3.2942 \(\int x^8 \sqrt{a+b (c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=113 \[ \frac{2 a^2 x^9 \left (a+b \left (c x^2\right )^{3/2}\right )^{3/2}}{9 b^3 \left (c x^2\right )^{9/2}}+\frac{2 x^9 \left (a+b \left (c x^2\right )^{3/2}\right )^{7/2}}{21 b^3 \left (c x^2\right )^{9/2}}-\frac{4 a x^9 \left (a+b \left (c x^2\right )^{3/2}\right )^{5/2}}{15 b^3 \left (c x^2\right )^{9/2}} \]

[Out]

(2*a^2*x^9*(a + b*(c*x^2)^(3/2))^(3/2))/(9*b^3*(c*x^2)^(9/2)) - (4*a*x^9*(a + b*(c*x^2)^(3/2))^(5/2))/(15*b^3*
(c*x^2)^(9/2)) + (2*x^9*(a + b*(c*x^2)^(3/2))^(7/2))/(21*b^3*(c*x^2)^(9/2))

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Rubi [A]  time = 0.0669988, antiderivative size = 113, normalized size of antiderivative = 1., 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 = {368, 266, 43} \[ \frac{2 a^2 x^9 \left (a+b \left (c x^2\right )^{3/2}\right )^{3/2}}{9 b^3 \left (c x^2\right )^{9/2}}+\frac{2 x^9 \left (a+b \left (c x^2\right )^{3/2}\right )^{7/2}}{21 b^3 \left (c x^2\right )^{9/2}}-\frac{4 a x^9 \left (a+b \left (c x^2\right )^{3/2}\right )^{5/2}}{15 b^3 \left (c x^2\right )^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^8*Sqrt[a + b*(c*x^2)^(3/2)],x]

[Out]

(2*a^2*x^9*(a + b*(c*x^2)^(3/2))^(3/2))/(9*b^3*(c*x^2)^(9/2)) - (4*a*x^9*(a + b*(c*x^2)^(3/2))^(5/2))/(15*b^3*
(c*x^2)^(9/2)) + (2*x^9*(a + b*(c*x^2)^(3/2))^(7/2))/(21*b^3*(c*x^2)^(9/2))

Rule 368

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*((c*x^q
)^(1/q))^(m + 1)), Subst[Int[x^m*(a + b*x^(n*q))^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, d, m, n, p, q
}, x] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]

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 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])

Rubi steps

\begin{align*} \int x^8 \sqrt{a+b \left (c x^2\right )^{3/2}} \, dx &=\frac{x^9 \operatorname{Subst}\left (\int x^8 \sqrt{a+b x^3} \, dx,x,\sqrt{c x^2}\right )}{\left (c x^2\right )^{9/2}}\\ &=\frac{x^9 \operatorname{Subst}\left (\int x^2 \sqrt{a+b x} \, dx,x,\left (c x^2\right )^{3/2}\right )}{3 \left (c x^2\right )^{9/2}}\\ &=\frac{x^9 \operatorname{Subst}\left (\int \left (\frac{a^2 \sqrt{a+b x}}{b^2}-\frac{2 a (a+b x)^{3/2}}{b^2}+\frac{(a+b x)^{5/2}}{b^2}\right ) \, dx,x,\left (c x^2\right )^{3/2}\right )}{3 \left (c x^2\right )^{9/2}}\\ &=\frac{2 a^2 x^9 \left (a+b \left (c x^2\right )^{3/2}\right )^{3/2}}{9 b^3 \left (c x^2\right )^{9/2}}-\frac{4 a x^9 \left (a+b \left (c x^2\right )^{3/2}\right )^{5/2}}{15 b^3 \left (c x^2\right )^{9/2}}+\frac{2 x^9 \left (a+b \left (c x^2\right )^{3/2}\right )^{7/2}}{21 b^3 \left (c x^2\right )^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.0389263, size = 67, normalized size = 0.59 \[ \frac{2 x \left (a+b \left (c x^2\right )^{3/2}\right )^{3/2} \left (8 a^2-12 a b \left (c x^2\right )^{3/2}+15 b^2 c^3 x^6\right )}{315 b^3 c^4 \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(a + b*(c*x^2)^(3/2))^(3/2)*(8*a^2 + 15*b^2*c^3*x^6 - 12*a*b*(c*x^2)^(3/2)))/(315*b^3*c^4*Sqrt[c*x^2])

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Maple [F]  time = 0.051, size = 0, normalized size = 0. \begin{align*} \int{x}^{8}\sqrt{a+b \left ( c{x}^{2} \right ) ^{{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((c^7 + 3*c^6 + 2*c^5)*b^3*x^9 + (c^5 + c^4)*a*b^2*sqrt(c)*x^6 - 2*a^2*b*c^3*x^3 + 2*a^3*sqrt(c))*sqrt(b*c
^(3/2)*x^3 + a)/((c^8 + 6*c^7 + 11*c^6 + 6*c^5)*b^3) + integrate(-((c^5 + 3*c^4 + 2*c^3 - (c^5 + 3*c^4 + 2*c^3
)*sqrt(c))*b^2*x^8 + 2*(c^3 + c^2 - (c^2 + c)*sqrt(c))*a*b*x^5 - 2*a^2*x^2*(sqrt(c) - 1))*sqrt(b*c^(3/2)*x^3 +
a), x)/((c^5 + 6*c^4 + 11*c^3 + 6*c^2)*b^2*sqrt(c))

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Fricas [A]  time = 1.34476, size = 169, normalized size = 1.5 \begin{align*} \frac{2 \,{\left (15 \, b^{3} c^{5} x^{10} - 4 \, a^{2} b c^{2} x^{4} +{\left (3 \, a b^{2} c^{3} x^{6} + 8 \, a^{3}\right )} \sqrt{c x^{2}}\right )} \sqrt{\sqrt{c x^{2}} b c x^{2} + a}}{315 \, b^{3} c^{5} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(15*b^3*c^5*x^10 - 4*a^2*b*c^2*x^4 + (3*a*b^2*c^3*x^6 + 8*a^3)*sqrt(c*x^2))*sqrt(sqrt(c*x^2)*b*c*x^2 + a
)/(b^3*c^5*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{8} \sqrt{a + b \left (c x^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18691, size = 74, normalized size = 0.65 \begin{align*} \frac{2 \,{\left (15 \,{\left (b c^{\frac{3}{2}} x^{3} + a\right )}^{\frac{7}{2}} - 42 \,{\left (b c^{\frac{3}{2}} x^{3} + a\right )}^{\frac{5}{2}} a + 35 \,{\left (b c^{\frac{3}{2}} x^{3} + a\right )}^{\frac{3}{2}} a^{2}\right )}}{315 \, b^{3} c^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(15*(b*c^(3/2)*x^3 + a)^(7/2) - 42*(b*c^(3/2)*x^3 + a)^(5/2)*a + 35*(b*c^(3/2)*x^3 + a)^(3/2)*a^2)/(b^3*
c^(9/2))