\,\sqrt {b\,x+2}+2\,b^3\,x^3\,\sqrt {b\,x+2}}{x^{3/2}\,\left (x\,\left (3\,x\,b^2+12\,b\right )+12\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/2)*(b*x + 2)^(5/2)),x)

[Out]

(3*b*x*(b*x + 2)^(1/2) - (b*x + 2)^(1/2) + 6*b^2*x^2*(b*x + 2)^(1/2) + 2*b^3*x^3*(b*x + 2)^(1/2))/(x^(3/2)*(x*
(12*b + 3*b^2*x) + 12))

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sympy [B]  time = 6.85, size = 257, normalized size = 3.62 \begin {gather*} \frac {2 b^{\frac {27}{2}} x^{4} \sqrt {1 + \frac {2}{b x}}}{3 b^{12} x^{4} + 18 b^{11} x^{3} + 36 b^{10} x^{2} + 24 b^{9} x} + \frac {10 b^{\frac {25}{2}} x^{3} \sqrt {1 + \frac {2}{b x}}}{3 b^{12} x^{4} + 18 b^{11} x^{3} + 36 b^{10} x^{2} + 24 b^{9} x} + \frac {15 b^{\frac {23}{2}} x^{2} \sqrt {1 + \frac {2}{b x}}}{3 b^{12} x^{4} + 18 b^{11} x^{3} + 36 b^{10} x^{2} + 24 b^{9} x} + \frac {5 b^{\frac {21}{2}} x \sqrt {1 + \frac {2}{b x}}}{3 b^{12} x^{4} + 18 b^{11} x^{3} + 36 b^{10} x^{2} + 24 b^{9} x} - \frac {2 b^{\frac {19}{2}} \sqrt {1 + \frac {2}{b x}}}{3 b^{12} x^{4} + 18 b^{11} x^{3} + 36 b^{10} x^{2} + 24 b^{9} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(5/2)/(b*x+2)**(5/2),x)

[Out]

2*b**(27/2)*x**4*sqrt(1 + 2/(b*x))/(3*b**12*x**4 + 18*b**11*x**3 + 36*b**10*x**2 + 24*b**9*x) + 10*b**(25/2)*x
**3*sqrt(1 + 2/(b*x))/(3*b**12*x**4 + 18*b**11*x**3 + 36*b**10*x**2 + 24*b**9*x) + 15*b**(23/2)*x**2*sqrt(1 +
2/(b*x))/(3*b**12*x**4 + 18*b**11*x**3 + 36*b**10*x**2 + 24*b**9*x) + 5*b**(21/2)*x*sqrt(1 + 2/(b*x))/(3*b**12
*x**4 + 18*b**11*x**3 + 36*b**10*x**2 + 24*b**9*x) - 2*b**(19/2)*sqrt(1 + 2/(b*x))/(3*b**12*x**4 + 18*b**11*x*
*3 + 36*b**10*x**2 + 24*b**9*x)

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