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

Optimal. Leaf size=86 \[ -\frac{16 b^2 \sqrt{a x^2+b x^3}}{15 a^3 x^{3/2}}+\frac{8 b \sqrt{a x^2+b x^3}}{15 a^2 x^{5/2}}-\frac{2 \sqrt{a x^2+b x^3}}{5 a x^{7/2}} \]

[Out]

(-2*Sqrt[a*x^2 + b*x^3])/(5*a*x^(7/2)) + (8*b*Sqrt[a*x^2 + b*x^3])/(15*a^2*x^(5/2)) - (16*b^2*Sqrt[a*x^2 + b*x
^3])/(15*a^3*x^(3/2))

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Rubi [A]  time = 0.1178, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2016, 2014} \[ -\frac{16 b^2 \sqrt{a x^2+b x^3}}{15 a^3 x^{3/2}}+\frac{8 b \sqrt{a x^2+b x^3}}{15 a^2 x^{5/2}}-\frac{2 \sqrt{a x^2+b x^3}}{5 a x^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[a*x^2 + b*x^3])/(5*a*x^(7/2)) + (8*b*Sqrt[a*x^2 + b*x^3])/(15*a^2*x^(5/2)) - (16*b^2*Sqrt[a*x^2 + b*x
^3])/(15*a^3*x^(3/2))

Rule 2016

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \frac{1}{x^{5/2} \sqrt{a x^2+b x^3}} \, dx &=-\frac{2 \sqrt{a x^2+b x^3}}{5 a x^{7/2}}-\frac{(4 b) \int \frac{1}{x^{3/2} \sqrt{a x^2+b x^3}} \, dx}{5 a}\\ &=-\frac{2 \sqrt{a x^2+b x^3}}{5 a x^{7/2}}+\frac{8 b \sqrt{a x^2+b x^3}}{15 a^2 x^{5/2}}+\frac{\left (8 b^2\right ) \int \frac{1}{\sqrt{x} \sqrt{a x^2+b x^3}} \, dx}{15 a^2}\\ &=-\frac{2 \sqrt{a x^2+b x^3}}{5 a x^{7/2}}+\frac{8 b \sqrt{a x^2+b x^3}}{15 a^2 x^{5/2}}-\frac{16 b^2 \sqrt{a x^2+b x^3}}{15 a^3 x^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0152781, size = 44, normalized size = 0.51 \[ -\frac{2 \sqrt{x^2 (a+b x)} \left (3 a^2-4 a b x+8 b^2 x^2\right )}{15 a^3 x^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[x^2*(a + b*x)]*(3*a^2 - 4*a*b*x + 8*b^2*x^2))/(15*a^3*x^(7/2))

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Maple [A]  time = 0.004, size = 46, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 2\,bx+2\,a \right ) \left ( 8\,{b}^{2}{x}^{2}-4\,abx+3\,{a}^{2} \right ) }{15\,{a}^{3}}{x}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{b{x}^{3}+a{x}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(b*x+a)*(8*b^2*x^2-4*a*b*x+3*a^2)/x^(3/2)/a^3/(b*x^3+a*x^2)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b x^{3} + a x^{2}} x^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.789147, size = 96, normalized size = 1.12 \begin{align*} -\frac{2 \,{\left (8 \, b^{2} x^{2} - 4 \, a b x + 3 \, a^{2}\right )} \sqrt{b x^{3} + a x^{2}}}{15 \, a^{3} x^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(8*b^2*x^2 - 4*a*b*x + 3*a^2)*sqrt(b*x^3 + a*x^2)/(a^3*x^(7/2))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{\frac{5}{2}} \sqrt{x^{2} \left (a + b x\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**(5/2)*sqrt(x**2*(a + b*x))), x)

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Giac [A]  time = 1.15967, size = 104, normalized size = 1.21 \begin{align*} \frac{32 \,{\left (10 \,{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{4} - 5 \, a{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} + a^{2}\right )} b^{\frac{5}{2}}}{15 \,{\left ({\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} - a\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/15*(10*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^4 - 5*a*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 + a^2)*b^(5/2)/((sqrt(
b)*sqrt(x) - sqrt(b*x + a))^2 - a)^5