3.1176 \(\int x^{15} \sqrt [4]{a-b x^4} \, dx\)

Optimal. Leaf size=84 \[ -\frac {a^3 \left (a-b x^4\right )^{5/4}}{5 b^4}+\frac {a^2 \left (a-b x^4\right )^{9/4}}{3 b^4}+\frac {\left (a-b x^4\right )^{17/4}}{17 b^4}-\frac {3 a \left (a-b x^4\right )^{13/4}}{13 b^4} \]

[Out]

-1/5*a^3*(-b*x^4+a)^(5/4)/b^4+1/3*a^2*(-b*x^4+a)^(9/4)/b^4-3/13*a*(-b*x^4+a)^(13/4)/b^4+1/17*(-b*x^4+a)^(17/4)
/b^4

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Rubi [A]  time = 0.05, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {266, 43} \[ \frac {a^2 \left (a-b x^4\right )^{9/4}}{3 b^4}-\frac {a^3 \left (a-b x^4\right )^{5/4}}{5 b^4}+\frac {\left (a-b x^4\right )^{17/4}}{17 b^4}-\frac {3 a \left (a-b x^4\right )^{13/4}}{13 b^4} \]

Antiderivative was successfully verified.

[In]

Int[x^15*(a - b*x^4)^(1/4),x]

[Out]

-(a^3*(a - b*x^4)^(5/4))/(5*b^4) + (a^2*(a - b*x^4)^(9/4))/(3*b^4) - (3*a*(a - b*x^4)^(13/4))/(13*b^4) + (a -
b*x^4)^(17/4)/(17*b^4)

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]]

Rubi steps

\begin {align*} \int x^{15} \sqrt [4]{a-b x^4} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int x^3 \sqrt [4]{a-b x} \, dx,x,x^4\right )\\ &=\frac {1}{4} \operatorname {Subst}\left (\int \left (\frac {a^3 \sqrt [4]{a-b x}}{b^3}-\frac {3 a^2 (a-b x)^{5/4}}{b^3}+\frac {3 a (a-b x)^{9/4}}{b^3}-\frac {(a-b x)^{13/4}}{b^3}\right ) \, dx,x,x^4\right )\\ &=-\frac {a^3 \left (a-b x^4\right )^{5/4}}{5 b^4}+\frac {a^2 \left (a-b x^4\right )^{9/4}}{3 b^4}-\frac {3 a \left (a-b x^4\right )^{13/4}}{13 b^4}+\frac {\left (a-b x^4\right )^{17/4}}{17 b^4}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 51, normalized size = 0.61 \[ -\frac {\left (a-b x^4\right )^{5/4} \left (128 a^3+160 a^2 b x^4+180 a b^2 x^8+195 b^3 x^{12}\right )}{3315 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^15*(a - b*x^4)^(1/4),x]

[Out]

-1/3315*((a - b*x^4)^(5/4)*(128*a^3 + 160*a^2*b*x^4 + 180*a*b^2*x^8 + 195*b^3*x^12))/b^4

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fricas [A]  time = 0.87, size = 58, normalized size = 0.69 \[ \frac {{\left (195 \, b^{4} x^{16} - 15 \, a b^{3} x^{12} - 20 \, a^{2} b^{2} x^{8} - 32 \, a^{3} b x^{4} - 128 \, a^{4}\right )} {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{3315 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^15*(-b*x^4+a)^(1/4),x, algorithm="fricas")

[Out]

1/3315*(195*b^4*x^16 - 15*a*b^3*x^12 - 20*a^2*b^2*x^8 - 32*a^3*b*x^4 - 128*a^4)*(-b*x^4 + a)^(1/4)/b^4

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giac [A]  time = 0.16, size = 94, normalized size = 1.12 \[ \frac {195 \, {\left (b x^{4} - a\right )}^{4} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} + 765 \, {\left (b x^{4} - a\right )}^{3} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} a + 1105 \, {\left (b x^{4} - a\right )}^{2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} a^{2} - 663 \, {\left (-b x^{4} + a\right )}^{\frac {5}{4}} a^{3}}{3315 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^15*(-b*x^4+a)^(1/4),x, algorithm="giac")

