3.2970 \(\int x^{17} \sqrt{a+b (c x^3)^{3/2}} \, dx\)

Optimal. Leaf size=116 \[ \frac{4 a^2 \left (a+b \left (c x^3\right )^{3/2}\right )^{5/2}}{15 b^4 c^6}-\frac{4 a^3 \left (a+b \left (c x^3\right )^{3/2}\right )^{3/2}}{27 b^4 c^6}+\frac{4 \left (a+b \left (c x^3\right )^{3/2}\right )^{9/2}}{81 b^4 c^6}-\frac{4 a \left (a+b \left (c x^3\right )^{3/2}\right )^{7/2}}{21 b^4 c^6} \]

[Out]

(-4*a^3*(a + b*(c*x^3)^(3/2))^(3/2))/(27*b^4*c^6) + (4*a^2*(a + b*(c*x^3)^(3/2))^(5/2))/(15*b^4*c^6) - (4*a*(a
+ b*(c*x^3)^(3/2))^(7/2))/(21*b^4*c^6) + (4*(a + b*(c*x^3)^(3/2))^(9/2))/(81*b^4*c^6)

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Rubi [A]  time = 0.0764661, antiderivative size = 116, 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 = {369, 266, 43} \[ \frac{4 a^2 \left (a+b \left (c x^3\right )^{3/2}\right )^{5/2}}{15 b^4 c^6}-\frac{4 a^3 \left (a+b \left (c x^3\right )^{3/2}\right )^{3/2}}{27 b^4 c^6}+\frac{4 \left (a+b \left (c x^3\right )^{3/2}\right )^{9/2}}{81 b^4 c^6}-\frac{4 a \left (a+b \left (c x^3\right )^{3/2}\right )^{7/2}}{21 b^4 c^6} \]

Antiderivative was successfully verified.

[In]

Int[x^17*Sqrt[a + b*(c*x^3)^(3/2)],x]

[Out]

(-4*a^3*(a + b*(c*x^3)^(3/2))^(3/2))/(27*b^4*c^6) + (4*a^2*(a + b*(c*x^3)^(3/2))^(5/2))/(15*b^4*c^6) - (4*a*(a
+ b*(c*x^3)^(3/2))^(7/2))/(21*b^4*c^6) + (4*(a + b*(c*x^3)^(3/2))^(9/2))/(81*b^4*c^6)

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]

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^{17} \sqrt{a+b \left (c x^3\right )^{3/2}} \, dx &=\operatorname{Subst}\left (\int x^{17} \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^3 \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^3 \sqrt{a+b c^{3/2} x}}{b^3 c^{9/2}}+\frac{3 a^2 \left (a+b c^{3/2} x\right )^{3/2}}{b^3 c^{9/2}}-\frac{3 a \left (a+b c^{3/2} x\right )^{5/2}}{b^3 c^{9/2}}+\frac{\left (a+b c^{3/2} x\right )^{7/2}}{b^3 c^{9/2}}\right ) \, dx,x,x^{9/2}\right ),\sqrt{x},\frac{\sqrt{c x^3}}{\sqrt{c} x}\right )\\ &=-\frac{4 a^3 \left (a+b \left (c x^3\right )^{3/2}\right )^{3/2}}{27 b^4 c^6}+\frac{4 a^2 \left (a+b \left (c x^3\right )^{3/2}\right )^{5/2}}{15 b^4 c^6}-\frac{4 a \left (a+b \left (c x^3\right )^{3/2}\right )^{7/2}}{21 b^4 c^6}+\frac{4 \left (a+b \left (c x^3\right )^{3/2}\right )^{9/2}}{81 b^4 c^6}\\ \end{align*}

Mathematica [A]  time = 0.108308, size = 80, normalized size = 0.69 \[ \frac{4 \left (a+b \left (c x^3\right )^{3/2}\right )^{3/2} \left (24 a^2 b \left (c x^3\right )^{3/2}-16 a^3-30 a b^2 c^3 x^9+35 b^3 c^3 x^9 \left (c x^3\right )^{3/2}\right )}{2835 b^4 c^6} \]

Antiderivative was successfully verified.

