3.1099 \(\int \frac{1}{x^8 \sqrt [4]{a+b x^4}} \, dx\)

Optimal. Leaf size=44 \[ \frac{4 b \left (a+b x^4\right )^{3/4}}{21 a^2 x^3}-\frac{\left (a+b x^4\right )^{3/4}}{7 a x^7} \]

[Out]

-(a + b*x^4)^(3/4)/(7*a*x^7) + (4*b*(a + b*x^4)^(3/4))/(21*a^2*x^3)

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Rubi [A]  time = 0.0101368, 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} \[ \frac{4 b \left (a+b x^4\right )^{3/4}}{21 a^2 x^3}-\frac{\left (a+b x^4\right )^{3/4}}{7 a x^7} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^8*(a + b*x^4)^(1/4)),x]

[Out]

-(a + b*x^4)^(3/4)/(7*a*x^7) + (4*b*(a + b*x^4)^(3/4))/(21*a^2*x^3)

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^8 \sqrt [4]{a+b x^4}} \, dx &=-\frac{\left (a+b x^4\right )^{3/4}}{7 a x^7}-\frac{(4 b) \int \frac{1}{x^4 \sqrt [4]{a+b x^4}} \, dx}{7 a}\\ &=-\frac{\left (a+b x^4\right )^{3/4}}{7 a x^7}+\frac{4 b \left (a+b x^4\right )^{3/4}}{21 a^2 x^3}\\ \end{align*}

Mathematica [A]  time = 0.01283, size = 31, normalized size = 0.7 \[ \frac{\left (a+b x^4\right )^{3/4} \left (4 b x^4-3 a\right )}{21 a^2 x^7} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^4)^(3/4)*(-3*a + 4*b*x^4))/(21*a^2*x^7)

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Maple [A]  time = 0.004, size = 28, normalized size = 0.6 \begin{align*} -{\frac{-4\,b{x}^{4}+3\,a}{21\,{a}^{2}{x}^{7}} \left ( b{x}^{4}+a \right ) ^{{\frac{3}{4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/21*(b*x^4+a)^(3/4)*(-4*b*x^4+3*a)/a^2/x^7

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Maxima [A]  time = 1.01794, size = 47, normalized size = 1.07 \begin{align*} \frac{\frac{7 \,{\left (b x^{4} + a\right )}^{\frac{3}{4}} b}{x^{3}} - \frac{3 \,{\left (b x^{4} + a\right )}^{\frac{7}{4}}}{x^{7}}}{21 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21*(7*(b*x^4 + a)^(3/4)*b/x^3 - 3*(b*x^4 + a)^(7/4)/x^7)/a^2

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Fricas [A]  time = 1.51174, size = 68, normalized size = 1.55 \begin{align*} \frac{{\left (4 \, b x^{4} - 3 \, a\right )}{\left (b x^{4} + a\right )}^{\frac{3}{4}}}{21 \, a^{2} x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21*(4*b*x^4 - 3*a)*(b*x^4 + a)^(3/4)/(a^2*x^7)

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Sympy [A]  time = 1.85342, size = 70, normalized size = 1.59 \begin{align*} - \frac{3 b^{\frac{3}{4}} \left (\frac{a}{b x^{4}} + 1\right )^{\frac{3}{4}} \Gamma \left (- \frac{7}{4}\right )}{16 a x^{4} \Gamma \left (\frac{1}{4}\right )} + \frac{b^{\frac{7}{4}} \left (\frac{a}{b x^{4}} + 1\right )^{\frac{3}{4}} \Gamma \left (- \frac{7}{4}\right )}{4 a^{2} \Gamma \left (\frac{1}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*b**(3/4)*(a/(b*x**4) + 1)**(3/4)*gamma(-7/4)/(16*a*x**4*gamma(1/4)) + b**(7/4)*(a/(b*x**4) + 1)**(3/4)*gamm
a(-7/4)/(4*a**2*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{1}{4}} x^{8}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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