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

Optimal. Leaf size=80 \[ -\frac {a^3 \left (a+b x^4\right )^{5/4}}{5 b^4}+\frac {a^2 \left (a+b x^4\right )^{9/4}}{3 b^4}+\frac {\left (a+b x^4\right )^{17/4}}{17 b^4}-\frac {3 a \left (a+b x^4\right )^{13/4}}{13 b^4} \]

[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 = 0.04, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, 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 = 0.04, size = 50, normalized size = 0.62 \[ \frac {\left (a+b x^4\right )^{5/4} \left (-128 a^3+160 a^2 b x^4-180 a b^2 x^8+195 b^3 x^{12}\right )}{3315 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^15*(a + b*x^4)^(1/4),x]

[Out]

((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))/(3315*b^4)

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fricas [A]  time = 0.85, size = 57, normalized size = 0.71 \[ \frac {{\left (195 \, b^{4} x^{16} + 15 \, a b^{3} x^{12} - 20 \, a^{2} b^{2} x^{8} + 32 \, a^{3} b x^{4} - 128 \, a^{4}\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{3315 \, b^{4}} \]

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 = 57, normalized size = 0.71 \[ \frac {195 \, {\left (b x^{4} + a\right )}^{\frac {17}{4}} - 765 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}} a + 1105 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}} a^{2} - 663 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{3}}{3315 \, b^{4}} \]

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)^(17/4) - 765*(b*x^4 + a)^(13/4)*a + 1105*(b*x^4 + a)^(9/4)*a^2 - 663*(b*x^4 + a)^(5/4)
*a^3)/b^4

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maple [A]  time = 0.01, size = 47, normalized size = 0.59 \[ -\frac {\left (b \,x^{4}+a \right )^{\frac {5}{4}} \left (-195 b^{3} x^{12}+180 a \,b^{2} x^{8}-160 a^{2} b \,x^{4}+128 a^{3}\right )}{3315 b^{4}} \]

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 = 1.30, size = 64, normalized size = 0.80 \[ \frac {{\left (b x^{4} + a\right )}^{\frac {17}{4}}}{17 \, b^{4}} - \frac {3 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}} a}{13 \, b^{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {9}{4}} a^{2}}{3 \, b^{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{3}}{5 \, b^{4}} \]

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 = 1.13, size = 55, normalized size = 0.69 \[ {\left (b\,x^4+a\right )}^{1/4}\,\left (\frac {x^{16}}{17}-\frac {128\,a^4}{3315\,b^4}+\frac {a\,x^{12}}{221\,b}+\frac {32\,a^3\,x^4}{3315\,b^3}-\frac {4\,a^2\,x^8}{663\,b^2}\right ) \]

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)*(x^16/17 - (128*a^4)/(3315*b^4) + (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 = 15.77, size = 110, normalized size = 1.38 \[ \begin {cases} - \frac {128 a^{4} \sqrt [4]{a + b x^{4}}}{3315 b^{4}} + \frac {32 a^{3} x^{4} \sqrt [4]{a + b x^{4}}}{3315 b^{3}} - \frac {4 a^{2} x^{8} \sqrt [4]{a + b x^{4}}}{663 b^{2}} + \frac {a x^{12} \sqrt [4]{a + b x^{4}}}{221 b} + \frac {x^{16} \sqrt [4]{a + b x^{4}}}{17} & \text {for}\: b \neq 0 \\\frac {\sqrt [4]{a} x^{16}}{16} & \text {otherwise} \end {cases} \]

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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