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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 116, 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} \[ \frac {4 a^2 \left (a+b \left (c x^3\right )^{3/2}\right )^{5/2}}{15 b^4 c^6}-\frac {4 a^3 \left (a+b \left (c x^3\right )^{3/2}\right )^{3/2}}{27 b^4 c^6}+\frac {4 \left (a+b \left (c x^3\right )^{3/2}\right )^{9/2}}{81 b^4 c^6}-\frac {4 a \left (a+b \left (c x^3\right )^{3/2}\right )^{7/2}}{21 b^4 c^6} \]

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 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^{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 = 0.15, size = 80, normalized size = 0.69 \[ \frac {4 \left (a+b \left (c x^3\right )^{3/2}\right )^{3/2} \left (-16 a^3+24 a^2 b \left (c x^3\right )^{3/2}-30 a b^2 c^3 x^9+35 b^3 c^3 x^9 \left (c x^3\right )^{3/2}\right )}{2835 b^4 c^6} \]

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)

________________________________________________________________________________________

fricas [A]  time = 5.27, size = 87, normalized size = 0.75 \[ \frac {4 \, {\left (35 \, b^{4} c^{6} x^{18} - 6 \, a^{2} b^{2} c^{3} x^{9} - 16 \, a^{4} + {\left (5 \, a b^{3} c^{4} x^{12} + 8 \, a^{3} b c x^{3}\right )} \sqrt {c x^{3}}\right )} \sqrt {\sqrt {c x^{3}} b c x^{3} + a}}{2835 \, b^{4} c^{6}} \]

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)

________________________________________________________________________________________

giac [A]  time = 0.28, size = 121, normalized size = 1.04 \[ -\frac {4 \, {\left (105 \, {\left (\sqrt {c x} b c^{5} x^{4} + a c^{4}\right )}^{\frac {3}{2}} a^{3} c^{12} - 189 \, {\left (\sqrt {c x} b c^{5} x^{4} + a c^{4}\right )}^{\frac {5}{2}} a^{2} c^{8} + 135 \, {\left (\sqrt {c x} b c^{5} x^{4} + a c^{4}\right )}^{\frac {7}{2}} a c^{4} - 35 \, {\left (\sqrt {c x} b c^{5} x^{4} + a c^{4}\right )}^{\frac {9}{2}}\right )} {\left | c \right |}^{2}}{2835 \, b^{4} c^{26}} \]

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*(105*(sqrt(c*x)*b*c^5*x^4 + a*c^4)^(3/2)*a^3*c^12 - 189*(sqrt(c*x)*b*c^5*x^4 + a*c^4)^(5/2)*a^2*c^8 +
135*(sqrt(c*x)*b*c^5*x^4 + a*c^4)^(7/2)*a*c^4 - 35*(sqrt(c*x)*b*c^5*x^4 + a*c^4)^(9/2))*abs(c)^2/(b^4*c^26)

________________________________________________________________________________________

maple [F]  time = 0.25, size = 0, normalized size = 0.00 \[ \int \sqrt {a +\left (c \,x^{3}\right )^{\frac {3}{2}} b}\, x^{17}\, dx \]

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)

________________________________________________________________________________________

maxima [A]  time = 0.60, size = 85, normalized size = 0.73 \[ \frac {4 \, {\left (\frac {35 \, {\left (\left (c x^{3}\right )^{\frac {3}{2}} b + a\right )}^{\frac {9}{2}}}{b^{4}} - \frac {135 \, {\left (\left (c x^{3}\right )^{\frac {3}{2}} b + a\right )}^{\frac {7}{2}} a}{b^{4}} + \frac {189 \, {\left (\left (c x^{3}\right )^{\frac {3}{2}} b + a\right )}^{\frac {5}{2}} a^{2}}{b^{4}} - \frac {105 \, {\left (\left (c x^{3}\right )^{\frac {3}{2}} b + a\right )}^{\frac {3}{2}} a^{3}}{b^{4}}\right )}}{2835 \, c^{6}} \]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^{17}\,\sqrt {a+b\,{\left (c\,x^3\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{17} \sqrt {a + b \left (c x^{3}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**17*sqrt(a + b*(c*x**3)**(3/2)), x)

________________________________________________________________________________________