3.1015 \(\int \frac {\sqrt [4]{a+b x^4}}{x^{12}} \, dx\)

Optimal. Leaf size=128 \[ -\frac {4 b^{7/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{77 a^{5/2} \left (a+b x^4\right )^{3/4}}+\frac {2 b^2 \sqrt [4]{a+b x^4}}{77 a^2 x^3}-\frac {\sqrt [4]{a+b x^4}}{11 x^{11}}-\frac {b \sqrt [4]{a+b x^4}}{77 a x^7} \]

[Out]

-1/11*(b*x^4+a)^(1/4)/x^11-1/77*b*(b*x^4+a)^(1/4)/a/x^7+2/77*b^2*(b*x^4+a)^(1/4)/a^2/x^3-4/77*b^(7/2)*(1+a/b/x
^4)^(3/4)*x^3*(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)))*EllipticF(si
n(1/2*arccot(x^2*b^(1/2)/a^(1/2))),2^(1/2))/a^(5/2)/(b*x^4+a)^(3/4)

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Rubi [A]  time = 0.06, antiderivative size = 128, 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 = {277, 325, 237, 335, 275, 231} \[ \frac {2 b^2 \sqrt [4]{a+b x^4}}{77 a^2 x^3}-\frac {4 b^{7/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{77 a^{5/2} \left (a+b x^4\right )^{3/4}}-\frac {b \sqrt [4]{a+b x^4}}{77 a x^7}-\frac {\sqrt [4]{a+b x^4}}{11 x^{11}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(1/4)/x^12,x]

[Out]

-(a + b*x^4)^(1/4)/(11*x^11) - (b*(a + b*x^4)^(1/4))/(77*a*x^7) + (2*b^2*(a + b*x^4)^(1/4))/(77*a^2*x^3) - (4*
b^(7/2)*(1 + a/(b*x^4))^(3/4)*x^3*EllipticF[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(77*a^(5/2)*(a + b*x^4)^(3/4)
)

Rule 231

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

Rule 237

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

Rule 275

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 277

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 325

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 335

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 {\sqrt [4]{a+b x^4}}{x^{12}} \, dx &=-\frac {\sqrt [4]{a+b x^4}}{11 x^{11}}+\frac {1}{11} b \int \frac {1}{x^8 \left (a+b x^4\right )^{3/4}} \, dx\\ &=-\frac {\sqrt [4]{a+b x^4}}{11 x^{11}}-\frac {b \sqrt [4]{a+b x^4}}{77 a x^7}-\frac {\left (6 b^2\right ) \int \frac {1}{x^4 \left (a+b x^4\right )^{3/4}} \, dx}{77 a}\\ &=-\frac {\sqrt [4]{a+b x^4}}{11 x^{11}}-\frac {b \sqrt [4]{a+b x^4}}{77 a x^7}+\frac {2 b^2 \sqrt [4]{a+b x^4}}{77 a^2 x^3}+\frac {\left (4 b^3\right ) \int \frac {1}{\left (a+b x^4\right )^{3/4}} \, dx}{77 a^2}\\ &=-\frac {\sqrt [4]{a+b x^4}}{11 x^{11}}-\frac {b \sqrt [4]{a+b x^4}}{77 a x^7}+\frac {2 b^2 \sqrt [4]{a+b x^4}}{77 a^2 x^3}+\frac {\left (4 b^3 \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \int \frac {1}{\left (1+\frac {a}{b x^4}\right )^{3/4} x^3} \, dx}{77 a^2 \left (a+b x^4\right )^{3/4}}\\ &=-\frac {\sqrt [4]{a+b x^4}}{11 x^{11}}-\frac {b \sqrt [4]{a+b x^4}}{77 a x^7}+\frac {2 b^2 \sqrt [4]{a+b x^4}}{77 a^2 x^3}-\frac {\left (4 b^3 \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \operatorname {Subst}\left (\int \frac {x}{\left (1+\frac {a x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{x}\right )}{77 a^2 \left (a+b x^4\right )^{3/4}}\\ &=-\frac {\sqrt [4]{a+b x^4}}{11 x^{11}}-\frac {b \sqrt [4]{a+b x^4}}{77 a x^7}+\frac {2 b^2 \sqrt [4]{a+b x^4}}{77 a^2 x^3}-\frac {\left (2 b^3 \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{x^2}\right )}{77 a^2 \left (a+b x^4\right )^{3/4}}\\ &=-\frac {\sqrt [4]{a+b x^4}}{11 x^{11}}-\frac {b \sqrt [4]{a+b x^4}}{77 a x^7}+\frac {2 b^2 \sqrt [4]{a+b x^4}}{77 a^2 x^3}-\frac {4 b^{7/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{77 a^{5/2} \left (a+b x^4\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 51, normalized size = 0.40 \[ -\frac {\sqrt [4]{a+b x^4} \, _2F_1\left (-\frac {11}{4},-\frac {1}{4};-\frac {7}{4};-\frac {b x^4}{a}\right )}{11 x^{11} \sqrt [4]{\frac {b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(1/4)/x^12,x]

[Out]

-1/11*((a + b*x^4)^(1/4)*Hypergeometric2F1[-11/4, -1/4, -7/4, -((b*x^4)/a)])/(x^11*(1 + (b*x^4)/a)^(1/4))

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x^{12}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(1/4)/x^12,x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(1/4)/x^12, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x^{12}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(1/4)/x^12,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(1/4)/x^12, x)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{x^{12}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^4+a)^(1/4)/x^12,x)

[Out]

int((b*x^4+a)^(1/4)/x^12,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x^{12}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(1/4)/x^12,x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^(1/4)/x^12, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,x^4+a\right )}^{1/4}}{x^{12}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^(1/4)/x^12,x)

[Out]

int((a + b*x^4)^(1/4)/x^12, x)

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sympy [C]  time = 2.84, size = 46, normalized size = 0.36 \[ \frac {\sqrt [4]{a} \Gamma \left (- \frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {11}{4}, - \frac {1}{4} \\ - \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{11} \Gamma \left (- \frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**4+a)**(1/4)/x**12,x)

[Out]

a**(1/4)*gamma(-11/4)*hyper((-11/4, -1/4), (-7/4,), b*x**4*exp_polar(I*pi)/a)/(4*x**11*gamma(-7/4))

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