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*(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)

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Rubi [A]  time = 0.0566138, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, 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 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])

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]

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*}

Mathematica [A]  time = 0.013138, size = 46, normalized size = 0.6 \[ \frac{-a^2+6 a b x^3+9 b^2 x^6}{4 a^3 x^2 \left (x^3 \left (a+b x^3\right )\right )^{2/3}} \]

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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Maple [A]  time = 0.006, size = 46, normalized size = 0.6 \begin{align*} -{\frac{x \left ( b{x}^{3}+a \right ) \left ( -9\,{b}^{2}{x}^{6}-6\,b{x}^{3}a+{a}^{2} \right ) }{4\,{a}^{3}} \left ( b{x}^{6}+a{x}^{3} \right ) ^{-{\frac{5}{3}}}} \end{align*}

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 = 1.04392, size = 51, normalized size = 0.66 \begin{align*} \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}} \end{align*}

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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Fricas [A]  time = 1.99237, size = 107, normalized size = 1.39 \begin{align*} \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 )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a x^{3} + b x^{6}\right )^{\frac{5}{3}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{6} + a x^{3}\right )}^{\frac{5}{3}}}\,{d x} \end{align*}

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]

integrate((b*x^6 + a*x^3)^(-5/3), x)