\(\int x (a+b x^4)^{5/4} \, dx\) [1057]

Optimal result

Integrand size = 13, antiderivative size = 98 \[ \int x \left (a+b x^4\right )^{5/4} \, dx=\frac {5}{21} a x^2 \sqrt [4]{a+b x^4}+\frac {1}{7} x^2 \left (a+b x^4\right )^{5/4}+\frac {5 a^{5/2} \left (1+\frac {b x^4}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{21 \sqrt {b} \left (a+b x^4\right )^{3/4}} \]

[Out]

5/21*a*x^2*(b*x^4+a)^(1/4)+1/7*x^2*(b*x^4+a)^(5/4)+5/21*a^(5/2)*(1+b*x^4/a)^(3/4)*(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)))*EllipticF(sin(1/2*arctan(x^2*b^(1/2)/a^(1/2))),2^(1/2)
)/(b*x^4+a)^(3/4)/b^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {281, 201, 239, 237} \[ \int x \left (a+b x^4\right )^{5/4} \, dx=\frac {5 a^{5/2} \left (\frac {b x^4}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{21 \sqrt {b} \left (a+b x^4\right )^{3/4}}+\frac {5}{21} a x^2 \sqrt [4]{a+b x^4}+\frac {1}{7} x^2 \left (a+b x^4\right )^{5/4} \]

[In]

Int[x*(a + b*x^4)^(5/4),x]

[Out]

(5*a*x^2*(a + b*x^4)^(1/4))/21 + (x^2*(a + b*x^4)^(5/4))/7 + (5*a^(5/2)*(1 + (b*x^4)/a)^(3/4)*EllipticF[ArcTan
[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(21*Sqrt[b]*(a + b*x^4)^(3/4))

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 237

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]))*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 239

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \left (a+b x^2\right )^{5/4} \, dx,x,x^2\right ) \\ & = \frac {1}{7} x^2 \left (a+b x^4\right )^{5/4}+\frac {1}{14} (5 a) \text {Subst}\left (\int \sqrt [4]{a+b x^2} \, dx,x,x^2\right ) \\ & = \frac {5}{21} a x^2 \sqrt [4]{a+b x^4}+\frac {1}{7} x^2 \left (a+b x^4\right )^{5/4}+\frac {1}{42} \left (5 a^2\right ) \text {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^{3/4}} \, dx,x,x^2\right ) \\ & = \frac {5}{21} a x^2 \sqrt [4]{a+b x^4}+\frac {1}{7} x^2 \left (a+b x^4\right )^{5/4}+\frac {\left (5 a^2 \left (1+\frac {b x^4}{a}\right )^{3/4}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{3/4}} \, dx,x,x^2\right )}{42 \left (a+b x^4\right )^{3/4}} \\ & = \frac {5}{21} a x^2 \sqrt [4]{a+b x^4}+\frac {1}{7} x^2 \left (a+b x^4\right )^{5/4}+\frac {5 a^{5/2} \left (1+\frac {b x^4}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{21 \sqrt {b} \left (a+b x^4\right )^{3/4}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 8.32 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.53 \[ \int x \left (a+b x^4\right )^{5/4} \, dx=\frac {a x^2 \sqrt [4]{a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^4}{a}\right )}{2 \sqrt [4]{1+\frac {b x^4}{a}}} \]

[In]

Integrate[x*(a + b*x^4)^(5/4),x]

[Out]

(a*x^2*(a + b*x^4)^(1/4)*Hypergeometric2F1[-5/4, 1/2, 3/2, -((b*x^4)/a)])/(2*(1 + (b*x^4)/a)^(1/4))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((b*x^5 + a*x)*(b*x^4 + a)^(1/4), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.65 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.30 \[ \int x \left (a+b x^4\right )^{5/4} \, dx=\frac {a^{\frac {5}{4}} x^{2} {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2} \]

[In]

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

[Out]

a**(5/4)*x**2*hyper((-5/4, 1/2), (3/2,), b*x**4*exp_polar(I*pi)/a)/2

Maxima [F]

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

[In]

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

[Out]

integrate((b*x^4 + a)^(5/4)*x, x)

Giac [F]

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

[In]

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

[Out]

integrate((b*x^4 + a)^(5/4)*x, x)

Mupad [F(-1)]

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

[In]

int(x*(a + b*x^4)^(5/4),x)

[Out]

int(x*(a + b*x^4)^(5/4), x)