3.1253 \(\int \frac {1}{x^6 (a-b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=46 \[ -\frac {4 b \sqrt [4]{a-b x^4}}{5 a^2 x}-\frac {\sqrt [4]{a-b x^4}}{5 a x^5} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 46, normalized size of antiderivative = 1.00, 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} \[ -\frac {4 b \sqrt [4]{a-b x^4}}{5 a^2 x}-\frac {\sqrt [4]{a-b x^4}}{5 a x^5} \]

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 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]

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]

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*}

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Mathematica [A]  time = 0.01, size = 30, normalized size = 0.65 \[ -\frac {\sqrt [4]{a-b x^4} \left (a+4 b x^4\right )}{5 a^2 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.64, size = 26, normalized size = 0.57 \[ -\frac {{\left (4 \, b x^{4} + a\right )} {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{5 \, a^{2} x^{5}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{4} + a\right )}^{\frac {3}{4}} x^{6}}\,{d x} \]

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)

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maple [A]  time = 0.00, size = 27, normalized size = 0.59 \[ -\frac {\left (-b \,x^{4}+a \right )^{\frac {1}{4}} \left (4 b \,x^{4}+a \right )}{5 a^{2} x^{5}} \]

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)/a^2/x^5

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maxima [A]  time = 1.15, size = 36, normalized size = 0.78 \[ -\frac {\frac {5 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b}{x} + \frac {{\left (-b x^{4} + a\right )}^{\frac {5}{4}}}{x^{5}}}{5 \, a^{2}} \]

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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mupad [B]  time = 1.19, size = 26, normalized size = 0.57 \[ -\frac {{\left (a-b\,x^4\right )}^{1/4}\,\left (4\,b\,x^4+a\right )}{5\,a^2\,x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(a - b*x^4)^(3/4)),x)

[Out]

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

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sympy [A]  time = 2.35, size = 311, normalized size = 6.76 \[ \begin {cases} - \frac {\sqrt [4]{b} \sqrt [4]{\frac {a}{b x^{4}} - 1} \Gamma \left (- \frac {5}{4}\right )}{16 a x^{4} \Gamma \left (\frac {3}{4}\right )} - \frac {b^{\frac {5}{4}} \sqrt [4]{\frac {a}{b x^{4}} - 1} \Gamma \left (- \frac {5}{4}\right )}{4 a^{2} \Gamma \left (\frac {3}{4}\right )} & \text {for}\: \left |{\frac {a}{b x^{4}}}\right | > 1 \\- \frac {a^{2} b^{\frac {5}{4}} \sqrt [4]{- \frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {5}{4}\right )}{- 16 a^{3} b x^{4} e^{\frac {3 i \pi }{4}} \Gamma \left (\frac {3}{4}\right ) + 16 a^{2} b^{2} x^{8} e^{\frac {3 i \pi }{4}} \Gamma \left (\frac {3}{4}\right )} - \frac {3 a b^{\frac {9}{4}} x^{4} \sqrt [4]{- \frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {5}{4}\right )}{- 16 a^{3} b x^{4} e^{\frac {3 i \pi }{4}} \Gamma \left (\frac {3}{4}\right ) + 16 a^{2} b^{2} x^{8} e^{\frac {3 i \pi }{4}} \Gamma \left (\frac {3}{4}\right )} + \frac {4 b^{\frac {13}{4}} x^{8} \sqrt [4]{- \frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {5}{4}\right )}{- 16 a^{3} b x^{4} e^{\frac {3 i \pi }{4}} \Gamma \left (\frac {3}{4}\right ) + 16 a^{2} b^{2} x^{8} e^{\frac {3 i \pi }{4}} \Gamma \left (\frac {3}{4}\right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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