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

Optimal. Leaf size=49 \[ -\frac{2 a^2}{3 b^3 (a+b x)^{3/2}}+\frac{4 a}{b^3 \sqrt{a+b x}}+\frac{2 \sqrt{a+b x}}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.0130616, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {43} \[ -\frac{2 a^2}{3 b^3 (a+b x)^{3/2}}+\frac{4 a}{b^3 \sqrt{a+b x}}+\frac{2 \sqrt{a+b x}}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*x)^(5/2),x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0176182, size = 35, normalized size = 0.71 \[ \frac{2 \left (8 a^2+12 a b x+3 b^2 x^2\right )}{3 b^3 (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*x)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.006, size = 32, normalized size = 0.7 \begin{align*}{\frac{6\,{b}^{2}{x}^{2}+24\,abx+16\,{a}^{2}}{3\,{b}^{3}} \left ( bx+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.14883, size = 55, normalized size = 1.12 \begin{align*} \frac{2 \, \sqrt{b x + a}}{b^{3}} + \frac{4 \, a}{\sqrt{b x + a} b^{3}} - \frac{2 \, a^{2}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.55674, size = 111, normalized size = 2.27 \begin{align*} \frac{2 \,{\left (3 \, b^{2} x^{2} + 12 \, a b x + 8 \, a^{2}\right )} \sqrt{b x + a}}{3 \,{\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.54181, size = 121, normalized size = 2.47 \begin{align*} \begin{cases} \frac{16 a^{2}}{3 a b^{3} \sqrt{a + b x} + 3 b^{4} x \sqrt{a + b x}} + \frac{24 a b x}{3 a b^{3} \sqrt{a + b x} + 3 b^{4} x \sqrt{a + b x}} + \frac{6 b^{2} x^{2}}{3 a b^{3} \sqrt{a + b x} + 3 b^{4} x \sqrt{a + b x}} & \text{for}\: b \neq 0 \\\frac{x^{3}}{3 a^{\frac{5}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((16*a**2/(3*a*b**3*sqrt(a + b*x) + 3*b**4*x*sqrt(a + b*x)) + 24*a*b*x/(3*a*b**3*sqrt(a + b*x) + 3*b*
*4*x*sqrt(a + b*x)) + 6*b**2*x**2/(3*a*b**3*sqrt(a + b*x) + 3*b**4*x*sqrt(a + b*x)), Ne(b, 0)), (x**3/(3*a**(5
/2)), True))

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Giac [A]  time = 1.15255, size = 53, normalized size = 1.08 \begin{align*} \frac{2 \, \sqrt{b x + a}}{b^{3}} + \frac{2 \,{\left (6 \,{\left (b x + a\right )} a - a^{2}\right )}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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