3.582 \(\int \frac {1}{x^{5/2} (a+b x)^{3/2}} \, dx\)

Optimal. Leaf size=63 \[ \frac {16 b \sqrt {a+b x}}{3 a^3 \sqrt {x}}-\frac {8 \sqrt {a+b x}}{3 a^2 x^{3/2}}+\frac {2}{a x^{3/2} \sqrt {a+b x}} \]

[Out]

2/a/x^(3/2)/(b*x+a)^(1/2)-8/3*(b*x+a)^(1/2)/a^2/x^(3/2)+16/3*b*(b*x+a)^(1/2)/a^3/x^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 63, 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} \[ -\frac {8 \sqrt {a+b x}}{3 a^2 x^{3/2}}+\frac {16 b \sqrt {a+b x}}{3 a^3 \sqrt {x}}+\frac {2}{a x^{3/2} \sqrt {a+b x}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*(a + b*x)^(3/2)),x]

[Out]

2/(a*x^(3/2)*Sqrt[a + b*x]) - (8*Sqrt[a + b*x])/(3*a^2*x^(3/2)) + (16*b*Sqrt[a + b*x])/(3*a^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} (a+b x)^{3/2}} \, dx &=\frac {2}{a x^{3/2} \sqrt {a+b x}}+\frac {4 \int \frac {1}{x^{5/2} \sqrt {a+b x}} \, dx}{a}\\ &=\frac {2}{a x^{3/2} \sqrt {a+b x}}-\frac {8 \sqrt {a+b x}}{3 a^2 x^{3/2}}-\frac {(8 b) \int \frac {1}{x^{3/2} \sqrt {a+b x}} \, dx}{3 a^2}\\ &=\frac {2}{a x^{3/2} \sqrt {a+b x}}-\frac {8 \sqrt {a+b x}}{3 a^2 x^{3/2}}+\frac {16 b \sqrt {a+b x}}{3 a^3 \sqrt {x}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 38, normalized size = 0.60 \[ -\frac {2 \left (a^2-4 a b x-8 b^2 x^2\right )}{3 a^3 x^{3/2} \sqrt {a+b x}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*(a + b*x)^(3/2)),x]

[Out]

(-2*(a^2 - 4*a*b*x - 8*b^2*x^2))/(3*a^3*x^(3/2)*Sqrt[a + b*x])

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fricas [A]  time = 0.43, size = 49, normalized size = 0.78 \[ \frac {2 \, {\left (8 \, b^{2} x^{2} + 4 \, a b x - a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{3 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

2/3*(8*b^2*x^2 + 4*a*b*x - a^2)*sqrt(b*x + a)*sqrt(x)/(a^3*b*x^3 + a^4*x^2)

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giac [B]  time = 1.20, size = 98, normalized size = 1.56 \[ \frac {2 \, \sqrt {b x + a} {\left (\frac {5 \, {\left (b x + a\right )} b^{2} {\left | b \right |}}{a^{3}} - \frac {6 \, b^{2} {\left | b \right |}}{a^{2}}\right )}}{3 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {3}{2}}} + \frac {4 \, b^{\frac {7}{2}}}{{\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )} a^{2} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x+a)^(3/2),x, algorithm="giac")

[Out]

2/3*sqrt(b*x + a)*(5*(b*x + a)*b^2*abs(b)/a^3 - 6*b^2*abs(b)/a^2)/((b*x + a)*b - a*b)^(3/2) + 4*b^(7/2)/(((sqr
t(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b)*a^2*abs(b))

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maple [A]  time = 0.00, size = 33, normalized size = 0.52 \[ -\frac {2 \left (-8 b^{2} x^{2}-4 a b x +a^{2}\right )}{3 \sqrt {b x +a}\, a^{3} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/2)/(b*x+a)^(3/2),x)

[Out]

-2/3*(-8*b^2*x^2-4*a*b*x+a^2)/(b*x+a)^(1/2)/x^(3/2)/a^3

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maxima [A]  time = 1.31, size = 50, normalized size = 0.79 \[ \frac {2 \, b^{2} \sqrt {x}}{\sqrt {b x + a} a^{3}} + \frac {2 \, {\left (\frac {6 \, \sqrt {b x + a} b}{\sqrt {x}} - \frac {{\left (b x + a\right )}^{\frac {3}{2}}}{x^{\frac {3}{2}}}\right )}}{3 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.41, size = 46, normalized size = 0.73 \[ \frac {\sqrt {a+b\,x}\,\left (\frac {8\,x}{3\,a^2}-\frac {2}{3\,a\,b}+\frac {16\,b\,x^2}{3\,a^3}\right )}{x^{5/2}+\frac {a\,x^{3/2}}{b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/2)*(a + b*x)^(3/2)),x)

[Out]

((a + b*x)^(1/2)*((8*x)/(3*a^2) - 2/(3*a*b) + (16*b*x^2)/(3*a^3)))/(x^(5/2) + (a*x^(3/2))/b)

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sympy [B]  time = 3.98, size = 219, normalized size = 3.48 \[ - \frac {2 a^{3} b^{\frac {9}{2}} \sqrt {\frac {a}{b x} + 1}}{3 a^{5} b^{4} x + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{3}} + \frac {6 a^{2} b^{\frac {11}{2}} x \sqrt {\frac {a}{b x} + 1}}{3 a^{5} b^{4} x + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{3}} + \frac {24 a b^{\frac {13}{2}} x^{2} \sqrt {\frac {a}{b x} + 1}}{3 a^{5} b^{4} x + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{3}} + \frac {16 b^{\frac {15}{2}} x^{3} \sqrt {\frac {a}{b x} + 1}}{3 a^{5} b^{4} x + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(5/2)/(b*x+a)**(3/2),x)

[Out]

-2*a**3*b**(9/2)*sqrt(a/(b*x) + 1)/(3*a**5*b**4*x + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**3) + 6*a**2*b**(11/2)*x*
sqrt(a/(b*x) + 1)/(3*a**5*b**4*x + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**3) + 24*a*b**(13/2)*x**2*sqrt(a/(b*x) + 1
)/(3*a**5*b**4*x + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**3) + 16*b**(15/2)*x**3*sqrt(a/(b*x) + 1)/(3*a**5*b**4*x +
6*a**4*b**5*x**2 + 3*a**3*b**6*x**3)

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