\(\int \frac {x^5}{(a+b x^4)^{5/4}} \, dx\) [1149]

Optimal result

Integrand size = 15, antiderivative size = 77 \[ \int \frac {x^5}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {x^2}{b \sqrt [4]{a+b x^4}}-\frac {2 \sqrt {a} \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{b^{3/2} \sqrt [4]{a+b x^4}} \]

[Out]

x^2/b/(b*x^4+a)^(1/4)-2*(1+b*x^4/a)^(1/4)*(cos(1/2*arctan(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x^2*b^
(1/2)/a^(1/2)))*EllipticE(sin(1/2*arctan(x^2*b^(1/2)/a^(1/2))),2^(1/2))*a^(1/2)/b^(3/2)/(b*x^4+a)^(1/4)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 77, 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 = {281, 291, 203, 202} \[ \int \frac {x^5}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {x^2}{b \sqrt [4]{a+b x^4}}-\frac {2 \sqrt {a} \sqrt [4]{\frac {b x^4}{a}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{b^{3/2} \sqrt [4]{a+b x^4}} \]

[In]

Int[x^5/(a + b*x^4)^(5/4),x]

[Out]

x^2/(b*(a + b*x^4)^(1/4)) - (2*Sqrt[a]*(1 + (b*x^4)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(b
^(3/2)*(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 203

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Dist[(1 + b*(x^2/a))^(1/4)/(a*(a + b*x^2)^(1/4)), Int[1/(1 + b*
(x^2/a))^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a] && 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 291

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x^2}{\left (a+b x^2\right )^{5/4}} \, dx,x,x^2\right ) \\ & = \frac {x^2}{b \sqrt [4]{a+b x^4}}-\frac {a \text {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^{5/4}} \, dx,x,x^2\right )}{b} \\ & = \frac {x^2}{b \sqrt [4]{a+b x^4}}-\frac {\sqrt [4]{1+\frac {b x^4}{a}} \text {Subst}\left (\int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx,x,x^2\right )}{b \sqrt [4]{a+b x^4}} \\ & = \frac {x^2}{b \sqrt [4]{a+b x^4}}-\frac {2 \sqrt {a} \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{b^{3/2} \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 = 8.67 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.70 \[ \int \frac {x^5}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {x^2 \left (-1+\sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^4}{a}\right )\right )}{b \sqrt [4]{a+b x^4}} \]

[In]

Integrate[x^5/(a + b*x^4)^(5/4),x]

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((b*x^4 + a)^(3/4)*x^5/(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 = 27, normalized size of antiderivative = 0.35 \[ \int \frac {x^5}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {x^{6} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{6 a^{\frac {5}{4}}} \]

[In]

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

[Out]

x**6*hyper((5/4, 3/2), (5/2,), b*x**4*exp_polar(I*pi)/a)/(6*a**(5/4))

Maxima [F]

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

[In]

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

[Out]

integrate(x^5/(b*x^4 + a)^(5/4), x)

Giac [F]

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

[In]

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

[Out]

integrate(x^5/(b*x^4 + a)^(5/4), x)

Mupad [F(-1)]

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

[In]

int(x^5/(a + b*x^4)^(5/4),x)

[Out]

int(x^5/(a + b*x^4)^(5/4), x)