\(\int \frac {x^{15}}{\sqrt [4]{a-b x^4}} \, dx\) [1205]

Optimal result

Integrand size = 16, antiderivative size = 84 \[ \int \frac {x^{15}}{\sqrt [4]{a-b x^4}} \, dx=-\frac {a^3 \left (a-b x^4\right )^{3/4}}{3 b^4}+\frac {3 a^2 \left (a-b x^4\right )^{7/4}}{7 b^4}-\frac {3 a \left (a-b x^4\right )^{11/4}}{11 b^4}+\frac {\left (a-b x^4\right )^{15/4}}{15 b^4} \]

[Out]

-1/3*a^3*(-b*x^4+a)^(3/4)/b^4+3/7*a^2*(-b*x^4+a)^(7/4)/b^4-3/11*a*(-b*x^4+a)^(11/4)/b^4+1/15*(-b*x^4+a)^(15/4)
/b^4

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {272, 45} \[ \int \frac {x^{15}}{\sqrt [4]{a-b x^4}} \, dx=-\frac {a^3 \left (a-b x^4\right )^{3/4}}{3 b^4}+\frac {3 a^2 \left (a-b x^4\right )^{7/4}}{7 b^4}+\frac {\left (a-b x^4\right )^{15/4}}{15 b^4}-\frac {3 a \left (a-b x^4\right )^{11/4}}{11 b^4} \]

[In]

Int[x^15/(a - b*x^4)^(1/4),x]

[Out]

-1/3*(a^3*(a - b*x^4)^(3/4))/b^4 + (3*a^2*(a - b*x^4)^(7/4))/(7*b^4) - (3*a*(a - b*x^4)^(11/4))/(11*b^4) + (a
- b*x^4)^(15/4)/(15*b^4)

Rule 45

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 272

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*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \frac {x^3}{\sqrt [4]{a-b x}} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \left (\frac {a^3}{b^3 \sqrt [4]{a-b x}}-\frac {3 a^2 (a-b x)^{3/4}}{b^3}+\frac {3 a (a-b x)^{7/4}}{b^3}-\frac {(a-b x)^{11/4}}{b^3}\right ) \, dx,x,x^4\right ) \\ & = -\frac {a^3 \left (a-b x^4\right )^{3/4}}{3 b^4}+\frac {3 a^2 \left (a-b x^4\right )^{7/4}}{7 b^4}-\frac {3 a \left (a-b x^4\right )^{11/4}}{11 b^4}+\frac {\left (a-b x^4\right )^{15/4}}{15 b^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.61 \[ \int \frac {x^{15}}{\sqrt [4]{a-b x^4}} \, dx=\frac {\left (a-b x^4\right )^{3/4} \left (-128 a^3-96 a^2 b x^4-84 a b^2 x^8-77 b^3 x^{12}\right )}{1155 b^4} \]

[In]

Integrate[x^15/(a - b*x^4)^(1/4),x]

[Out]

((a - b*x^4)^(3/4)*(-128*a^3 - 96*a^2*b*x^4 - 84*a*b^2*x^8 - 77*b^3*x^12))/(1155*b^4)

Maple [A] (verified)

Time = 4.38 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.57

method	result	size
gosper	\(-\frac {\left (-b \,x^{4}+a \right )^{\frac {3}{4}} \left (77 b^{3} x^{12}+84 a \,b^{2} x^{8}+96 a^{2} b \,x^{4}+128 a^{3}\right )}{1155 b^{4}}\)	\(48\)
trager	\(-\frac {\left (-b \,x^{4}+a \right )^{\frac {3}{4}} \left (77 b^{3} x^{12}+84 a \,b^{2} x^{8}+96 a^{2} b \,x^{4}+128 a^{3}\right )}{1155 b^{4}}\)	\(48\)
risch	\(-\frac {\left (-b \,x^{4}+a \right )^{\frac {3}{4}} \left (77 b^{3} x^{12}+84 a \,b^{2} x^{8}+96 a^{2} b \,x^{4}+128 a^{3}\right )}{1155 b^{4}}\)	\(48\)
pseudoelliptic	\(-\frac {\left (-b \,x^{4}+a \right )^{\frac {3}{4}} \left (77 b^{3} x^{12}+84 a \,b^{2} x^{8}+96 a^{2} b \,x^{4}+128 a^{3}\right )}{1155 b^{4}}\)	\(48\)

[In]

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

[Out]

