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

Optimal. Leaf size=54 \[ \frac{16 b (a+2 b x)}{3 a^4 \sqrt{a x+b x^2}}-\frac{2 (a+2 b x)}{3 a^2 \left (a x+b x^2\right )^{3/2}} \]

[Out]

(-2*(a + 2*b*x))/(3*a^2*(a*x + b*x^2)^(3/2)) + (16*b*(a + 2*b*x))/(3*a^4*Sqrt[a*x + b*x^2])

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Rubi [A]  time = 0.0090646, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {614, 613} \[ \frac{16 b (a+2 b x)}{3 a^4 \sqrt{a x+b x^2}}-\frac{2 (a+2 b x)}{3 a^2 \left (a x+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^2)^(-5/2),x]

[Out]

(-2*(a + 2*b*x))/(3*a^2*(a*x + b*x^2)^(3/2)) + (16*b*(a + 2*b*x))/(3*a^4*Sqrt[a*x + b*x^2])

Rule 614

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

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

Mathematica [A]  time = 0.0139228, size = 48, normalized size = 0.89 \[ \frac{12 a^2 b x-2 a^3+48 a b^2 x^2+32 b^3 x^3}{3 a^4 (x (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^2)^(-5/2),x]

[Out]

(-2*a^3 + 12*a^2*b*x + 48*a*b^2*x^2 + 32*b^3*x^3)/(3*a^4*(x*(a + b*x))^(3/2))

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Maple [A]  time = 0.045, size = 51, normalized size = 0.9 \begin{align*} -{\frac{2\,x \left ( bx+a \right ) \left ( -16\,{b}^{3}{x}^{3}-24\,a{b}^{2}{x}^{2}-6\,bx{a}^{2}+{a}^{3} \right ) }{3\,{a}^{4}} \left ( b{x}^{2}+ax \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*x*(b*x+a)*(-16*b^3*x^3-24*a*b^2*x^2-6*a^2*b*x+a^3)/a^4/(b*x^2+a*x)^(5/2)

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Maxima [A]  time = 1.18368, size = 97, normalized size = 1.8 \begin{align*} -\frac{4 \, b x}{3 \,{\left (b x^{2} + a x\right )}^{\frac{3}{2}} a^{2}} + \frac{32 \, b^{2} x}{3 \, \sqrt{b x^{2} + a x} a^{4}} - \frac{2}{3 \,{\left (b x^{2} + a x\right )}^{\frac{3}{2}} a} + \frac{16 \, b}{3 \, \sqrt{b x^{2} + a x} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3*b*x/((b*x^2 + a*x)^(3/2)*a^2) + 32/3*b^2*x/(sqrt(b*x^2 + a*x)*a^4) - 2/3/((b*x^2 + a*x)^(3/2)*a) + 16/3*b
/(sqrt(b*x^2 + a*x)*a^3)

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Fricas [A]  time = 2.01023, size = 144, normalized size = 2.67 \begin{align*} \frac{2 \,{\left (16 \, b^{3} x^{3} + 24 \, a b^{2} x^{2} + 6 \, a^{2} b x - a^{3}\right )} \sqrt{b x^{2} + a x}}{3 \,{\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(16*b^3*x^3 + 24*a*b^2*x^2 + 6*a^2*b*x - a^3)*sqrt(b*x^2 + a*x)/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*x + b*x**2)**(-5/2), x)

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Giac [A]  time = 1.17887, size = 68, normalized size = 1.26 \begin{align*} \frac{2 \,{\left (2 \,{\left (4 \, x{\left (\frac{2 \, b^{3} x}{a^{4}} + \frac{3 \, b^{2}}{a^{3}}\right )} + \frac{3 \, b}{a^{2}}\right )} x - \frac{1}{a}\right )}}{3 \,{\left (b x^{2} + a x\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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