[In]

Integrate[x^17*Sqrt[a + b*(c*x^3)^(3/2)],x]

[Out]

(4*(a + b*(c*x^3)^(3/2))^(3/2)*(-16*a^3 - 30*a*b^2*c^3*x^9 + 24*a^2*b*(c*x^3)^(3/2) + 35*b^3*c^3*x^9*(c*x^3)^(
3/2)))/(2835*b^4*c^6)

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Maple [F]  time = 0.061, size = 0, normalized size = 0. \begin{align*} \int{x}^{17}\sqrt{a+b \left ( c{x}^{3} \right ) ^{{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^17*(a+b*(c*x^3)^(3/2))^(1/2),x)

[Out]

int(x^17*(a+b*(c*x^3)^(3/2))^(1/2),x)

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Maxima [A]  time = 0.95555, size = 115, normalized size = 0.99 \begin{align*} \frac{4 \,{\left (\frac{35 \,{\left (\left (c x^{3}\right )^{\frac{3}{2}} b + a\right )}^{\frac{9}{2}}}{b^{4}} - \frac{135 \,{\left (\left (c x^{3}\right )^{\frac{3}{2}} b + a\right )}^{\frac{7}{2}} a}{b^{4}} + \frac{189 \,{\left (\left (c x^{3}\right )^{\frac{3}{2}} b + a\right )}^{\frac{5}{2}} a^{2}}{b^{4}} - \frac{105 \,{\left (\left (c x^{3}\right )^{\frac{3}{2}} b + a\right )}^{\frac{3}{2}} a^{3}}{b^{4}}\right )}}{2835 \, c^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^17*(a+b*(c*x^3)^(3/2))^(1/2),x, algorithm="maxima")

[Out]

4/2835*(35*((c*x^3)^(3/2)*b + a)^(9/2)/b^4 - 135*((c*x^3)^(3/2)*b + a)^(7/2)*a/b^4 + 189*((c*x^3)^(3/2)*b + a)
^(5/2)*a^2/b^4 - 105*((c*x^3)^(3/2)*b + a)^(3/2)*a^3/b^4)/c^6

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Fricas [A]  time = 19.521, size = 194, normalized size = 1.67 \begin{align*} \frac{4 \,{\left (35 \, b^{4} c^{6} x^{18} - 6 \, a^{2} b^{2} c^{3} x^{9} - 16 \, a^{4} +{\left (5 \, a b^{3} c^{4} x^{12} + 8 \, a^{3} b c x^{3}\right )} \sqrt{c x^{3}}\right )} \sqrt{\sqrt{c x^{3}} b c x^{3} + a}}{2835 \, b^{4} c^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^17*(a+b*(c*x^3)^(3/2))^(1/2),x, algorithm="fricas")

[Out]

4/2835*(35*b^4*c^6*x^18 - 6*a^2*b^2*c^3*x^9 - 16*a^4 + (5*a*b^3*c^4*x^12 + 8*a^3*b*c*x^3)*sqrt(c*x^3))*sqrt(sq
rt(c*x^3)*b*c*x^3 + a)/(b^4*c^6)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**17*(a+b*(c*x**3)**(3/2))**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.21507, size = 193, normalized size = 1.66 \begin{align*} \frac{4 \,{\left (\frac{16 \, \sqrt{a c^{3}} a^{4}}{b^{4} c^{5}} - \frac{105 \,{\left (\sqrt{c x} b c^{4} x^{4} + a c^{3}\right )}^{\frac{3}{2}} a^{3} c^{9} - 189 \,{\left (\sqrt{c x} b c^{4} x^{4} + a c^{3}\right )}^{\frac{5}{2}} a^{2} c^{6} + 135 \,{\left (\sqrt{c x} b c^{4} x^{4} + a c^{3}\right )}^{\frac{7}{2}} a c^{3} - 35 \,{\left (\sqrt{c x} b c^{4} x^{4} + a c^{3}\right )}^{\frac{9}{2}}}{b^{4} c^{17}}\right )}{\left | c \right |}}{2835 \, c^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^17*(a+b*(c*x^3)^(3/2))^(1/2),x, algorithm="giac")

[Out]

4/2835*(16*sqrt(a*c^3)*a^4/(b^4*c^5) - (105*(sqrt(c*x)*b*c^4*x^4 + a*c^3)^(3/2)*a^3*c^9 - 189*(sqrt(c*x)*b*c^4
*x^4 + a*c^3)^(5/2)*a^2*c^6 + 135*(sqrt(c*x)*b*c^4*x^4 + a*c^3)^(7/2)*a*c^3 - 35*(sqrt(c*x)*b*c^4*x^4 + a*c^3)
^(9/2))/(b^4*c^17))*abs(c)/c^(7/2)