\(\int \frac {1}{x^8 (a+b x^4)^{3/4}} \, dx\) [1137]

Optimal result

Integrand size = 15, antiderivative size = 107 \[ \int \frac {1}{x^8 \left (a+b x^4\right )^{3/4}} \, dx=-\frac {\sqrt [4]{a+b x^4}}{7 a x^7}+\frac {2 b \sqrt [4]{a+b x^4}}{7 a^2 x^3}-\frac {4 b^{5/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{7 a^{5/2} \left (a+b x^4\right )^{3/4}} \]

[Out]

-1/7*(b*x^4+a)^(1/4)/a/x^7+2/7*b*(b*x^4+a)^(1/4)/a^2/x^3-4/7*b^(5/2)*(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))/a^(5/2)/(b*x^4+a)^(3/4)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {331, 243, 342, 281, 237} \[ \int \frac {1}{x^8 \left (a+b x^4\right )^{3/4}} \, dx=-\frac {4 b^{5/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{7 a^{5/2} \left (a+b x^4\right )^{3/4}}+\frac {2 b \sqrt [4]{a+b x^4}}{7 a^2 x^3}-\frac {\sqrt [4]{a+b x^4}}{7 a x^7} \]

[In]

Int[1/(x^8*(a + b*x^4)^(3/4)),x]

[Out]

-1/7*(a + b*x^4)^(1/4)/(a*x^7) + (2*b*(a + b*x^4)^(1/4))/(7*a^2*x^3) - (4*b^(5/2)*(1 + a/(b*x^4))^(3/4)*x^3*El
lipticF[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(7*a^(5/2)*(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 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]

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 {\sqrt [4]{a+b x^4}}{7 a x^7}-\frac {(6 b) \int \frac {1}{x^4 \left (a+b x^4\right )^{3/4}} \, dx}{7 a} \\ & = -\frac {\sqrt [4]{a+b x^4}}{7 a x^7}+\frac {2 b \sqrt [4]{a+b x^4}}{7 a^2 x^3}+\frac {\left (4 b^2\right ) \int \frac {1}{\left (a+b x^4\right )^{3/4}} \, dx}{7 a^2} \\ & = -\frac {\sqrt [4]{a+b x^4}}{7 a x^7}+\frac {2 b \sqrt [4]{a+b x^4}}{7 a^2 x^3}+\frac {\left (4 b^2 \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}{7 a^2 \left (a+b x^4\right )^{3/4}} \\ & = -\frac {\sqrt [4]{a+b x^4}}{7 a x^7}+\frac {2 b \sqrt [4]{a+b x^4}}{7 a^2 x^3}-\frac {\left (4 b^2 \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 )}{7 a^2 \left (a+b x^4\right )^{3/4}} \\ & = -\frac {\sqrt [4]{a+b x^4}}{7 a x^7}+\frac {2 b \sqrt [4]{a+b x^4}}{7 a^2 x^3}-\frac {\left (2 b^2 \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 )}{7 a^2 \left (a+b x^4\right )^{3/4}} \\ & = -\frac {\sqrt [4]{a+b x^4}}{7 a x^7}+\frac {2 b \sqrt [4]{a+b x^4}}{7 a^2 x^3}-\frac {4 b^{5/2} \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 )}{7 a^{5/2} \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 = 10.02 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.48 \[ \int \frac {1}{x^8 \left (a+b x^4\right )^{3/4}} \, dx=-\frac {\left (1+\frac {b x^4}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},\frac {3}{4},-\frac {3}{4},-\frac {b x^4}{a}\right )}{7 x^7 \left (a+b x^4\right )^{3/4}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.68 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.41 \[ \int \frac {1}{x^8 \left (a+b x^4\right )^{3/4}} \, dx=\frac {\Gamma \left (- \frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{4}, \frac {3}{4} \\ - \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{4}} x^{7} \Gamma \left (- \frac {3}{4}\right )} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(x^8*(a + b*x^4)^(3/4)),x)

[Out]

int(1/(x^8*(a + b*x^4)^(3/4)), x)