[Out]

1/3315*(195*(b*x^4 - a)^4*(-b*x^4 + a)^(1/4) + 765*(b*x^4 - a)^3*(-b*x^4 + a)^(1/4)*a + 1105*(b*x^4 - a)^2*(-b
*x^4 + a)^(1/4)*a^2 - 663*(-b*x^4 + a)^(5/4)*a^3)/b^4

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maple [A]  time = 0.01, size = 48, normalized size = 0.57 \[ -\frac {\left (-b \,x^{4}+a \right )^{\frac {5}{4}} \left (195 b^{3} x^{12}+180 a \,b^{2} x^{8}+160 a^{2} b \,x^{4}+128 a^{3}\right )}{3315 b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^15*(-b*x^4+a)^(1/4),x)

[Out]

-1/3315*(-b*x^4+a)^(5/4)*(195*b^3*x^12+180*a*b^2*x^8+160*a^2*b*x^4+128*a^3)/b^4

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maxima [A]  time = 1.41, size = 68, normalized size = 0.81 \[ \frac {{\left (-b x^{4} + a\right )}^{\frac {17}{4}}}{17 \, b^{4}} - \frac {3 \, {\left (-b x^{4} + a\right )}^{\frac {13}{4}} a}{13 \, b^{4}} + \frac {{\left (-b x^{4} + a\right )}^{\frac {9}{4}} a^{2}}{3 \, b^{4}} - \frac {{\left (-b x^{4} + a\right )}^{\frac {5}{4}} a^{3}}{5 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^15*(-b*x^4+a)^(1/4),x, algorithm="maxima")

[Out]

1/17*(-b*x^4 + a)^(17/4)/b^4 - 3/13*(-b*x^4 + a)^(13/4)*a/b^4 + 1/3*(-b*x^4 + a)^(9/4)*a^2/b^4 - 1/5*(-b*x^4 +
a)^(5/4)*a^3/b^4

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mupad [B]  time = 1.12, size = 57, normalized size = 0.68 \[ -{\left (a-b\,x^4\right )}^{1/4}\,\left (\frac {128\,a^4}{3315\,b^4}-\frac {x^{16}}{17}+\frac {a\,x^{12}}{221\,b}+\frac {32\,a^3\,x^4}{3315\,b^3}+\frac {4\,a^2\,x^8}{663\,b^2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^15*(a - b*x^4)^(1/4),x)

[Out]

-(a - b*x^4)^(1/4)*((128*a^4)/(3315*b^4) - x^16/17 + (a*x^12)/(221*b) + (32*a^3*x^4)/(3315*b^3) + (4*a^2*x^8)/
(663*b^2))

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sympy [A]  time = 13.33, size = 110, normalized size = 1.31 \[ \begin {cases} - \frac {128 a^{4} \sqrt [4]{a - b x^{4}}}{3315 b^{4}} - \frac {32 a^{3} x^{4} \sqrt [4]{a - b x^{4}}}{3315 b^{3}} - \frac {4 a^{2} x^{8} \sqrt [4]{a - b x^{4}}}{663 b^{2}} - \frac {a x^{12} \sqrt [4]{a - b x^{4}}}{221 b} + \frac {x^{16} \sqrt [4]{a - b x^{4}}}{17} & \text {for}\: b \neq 0 \\\frac {\sqrt [4]{a} x^{16}}{16} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**15*(-b*x**4+a)**(1/4),x)

[Out]

Piecewise((-128*a**4*(a - b*x**4)**(1/4)/(3315*b**4) - 32*a**3*x**4*(a - b*x**4)**(1/4)/(3315*b**3) - 4*a**2*x
**8*(a - b*x**4)**(1/4)/(663*b**2) - a*x**12*(a - b*x**4)**(1/4)/(221*b) + x**16*(a - b*x**4)**(1/4)/17, Ne(b,
0)), (a**(1/4)*x**16/16, True))

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