3.1193 \(\int \frac{\sqrt [4]{a-b x^4}}{x^{10}} \, dx\)

Optimal. Leaf size=46 \[ -\frac{4 b \left (a-b x^4\right )^{5/4}}{45 a^2 x^5}-\frac{\left (a-b x^4\right )^{5/4}}{9 a x^9} \]

[Out]

-(a - b*x^4)^(5/4)/(9*a*x^9) - (4*b*(a - b*x^4)^(5/4))/(45*a^2*x^5)

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Rubi [A]  time = 0.0107183, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {271, 264} \[ -\frac{4 b \left (a-b x^4\right )^{5/4}}{45 a^2 x^5}-\frac{\left (a-b x^4\right )^{5/4}}{9 a x^9} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x^4)^(1/4)/x^10,x]

[Out]

-(a - b*x^4)^(5/4)/(9*a*x^9) - (4*b*(a - b*x^4)^(5/4))/(45*a^2*x^5)

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{\sqrt [4]{a-b x^4}}{x^{10}} \, dx &=-\frac{\left (a-b x^4\right )^{5/4}}{9 a x^9}+\frac{(4 b) \int \frac{\sqrt [4]{a-b x^4}}{x^6} \, dx}{9 a}\\ &=-\frac{\left (a-b x^4\right )^{5/4}}{9 a x^9}-\frac{4 b \left (a-b x^4\right )^{5/4}}{45 a^2 x^5}\\ \end{align*}

Mathematica [A]  time = 0.007507, size = 32, normalized size = 0.7 \[ -\frac{\left (a-b x^4\right )^{5/4} \left (5 a+4 b x^4\right )}{45 a^2 x^9} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^4)^(1/4)/x^10,x]

[Out]

-((a - b*x^4)^(5/4)*(5*a + 4*b*x^4))/(45*a^2*x^9)

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Maple [A]  time = 0.004, size = 29, normalized size = 0.6 \begin{align*} -{\frac{4\,b{x}^{4}+5\,a}{45\,{a}^{2}{x}^{9}} \left ( -b{x}^{4}+a \right ) ^{{\frac{5}{4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-b*x^4+a)^(1/4)/x^10,x)

[Out]

-1/45*(-b*x^4+a)^(5/4)*(4*b*x^4+5*a)/a^2/x^9

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Maxima [A]  time = 0.962667, size = 50, normalized size = 1.09 \begin{align*} -\frac{\frac{9 \,{\left (-b x^{4} + a\right )}^{\frac{5}{4}} b}{x^{5}} + \frac{5 \,{\left (-b x^{4} + a\right )}^{\frac{9}{4}}}{x^{9}}}{45 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^4+a)^(1/4)/x^10,x, algorithm="maxima")

[Out]

-1/45*(9*(-b*x^4 + a)^(5/4)*b/x^5 + 5*(-b*x^4 + a)^(9/4)/x^9)/a^2

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Fricas [A]  time = 1.78002, size = 88, normalized size = 1.91 \begin{align*} \frac{{\left (4 \, b^{2} x^{8} + a b x^{4} - 5 \, a^{2}\right )}{\left (-b x^{4} + a\right )}^{\frac{1}{4}}}{45 \, a^{2} x^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^4+a)^(1/4)/x^10,x, algorithm="fricas")

[Out]

1/45*(4*b^2*x^8 + a*b*x^4 - 5*a^2)*(-b*x^4 + a)^(1/4)/(a^2*x^9)

