3.11.1 \(\int \frac {\sqrt [4]{a+b x^4}}{x^{11}} \, dx\) [1001]

Optimal. Leaf size=125 \[ -\frac {\sqrt [4]{a+b x^4}}{10 x^{10}}-\frac {b \sqrt [4]{a+b x^4}}{60 a x^6}+\frac {b^2 \sqrt [4]{a+b x^4}}{24 a^2 x^2}+\frac {b^{5/2} \left (1+\frac {b x^4}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{24 a^{3/2} \left (a+b x^4\right )^{3/4}} \]

[Out]

-1/10*(b*x^4+a)^(1/4)/x^10-1/60*b*(b*x^4+a)^(1/4)/a/x^6+1/24*b^2*(b*x^4+a)^(1/4)/a^2/x^2+1/24*b^(5/2)*(1+b*x^4
/a)^(3/4)*(cos(1/2*arctan(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x^2*b^(1/2)/a^(1/2)))*EllipticF(sin(1/
2*arctan(x^2*b^(1/2)/a^(1/2))),2^(1/2))/a^(3/2)/(b*x^4+a)^(3/4)

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Rubi [A]
time = 0.06, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {281, 283, 331, 239, 237} \begin {gather*} \frac {b^{5/2} \left (\frac {b x^4}{a}+1\right )^{3/4} F\left (\left .\frac {1}{2} \text {ArcTan}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{24 a^{3/2} \left (a+b x^4\right )^{3/4}}+\frac {b^2 \sqrt [4]{a+b x^4}}{24 a^2 x^2}-\frac {\sqrt [4]{a+b x^4}}{10 x^{10}}-\frac {b \sqrt [4]{a+b x^4}}{60 a x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(1/4)/x^11,x]

[Out]

-1/10*(a + b*x^4)^(1/4)/x^10 - (b*(a + b*x^4)^(1/4))/(60*a*x^6) + (b^2*(a + b*x^4)^(1/4))/(24*a^2*x^2) + (b^(5
/2)*(1 + (b*x^4)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(24*a^(3/2)*(a + b*x^4)^(3/4))

Rule 237

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]))*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 239

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[(1 + b*(x^2/a))^(3/4)/(a + b*x^2)^(3/4), Int[1/(1 + b*(x^2
/a))^(3/4), x], x] /; FreeQ[{a, b}, x] && PosQ[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 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

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]

Rubi steps

\begin {align*} \int \frac {\sqrt [4]{a+b x^4}}{x^{11}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {\sqrt [4]{a+b x^2}}{x^6} \, dx,x,x^2\right )\\ &=-\frac {\sqrt [4]{a+b x^4}}{10 x^{10}}+\frac {1}{20} b \text {Subst}\left (\int \frac {1}{x^4 \left (a+b x^2\right )^{3/4}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt [4]{a+b x^4}}{10 x^{10}}-\frac {b \sqrt [4]{a+b x^4}}{60 a x^6}-\frac {b^2 \text {Subst}\left (\int \frac {1}{x^2 \left (a+b x^2\right )^{3/4}} \, dx,x,x^2\right )}{24 a}\\ &=-\frac {\sqrt [4]{a+b x^4}}{10 x^{10}}-\frac {b \sqrt [4]{a+b x^4}}{60 a x^6}+\frac {b^2 \sqrt [4]{a+b x^4}}{24 a^2 x^2}+\frac {b^3 \text {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^{3/4}} \, dx,x,x^2\right )}{48 a^2}\\ &=-\frac {\sqrt [4]{a+b x^4}}{10 x^{10}}-\frac {b \sqrt [4]{a+b x^4}}{60 a x^6}+\frac {b^2 \sqrt [4]{a+b x^4}}{24 a^2 x^2}+\frac {\left (b^3 \left (1+\frac {b x^4}{a}\right )^{3/4}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{3/4}} \, dx,x,x^2\right )}{48 a^2 \left (a+b x^4\right )^{3/4}}\\ &=-\frac {\sqrt [4]{a+b x^4}}{10 x^{10}}-\frac {b \sqrt [4]{a+b x^4}}{60 a x^6}+\frac {b^2 \sqrt [4]{a+b x^4}}{24 a^2 x^2}+\frac {b^{5/2} \left (1+\frac {b x^4}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{24 a^{3/2} \left (a+b x^4\right )^{3/4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.01, size = 51, normalized size = 0.41 \begin {gather*} -\frac {\sqrt [4]{a+b x^4} \, _2F_1\left (-\frac {5}{2},-\frac {1}{4};-\frac {3}{2};-\frac {b x^4}{a}\right )}{10 x^{10} \sqrt [4]{1+\frac {b x^4}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(1/4)/x^11,x]

[Out]

-1/10*((a + b*x^4)^(1/4)*Hypergeometric2F1[-5/2, -1/4, -3/2, -((b*x^4)/a)])/(x^10*(1 + (b*x^4)/a)^(1/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{x^{11}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^4+a)^(1/4)/x^11,x)

[Out]

int((b*x^4+a)^(1/4)/x^11,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((b*x^4+a)^(1/4)/x^11,x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^(1/4)/x^11, x)

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Fricas [F]
time = 0.07, size = 15, normalized size = 0.12 \begin {gather*} {\rm integral}\left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x^{11}}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(1/4)/x^11,x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(1/4)/x^11, x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.77, size = 34, normalized size = 0.27 \begin {gather*} - \frac {\sqrt [4]{a} {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, - \frac {1}{4} \\ - \frac {3}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{10 x^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**4+a)**(1/4)/x**11,x)

[Out]

-a**(1/4)*hyper((-5/2, -1/4), (-3/2,), b*x**4*exp_polar(I*pi)/a)/(10*x**10)

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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((b*x^4+a)^(1/4)/x^11,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(1/4)/x^11, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^4+a\right )}^{1/4}}{x^{11}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^(1/4)/x^11,x)

[Out]

int((a + b*x^4)^(1/4)/x^11, x)

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