3.1129 \(\int \frac{1}{x^6 (a+b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=44 \[ \frac{4 b \sqrt [4]{a+b x^4}}{5 a^2 x}-\frac{\sqrt [4]{a+b x^4}}{5 a x^5} \]

[Out]

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

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Rubi [A]  time = 0.0118411, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {271, 264} \[ \frac{4 b \sqrt [4]{a+b x^4}}{5 a^2 x}-\frac{\sqrt [4]{a+b x^4}}{5 a x^5} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(a + b*x^4)^(3/4)),x]

[Out]

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

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 264

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] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{1}{x^6 \left (a+b x^4\right )^{3/4}} \, dx &=-\frac{\sqrt [4]{a+b x^4}}{5 a x^5}-\frac{(4 b) \int \frac{1}{x^2 \left (a+b x^4\right )^{3/4}} \, dx}{5 a}\\ &=-\frac{\sqrt [4]{a+b x^4}}{5 a x^5}+\frac{4 b \sqrt [4]{a+b x^4}}{5 a^2 x}\\ \end{align*}

Mathematica [A]  time = 0.0094761, size = 29, normalized size = 0.66 \[ -\frac{\left (a-4 b x^4\right ) \sqrt [4]{a+b x^4}}{5 a^2 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^6*(a + b*x^4)^(3/4)),x]

[Out]

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

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Maple [A]  time = 0.005, size = 26, normalized size = 0.6 \begin{align*} -{\frac{-4\,b{x}^{4}+a}{5\,{x}^{5}{a}^{2}}\sqrt [4]{b{x}^{4}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^6/(b*x^4+a)^(3/4),x)

[Out]

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

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Maxima [A]  time = 0.999628, size = 47, normalized size = 1.07 \begin{align*} \frac{\frac{5 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}} b}{x} - \frac{{\left (b x^{4} + a\right )}^{\frac{5}{4}}}{x^{5}}}{5 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.48579, size = 63, normalized size = 1.43 \begin{align*} \frac{{\left (4 \, b x^{4} - a\right )}{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{5 \, a^{2} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.58591, size = 68, normalized size = 1.55 \begin{align*} - \frac{\sqrt [4]{b} \sqrt [4]{\frac{a}{b x^{4}} + 1} \Gamma \left (- \frac{5}{4}\right )}{16 a x^{4} \Gamma \left (\frac{3}{4}\right )} + \frac{b^{\frac{5}{4}} \sqrt [4]{\frac{a}{b x^{4}} + 1} \Gamma \left (- \frac{5}{4}\right )}{4 a^{2} \Gamma \left (\frac{3}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**6/(b*x**4+a)**(3/4),x)

[Out]

-b**(1/4)*(a/(b*x**4) + 1)**(1/4)*gamma(-5/4)/(16*a*x**4*gamma(3/4)) + b**(5/4)*(a/(b*x**4) + 1)**(1/4)*gamma(
-5/4)/(4*a**2*gamma(3/4))

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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^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(3/4)*x^6), x)