3.2971 \(\int x^8 \sqrt {a+b (c x^3)^{3/2}} \, dx\)

Optimal. Leaf size=56 \[ \frac {4 \left (a+b \left (c x^3\right )^{3/2}\right )^{5/2}}{45 b^2 c^3}-\frac {4 a \left (a+b \left (c x^3\right )^{3/2}\right )^{3/2}}{27 b^2 c^3} \]

[Out]

-4/27*a*(a+b*(c*x^3)^(3/2))^(3/2)/b^2/c^3+4/45*(a+b*(c*x^3)^(3/2))^(5/2)/b^2/c^3

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Rubi [A]  time = 0.04, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {369, 266, 43} \[ \frac {4 \left (a+b \left (c x^3\right )^{3/2}\right )^{5/2}}{45 b^2 c^3}-\frac {4 a \left (a+b \left (c x^3\right )^{3/2}\right )^{3/2}}{27 b^2 c^3} \]

Antiderivative was successfully verified.

[In]

Int[x^8*Sqrt[a + b*(c*x^3)^(3/2)],x]

[Out]

(-4*a*(a + b*(c*x^3)^(3/2))^(3/2))/(27*b^2*c^3) + (4*(a + b*(c*x^3)^(3/2))^(5/2))/(45*b^2*c^3)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 369

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> With[{k = Denominator[n]}, Su
bst[Int[(d*x)^m*(a + b*c^n*x^(n*q))^p, x], x^(1/k), (c*x^q)^(1/k)/(c^(1/k)*(x^(1/k))^(q - 1))]] /; FreeQ[{a, b
, c, d, m, p, q}, x] && FractionQ[n]

Rubi steps

\begin {align*} \int x^8 \sqrt {a+b \left (c x^3\right )^{3/2}} \, dx &=\operatorname {Subst}\left (\int x^8 \sqrt {a+b c^{3/2} x^{9/2}} \, dx,\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right )\\ &=\operatorname {Subst}\left (\frac {2}{9} \operatorname {Subst}\left (\int x \sqrt {a+b c^{3/2} x} \, dx,x,x^{9/2}\right ),\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right )\\ &=\operatorname {Subst}\left (\frac {2}{9} \operatorname {Subst}\left (\int \left (-\frac {a \sqrt {a+b c^{3/2} x}}{b c^{3/2}}+\frac {\left (a+b c^{3/2} x\right )^{3/2}}{b c^{3/2}}\right ) \, dx,x,x^{9/2}\right ),\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right )\\ &=-\frac {4 a \left (a+b \left (c x^3\right )^{3/2}\right )^{3/2}}{27 b^2 c^3}+\frac {4 \left (a+b \left (c x^3\right )^{3/2}\right )^{5/2}}{45 b^2 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 43, normalized size = 0.77 \[ \frac {4 \left (a+b \left (c x^3\right )^{3/2}\right )^{3/2} \left (3 b \left (c x^3\right )^{3/2}-2 a\right )}{135 b^2 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8*Sqrt[a + b*(c*x^3)^(3/2)],x]

[Out]

(4*(a + b*(c*x^3)^(3/2))^(3/2)*(-2*a + 3*b*(c*x^3)^(3/2)))/(135*b^2*c^3)

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fricas [A]  time = 2.86, size = 56, normalized size = 1.00 \[ \frac {4 \, {\left (3 \, b^{2} c^{3} x^{9} + \sqrt {c x^{3}} a b c x^{3} - 2 \, a^{2}\right )} \sqrt {\sqrt {c x^{3}} b c x^{3} + a}}{135 \, b^{2} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(a+b*(c*x^3)^(3/2))^(1/2),x, algorithm="fricas")

[Out]

4/135*(3*b^2*c^3*x^9 + sqrt(c*x^3)*a*b*c*x^3 - 2*a^2)*sqrt(sqrt(c*x^3)*b*c*x^3 + a)/(b^2*c^3)

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giac [A]  time = 0.20, size = 63, normalized size = 1.12 \[ -\frac {4 \, {\left (5 \, {\left (\sqrt {c x} b c^{5} x^{4} + a c^{4}\right )}^{\frac {3}{2}} a c^{4} - 3 \, {\left (\sqrt {c x} b c^{5} x^{4} + a c^{4}\right )}^{\frac {5}{2}}\right )} {\left | c \right |}^{2}}{135 \, b^{2} c^{15}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(a+b*(c*x^3)^(3/2))^(1/2),x, algorithm="giac")

[Out]

-4/135*(5*(sqrt(c*x)*b*c^5*x^4 + a*c^4)^(3/2)*a*c^4 - 3*(sqrt(c*x)*b*c^5*x^4 + a*c^4)^(5/2))*abs(c)^2/(b^2*c^1
5)

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maple [F]  time = 0.26, size = 0, normalized size = 0.00 \[ \int \sqrt {a +\left (c \,x^{3}\right )^{\frac {3}{2}} b}\, x^{8}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(a+(c*x^3)^(3/2)*b)^(1/2),x)

[Out]

int(x^8*(a+(c*x^3)^(3/2)*b)^(1/2),x)

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maxima [A]  time = 0.62, size = 43, normalized size = 0.77 \[ \frac {4 \, {\left (\frac {3 \, {\left (\left (c x^{3}\right )^{\frac {3}{2}} b + a\right )}^{\frac {5}{2}}}{b^{2}} - \frac {5 \, {\left (\left (c x^{3}\right )^{\frac {3}{2}} b + a\right )}^{\frac {3}{2}} a}{b^{2}}\right )}}{135 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(a+b*(c*x^3)^(3/2))^(1/2),x, algorithm="maxima")

[Out]

4/135*(3*((c*x^3)^(3/2)*b + a)^(5/2)/b^2 - 5*((c*x^3)^(3/2)*b + a)^(3/2)*a/b^2)/c^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x^8\,\sqrt {a+b\,{\left (c\,x^3\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(a + b*(c*x^3)^(3/2))^(1/2),x)

[Out]

int(x^8*(a + b*(c*x^3)^(3/2))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{8} \sqrt {a + b \left (c x^{3}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*(a+b*(c*x**3)**(3/2))**(1/2),x)

[Out]

Integral(x**8*sqrt(a + b*(c*x**3)**(3/2)), x)

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