\(\int \frac {1}{x^6 (a+b x^4)^{5/4}} \, dx\) [1167]

Optimal result

Integrand size = 15, antiderivative size = 105 \[ \int \frac {1}{x^6 \left (a+b x^4\right )^{5/4}} \, dx=-\frac {1}{5 a x^5 \sqrt [4]{a+b x^4}}+\frac {6 b}{5 a^2 x \sqrt [4]{a+b x^4}}-\frac {12 b^{3/2} \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{5 a^{5/2} \sqrt [4]{a+b x^4}} \]

[Out]

-1/5/a/x^5/(b*x^4+a)^(1/4)+6/5*b/a^2/x/(b*x^4+a)^(1/4)-12/5*b^(3/2)*(1+a/b/x^4)^(1/4)*x*(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)))*EllipticE(sin(1/2*arccot(x^2*b^(1/2)/a^(1/2))),2
^(1/2))/a^(5/2)/(b*x^4+a)^(1/4)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 105, 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 = {289, 287, 342, 281, 202} \[ \int \frac {1}{x^6 \left (a+b x^4\right )^{5/4}} \, dx=-\frac {12 b^{3/2} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{5 a^{5/2} \sqrt [4]{a+b x^4}}+\frac {6 b}{5 a^2 x \sqrt [4]{a+b x^4}}-\frac {1}{5 a x^5 \sqrt [4]{a+b x^4}} \]

[In]

Int[1/(x^6*(a + b*x^4)^(5/4)),x]

[Out]

-1/5*1/(a*x^5*(a + b*x^4)^(1/4)) + (6*b)/(5*a^2*x*(a + b*x^4)^(1/4)) - (12*b^(3/2)*(1 + a/(b*x^4))^(1/4)*x*Ell
ipticE[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(5*a^(5/2)*(a + b*x^4)^(1/4))

Rule 202

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]))*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/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]

Rule 287

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Dist[x*((1 + a/(b*x^4))^(1/4)/(b*(a + b*x^4)^(1/4))), Int
[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 289

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x^(m + 1)/(a*(m + 1)*(a + b*x^4)^(1/4)), x] - Dis
t[b*(m/(a*(m + 1))), Int[x^(m + 4)/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a] && ILtQ[(m - 2)/
4, 0]

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}{5 a x^5 \sqrt [4]{a+b x^4}}-\frac {(6 b) \int \frac {1}{x^2 \left (a+b x^4\right )^{5/4}} \, dx}{5 a} \\ & = -\frac {1}{5 a x^5 \sqrt [4]{a+b x^4}}+\frac {6 b}{5 a^2 x \sqrt [4]{a+b x^4}}+\frac {\left (12 b^2\right ) \int \frac {x^2}{\left (a+b x^4\right )^{5/4}} \, dx}{5 a^2} \\ & = -\frac {1}{5 a x^5 \sqrt [4]{a+b x^4}}+\frac {6 b}{5 a^2 x \sqrt [4]{a+b x^4}}+\frac {\left (12 b \sqrt [4]{1+\frac {a}{b x^4}} x\right ) \int \frac {1}{\left (1+\frac {a}{b x^4}\right )^{5/4} x^3} \, dx}{5 a^2 \sqrt [4]{a+b x^4}} \\ & = -\frac {1}{5 a x^5 \sqrt [4]{a+b x^4}}+\frac {6 b}{5 a^2 x \sqrt [4]{a+b x^4}}-\frac {\left (12 b \sqrt [4]{1+\frac {a}{b x^4}} x\right ) \text {Subst}\left (\int \frac {x}{\left (1+\frac {a x^4}{b}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )}{5 a^2 \sqrt [4]{a+b x^4}} \\ & = -\frac {1}{5 a x^5 \sqrt [4]{a+b x^4}}+\frac {6 b}{5 a^2 x \sqrt [4]{a+b x^4}}-\frac {\left (6 b \sqrt [4]{1+\frac {a}{b x^4}} x\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{5/4}} \, dx,x,\frac {1}{x^2}\right )}{5 a^2 \sqrt [4]{a+b x^4}} \\ & = -\frac {1}{5 a x^5 \sqrt [4]{a+b x^4}}+\frac {6 b}{5 a^2 x \sqrt [4]{a+b x^4}}-\frac {12 b^{3/2} \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{5 a^{5/2} \sqrt [4]{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 = 10.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.51 \[ \int \frac {1}{x^6 \left (a+b x^4\right )^{5/4}} \, dx=-\frac {\sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {5}{4},-\frac {1}{4},-\frac {b x^4}{a}\right )}{5 a x^5 \sqrt [4]{a+b x^4}} \]

[In]

Integrate[1/(x^6*(a + b*x^4)^(5/4)),x]

[Out]

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

Maple [F]

\[\int \frac {1}{x^{6} \left (b \,x^{4}+a \right )^{\frac {5}{4}}}d x\]

[In]

int(1/x^6/(b*x^4+a)^(5/4),x)

[Out]

int(1/x^6/(b*x^4+a)^(5/4),x)

Fricas [F]

\[ \int \frac {1}{x^6 \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{6}} \,d x } \]

[In]

integrate(1/x^6/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(3/4)/(b^2*x^14 + 2*a*b*x^10 + a^2*x^6), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.68 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.42 \[ \int \frac {1}{x^6 \left (a+b x^4\right )^{5/4}} \, dx=\frac {\Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {5}{4} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} x^{5} \Gamma \left (- \frac {1}{4}\right )} \]

[In]

integrate(1/x**6/(b*x**4+a)**(5/4),x)

[Out]

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

Maxima [F]

\[ \int \frac {1}{x^6 \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{6}} \,d x } \]

[In]

integrate(1/x^6/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)^(5/4)*x^6), x)

Giac [F]

\[ \int \frac {1}{x^6 \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{6}} \,d x } \]

[In]

integrate(1/x^6/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(5/4)*x^6), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^6 \left (a+b x^4\right )^{5/4}} \, dx=\int \frac {1}{x^6\,{\left (b\,x^4+a\right )}^{5/4}} \,d x \]

[In]

int(1/(x^6*(a + b*x^4)^(5/4)),x)

[Out]

int(1/(x^6*(a + b*x^4)^(5/4)), x)