3.12.29 \(\int \frac {1}{x^6 (a+b x^4)^{3/4}} \, dx\) [1129]

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

[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 = 44, normalized size of antiderivative = 1.00, 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 = {277, 270} \begin {gather*} \frac {4 b \sqrt [4]{a+b x^4}}{5 a^2 x}-\frac {\sqrt [4]{a+b x^4}}{5 a x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(a + b*x^4)^(3/4)),x]

[Out]

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

Rule 270

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 277

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 = 0.21, size = 31, normalized size = 0.70 \begin {gather*} \frac {\sqrt [4]{a+b x^4} \left (-a+4 b x^4\right )}{5 a^2 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[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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Maple [A]
time = 0.16, size = 26, normalized size = 0.59

method	result	size
gosper	\(-\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-4 b \,x^{4}+a \right )}{5 a^{2} x^{5}}\)	\(26\)
trager	\(-\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-4 b \,x^{4}+a \right )}{5 a^{2} x^{5}}\)	\(26\)
risch	\(-\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-4 b \,x^{4}+a \right )}{5 a^{2} x^{5}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.29, size = 35, normalized size = 0.80 \begin {gather*} \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 {gather*}

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 = 0.38, size = 27, normalized size = 0.61 \begin {gather*} \frac {{\left (4 \, b x^{4} - a\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{5 \, a^{2} x^{5}} \end {gather*}

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 = 0.52, size = 68, normalized size = 1.55 \begin {gather*} - \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [B]
time = 1.17, size = 25, normalized size = 0.57 \begin {gather*} -\frac {{\left (b\,x^4+a\right )}^{1/4}\,\left (a-4\,b\,x^4\right )}{5\,a^2\,x^5} \end {gather*}

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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