\(\int \frac {x^8}{(a+b x^4)^{3/2}} \, dx\) [862]

Optimal result

Integrand size = 15, antiderivative size = 129 \[ \int \frac {x^8}{\left (a+b x^4\right )^{3/2}} \, dx=-\frac {x^5}{2 b \sqrt {a+b x^4}}+\frac {5 x \sqrt {a+b x^4}}{6 b^2}-\frac {5 a^{3/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{12 b^{9/4} \sqrt {a+b x^4}} \]

[Out]

-1/2*x^5/b/(b*x^4+a)^(1/2)+5/6*x*(b*x^4+a)^(1/2)/b^2-5/12*a^(3/4)*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/c
os(2*arctan(b^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2))*
((b*x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/b^(9/4)/(b*x^4+a)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {294, 327, 226} \[ \int \frac {x^8}{\left (a+b x^4\right )^{3/2}} \, dx=-\frac {5 a^{3/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{12 b^{9/4} \sqrt {a+b x^4}}+\frac {5 x \sqrt {a+b x^4}}{6 b^2}-\frac {x^5}{2 b \sqrt {a+b x^4}} \]

[In]

Int[x^8/(a + b*x^4)^(3/2),x]

[Out]

-1/2*x^5/(b*Sqrt[a + b*x^4]) + (5*x*Sqrt[a + b*x^4])/(6*b^2) - (5*a^(3/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*
x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(12*b^(9/4)*Sqrt[a + b*x^4])

Rule 226

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

Rule 294

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*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 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]

Rubi steps \begin{align*} \text {integral}& = -\frac {x^5}{2 b \sqrt {a+b x^4}}+\frac {5 \int \frac {x^4}{\sqrt {a+b x^4}} \, dx}{2 b} \\ & = -\frac {x^5}{2 b \sqrt {a+b x^4}}+\frac {5 x \sqrt {a+b x^4}}{6 b^2}-\frac {(5 a) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{6 b^2} \\ & = -\frac {x^5}{2 b \sqrt {a+b x^4}}+\frac {5 x \sqrt {a+b x^4}}{6 b^2}-\frac {5 a^{3/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{12 b^{9/4} \sqrt {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 = 10.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.51 \[ \int \frac {x^8}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {5 a x+2 b x^5-5 a x \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )}{6 b^2 \sqrt {a+b x^4}} \]

[In]

Integrate[x^8/(a + b*x^4)^(3/2),x]

[Out]

(5*a*x + 2*b*x^5 - 5*a*x*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)])/(6*b^2*Sqrt[a + b
*x^4])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 5.06 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.86

method	result	size
default	\(\frac {a x}{2 b^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {x \sqrt {b \,x^{4}+a}}{3 b^{2}}-\frac {5 a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{6 b^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(111\)
elliptic	\(\frac {a x}{2 b^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {x \sqrt {b \,x^{4}+a}}{3 b^{2}}-\frac {5 a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{6 b^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(111\)
risch	\(\frac {x \sqrt {b \,x^{4}+a}}{3 b^{2}}-\frac {a \left (a \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+4 b \left (-\frac {x}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\right )}{3 b^{2}}\)	\(215\)

[In]

int(x^8/(b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/2/b^2*a*x/((x^4+a/b)*b)^(1/2)+1/3*x*(b*x^4+a)^(1/2)/b^2-5/6*a/b^2/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(
1/2)*x^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*x^2)^(1/2)/(b*x^4+a)^(1/2)*EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.58 \[ \int \frac {x^8}{\left (a+b x^4\right )^{3/2}} \, dx=-\frac {5 \, {\left (b x^{4} + a\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - {\left (2 \, b x^{5} + 5 \, a x\right )} \sqrt {b x^{4} + a}}{6 \, {\left (b^{3} x^{4} + a b^{2}\right )}} \]

[In]

integrate(x^8/(b*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

-1/6*(5*(b*x^4 + a)*sqrt(b)*(-a/b)^(3/4)*elliptic_f(arcsin((-a/b)^(1/4)/x), -1) - (2*b*x^5 + 5*a*x)*sqrt(b*x^4
+ a))/(b^3*x^4 + a*b^2)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.53 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.29 \[ \int \frac {x^8}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {13}{4}\right )} \]

[In]

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

[Out]

x**9*gamma(9/4)*hyper((3/2, 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(13/4))

Maxima [F]

\[ \int \frac {x^8}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {x^{8}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(x^8/(b*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^8/(b*x^4 + a)^(3/2), x)

Giac [F]

\[ \int \frac {x^8}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {x^{8}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(x^8/(b*x^4+a)^(3/2),x, algorithm="giac")

[Out]

integrate(x^8/(b*x^4 + a)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^8}{\left (a+b x^4\right )^{3/2}} \, dx=\int \frac {x^8}{{\left (b\,x^4+a\right )}^{3/2}} \,d x \]

[In]

int(x^8/(a + b*x^4)^(3/2),x)

[Out]

int(x^8/(a + b*x^4)^(3/2), x)