3.12.8 \(\int \frac {1}{x^{10} \sqrt [4]{a+b x^4}} \, dx\) [1108]

Optimal. Leaf size=129 \[ -\frac {4 b^2}{15 a^2 x \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{9 a x^9}+\frac {2 b \left (a+b x^4\right )^{3/4}}{15 a^2 x^5}+\frac {4 b^{5/2} \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{15 a^{5/2} \sqrt [4]{a+b x^4}} \]

[Out]

-4/15*b^2/a^2/x/(b*x^4+a)^(1/4)-1/9*(b*x^4+a)^(3/4)/a/x^9+2/15*b*(b*x^4+a)^(3/4)/a^2/x^5+4/15*b^(5/2)*(1+a/b/x
^4)^(1/4)*x*(cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))*EllipticE(sin(
1/2*arccot(x^2*b^(1/2)/a^(1/2))),2^(1/2))/a^(5/2)/(b*x^4+a)^(1/4)

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Rubi [A]
time = 0.04, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {331, 318, 287, 342, 281, 202} \begin {gather*} \frac {4 b^{5/2} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{15 a^{5/2} \sqrt [4]{a+b x^4}}-\frac {4 b^2}{15 a^2 x \sqrt [4]{a+b x^4}}+\frac {2 b \left (a+b x^4\right )^{3/4}}{15 a^2 x^5}-\frac {\left (a+b x^4\right )^{3/4}}{9 a x^9} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^10*(a + b*x^4)^(1/4)),x]

[Out]

(-4*b^2)/(15*a^2*x*(a + b*x^4)^(1/4)) - (a + b*x^4)^(3/4)/(9*a*x^9) + (2*b*(a + b*x^4)^(3/4))/(15*a^2*x^5) + (
4*b^(5/2)*(1 + a/(b*x^4))^(1/4)*x*EllipticE[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(15*a^(5/2)*(a + b*x^4)^(1/4)
)

Rule 202

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]))*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 287

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Dist[x*((1 + a/(b*x^4))^(1/4)/(b*(a + b*x^4)^(1/4))), Int
[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 318

Int[1/((x_)^2*((a_) + (b_.)*(x_)^4)^(1/4)), x_Symbol] :> -Simp[(x*(a + b*x^4)^(1/4))^(-1), x] - Dist[b, Int[x^
2/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 331

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] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {1}{x^{10} \sqrt [4]{a+b x^4}} \, dx &=-\frac {\left (a+b x^4\right )^{3/4}}{9 a x^9}-\frac {(2 b) \int \frac {1}{x^6 \sqrt [4]{a+b x^4}} \, dx}{3 a}\\ &=-\frac {\left (a+b x^4\right )^{3/4}}{9 a x^9}+\frac {2 b \left (a+b x^4\right )^{3/4}}{15 a^2 x^5}+\frac {\left (4 b^2\right ) \int \frac {1}{x^2 \sqrt [4]{a+b x^4}} \, dx}{15 a^2}\\ &=-\frac {4 b^2}{15 a^2 x \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{9 a x^9}+\frac {2 b \left (a+b x^4\right )^{3/4}}{15 a^2 x^5}-\frac {\left (4 b^3\right ) \int \frac {x^2}{\left (a+b x^4\right )^{5/4}} \, dx}{15 a^2}\\ &=-\frac {4 b^2}{15 a^2 x \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{9 a x^9}+\frac {2 b \left (a+b x^4\right )^{3/4}}{15 a^2 x^5}-\frac {\left (4 b^2 \sqrt [4]{1+\frac {a}{b x^4}} x\right ) \int \frac {1}{\left (1+\frac {a}{b x^4}\right )^{5/4} x^3} \, dx}{15 a^2 \sqrt [4]{a+b x^4}}\\ &=-\frac {4 b^2}{15 a^2 x \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{9 a x^9}+\frac {2 b \left (a+b x^4\right )^{3/4}}{15 a^2 x^5}+\frac {\left (4 b^2 \sqrt [4]{1+\frac {a}{b x^4}} x\right ) \text {Subst}\left (\int \frac {x}{\left (1+\frac {a x^4}{b}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )}{15 a^2 \sqrt [4]{a+b x^4}}\\ &=-\frac {4 b^2}{15 a^2 x \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{9 a x^9}+\frac {2 b \left (a+b x^4\right )^{3/4}}{15 a^2 x^5}+\frac {\left (2 b^2 \sqrt [4]{1+\frac {a}{b x^4}} x\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{5/4}} \, dx,x,\frac {1}{x^2}\right )}{15 a^2 \sqrt [4]{a+b x^4}}\\ &=-\frac {4 b^2}{15 a^2 x \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{9 a x^9}+\frac {2 b \left (a+b x^4\right )^{3/4}}{15 a^2 x^5}+\frac {4 b^{5/2} \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{15 a^{5/2} \sqrt [4]{a+b x^4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.02, size = 51, normalized size = 0.40 \begin {gather*} -\frac {\sqrt [4]{1+\frac {b x^4}{a}} \, _2F_1\left (-\frac {9}{4},\frac {1}{4};-\frac {5}{4};-\frac {b x^4}{a}\right )}{9 x^9 \sqrt [4]{a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^10*(a + b*x^4)^(1/4)),x]

[Out]

-1/9*((1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[-9/4, 1/4, -5/4, -((b*x^4)/a)])/(x^9*(a + b*x^4)^(1/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{10} \left (b \,x^{4}+a \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.07, size = 25, normalized size = 0.19 \begin {gather*} {\rm integral}\left (\frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{b x^{14} + a x^{10}}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^4 + a)^(3/4)/(b*x^14 + a*x^10), x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.69, size = 44, normalized size = 0.34 \begin {gather*} \frac {\Gamma \left (- \frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {9}{4}, \frac {1}{4} \\ - \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [4]{a} x^{9} \Gamma \left (- \frac {5}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

gamma(-9/4)*hyper((-9/4, 1/4), (-5/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(1/4)*x**9*gamma(-5/4))

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^{10}\,{\left (b\,x^4+a\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^10*(a + b*x^4)^(1/4)),x)

[Out]

int(1/(x^10*(a + b*x^4)^(1/4)), x)

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