-1/1155*(-b*x^4+a)^(3/4)*(77*b^3*x^12+84*a*b^2*x^8+96*a^2*b*x^4+128*a^3)/b^4

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.56 \[ \int \frac {x^{15}}{\sqrt [4]{a-b x^4}} \, dx=-\frac {{\left (77 \, b^{3} x^{12} + 84 \, a b^{2} x^{8} + 96 \, a^{2} b x^{4} + 128 \, a^{3}\right )} {\left (-b x^{4} + a\right )}^{\frac {3}{4}}}{1155 \, b^{4}} \]

[In]

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

[Out]

-1/1155*(77*b^3*x^12 + 84*a*b^2*x^8 + 96*a^2*b*x^4 + 128*a^3)*(-b*x^4 + a)^(3/4)/b^4

Sympy [A] (verification not implemented)

Time = 0.59 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.12 \[ \int \frac {x^{15}}{\sqrt [4]{a-b x^4}} \, dx=\begin {cases} - \frac {128 a^{3} \left (a - b x^{4}\right )^{\frac {3}{4}}}{1155 b^{4}} - \frac {32 a^{2} x^{4} \left (a - b x^{4}\right )^{\frac {3}{4}}}{385 b^{3}} - \frac {4 a x^{8} \left (a - b x^{4}\right )^{\frac {3}{4}}}{55 b^{2}} - \frac {x^{12} \left (a - b x^{4}\right )^{\frac {3}{4}}}{15 b} & \text {for}\: b \neq 0 \\\frac {x^{16}}{16 \sqrt [4]{a}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-128*a**3*(a - b*x**4)**(3/4)/(1155*b**4) - 32*a**2*x**4*(a - b*x**4)**(3/4)/(385*b**3) - 4*a*x**8*
(a - b*x**4)**(3/4)/(55*b**2) - x**12*(a - b*x**4)**(3/4)/(15*b), Ne(b, 0)), (x**16/(16*a**(1/4)), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.81 \[ \int \frac {x^{15}}{\sqrt [4]{a-b x^4}} \, dx=\frac {{\left (-b x^{4} + a\right )}^{\frac {15}{4}}}{15 \, b^{4}} - \frac {3 \, {\left (-b x^{4} + a\right )}^{\frac {11}{4}} a}{11 \, b^{4}} + \frac {3 \, {\left (-b x^{4} + a\right )}^{\frac {7}{4}} a^{2}}{7 \, b^{4}} - \frac {{\left (-b x^{4} + a\right )}^{\frac {3}{4}} a^{3}}{3 \, b^{4}} \]

[In]

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

[Out]

1/15*(-b*x^4 + a)^(15/4)/b^4 - 3/11*(-b*x^4 + a)^(11/4)*a/b^4 + 3/7*(-b*x^4 + a)^(7/4)*a^2/b^4 - 1/3*(-b*x^4 +
a)^(3/4)*a^3/b^4

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.99 \[ \int \frac {x^{15}}{\sqrt [4]{a-b x^4}} \, dx=-\frac {77 \, {\left (b x^{4} - a\right )}^{3} {\left (-b x^{4} + a\right )}^{\frac {3}{4}} + 315 \, {\left (b x^{4} - a\right )}^{2} {\left (-b x^{4} + a\right )}^{\frac {3}{4}} a - 495 \, {\left (-b x^{4} + a\right )}^{\frac {7}{4}} a^{2} + 385 \, {\left (-b x^{4} + a\right )}^{\frac {3}{4}} a^{3}}{1155 \, b^{4}} \]

[In]

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

[Out]

-1/1155*(77*(b*x^4 - a)^3*(-b*x^4 + a)^(3/4) + 315*(b*x^4 - a)^2*(-b*x^4 + a)^(3/4)*a - 495*(-b*x^4 + a)^(7/4)
*a^2 + 385*(-b*x^4 + a)^(3/4)*a^3)/b^4

Mupad [B] (verification not implemented)

Time = 5.51 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.58 \[ \int \frac {x^{15}}{\sqrt [4]{a-b x^4}} \, dx=-{\left (a-b\,x^4\right )}^{3/4}\,\left (\frac {128\,a^3}{1155\,b^4}+\frac {x^{12}}{15\,b}+\frac {4\,a\,x^8}{55\,b^2}+\frac {32\,a^2\,x^4}{385\,b^3}\right ) \]

[In]

int(x^15/(a - b*x^4)^(1/4),x)

[Out]

-(a - b*x^4)^(3/4)*((128*a^3)/(1155*b^4) + x^12/(15*b) + (4*a*x^8)/(55*b^2) + (32*a^2*x^4)/(385*b^3))