\(\int x^8 (a+b x^4)^{5/4} \, dx\) [1072]

Optimal result

Integrand size = 15, antiderivative size = 147 \[ \int x^8 \left (a+b x^4\right )^{5/4} \, dx=-\frac {5 a^3 x \sqrt [4]{a+b x^4}}{336 b^2}+\frac {a^2 x^5 \sqrt [4]{a+b x^4}}{168 b}+\frac {1}{28} a x^9 \sqrt [4]{a+b x^4}+\frac {1}{14} x^9 \left (a+b x^4\right )^{5/4}-\frac {5 a^{7/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{336 b^{3/2} \left (a+b x^4\right )^{3/4}} \]

[Out]

-5/336*a^3*x*(b*x^4+a)^(1/4)/b^2+1/168*a^2*x^5*(b*x^4+a)^(1/4)/b+1/28*a*x^9*(b*x^4+a)^(1/4)+1/14*x^9*(b*x^4+a)
^(5/4)-5/336*a^(7/2)*(1+a/b/x^4)^(3/4)*x^3*(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)))*EllipticF(sin(1/2*arccot(x^2*b^(1/2)/a^(1/2))),2^(1/2))/b^(3/2)/(b*x^4+a)^(3/4)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {285, 327, 243, 342, 281, 237} \[ \int x^8 \left (a+b x^4\right )^{5/4} \, dx=-\frac {5 a^{7/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{336 b^{3/2} \left (a+b x^4\right )^{3/4}}-\frac {5 a^3 x \sqrt [4]{a+b x^4}}{336 b^2}+\frac {a^2 x^5 \sqrt [4]{a+b x^4}}{168 b}+\frac {1}{14} x^9 \left (a+b x^4\right )^{5/4}+\frac {1}{28} a x^9 \sqrt [4]{a+b x^4} \]

[In]

Int[x^8*(a + b*x^4)^(5/4),x]

[Out]

(-5*a^3*x*(a + b*x^4)^(1/4))/(336*b^2) + (a^2*x^5*(a + b*x^4)^(1/4))/(168*b) + (a*x^9*(a + b*x^4)^(1/4))/28 +
(x^9*(a + b*x^4)^(5/4))/14 - (5*a^(7/2)*(1 + a/(b*x^4))^(3/4)*x^3*EllipticF[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2
])/(336*b^(3/2)*(a + b*x^4)^(3/4))

