3.4 \(\int \frac {1}{(a x^3+b x^6)^{5/3}} \, dx\)

Optimal. Leaf size=77 \[ \frac {9 b \sqrt [3]{a x^3+b x^6}}{4 a^3 x^2}-\frac {3 \sqrt [3]{a x^3+b x^6}}{4 a^2 x^5}+\frac {1}{2 a x^2 \left (a x^3+b x^6\right )^{2/3}} \]

[Out]

1/2/a/x^2/(b*x^6+a*x^3)^(2/3)-3/4*(b*x^6+a*x^3)^(1/3)/a^2/x^5+9/4*b*(b*x^6+a*x^3)^(1/3)/a^3/x^2

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Rubi [A]  time = 0.06, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2001, 2016, 2000} \[ \frac {9 b \sqrt [3]{a x^3+b x^6}}{4 a^3 x^2}-\frac {3 \sqrt [3]{a x^3+b x^6}}{4 a^2 x^5}+\frac {1}{2 a x^2 \left (a x^3+b x^6\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a*x^3 + b*x^6)^(-5/3),x]

[Out]

1/(2*a*x^2*(a*x^3 + b*x^6)^(2/3)) - (3*(a*x^3 + b*x^6)^(1/3))/(4*a^2*x^5) + (9*b*(a*x^3 + b*x^6)^(1/3))/(4*a^3
*x^2)

Rule 2000

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(b*(n - j)*(p + 1)*x
^(n - 1)), x] /; FreeQ[{a, b, j, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && EqQ[j*p - n + j + 1, 0]

Rule 2001

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(a*x^j + b*x^n)^(p + 1)/(a*(n - j)*(p + 1)*
x^(j - 1)), x] + Dist[(n*p + n - j + 1)/(a*(n - j)*(p + 1)), Int[(a*x^j + b*x^n)^(p + 1)/x^j, x], x] /; FreeQ[
{a, b, j, n}, x] &&  !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(n*p + n - j + 1)/(n - j)], 0] && LtQ[p, -1]

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 {1}{\left (a x^3+b x^6\right )^{5/3}} \, dx &=\frac {1}{2 a x^2 \left (a x^3+b x^6\right )^{2/3}}+\frac {3 \int \frac {1}{x^3 \left (a x^3+b x^6\right )^{2/3}} \, dx}{a}\\ &=\frac {1}{2 a x^2 \left (a x^3+b x^6\right )^{2/3}}-\frac {3 \sqrt [3]{a x^3+b x^6}}{4 a^2 x^5}-\frac {(9 b) \int \frac {1}{\left (a x^3+b x^6\right )^{2/3}} \, dx}{4 a^2}\\ &=\frac {1}{2 a x^2 \left (a x^3+b x^6\right )^{2/3}}-\frac {3 \sqrt [3]{a x^3+b x^6}}{4 a^2 x^5}+\frac {9 b \sqrt [3]{a x^3+b x^6}}{4 a^3 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 46, normalized size = 0.60 \[ \frac {-a^2+6 a b x^3+9 b^2 x^6}{4 a^3 x^2 \left (x^3 \left (a+b x^3\right )\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x^3 + b*x^6)^(-5/3),x]

[Out]

(-a^2 + 6*a*b*x^3 + 9*b^2*x^6)/(4*a^3*x^2*(x^3*(a + b*x^3))^(2/3))

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fricas [A]  time = 0.84, size = 54, normalized size = 0.70 \[ \frac {{\left (9 \, b^{2} x^{6} + 6 \, a b x^{3} - a^{2}\right )} {\left (b x^{6} + a x^{3}\right )}^{\frac {1}{3}}}{4 \, {\left (a^{3} b x^{8} + a^{4} x^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^6+a*x^3)^(5/3),x, algorithm="fricas")

[Out]

1/4*(9*b^2*x^6 + 6*a*b*x^3 - a^2)*(b*x^6 + a*x^3)^(1/3)/(a^3*b*x^8 + a^4*x^5)

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giac [A]  time = 23.79, size = 52, normalized size = 0.68 \[ \frac {b^{2}}{2 \, a^{3} {\left (b + \frac {a}{x^{3}}\right )}^{\frac {2}{3}}} - \frac {a^{9} {\left (b + \frac {a}{x^{3}}\right )}^{\frac {4}{3}} - 8 \, a^{9} {\left (b + \frac {a}{x^{3}}\right )}^{\frac {1}{3}} b}{4 \, a^{12}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^6+a*x^3)^(5/3),x, algorithm="giac")

[Out]

1/2*b^2/(a^3*(b + a/x^3)^(2/3)) - 1/4*(a^9*(b + a/x^3)^(4/3) - 8*a^9*(b + a/x^3)^(1/3)*b)/a^12

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maple [A]  time = 0.00, size = 46, normalized size = 0.60 \[ -\frac {\left (b \,x^{3}+a \right ) \left (-9 b^{2} x^{6}-6 b \,x^{3} a +a^{2}\right ) x}{4 \left (b \,x^{6}+a \,x^{3}\right )^{\frac {5}{3}} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^6+a*x^3)^(5/3),x)

[Out]

-1/4*x*(b*x^3+a)*(-9*b^2*x^6-6*a*b*x^3+a^2)/a^3/(b*x^6+a*x^3)^(5/3)

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maxima [A]  time = 0.47, size = 38, normalized size = 0.49 \[ \frac {9 \, b^{2} x^{6} + 6 \, a b x^{3} - a^{2}}{4 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{3} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^6+a*x^3)^(5/3),x, algorithm="maxima")

[Out]

1/4*(9*b^2*x^6 + 6*a*b*x^3 - a^2)/((b*x^3 + a)^(2/3)*a^3*x^4)

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mupad [B]  time = 1.28, size = 51, normalized size = 0.66 \[ \frac {{\left (b\,x^6+a\,x^3\right )}^{1/3}\,\left (-a^2+6\,a\,b\,x^3+9\,b^2\,x^6\right )}{4\,a^3\,x^5\,\left (b\,x^3+a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x^3 + b*x^6)^(5/3),x)

[Out]

((a*x^3 + b*x^6)^(1/3)*(9*b^2*x^6 - a^2 + 6*a*b*x^3))/(4*a^3*x^5*(a + b*x^3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a x^{3} + b x^{6}\right )^{\frac {5}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**6+a*x**3)**(5/3),x)

[Out]

Integral((a*x**3 + b*x**6)**(-5/3), x)

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