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

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

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Rubi [A]  time = 0.11, antiderivative size = 72, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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]

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])

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*}

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Mathematica [A]  time = 0.02, size = 39, normalized size = 0.54 \begin {gather*} \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)}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 3.92, size = 42, normalized size = 0.58 \begin {gather*} \frac {2 x \left (-3 a^2-6 a (a+b x)+(a+b x)^2\right )}{3 b^3 \sqrt {x^2 (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(-3*a^2 - 6*a*(a + b*x) + (a + b*x)^2))/(3*b^3*Sqrt[x^2*(a + b*x)])

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fricas [A]  time = 0.38, size = 49, normalized size = 0.68 \begin {gather*} \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 {gather*}

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5}}{{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}}}\,{d x} \end {gather*}

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)

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maple [A]  time = 0.05, size = 46, normalized size = 0.64 \begin {gather*} -\frac {2 \left (b x +a \right ) \left (-b^{2} x^{2}+4 a b x +8 a^{2}\right ) x^{3}}{3 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} b^{3}} \end {gather*}

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 = 1.44, size = 30, normalized size = 0.42 \begin {gather*} \frac {2 \, {\left (b^{2} x^{2} - 4 \, a b x - 8 \, a^{2}\right )}}{3 \, \sqrt {b x + a} b^{3}} \end {gather*}

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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mupad [B]  time = 5.22, size = 47, normalized size = 0.65 \begin {gather*} -\frac {2\,\sqrt {b\,x^3+a\,x^2}\,\left (8\,a^2+4\,a\,b\,x-b^2\,x^2\right )}{3\,b^3\,x\,\left (a+b\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a*x^2 + b*x^3)^(3/2),x)

[Out]

-(2*(a*x^2 + b*x^3)^(1/2)*(8*a^2 - b^2*x^2 + 4*a*b*x))/(3*b^3*x*(a + b*x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5}}{\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

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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