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 243

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

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 285

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

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 {1}{14} x^9 \left (a+b x^4\right )^{5/4}+\frac {1}{14} (5 a) \int x^8 \sqrt [4]{a+b x^4} \, dx \\ & = \frac {1}{28} a x^9 \sqrt [4]{a+b x^4}+\frac {1}{14} x^9 \left (a+b x^4\right )^{5/4}+\frac {1}{28} a^2 \int \frac {x^8}{\left (a+b x^4\right )^{3/4}} \, dx \\ & = \frac {a^2 x^5 \sqrt [4]{a+b x^4}}{168 b}+\frac {1}{28} a x^9 \sqrt [4]{a+b x^4}+\frac {1}{14} x^9 \left (a+b x^4\right )^{5/4}-\frac {\left (5 a^3\right ) \int \frac {x^4}{\left (a+b x^4\right )^{3/4}} \, dx}{168 b} \\ & = -\frac {5 a^3 x \sqrt [4]{a+b x^4}}{336 b^2}+\frac {a^2 x^5 \sqrt [4]{a+b x^4}}{168 b}+\frac {1}{28} a x^9 \sqrt [4]{a+b x^4}+\frac {1}{14} x^9 \left (a+b x^4\right )^{5/4}+\frac {\left (5 a^4\right ) \int \frac {1}{\left (a+b x^4\right )^{3/4}} \, dx}{336 b^2} \\ & = -\frac {5 a^3 x \sqrt [4]{a+b x^4}}{336 b^2}+\frac {a^2 x^5 \sqrt [4]{a+b x^4}}{168 b}+\frac {1}{28} a x^9 \sqrt [4]{a+b x^4}+\frac {1}{14} x^9 \left (a+b x^4\right )^{5/4}+\frac {\left (5 a^4 \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \int \frac {1}{\left (1+\frac {a}{b x^4}\right )^{3/4} x^3} \, dx}{336 b^2 \left (a+b x^4\right )^{3/4}} \\ & = -\frac {5 a^3 x \sqrt [4]{a+b x^4}}{336 b^2}+\frac {a^2 x^5 \sqrt [4]{a+b x^4}}{168 b}+\frac {1}{28} a x^9 \sqrt [4]{a+b x^4}+\frac {1}{14} x^9 \left (a+b x^4\right )^{5/4}-\frac {\left (5 a^4 \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \text {Subst}\left (\int \frac {x}{\left (1+\frac {a x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{x}\right )}{336 b^2 \left (a+b x^4\right )^{3/4}} \\ & = -\frac {5 a^3 x \sqrt [4]{a+b x^4}}{336 b^2}+\frac {a^2 x^5 \sqrt [4]{a+b x^4}}{168 b}+\frac {1}{28} a x^9 \sqrt [4]{a+b x^4}+\frac {1}{14} x^9 \left (a+b x^4\right )^{5/4}-\frac {\left (5 a^4 \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{x^2}\right )}{672 b^2 \left (a+b x^4\right )^{3/4}} \\ & = -\frac {5 a^3 x \sqrt [4]{a+b x^4}}{336 b^2}+\frac {a^2 x^5 \sqrt [4]{a+b x^4}}{168 b}+\frac {1}{28} a x^9 \sqrt [4]{a+b x^4}+\frac {1}{14} x^9 \left (a+b x^4\right )^{5/4}-\frac {5 a^{7/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{336 b^{3/2} \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 = 9.37 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.52 \[ \int x^8 \left (a+b x^4\right )^{5/4} \, dx=\frac {x \sqrt [4]{a+b x^4} \left (-\left (\left (a-2 b x^4\right ) \left (a+b x^4\right )^2\right )+\frac {a^3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{4},\frac {5}{4},-\frac {b x^4}{a}\right )}{\sqrt [4]{1+\frac {b x^4}{a}}}\right )}{28 b^2} \]

[In]

Integrate[x^8*(a + b*x^4)^(5/4),x]

[Out]

(x*(a + b*x^4)^(1/4)*(-((a - 2*b*x^4)*(a + b*x^4)^2) + (a^3*Hypergeometric2F1[-5/4, 1/4, 5/4, -((b*x^4)/a)])/(
1 + (b*x^4)/a)^(1/4)))/(28*b^2)

Maple [F]

\[\int x^{8} \left (b \,x^{4}+a \right )^{\frac {5}{4}}d x\]

[In]

int(x^8*(b*x^4+a)^(5/4),x)

[Out]

int(x^8*(b*x^4+a)^(5/4),x)

Fricas [F]

\[ \int x^8 \left (a+b x^4\right )^{5/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{8} \,d x } \]

[In]

integrate(x^8*(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

integral((b*x^12 + a*x^8)*(b*x^4 + a)^(1/4), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.94 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.27 \[ \int x^8 \left (a+b x^4\right )^{5/4} \, dx=\frac {a^{\frac {5}{4}} x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {13}{4}\right )} \]

[In]

integrate(x**8*(b*x**4+a)**(5/4),x)

[Out]

a**(5/4)*x**9*gamma(9/4)*hyper((-5/4, 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(13/4))

Maxima [F]

\[ \int x^8 \left (a+b x^4\right )^{5/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{8} \,d x } \]

[In]

integrate(x^8*(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^(5/4)*x^8, x)

Giac [F]

\[ \int x^8 \left (a+b x^4\right )^{5/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{8} \,d x } \]

[In]

integrate(x^8*(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(5/4)*x^8, x)

Mupad [F(-1)]

Timed out. \[ \int x^8 \left (a+b x^4\right )^{5/4} \, dx=\int x^8\,{\left (b\,x^4+a\right )}^{5/4} \,d x \]

[In]

int(x^8*(a + b*x^4)^(5/4),x)

[Out]

int(x^8*(a + b*x^4)^(5/4), x)