\(\int \frac {x^{10}}{\sqrt [4]{a-b x^4}} \, dx\) [1227]

Optimal result

Integrand size = 16, antiderivative size = 134 \[ \int \frac {x^{10}}{\sqrt [4]{a-b x^4}} \, dx=-\frac {7 a^2 \left (a-b x^4\right )^{3/4}}{40 b^3 x}-\frac {7 a x^3 \left (a-b x^4\right )^{3/4}}{60 b^2}-\frac {x^7 \left (a-b x^4\right )^{3/4}}{10 b}+\frac {7 a^{5/2} \sqrt [4]{1-\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{40 b^{5/2} \sqrt [4]{a-b x^4}} \]

[Out]

-7/40*a^2*(-b*x^4+a)^(3/4)/b^3/x-7/60*a*x^3*(-b*x^4+a)^(3/4)/b^2-1/10*x^7*(-b*x^4+a)^(3/4)/b+7/40*a^(5/2)*(1-a
/b/x^4)^(1/4)*x*(cos(1/2*arccsc(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccsc(x^2*b^(1/2)/a^(1/2)))*EllipticE(
sin(1/2*arccsc(x^2*b^(1/2)/a^(1/2))),2^(1/2))/b^(5/2)/(-b*x^4+a)^(1/4)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {327, 317, 319, 342, 281, 234} \[ \int \frac {x^{10}}{\sqrt [4]{a-b x^4}} \, dx=\frac {7 a^{5/2} x \sqrt [4]{1-\frac {a}{b x^4}} E\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{40 b^{5/2} \sqrt [4]{a-b x^4}}-\frac {7 a^2 \left (a-b x^4\right )^{3/4}}{40 b^3 x}-\frac {7 a x^3 \left (a-b x^4\right )^{3/4}}{60 b^2}-\frac {x^7 \left (a-b x^4\right )^{3/4}}{10 b} \]

[In]

Int[x^10/(a - b*x^4)^(1/4),x]

[Out]

(-7*a^2*(a - b*x^4)^(3/4))/(40*b^3*x) - (7*a*x^3*(a - b*x^4)^(3/4))/(60*b^2) - (x^7*(a - b*x^4)^(3/4))/(10*b)
+ (7*a^(5/2)*(1 - a/(b*x^4))^(1/4)*x*EllipticE[ArcCsc[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(40*b^(5/2)*(a - b*x^4)^(1
/4))

Rule 234

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2/(a^(1/4)*Rt[-b/a, 2]))*EllipticE[(1/2)*ArcSin[Rt[-b/a,
2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 317

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(1/4), x_Symbol] :> Simp[(a + b*x^4)^(3/4)/(2*b*x), x] + Dist[a/(2*b), Int[1/
(x^2*(a + b*x^4)^(1/4)), x], x] /; FreeQ[{a, b}, x] && NegQ[b/a]

