\(\int \frac {1}{x^8 (a+b x^4)^{3/2}} \, dx\) [866]

Optimal result

Integrand size = 15, antiderivative size = 153 \[ \int \frac {1}{x^8 \left (a+b x^4\right )^{3/2}} \, dx=\frac {1}{2 a x^7 \sqrt {a+b x^4}}-\frac {9 \sqrt {a+b x^4}}{14 a^2 x^7}+\frac {15 b \sqrt {a+b x^4}}{14 a^3 x^3}+\frac {15 b^{7/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{28 a^{13/4} \sqrt {a+b x^4}} \]

[Out]

1/2/a/x^7/(b*x^4+a)^(1/2)-9/14*(b*x^4+a)^(1/2)/a^2/x^7+15/14*b*(b*x^4+a)^(1/2)/a^3/x^3+15/28*b^(7/4)*(cos(2*ar
ctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(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)/a^(13/4)/(b*x^4+a)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {296, 331, 226} \[ \int \frac {1}{x^8 \left (a+b x^4\right )^{3/2}} \, dx=\frac {15 b^{7/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{28 a^{13/4} \sqrt {a+b x^4}}+\frac {15 b \sqrt {a+b x^4}}{14 a^3 x^3}-\frac {9 \sqrt {a+b x^4}}{14 a^2 x^7}+\frac {1}{2 a x^7 \sqrt {a+b x^4}} \]

[In]

Int[1/(x^8*(a + b*x^4)^(3/2)),x]

[Out]

1/(2*a*x^7*Sqrt[a + b*x^4]) - (9*Sqrt[a + b*x^4])/(14*a^2*x^7) + (15*b*Sqrt[a + b*x^4])/(14*a^3*x^3) + (15*b^(
7/4)*(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])/(28*a^(13/4)*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 296

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

Rule 331

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2 a x^7 \sqrt {a+b x^4}}+\frac {9 \int \frac {1}{x^8 \sqrt {a+b x^4}} \, dx}{2 a} \\ & = \frac {1}{2 a x^7 \sqrt {a+b x^4}}-\frac {9 \sqrt {a+b x^4}}{14 a^2 x^7}-\frac {(45 b) \int \frac {1}{x^4 \sqrt {a+b x^4}} \, dx}{14 a^2} \\ & = \frac {1}{2 a x^7 \sqrt {a+b x^4}}-\frac {9 \sqrt {a+b x^4}}{14 a^2 x^7}+\frac {15 b \sqrt {a+b x^4}}{14 a^3 x^3}+\frac {\left (15 b^2\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{14 a^3} \\ & = \frac {1}{2 a x^7 \sqrt {a+b x^4}}-\frac {9 \sqrt {a+b x^4}}{14 a^2 x^7}+\frac {15 b \sqrt {a+b x^4}}{14 a^3 x^3}+\frac {15 b^{7/4} \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 )}{28 a^{13/4} \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 = 10.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.35 \[ \int \frac {1}{x^8 \left (a+b x^4\right )^{3/2}} \, dx=-\frac {\sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},\frac {3}{2},-\frac {3}{4},-\frac {b x^4}{a}\right )}{7 a x^7 \sqrt {a+b x^4}} \]

[In]

Integrate[1/(x^8*(a + b*x^4)^(3/2)),x]

[Out]

-1/7*(Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[-7/4, 3/2, -3/4, -((b*x^4)/a)])/(a*x^7*Sqrt[a + b*x^4])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 5.19 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.88

method	result	size
default	\(\frac {b^{2} x}{2 a^{3} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {\sqrt {b \,x^{4}+a}}{7 a^{2} x^{7}}+\frac {4 b \sqrt {b \,x^{4}+a}}{7 a^{3} x^{3}}+\frac {15 b^{2} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{14 a^{3} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(135\)
elliptic	\(\frac {b^{2} x}{2 a^{3} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {\sqrt {b \,x^{4}+a}}{7 a^{2} x^{7}}+\frac {4 b \sqrt {b \,x^{4}+a}}{7 a^{3} x^{3}}+\frac {15 b^{2} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{14 a^{3} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(135\)
risch	\(-\frac {\sqrt {b \,x^{4}+a}\, \left (-4 b \,x^{4}+a \right )}{7 a^{3} x^{7}}+\frac {b^{2} \left (11 a \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+4 b \left (-\frac {x}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\right )}{7 a^{3}}\)	\(228\)

[In]

int(1/x^8/(b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/2*b^2/a^3*x/((x^4+a/b)*b)^(1/2)-1/7*(b*x^4+a)^(1/2)/a^2/x^7+4/7*b*(b*x^4+a)^(1/2)/a^3/x^3+15/14*b^2/a^3/(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)*Elliptic
F(x*(I/a^(1/2)*b^(1/2))^(1/2),I)

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.61 \[ \int \frac {1}{x^8 \left (a+b x^4\right )^{3/2}} \, dx=-\frac {15 \, {\left (b^{2} x^{11} + a b x^{7}\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - {\left (15 \, b^{2} x^{8} + 6 \, a b x^{4} - 2 \, a^{2}\right )} \sqrt {b x^{4} + a}}{14 \, {\left (a^{3} b x^{11} + a^{4} x^{7}\right )}} \]

[In]

integrate(1/x^8/(b*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

-1/14*(15*(b^2*x^11 + a*b*x^7)*sqrt(a)*(-b/a)^(3/4)*elliptic_f(arcsin(x*(-b/a)^(1/4)), -1) - (15*b^2*x^8 + 6*a
*b*x^4 - 2*a^2)*sqrt(b*x^4 + a))/(a^3*b*x^11 + a^4*x^7)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.73 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.29 \[ \int \frac {1}{x^8 \left (a+b x^4\right )^{3/2}} \, dx=\frac {\Gamma \left (- \frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{4}, \frac {3}{2} \\ - \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} x^{7} \Gamma \left (- \frac {3}{4}\right )} \]

[In]

integrate(1/x**8/(b*x**4+a)**(3/2),x)

[Out]

gamma(-7/4)*hyper((-7/4, 3/2), (-3/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*x**7*gamma(-3/4))

Maxima [F]

\[ \int \frac {1}{x^8 \left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{2}} x^{8}} \,d x } \]

[In]

integrate(1/x^8/(b*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)^(3/2)*x^8), x)

Giac [F]

\[ \int \frac {1}{x^8 \left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{2}} x^{8}} \,d x } \]

[In]

integrate(1/x^8/(b*x^4+a)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(3/2)*x^8), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^8 \left (a+b x^4\right )^{3/2}} \, dx=\int \frac {1}{x^8\,{\left (b\,x^4+a\right )}^{3/2}} \,d x \]

[In]

int(1/(x^8*(a + b*x^4)^(3/2)),x)

[Out]

int(1/(x^8*(a + b*x^4)^(3/2)), x)