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Sympy [B]  time = 2.41011, size = 410, normalized size = 8.91 \begin{align*} \begin{cases} - \frac{5 \sqrt [4]{b} \sqrt [4]{\frac{a}{b x^{4}} - 1} \Gamma \left (- \frac{9}{4}\right )}{16 x^{8} \Gamma \left (- \frac{1}{4}\right )} + \frac{b^{\frac{5}{4}} \sqrt [4]{\frac{a}{b x^{4}} - 1} \Gamma \left (- \frac{9}{4}\right )}{16 a x^{4} \Gamma \left (- \frac{1}{4}\right )} + \frac{b^{\frac{9}{4}} \sqrt [4]{\frac{a}{b x^{4}} - 1} \Gamma \left (- \frac{9}{4}\right )}{4 a^{2} \Gamma \left (- \frac{1}{4}\right )} & \text{for}\: \frac{\left |{a}\right |}{\left |{b}\right | \left |{x^{4}}\right |} > 1 \\\frac{5 a^{3} b^{\frac{5}{4}} \sqrt [4]{- \frac{a}{b x^{4}} + 1} e^{\frac{i \pi }{4}} \Gamma \left (- \frac{9}{4}\right )}{x^{4} \left (- 16 a^{3} b x^{4} \Gamma \left (- \frac{1}{4}\right ) + 16 a^{2} b^{2} x^{8} \Gamma \left (- \frac{1}{4}\right )\right )} - \frac{6 a^{2} b^{\frac{9}{4}} \sqrt [4]{- \frac{a}{b x^{4}} + 1} e^{\frac{i \pi }{4}} \Gamma \left (- \frac{9}{4}\right )}{- 16 a^{3} b x^{4} \Gamma \left (- \frac{1}{4}\right ) + 16 a^{2} b^{2} x^{8} \Gamma \left (- \frac{1}{4}\right )} - \frac{3 a b^{\frac{13}{4}} x^{4} \sqrt [4]{- \frac{a}{b x^{4}} + 1} e^{\frac{i \pi }{4}} \Gamma \left (- \frac{9}{4}\right )}{- 16 a^{3} b x^{4} \Gamma \left (- \frac{1}{4}\right ) + 16 a^{2} b^{2} x^{8} \Gamma \left (- \frac{1}{4}\right )} + \frac{4 b^{\frac{17}{4}} x^{8} \sqrt [4]{- \frac{a}{b x^{4}} + 1} e^{\frac{i \pi }{4}} \Gamma \left (- \frac{9}{4}\right )}{- 16 a^{3} b x^{4} \Gamma \left (- \frac{1}{4}\right ) + 16 a^{2} b^{2} x^{8} \Gamma \left (- \frac{1}{4}\right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x**4+a)**(1/4)/x**10,x)

[Out]

Piecewise((-5*b**(1/4)*(a/(b*x**4) - 1)**(1/4)*gamma(-9/4)/(16*x**8*gamma(-1/4)) + b**(5/4)*(a/(b*x**4) - 1)**
(1/4)*gamma(-9/4)/(16*a*x**4*gamma(-1/4)) + b**(9/4)*(a/(b*x**4) - 1)**(1/4)*gamma(-9/4)/(4*a**2*gamma(-1/4)),
Abs(a)/(Abs(b)*Abs(x**4)) > 1), (5*a**3*b**(5/4)*(-a/(b*x**4) + 1)**(1/4)*exp(I*pi/4)*gamma(-9/4)/(x**4*(-16*
a**3*b*x**4*gamma(-1/4) + 16*a**2*b**2*x**8*gamma(-1/4))) - 6*a**2*b**(9/4)*(-a/(b*x**4) + 1)**(1/4)*exp(I*pi/
4)*gamma(-9/4)/(-16*a**3*b*x**4*gamma(-1/4) + 16*a**2*b**2*x**8*gamma(-1/4)) - 3*a*b**(13/4)*x**4*(-a/(b*x**4)
+ 1)**(1/4)*exp(I*pi/4)*gamma(-9/4)/(-16*a**3*b*x**4*gamma(-1/4) + 16*a**2*b**2*x**8*gamma(-1/4)) + 4*b**(17/
4)*x**8*(-a/(b*x**4) + 1)**(1/4)*exp(I*pi/4)*gamma(-9/4)/(-16*a**3*b*x**4*gamma(-1/4) + 16*a**2*b**2*x**8*gamm
a(-1/4)), True))

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Giac [A]  time = 1.18235, size = 85, normalized size = 1.85 \begin{align*} \frac{\frac{9 \,{\left (-b x^{4} + a\right )}^{\frac{1}{4}}{\left (b - \frac{a}{x^{4}}\right )} b}{x} - \frac{5 \,{\left (b^{2} x^{8} - 2 \, a b x^{4} + a^{2}\right )}{\left (-b x^{4} + a\right )}^{\frac{1}{4}}}{x^{9}}}{45 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^4+a)^(1/4)/x^10,x, algorithm="giac")

[Out]

1/45*(9*(-b*x^4 + a)^(1/4)*(b - a/x^4)*b/x - 5*(b^2*x^8 - 2*a*b*x^4 + a^2)*(-b*x^4 + a)^(1/4)/x^9)/a^2