Rule 319

Int[1/((x_)^2*((a_) + (b_.)*(x_)^4)^(1/4)), x_Symbol] :> Dist[x*((1 + a/(b*x^4))^(1/4)/(a + b*x^4)^(1/4)), Int
[1/(x^3*(1 + a/(b*x^4))^(1/4)), x], x] /; FreeQ[{a, b}, x] && NegQ[b/a]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = -\frac {x^7 \left (a-b x^4\right )^{3/4}}{10 b}+\frac {(7 a) \int \frac {x^6}{\sqrt [4]{a-b x^4}} \, dx}{10 b} \\ & = -\frac {7 a x^3 \left (a-b x^4\right )^{3/4}}{60 b^2}-\frac {x^7 \left (a-b x^4\right )^{3/4}}{10 b}+\frac {\left (7 a^2\right ) \int \frac {x^2}{\sqrt [4]{a-b x^4}} \, dx}{20 b^2} \\ & = -\frac {7 a^2 \left (a-b x^4\right )^{3/4}}{40 b^3 x}-\frac {7 a x^3 \left (a-b x^4\right )^{3/4}}{60 b^2}-\frac {x^7 \left (a-b x^4\right )^{3/4}}{10 b}-\frac {\left (7 a^3\right ) \int \frac {1}{x^2 \sqrt [4]{a-b x^4}} \, dx}{40 b^3} \\ & = -\frac {7 a^2 \left (a-b x^4\right )^{3/4}}{40 b^3 x}-\frac {7 a x^3 \left (a-b x^4\right )^{3/4}}{60 b^2}-\frac {x^7 \left (a-b x^4\right )^{3/4}}{10 b}-\frac {\left (7 a^3 \sqrt [4]{1-\frac {a}{b x^4}} x\right ) \int \frac {1}{\sqrt [4]{1-\frac {a}{b x^4}} x^3} \, dx}{40 b^3 \sqrt [4]{a-b x^4}} \\ & = -\frac {7 a^2 \left (a-b x^4\right )^{3/4}}{40 b^3 x}-\frac {7 a x^3 \left (a-b x^4\right )^{3/4}}{60 b^2}-\frac {x^7 \left (a-b x^4\right )^{3/4}}{10 b}+\frac {\left (7 a^3 \sqrt [4]{1-\frac {a}{b x^4}} x\right ) \text {Subst}\left (\int \frac {x}{\sqrt [4]{1-\frac {a x^4}{b}}} \, dx,x,\frac {1}{x}\right )}{40 b^3 \sqrt [4]{a-b x^4}} \\ & = -\frac {7 a^2 \left (a-b x^4\right )^{3/4}}{40 b^3 x}-\frac {7 a x^3 \left (a-b x^4\right )^{3/4}}{60 b^2}-\frac {x^7 \left (a-b x^4\right )^{3/4}}{10 b}+\frac {\left (7 a^3 \sqrt [4]{1-\frac {a}{b x^4}} x\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1-\frac {a x^2}{b}}} \, dx,x,\frac {1}{x^2}\right )}{80 b^3 \sqrt [4]{a-b x^4}} \\ & = -\frac {7 a^2 \left (a-b x^4\right )^{3/4}}{40 b^3 x}-\frac {7 a x^3 \left (a-b x^4\right )^{3/4}}{60 b^2}-\frac {x^7 \left (a-b x^4\right )^{3/4}}{10 b}+\frac {7 a^{5/2} \sqrt [4]{1-\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{40 b^{5/2} \sqrt [4]{a-b x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 7.38 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.60 \[ \int \frac {x^{10}}{\sqrt [4]{a-b x^4}} \, dx=\frac {x^3 \left (-7 a^2+a b x^4+6 b^2 x^8+7 a^2 \sqrt [4]{1-\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {7}{4},\frac {b x^4}{a}\right )\right )}{60 b^2 \sqrt [4]{a-b x^4}} \]

[In]

Integrate[x^10/(a - b*x^4)^(1/4),x]

[Out]

(x^3*(-7*a^2 + a*b*x^4 + 6*b^2*x^8 + 7*a^2*(1 - (b*x^4)/a)^(1/4)*Hypergeometric2F1[1/4, 3/4, 7/4, (b*x^4)/a]))
/(60*b^2*(a - b*x^4)^(1/4))

Maple [F]

\[\int \frac {x^{10}}{\left (-b \,x^{4}+a \right )^{\frac {1}{4}}}d x\]

[In]

int(x^10/(-b*x^4+a)^(1/4),x)

[Out]

int(x^10/(-b*x^4+a)^(1/4),x)

Fricas [F]

\[ \int \frac {x^{10}}{\sqrt [4]{a-b x^4}} \, dx=\int { \frac {x^{10}}{{\left (-b x^{4} + a\right )}^{\frac {1}{4}}} \,d x } \]

[In]

integrate(x^10/(-b*x^4+a)^(1/4),x, algorithm="fricas")

[Out]

integral(-(-b*x^4 + a)^(3/4)*x^10/(b*x^4 - a), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.63 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.29 \[ \int \frac {x^{10}}{\sqrt [4]{a-b x^4}} \, dx=\frac {x^{11} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{4 \sqrt [4]{a} \Gamma \left (\frac {15}{4}\right )} \]

[In]

integrate(x**10/(-b*x**4+a)**(1/4),x)

[Out]

x**11*gamma(11/4)*hyper((1/4, 11/4), (15/4,), b*x**4*exp_polar(2*I*pi)/a)/(4*a**(1/4)*gamma(15/4))

Maxima [F]

\[ \int \frac {x^{10}}{\sqrt [4]{a-b x^4}} \, dx=\int { \frac {x^{10}}{{\left (-b x^{4} + a\right )}^{\frac {1}{4}}} \,d x } \]

[In]

integrate(x^10/(-b*x^4+a)^(1/4),x, algorithm="maxima")

[Out]

integrate(x^10/(-b*x^4 + a)^(1/4), x)

Giac [F]

\[ \int \frac {x^{10}}{\sqrt [4]{a-b x^4}} \, dx=\int { \frac {x^{10}}{{\left (-b x^{4} + a\right )}^{\frac {1}{4}}} \,d x } \]

[In]

integrate(x^10/(-b*x^4+a)^(1/4),x, algorithm="giac")

[Out]

integrate(x^10/(-b*x^4 + a)^(1/4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{10}}{\sqrt [4]{a-b x^4}} \, dx=\int \frac {x^{10}}{{\left (a-b\,x^4\right )}^{1/4}} \,d x \]

[In]

int(x^10/(a - b*x^4)^(1/4),x)

[Out]

int(x^10/(a - b*x^4)^(1/4), x)