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

Optimal result

Integrand size = 15, antiderivative size = 25 \[ \int \left (a x^3+b x^6\right )^{2/3} \, dx=\frac {\left (a x^3+b x^6\right )^{5/3}}{5 b x^5} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2025} \[ \int \left (a x^3+b x^6\right )^{2/3} \, dx=\frac {\left (a x^3+b x^6\right )^{5/3}}{5 b x^5} \]

[In]

Int[(a*x^3 + b*x^6)^(2/3),x]

[Out]

(a*x^3 + b*x^6)^(5/3)/(5*b*x^5)

Rule 2025

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*} \text {integral}& = \frac {\left (a x^3+b x^6\right )^{5/3}}{5 b x^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \left (a x^3+b x^6\right )^{2/3} \, dx=\frac {\left (x^3 \left (a+b x^3\right )\right )^{5/3}}{5 b x^5} \]

[In]

Integrate[(a*x^3 + b*x^6)^(2/3),x]

[Out]

(x^3*(a + b*x^3))^(5/3)/(5*b*x^5)

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16

method	result	size
gosper	\(\frac {\left (b \,x^{3}+a \right ) \left (b \,x^{6}+a \,x^{3}\right )^{\frac {2}{3}}}{5 b \,x^{2}}\)	\(29\)
trager	\(\frac {\left (b \,x^{3}+a \right ) \left (b \,x^{6}+a \,x^{3}\right )^{\frac {2}{3}}}{5 b \,x^{2}}\)	\(29\)
risch	\(\frac {\left (x^{3} \left (b \,x^{3}+a \right )\right )^{\frac {2}{3}} \left (b \,x^{3}+a \right )}{5 x^{2} b}\)	\(29\)

[In]

int((b*x^6+a*x^3)^(2/3),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \left (a x^3+b x^6\right )^{2/3} \, dx=\frac {{\left (b x^{6} + a x^{3}\right )}^{\frac {2}{3}} {\left (b x^{3} + a\right )}}{5 \, b x^{2}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \left (a x^3+b x^6\right )^{2/3} \, dx=\int \left (a x^{3} + b x^{6}\right )^{\frac {2}{3}}\, dx \]

[In]

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

[Out]

Integral((a*x**3 + b*x**6)**(2/3), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.56 \[ \int \left (a x^3+b x^6\right )^{2/3} \, dx=\frac {{\left (b x^{3} + a\right )}^{\frac {5}{3}}}{5 \, b} \]

[In]

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

[Out]

1/5*(b*x^3 + a)^(5/3)/b

Giac [F]

\[ \int \left (a x^3+b x^6\right )^{2/3} \, dx=\int { {\left (b x^{6} + a x^{3}\right )}^{\frac {2}{3}} \,d x } \]

[In]

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

[Out]

integrate((b*x^6 + a*x^3)^(2/3), x)

Mupad [B] (verification not implemented)

Time = 8.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \left (a x^3+b x^6\right )^{2/3} \, dx=\frac {\left (\frac {a}{5\,b}+\frac {x^3}{5}\right )\,{\left (b\,x^6+a\,x^3\right )}^{2/3}}{x^2} \]

[In]

int((a*x^3 + b*x^6)^(2/3),x)

[Out]

((a/(5*b) + x^3/5)*(a*x^3 + b*x^6)^(2/3))/x^2