3.8.91 \(\int \frac {\sqrt [4]{a+b x^2}}{x^4} \, dx\) [791]

Optimal. Leaf size=99 \[ -\frac {\sqrt [4]{a+b x^2}}{3 x^3}-\frac {b \sqrt [4]{a+b x^2}}{6 a x}-\frac {b^{3/2} \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{6 \sqrt {a} \left (a+b x^2\right )^{3/4}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {283, 331, 239, 237} \begin {gather*} -\frac {b^{3/2} \left (\frac {b x^2}{a}+1\right )^{3/4} F\left (\left .\frac {1}{2} \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{6 \sqrt {a} \left (a+b x^2\right )^{3/4}}-\frac {b \sqrt [4]{a+b x^2}}{6 a x}-\frac {\sqrt [4]{a+b x^2}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(1/4)/x^4,x]

[Out]

-1/3*(a + b*x^2)^(1/4)/x^3 - (b*(a + b*x^2)^(1/4))/(6*a*x) - (b^(3/2)*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(
Sqrt[b]*x)/Sqrt[a]]/2, 2])/(6*Sqrt[a]*(a + b*x^2)^(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 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^2}}{x^4} \, dx &=-\frac {\sqrt [4]{a+b x^2}}{3 x^3}+\frac {1}{6} b \int \frac {1}{x^2 \left (a+b x^2\right )^{3/4}} \, dx\\ &=-\frac {\sqrt [4]{a+b x^2}}{3 x^3}-\frac {b \sqrt [4]{a+b x^2}}{6 a x}-\frac {b^2 \int \frac {1}{\left (a+b x^2\right )^{3/4}} \, dx}{12 a}\\ &=-\frac {\sqrt [4]{a+b x^2}}{3 x^3}-\frac {b \sqrt [4]{a+b x^2}}{6 a x}-\frac {\left (b^2 \left (1+\frac {b x^2}{a}\right )^{3/4}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{3/4}} \, dx}{12 a \left (a+b x^2\right )^{3/4}}\\ &=-\frac {\sqrt [4]{a+b x^2}}{3 x^3}-\frac {b \sqrt [4]{a+b x^2}}{6 a x}-\frac {b^{3/2} \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{6 \sqrt {a} \left (a+b x^2\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 = 10.02, size = 51, normalized size = 0.52 \begin {gather*} -\frac {\sqrt [4]{a+b x^2} \, _2F_1\left (-\frac {3}{2},-\frac {1}{4};-\frac {1}{2};-\frac {b x^2}{a}\right )}{3 x^3 \sqrt [4]{1+\frac {b x^2}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(1/4)/x^4,x]

[Out]

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

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{x^{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(1/4)/x^4,x)

[Out]

int((b*x^2+a)^(1/4)/x^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((b*x^2+a)^(1/4)/x^4,x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(1/4)/x^4, x)

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Fricas [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^2+a)^(1/4)/x^4,x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(1/4)/x^4, x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.55, size = 34, normalized size = 0.34 \begin {gather*} - \frac {\sqrt [4]{a} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - \frac {1}{4} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(1/4)/x**4,x)

[Out]

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

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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^2+a)^(1/4)/x^4,x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(1/4)/x^4, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^{1/4}}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/4)/x^4,x)

[Out]

int((a + b*x^2)^(1/4)/x^4, x)

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