3.11.66 \(\int \frac {\sqrt [4]{b x^2+a x^4}}{x^2} \, dx\) [1066]

Optimal. Leaf size=80 \[ -\frac {2 \sqrt [4]{b x^2+a x^4}}{x}-\sqrt [4]{a} \text {ArcTan}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )+\sqrt [4]{a} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 142, normalized size of antiderivative = 1.78, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2045, 2057, 335, 338, 304, 209, 212} \begin {gather*} -\frac {\sqrt [4]{a} x^{3/2} \left (a x^2+b\right )^{3/4} \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{\left (a x^4+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a x^4+b x^2}}{x}+\frac {\sqrt [4]{a} x^{3/2} \left (a x^2+b\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{\left (a x^4+b x^2\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + a*x^4)^(1/4)/x^2,x]

[Out]

(-2*(b*x^2 + a*x^4)^(1/4))/x - (a^(1/4)*x^(3/2)*(b + a*x^2)^(3/4)*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])
/(b*x^2 + a*x^4)^(3/4) + (a^(1/4)*x^(3/2)*(b + a*x^2)^(3/4)*ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(b*x
^2 + a*x^4)^(3/4)

Rule 209

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 335

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

Rule 338

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

Rule 2045

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

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps

\begin {align*} \int \frac {\sqrt [4]{b x^2+a x^4}}{x^2} \, dx &=-\frac {2 \sqrt [4]{b x^2+a x^4}}{x}+a \int \frac {x^2}{\left (b x^2+a x^4\right )^{3/4}} \, dx\\ &=-\frac {2 \sqrt [4]{b x^2+a x^4}}{x}+\frac {\left (a x^{3/2} \left (b+a x^2\right )^{3/4}\right ) \int \frac {\sqrt {x}}{\left (b+a x^2\right )^{3/4}} \, dx}{\left (b x^2+a x^4\right )^{3/4}}\\ &=-\frac {2 \sqrt [4]{b x^2+a x^4}}{x}+\frac {\left (2 a x^{3/2} \left (b+a x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{\left (b+a x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\left (b x^2+a x^4\right )^{3/4}}\\ &=-\frac {2 \sqrt [4]{b x^2+a x^4}}{x}+\frac {\left (2 a x^{3/2} \left (b+a x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\left (b x^2+a x^4\right )^{3/4}}\\ &=-\frac {2 \sqrt [4]{b x^2+a x^4}}{x}+\frac {\left (\sqrt {a} x^{3/2} \left (b+a x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\left (b x^2+a x^4\right )^{3/4}}-\frac {\left (\sqrt {a} x^{3/2} \left (b+a x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\left (b x^2+a x^4\right )^{3/4}}\\ &=-\frac {2 \sqrt [4]{b x^2+a x^4}}{x}-\frac {\sqrt [4]{a} x^{3/2} \left (b+a x^2\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\left (b x^2+a x^4\right )^{3/4}}+\frac {\sqrt [4]{a} x^{3/2} \left (b+a x^2\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\left (b x^2+a x^4\right )^{3/4}}\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 111, normalized size = 1.39 \begin {gather*} \frac {x \left (b+a x^2\right )^{3/4} \left (-2 \sqrt [4]{b+a x^2}-\sqrt [4]{a} \sqrt {x} \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )+\sqrt [4]{a} \sqrt {x} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )\right )}{\left (x^2 \left (b+a x^2\right )\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + a*x^4)^(1/4)/x^2,x]

[Out]

(x*(b + a*x^2)^(3/4)*(-2*(b + a*x^2)^(1/4) - a^(1/4)*Sqrt[x]*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)] + a^(
1/4)*Sqrt[x]*ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)]))/(x^2*(b + a*x^2))^(3/4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [4]{x^{2} \left (a x^{2} + b\right )}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (66) = 132\).
time = 0.41, size = 185, normalized size = 2.31 \begin {gather*} \frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right ) - \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right ) - 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*(-a)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x^2)^(1/4))/(-a)^(1/4)) + 1/2*sqrt(2)
*(-a)^(1/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x^2)^(1/4))/(-a)^(1/4)) + 1/4*sqrt(2)*(-a)^(1/4
)*log(sqrt(2)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2)) - 1/4*sqrt(2)*(-a)^(1/4)*log(-sqrt(2)
*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2)) - 2*(a + b/x^2)^(1/4)

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Mupad [B]
time = 0.98, size = 44, normalized size = 0.55 \begin {gather*} -\frac {2\,{\left (a\,x^4+b\,x^2\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},-\frac {1}{4};\ \frac {3}{4};\ -\frac {a\,x^2}{b}\right )}{x\,{\left (\frac {a\,x^2}{b}+1\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^4 + b*x^2)^(1/4)/x^2,x)

[Out]

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

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