3.1136 \(\int \frac{1}{x^4 (a+b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=85 \[ \frac{2 b^{3/2} x^3 \left (\frac{a}{b x^4}+1\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right ),2\right )}{3 a^{3/2} \left (a+b x^4\right )^{3/4}}-\frac{\sqrt [4]{a+b x^4}}{3 a x^3} \]

[Out]

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

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Rubi [A]  time = 0.0339459, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {325, 237, 335, 275, 231} \[ \frac{2 b^{3/2} x^3 \left (\frac{a}{b x^4}+1\right )^{3/4} F\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{3 a^{3/2} \left (a+b x^4\right )^{3/4}}-\frac{\sqrt [4]{a+b x^4}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^4)^(3/4)),x]

[Out]

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

Rule 325

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 237

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 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 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(3/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rubi steps

\begin{align*} \int \frac{1}{x^4 \left (a+b x^4\right )^{3/4}} \, dx &=-\frac{\sqrt [4]{a+b x^4}}{3 a x^3}-\frac{(2 b) \int \frac{1}{\left (a+b x^4\right )^{3/4}} \, dx}{3 a}\\ &=-\frac{\sqrt [4]{a+b x^4}}{3 a x^3}-\frac{\left (2 b \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}{3 a \left (a+b x^4\right )^{3/4}}\\ &=-\frac{\sqrt [4]{a+b x^4}}{3 a x^3}+\frac{\left (2 b \left (1+\frac{a}{b x^4}\right )^{3/4} x^3\right ) \operatorname{Subst}\left (\int \frac{x}{\left (1+\frac{a x^4}{b}\right )^{3/4}} \, dx,x,\frac{1}{x}\right )}{3 a \left (a+b x^4\right )^{3/4}}\\ &=-\frac{\sqrt [4]{a+b x^4}}{3 a x^3}+\frac{\left (b \left (1+\frac{a}{b x^4}\right )^{3/4} x^3\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{a x^2}{b}\right )^{3/4}} \, dx,x,\frac{1}{x^2}\right )}{3 a \left (a+b x^4\right )^{3/4}}\\ &=-\frac{\sqrt [4]{a+b x^4}}{3 a x^3}+\frac{2 b^{3/2} \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 )}{3 a^{3/2} \left (a+b x^4\right )^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0087688, size = 51, normalized size = 0.6 \[ -\frac{\left (\frac{b x^4}{a}+1\right )^{3/4} \, _2F_1\left (-\frac{3}{4},\frac{3}{4};\frac{1}{4};-\frac{b x^4}{a}\right )}{3 x^3 \left (a+b x^4\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.03, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4}} \left ( b{x}^{4}+a \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/x^4/(b*x^4+a)^(3/4),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{3}{4}} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((b*x^4 + a)^(3/4)*x^4), 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{1}{4}}}{b x^{8} + a x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 0.978788, size = 41, normalized size = 0.48 \begin{align*} \frac{\Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, \frac{3}{4} \\ \frac{1}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac{3}{4}} x^{3} \Gamma \left (\frac{1}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

gamma(-3/4)*hyper((-3/4, 3/4), (1/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/4)*x**3*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{3}{4}} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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