\(\int \frac {x^{10}}{\sqrt [4]{a+b x^4}} \, dx\) [1103]

Optimal result

Integrand size = 15, antiderivative size = 129 \[ \int \frac {x^{10}}{\sqrt [4]{a+b x^4}} \, dx=\frac {7 a^2 x^3}{40 b^2 \sqrt [4]{a+b x^4}}-\frac {7 a x^3 \left (a+b x^4\right )^{3/4}}{60 b^2}+\frac {x^7 \left (a+b x^4\right )^{3/4}}{10 b}+\frac {7 a^{5/2} \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{40 b^{5/2} \sqrt [4]{a+b x^4}} \]

[Out]

7/40*a^2*x^3/b^2/(b*x^4+a)^(1/4)-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*arccot(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))*EllipticE(si
n(1/2*arccot(x^2*b^(1/2)/a^(1/2))),2^(1/2))/b^(5/2)/(b*x^4+a)^(1/4)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {327, 316, 287, 342, 281, 202} \[ \int \frac {x^{10}}{\sqrt [4]{a+b x^4}} \, dx=\frac {7 a^{5/2} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{40 b^{5/2} \sqrt [4]{a+b x^4}}+\frac {7 a^2 x^3}{40 b^2 \sqrt [4]{a+b x^4}}-\frac {7 a x^3 \left (a+b x^4\right )^{3/4}}{60 b^2}+\frac {x^7 \left (a+b x^4\right )^{3/4}}{10 b} \]

[In]

Int[x^10/(a + b*x^4)^(1/4),x]

[Out]

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

Rule 202

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]))*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[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 287

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Dist[x*((1 + a/(b*x^4))^(1/4)/(b*(a + b*x^4)^(1/4))), Int
[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 316

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(1/4), x_Symbol] :> Simp[x^3/(2*(a + b*x^4)^(1/4)), x] - Dist[a/2, Int[x^2/(a
+ b*x^4)^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[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 x^3}{40 b^2 \sqrt [4]{a+b x^4}}-\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 {x^2}{\left (a+b x^4\right )^{5/4}} \, dx}{40 b^2} \\ & = \frac {7 a^2 x^3}{40 b^2 \sqrt [4]{a+b x^4}}-\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}{\left (1+\frac {a}{b x^4}\right )^{5/4} x^3} \, dx}{40 b^3 \sqrt [4]{a+b x^4}} \\ & = \frac {7 a^2 x^3}{40 b^2 \sqrt [4]{a+b x^4}}-\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}{\left (1+\frac {a x^4}{b}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )}{40 b^3 \sqrt [4]{a+b x^4}} \\ & = \frac {7 a^2 x^3}{40 b^2 \sqrt [4]{a+b x^4}}-\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}{\left (1+\frac {a x^2}{b}\right )^{5/4}} \, dx,x,\frac {1}{x^2}\right )}{80 b^3 \sqrt [4]{a+b x^4}} \\ & = \frac {7 a^2 x^3}{40 b^2 \sqrt [4]{a+b x^4}}-\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} \cot ^{-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 = 7.48 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.62 \[ \int \frac {x^{10}}{\sqrt [4]{a+b x^4}} \, dx=\frac {x^3 \left (-7 a^2-a b x^4+6 b^2 x^8+7 a^2 \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )\right )}{60 b^2 \sqrt [4]{a+b x^4}} \]

[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(x^10/(b*x^4 + a)^(1/4), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.60 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.29 \[ \int \frac {x^{10}}{\sqrt [4]{a+b x^4}} \, dx=\frac {x^{11} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [4]{a} \Gamma \left (\frac {15}{4}\right )} \]

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

[In]

int(x^10/(a + b*x^4)^(1/4),x)

[Out]

int(x^10/(a + b*x^4)^(1/4), x)