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

Optimal. Leaf size=72 \[ \frac{8 \sqrt{a x^2+b x^3}}{3 b^2}-\frac{16 a \sqrt{a x^2+b x^3}}{3 b^3 x}-\frac{2 x^3}{b \sqrt{a x^2+b x^3}} \]

[Out]

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

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Rubi [A]  time = 0.105215, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2015, 2016, 1588} \[ \frac{8 \sqrt{a x^2+b x^3}}{3 b^2}-\frac{16 a \sqrt{a x^2+b x^3}}{3 b^3 x}-\frac{2 x^3}{b \sqrt{a x^2+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a*x^2 + b*x^3)^(3/2),x]

[Out]

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

Rule 2015

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] + Dist[(c^j*(m + n*p + n - j + 1))/(a*(n - j)*(p + 1)),
Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, j, m, n}, x] &&  !IntegerQ[p] && NeQ[n, j
] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && LtQ[p, -1] && (IntegerQ[j] || GtQ[c, 0])

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 1588

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*x^(p - q
+ 1)*Qq^(m + 1))/((p + m*q + 1)*Coeff[Qq, x, q]), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0172749, size = 39, normalized size = 0.54 \[ \frac{2 x \left (-8 a^2-4 a b x+b^2 x^2\right )}{3 b^3 \sqrt{x^2 (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 46, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2\,bx+2\,a \right ) \left ( -{b}^{2}{x}^{2}+4\,abx+8\,{a}^{2} \right ){x}^{3}}{3\,{b}^{3}} \left ( b{x}^{3}+a{x}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.14535, size = 41, normalized size = 0.57 \begin{align*} \frac{2 \,{\left (b^{2} x^{2} - 4 \, a b x - 8 \, a^{2}\right )}}{3 \, \sqrt{b x + a} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.829334, size = 99, normalized size = 1.38 \begin{align*} \frac{2 \,{\left (b^{2} x^{2} - 4 \, a b x - 8 \, a^{2}\right )} \sqrt{b x^{3} + a x^{2}}}{3 \,{\left (b^{4} x^{2} + a b^{3} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{\left (x^{2} \left (a + b x\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{{\left (b x^{3} + a x^{2}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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