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

Optimal. Leaf size=48 \[ \frac{2 x}{3 a \left (a x+b x^2\right )^{3/2}}-\frac{8 (a+2 b x)}{3 a^3 \sqrt{a x+b x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0111226, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {638, 613} \[ \frac{2 x}{3 a \left (a x+b x^2\right )^{3/2}}-\frac{8 (a+2 b x)}{3 a^3 \sqrt{a x+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a*x + b*x^2)^(5/2),x]

[Out]

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

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

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

Mathematica [A]  time = 0.0128015, size = 38, normalized size = 0.79 \[ -\frac{2 x \left (3 a^2+12 a b x+8 b^2 x^2\right )}{3 a^3 (x (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a*x + b*x^2)^(5/2),x]

[Out]

(-2*x*(3*a^2 + 12*a*b*x + 8*b^2*x^2))/(3*a^3*(x*(a + b*x))^(3/2))

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Maple [A]  time = 0.049, size = 44, normalized size = 0.9 \begin{align*} -{\frac{2\,{x}^{2} \left ( bx+a \right ) \left ( 8\,{b}^{2}{x}^{2}+12\,bxa+3\,{a}^{2} \right ) }{3\,{a}^{3}} \left ( b{x}^{2}+ax \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*x^2*(b*x+a)*(8*b^2*x^2+12*a*b*x+3*a^2)/a^3/(b*x^2+a*x)^(5/2)

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Maxima [A]  time = 1.04203, size = 70, normalized size = 1.46 \begin{align*} \frac{2 \, x}{3 \,{\left (b x^{2} + a x\right )}^{\frac{3}{2}} a} - \frac{16 \, b x}{3 \, \sqrt{b x^{2} + a x} a^{3}} - \frac{8}{3 \, \sqrt{b x^{2} + a x} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.86979, size = 123, normalized size = 2.56 \begin{align*} -\frac{2 \,{\left (8 \, b^{2} x^{2} + 12 \, a b x + 3 \, a^{2}\right )} \sqrt{b x^{2} + a x}}{3 \,{\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(8*b^2*x^2 + 12*a*b*x + 3*a^2)*sqrt(b*x^2 + a*x)/(a^3*b^2*x^3 + 2*a^4*b*x^2 + a^5*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/(x*(a + b*x))**(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(b*x^2 + a*x)^(5/2), x)