### 3.64 $$\int \frac{x^2}{(a x+b x^2)^{5/2}} \, dx$$

Optimal. Leaf size=51 $\frac{2 (a+2 b x)}{3 a^2 b \sqrt{a x+b x^2}}-\frac{2 x}{3 b \left (a x+b x^2\right )^{3/2}}$

[Out]

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

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Rubi [A]  time = 0.0134114, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.118, Rules used = {652, 613} $\frac{2 (a+2 b x)}{3 a^2 b \sqrt{a x+b x^2}}-\frac{2 x}{3 b \left (a x+b x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^2/(a*x + b*x^2)^(5/2),x]

[Out]

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

Rule 652

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

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

Mathematica [A]  time = 0.0110211, size = 29, normalized size = 0.57 $\frac{2 x^2 (3 a+2 b x)}{3 a^2 (x (a+b x))^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^2/(a*x + b*x^2)^(5/2),x]

[Out]

(2*x^2*(3*a + 2*b*x))/(3*a^2*(x*(a + b*x))^(3/2))

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Maple [A]  time = 0.045, size = 33, normalized size = 0.7 \begin{align*}{\frac{2\,{x}^{3} \left ( bx+a \right ) \left ( 2\,bx+3\,a \right ) }{3\,{a}^{2}} \left ( b{x}^{2}+ax \right ) ^{-{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.1577, size = 73, normalized size = 1.43 \begin{align*} \frac{4 \, x}{3 \, \sqrt{b x^{2} + a x} a^{2}} - \frac{2 \, x}{3 \,{\left (b x^{2} + a x\right )}^{\frac{3}{2}} b} + \frac{2}{3 \, \sqrt{b x^{2} + a x} a b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.07175, size = 93, normalized size = 1.82 \begin{align*} \frac{2 \, \sqrt{b x^{2} + a x}{\left (2 \, b x + 3 \, a\right )}}{3 \,{\left (a^{2} b^{2} x^{2} + 2 \, a^{3} b x + a^{4}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.21426, size = 82, normalized size = 1.61 \begin{align*} \frac{2 \,{\left (3 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )} b + 2 \, a \sqrt{b}\right )}}{3 \,{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )} \sqrt{b} + a\right )}^{3} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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