3.1167 \(\int \frac{1}{x^6 (a+b x^4)^{5/4}} \, dx\)

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

[Out]

-1/(5*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*Ellip
ticE[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(5*a^(5/2)*(a + b*x^4)^(1/4))

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Rubi [A]  time = 0.0484748, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {283, 281, 335, 275, 196} \[ -\frac{12 b^{3/2} x \sqrt [4]{\frac{a}{b x^4}+1} E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{5 a^{5/2} \sqrt [4]{a+b x^4}}+\frac{6 b}{5 a^2 x \sqrt [4]{a+b x^4}}-\frac{1}{5 a x^5 \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(a + b*x^4)^(5/4)),x]

[Out]

-1/(5*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*Ellip
ticE[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(5*a^(5/2)*(a + b*x^4)^(1/4))

Rule 283

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 281

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 335

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]

Rule 275

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 196

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

Rubi steps

\begin{align*} \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) \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 ) \operatorname{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 ) \operatorname{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]  time = 0.0099544, size = 54, normalized size = 0.51 \[ -\frac{\sqrt [4]{\frac{b x^4}{a}+1} \, _2F_1\left (-\frac{5}{4},\frac{5}{4};-\frac{1}{4};-\frac{b x^4}{a}\right )}{5 a x^5 \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.055, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{6}} \left ( b{x}^{4}+a \right ) ^{-{\frac{5}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{4} + a\right )}^{\frac{5}{4}} x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{4} + a\right )}^{\frac{3}{4}}}{b^{2} x^{14} + 2 \, a b x^{10} + a^{2} x^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [C]  time = 2.1668, size = 44, normalized size = 0.42 \begin{align*} \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{4} + a\right )}^{\frac{5}{4}} x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)