3.260 \(\int \frac{x^6}{(a x^2+b x^3)^{3/2}} \, dx\)

Optimal. Leaf size=98 \[ \frac{32 a^2 \sqrt{a x^2+b x^3}}{5 b^4 x}+\frac{12 x \sqrt{a x^2+b x^3}}{5 b^2}-\frac{16 a \sqrt{a x^2+b x^3}}{5 b^3}-\frac{2 x^4}{b \sqrt{a x^2+b x^3}} \]

[Out]

(-2*x^4)/(b*Sqrt[a*x^2 + b*x^3]) - (16*a*Sqrt[a*x^2 + b*x^3])/(5*b^3) + (32*a^2*Sqrt[a*x^2 + b*x^3])/(5*b^4*x)
+ (12*x*Sqrt[a*x^2 + b*x^3])/(5*b^2)

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Rubi [A]  time = 0.150882, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2015, 2016, 1588} \[ \frac{32 a^2 \sqrt{a x^2+b x^3}}{5 b^4 x}+\frac{12 x \sqrt{a x^2+b x^3}}{5 b^2}-\frac{16 a \sqrt{a x^2+b x^3}}{5 b^3}-\frac{2 x^4}{b \sqrt{a x^2+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(a*x^2 + b*x^3)^(3/2),x]

[Out]

(-2*x^4)/(b*Sqrt[a*x^2 + b*x^3]) - (16*a*Sqrt[a*x^2 + b*x^3])/(5*b^3) + (32*a^2*Sqrt[a*x^2 + b*x^3])/(5*b^4*x)
+ (12*x*Sqrt[a*x^2 + b*x^3])/(5*b^2)

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^6}{\left (a x^2+b x^3\right )^{3/2}} \, dx &=-\frac{2 x^4}{b \sqrt{a x^2+b x^3}}+\frac{6 \int \frac{x^3}{\sqrt{a x^2+b x^3}} \, dx}{b}\\ &=-\frac{2 x^4}{b \sqrt{a x^2+b x^3}}+\frac{12 x \sqrt{a x^2+b x^3}}{5 b^2}-\frac{(24 a) \int \frac{x^2}{\sqrt{a x^2+b x^3}} \, dx}{5 b^2}\\ &=-\frac{2 x^4}{b \sqrt{a x^2+b x^3}}-\frac{16 a \sqrt{a x^2+b x^3}}{5 b^3}+\frac{12 x \sqrt{a x^2+b x^3}}{5 b^2}+\frac{\left (16 a^2\right ) \int \frac{x}{\sqrt{a x^2+b x^3}} \, dx}{5 b^3}\\ &=-\frac{2 x^4}{b \sqrt{a x^2+b x^3}}-\frac{16 a \sqrt{a x^2+b x^3}}{5 b^3}+\frac{32 a^2 \sqrt{a x^2+b x^3}}{5 b^4 x}+\frac{12 x \sqrt{a x^2+b x^3}}{5 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0242158, size = 50, normalized size = 0.51 \[ \frac{2 x \left (8 a^2 b x+16 a^3-2 a b^2 x^2+b^3 x^3\right )}{5 b^4 \sqrt{x^2 (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a*x^2 + b*x^3)^(3/2),x]

[Out]

(2*x*(16*a^3 + 8*a^2*b*x - 2*a*b^2*x^2 + b^3*x^3))/(5*b^4*Sqrt[x^2*(a + b*x)])

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Maple [A]  time = 0.005, size = 56, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2\,bx+2\,a \right ) \left ({x}^{3}{b}^{3}-2\,a{b}^{2}{x}^{2}+8\,{a}^{2}xb+16\,{a}^{3} \right ){x}^{3}}{5\,{b}^{4}} \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^6/(b*x^3+a*x^2)^(3/2),x)

[Out]

2/5*(b*x+a)*(b^3*x^3-2*a*b^2*x^2+8*a^2*b*x+16*a^3)*x^3/b^4/(b*x^3+a*x^2)^(3/2)

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Maxima [A]  time = 1.15932, size = 55, normalized size = 0.56 \begin{align*} \frac{2 \,{\left (b^{3} x^{3} - 2 \, a b^{2} x^{2} + 8 \, a^{2} b x + 16 \, a^{3}\right )}}{5 \, \sqrt{b x + a} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^3+a*x^2)^(3/2),x, algorithm="maxima")

[Out]

2/5*(b^3*x^3 - 2*a*b^2*x^2 + 8*a^2*b*x + 16*a^3)/(sqrt(b*x + a)*b^4)

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Fricas [A]  time = 0.905986, size = 122, normalized size = 1.24 \begin{align*} \frac{2 \,{\left (b^{3} x^{3} - 2 \, a b^{2} x^{2} + 8 \, a^{2} b x + 16 \, a^{3}\right )} \sqrt{b x^{3} + a x^{2}}}{5 \,{\left (b^{5} x^{2} + a b^{4} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^3+a*x^2)^(3/2),x, algorithm="fricas")

[Out]

2/5*(b^3*x^3 - 2*a*b^2*x^2 + 8*a^2*b*x + 16*a^3)*sqrt(b*x^3 + a*x^2)/(b^5*x^2 + a*b^4*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{6}}{\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**6/(b*x**3+a*x**2)**(3/2),x)

[Out]

Integral(x**6/(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^{6}}{{\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^6/(b*x^3+a*x^2)^(3/2),x, algorithm="giac")

[Out]

integrate(x^6/(b*x^3 + a*x^2)^(3/2), x)