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

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

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 98, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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^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 = 0.03, size = 50, normalized size = 0.51 \begin {gather*} \frac {2 x \left (16 a^3+8 a^2 b x-2 a b^2 x^2+b^3 x^3\right )}{5 b^4 \sqrt {x^2 (a+b x)}} \end {gather*}

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)])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 3.87, size = 54, normalized size = 0.55 \begin {gather*} \frac {2 x \left (5 a^3+15 a^2 (a+b x)-5 a (a+b x)^2+(a+b x)^3\right )}{5 b^4 \sqrt {x^2 (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(5*a^3 + 15*a^2*(a + b*x) - 5*a*(a + b*x)^2 + (a + b*x)^3))/(5*b^4*Sqrt[x^2*(a + b*x)])

________________________________________________________________________________________

fricas [A]  time = 0.40, size = 60, normalized size = 0.61 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{6}}{{\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^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)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 56, normalized size = 0.57 \begin {gather*} \frac {2 \left (b x +a \right ) \left (b^{3} x^{3}-2 a \,b^{2} x^{2}+8 a^{2} b x +16 a^{3}\right ) x^{3}}{5 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} b^{4}} \end {gather*}

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)

________________________________________________________________________________________

maxima [A]  time = 1.48, size = 41, normalized size = 0.42 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

mupad [B]  time = 5.28, size = 57, normalized size = 0.58 \begin {gather*} \frac {2\,\sqrt {b\,x^3+a\,x^2}\,\left (16\,a^3+8\,a^2\,b\,x-2\,a\,b^2\,x^2+b^3\,x^3\right )}{5\,b^4\,x\,\left (a+b\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(a*x^2 + b*x^3)^(3/2),x)

[Out]

(2*(a*x^2 + b*x^3)^(1/2)*(16*a^3 + b^3*x^3 - 2*a*b^2*x^2 + 8*a^2*b*x))/(5*b^4*x*(a + b*x))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{6}}{\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**6/(b*x**3+a*x**2)**(3/2),x)

[Out]

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

________________________________________________________________________________________