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

Optimal result

Integrand size = 15, antiderivative size = 52 \[ \int \left (a x^3+b x^6\right )^{5/3} \, dx=-\frac {3 a \left (a x^3+b x^6\right )^{8/3}}{88 b^2 x^8}+\frac {\left (a x^3+b x^6\right )^{8/3}}{11 b x^5} \]

[Out]

-3/88*a*(b*x^6+a*x^3)^(8/3)/b^2/x^8+1/11*(b*x^6+a*x^3)^(8/3)/b/x^5

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2027, 2039} \[ \int \left (a x^3+b x^6\right )^{5/3} \, dx=\frac {\left (a x^3+b x^6\right )^{8/3}}{11 b x^5}-\frac {3 a \left (a x^3+b x^6\right )^{8/3}}{88 b^2 x^8} \]

[In]

Int[(a*x^3 + b*x^6)^(5/3),x]

[Out]

(-3*a*(a*x^3 + b*x^6)^(8/3))/(88*b^2*x^8) + (a*x^3 + b*x^6)^(8/3)/(11*b*x^5)

Rule 2027

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

Rule 2039

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] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] &&
NeQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a x^3+b x^6\right )^{8/3}}{11 b x^5}-\frac {(3 a) \int \frac {\left (a x^3+b x^6\right )^{5/3}}{x^3} \, dx}{11 b} \\ & = -\frac {3 a \left (a x^3+b x^6\right )^{8/3}}{88 b^2 x^8}+\frac {\left (a x^3+b x^6\right )^{8/3}}{11 b x^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.67 \[ \int \left (a x^3+b x^6\right )^{5/3} \, dx=\frac {\left (x^3 \left (a+b x^3\right )\right )^{8/3} \left (-3 a+8 b x^3\right )}{88 b^2 x^8} \]

[In]

Integrate[(a*x^3 + b*x^6)^(5/3),x]

[Out]

((x^3*(a + b*x^3))^(8/3)*(-3*a + 8*b*x^3))/(88*b^2*x^8)

Maple [A] (verified)

Time = 0.77 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.75

method	result	size
gosper	\(-\frac {\left (b \,x^{3}+a \right ) \left (-8 b \,x^{3}+3 a \right ) \left (b \,x^{6}+a \,x^{3}\right )^{\frac {5}{3}}}{88 b^{2} x^{5}}\)	\(39\)
trager	\(-\frac {\left (-8 b^{3} x^{9}-13 b^{2} x^{6} a -2 a^{2} b \,x^{3}+3 a^{3}\right ) \left (b \,x^{6}+a \,x^{3}\right )^{\frac {2}{3}}}{88 b^{2} x^{2}}\)	\(54\)
risch	\(-\frac {\left (x^{3} \left (b \,x^{3}+a \right )\right )^{\frac {2}{3}} \left (-8 b^{3} x^{9}-13 b^{2} x^{6} a -2 a^{2} b \,x^{3}+3 a^{3}\right )}{88 x^{2} b^{2}}\)	\(54\)

[In]

int((b*x^6+a*x^3)^(5/3),x,method=_RETURNVERBOSE)

[Out]

-1/88*(b*x^3+a)*(-8*b*x^3+3*a)*(b*x^6+a*x^3)^(5/3)/b^2/x^5

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.02 \[ \int \left (a x^3+b x^6\right )^{5/3} \, dx=\frac {{\left (8 \, b^{3} x^{9} + 13 \, a b^{2} x^{6} + 2 \, a^{2} b x^{3} - 3 \, a^{3}\right )} {\left (b x^{6} + a x^{3}\right )}^{\frac {2}{3}}}{88 \, b^{2} x^{2}} \]

[In]

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

[Out]

1/88*(8*b^3*x^9 + 13*a*b^2*x^6 + 2*a^2*b*x^3 - 3*a^3)*(b*x^6 + a*x^3)^(2/3)/(b^2*x^2)

Sympy [F]

\[ \int \left (a x^3+b x^6\right )^{5/3} \, dx=\int \left (a x^{3} + b x^{6}\right )^{\frac {5}{3}}\, dx \]

[In]

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

[Out]

Integral((a*x**3 + b*x**6)**(5/3), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.88 \[ \int \left (a x^3+b x^6\right )^{5/3} \, dx=\frac {{\left (8 \, b^{3} x^{9} + 13 \, a b^{2} x^{6} + 2 \, a^{2} b x^{3} - 3 \, a^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{88 \, b^{2}} \]

[In]

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

[Out]

1/88*(8*b^3*x^9 + 13*a*b^2*x^6 + 2*a^2*b*x^3 - 3*a^3)*(b*x^3 + a)^(2/3)/b^2

Giac [F]

\[ \int \left (a x^3+b x^6\right )^{5/3} \, dx=\int { {\left (b x^{6} + a x^{3}\right )}^{\frac {5}{3}} \,d x } \]

[In]

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

[Out]

integrate((b*x^6 + a*x^3)^(5/3), x)

Mupad [B] (verification not implemented)

Time = 8.33 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.77 \[ \int \left (a x^3+b x^6\right )^{5/3} \, dx=-\frac {{\left (b\,x^3+a\right )}^2\,{\left (b\,x^6+a\,x^3\right )}^{2/3}\,\left (3\,a-8\,b\,x^3\right )}{88\,b^2\,x^2} \]

[In]

int((a*x^3 + b*x^6)^(5/3),x)

[Out]

-((a + b*x^3)^2*(a*x^3 + b*x^6)^(2/3)*(3*a - 8*b*x^3))/(88*b^2*x^2)