\(\int \frac {x^2}{(a+b x^4)^{5/4}} \, dx\) [1165]

Optimal result

Integrand size = 15, antiderivative size = 59 \[ \int \frac {x^2}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {\sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} \sqrt {b} \sqrt [4]{a+b x^4}} \]

[Out]

-(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)))*Ellip
ticE(sin(1/2*arccot(x^2*b^(1/2)/a^(1/2))),2^(1/2))/(b*x^4+a)^(1/4)/a^(1/2)/b^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {287, 342, 281, 202} \[ \int \frac {x^2}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} \sqrt {b} \sqrt [4]{a+b x^4}} \]

[In]

Int[x^2/(a + b*x^4)^(5/4),x]

[Out]

-(((1 + a/(b*x^4))^(1/4)*x*EllipticE[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(Sqrt[a]*Sqrt[b]*(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 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 {\left (\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}{b \sqrt [4]{a+b x^4}} \\ & = -\frac {\left (\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 )}{b \sqrt [4]{a+b x^4}} \\ & = -\frac {\left (\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 )}{2 b \sqrt [4]{a+b x^4}} \\ & = -\frac {\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 )}{\sqrt {a} \sqrt {b} \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 = 8.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.92 \[ \int \frac {x^2}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {x^3 \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{3 a \sqrt [4]{a+b x^4}} \]

[In]

Integrate[x^2/(a + b*x^4)^(5/4),x]

[Out]

(x^3*(1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[3/4, 5/4, 7/4, -((b*x^4)/a)])/(3*a*(a + b*x^4)^(1/4))

Maple [F]

\[\int \frac {x^{2}}{\left (b \,x^{4}+a \right )^{\frac {5}{4}}}d x\]

[In]

int(x^2/(b*x^4+a)^(5/4),x)

[Out]

int(x^2/(b*x^4+a)^(5/4),x)

Fricas [F]

\[ \int \frac {x^2}{\left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {x^{2}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}}} \,d x } \]

[In]

integrate(x^2/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(3/4)*x^2/(b^2*x^8 + 2*a*b*x^4 + a^2), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.49 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.63 \[ \int \frac {x^2}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} \Gamma \left (\frac {7}{4}\right )} \]

[In]

integrate(x**2/(b*x**4+a)**(5/4),x)

[Out]

x**3*gamma(3/4)*hyper((3/4, 5/4), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(5/4)*gamma(7/4))

Maxima [F]

\[ \int \frac {x^2}{\left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {x^{2}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}}} \,d x } \]

[In]

integrate(x^2/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(x^2/(b*x^4 + a)^(5/4), x)

Giac [F]

\[ \int \frac {x^2}{\left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {x^{2}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}}} \,d x } \]

[In]

integrate(x^2/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate(x^2/(b*x^4 + a)^(5/4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\left (a+b x^4\right )^{5/4}} \, dx=\int \frac {x^2}{{\left (b\,x^4+a\right )}^{5/4}} \,d x \]

[In]

int(x^2/(a + b*x^4)^(5/4),x)

[Out]

int(x^2/(a + b*x^4)^(5/4), x)