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

Optimal. Leaf size=94 \[ a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^2}}\right )+a^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^2}}\right )-\frac {4 \left (a x^4+b x^2\right )^{3/4} \left (4 a x^2-3 b\right )}{21 b x^5} \]

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Rubi [A]  time = 0.29, antiderivative size = 169, normalized size of antiderivative = 1.80, number of steps used = 10, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {2052, 2011, 329, 240, 212, 206, 203, 2016, 2014} \begin {gather*} \frac {a^{3/4} \sqrt {x} \sqrt [4]{a x^2+b} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{\sqrt [4]{a x^4+b x^2}}+\frac {a^{3/4} \sqrt {x} \sqrt [4]{a x^2+b} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{\sqrt [4]{a x^4+b x^2}}+\frac {4 \left (a x^4+b x^2\right )^{3/4}}{7 x^5}-\frac {16 a \left (a x^4+b x^2\right )^{3/4}}{21 b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(b*x^2 + a*x^4)^(3/4))/(7*x^5) - (16*a*(b*x^2 + a*x^4)^(3/4))/(21*b*x^3) + (a^(3/4)*Sqrt[x]*(b + a*x^2)^(1/
4)*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(b*x^2 + a*x^4)^(1/4) + (a^(3/4)*Sqrt[x]*(b + a*x^2)^(1/4)*Arc
Tanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(b*x^2 + a*x^4)^(1/4)

Rule 203

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

Rule 206

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

Rule 212

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

Rule 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p
+ 1/n]

Rule 329

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 2011

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2014

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

Rule 2016

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

Rule 2052

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(c*x)
^m*Pq*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !In
tegerQ[p] && NeQ[n, j]

Rubi steps

\begin {align*} \int \frac {-2 b+a x^4}{x^4 \sqrt [4]{b x^2+a x^4}} \, dx &=\int \left (\frac {a}{\sqrt [4]{b x^2+a x^4}}-\frac {2 b}{x^4 \sqrt [4]{b x^2+a x^4}}\right ) \, dx\\ &=a \int \frac {1}{\sqrt [4]{b x^2+a x^4}} \, dx-(2 b) \int \frac {1}{x^4 \sqrt [4]{b x^2+a x^4}} \, dx\\ &=\frac {4 \left (b x^2+a x^4\right )^{3/4}}{7 x^5}+\frac {1}{7} (8 a) \int \frac {1}{x^2 \sqrt [4]{b x^2+a x^4}} \, dx+\frac {\left (a \sqrt {x} \sqrt [4]{b+a x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt [4]{b+a x^2}} \, dx}{\sqrt [4]{b x^2+a x^4}}\\ &=\frac {4 \left (b x^2+a x^4\right )^{3/4}}{7 x^5}-\frac {16 a \left (b x^2+a x^4\right )^{3/4}}{21 b x^3}+\frac {\left (2 a \sqrt {x} \sqrt [4]{b+a x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{b+a x^4}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^4}}\\ &=\frac {4 \left (b x^2+a x^4\right )^{3/4}}{7 x^5}-\frac {16 a \left (b x^2+a x^4\right )^{3/4}}{21 b x^3}+\frac {\left (2 a \sqrt {x} \sqrt [4]{b+a x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-a x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\sqrt [4]{b x^2+a x^4}}\\ &=\frac {4 \left (b x^2+a x^4\right )^{3/4}}{7 x^5}-\frac {16 a \left (b x^2+a x^4\right )^{3/4}}{21 b x^3}+\frac {\left (a \sqrt {x} \sqrt [4]{b+a x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\sqrt [4]{b x^2+a x^4}}+\frac {\left (a \sqrt {x} \sqrt [4]{b+a x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\sqrt [4]{b x^2+a x^4}}\\ &=\frac {4 \left (b x^2+a x^4\right )^{3/4}}{7 x^5}-\frac {16 a \left (b x^2+a x^4\right )^{3/4}}{21 b x^3}+\frac {a^{3/4} \sqrt {x} \sqrt [4]{b+a x^2} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\sqrt [4]{b x^2+a x^4}}+\frac {a^{3/4} \sqrt {x} \sqrt [4]{b+a x^2} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\sqrt [4]{b x^2+a x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 142, normalized size = 1.51 \begin {gather*} \frac {21 a^{3/4} b x^{7/2} \sqrt [4]{a x^2+b} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )+21 a^{3/4} b x^{7/2} \sqrt [4]{a x^2+b} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )-4 \left (4 a^2 x^4+a b x^2-3 b^2\right )}{21 b x^3 \sqrt [4]{x^2 \left (a x^2+b\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(-3*b^2 + a*b*x^2 + 4*a^2*x^4) + 21*a^(3/4)*b*x^(7/2)*(b + a*x^2)^(1/4)*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^
2)^(1/4)] + 21*a^(3/4)*b*x^(7/2)*(b + a*x^2)^(1/4)*ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(21*b*x^3*(x^
2*(b + a*x^2))^(1/4))

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IntegrateAlgebraic [A]  time = 0.53, size = 94, normalized size = 1.00 \begin {gather*} -\frac {4 \left (-3 b+4 a x^2\right ) \left (b x^2+a x^4\right )^{3/4}}{21 b x^5}+a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )+a^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(-3*b + 4*a*x^2)*(b*x^2 + a*x^4)^(3/4))/(21*b*x^5) + a^(3/4)*ArcTan[(a^(1/4)*x)/(b*x^2 + a*x^4)^(1/4)] + a
^(3/4)*ArcTanh[(a^(1/4)*x)/(b*x^2 + a*x^4)^(1/4)]

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fricas [F(-1)]  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^4-2*b)/x^4/(a*x^4+b*x^2)^(1/4),x, algorithm="fricas")

[Out]

Timed out

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giac [B]  time = 0.22, size = 209, normalized size = 2.22 \begin {gather*} \frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {3}{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 {3}{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 {3}{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 {3}{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 {4 \, {\left (3 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {7}{4}} b^{6} - 7 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{4}} a b^{6}\right )}}{21 \, b^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*(-a)^(3/4)*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)^(3/4)*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)^(3/4
)*log(sqrt(2)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2)) + 1/4*sqrt(2)*(-a)^(3/4)*log(-sqrt(2)
*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2)) + 4/21*(3*(a + b/x^2)^(7/4)*b^6 - 7*(a + b/x^2)^(3
/4)*a*b^6)/b^7

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {a \,x^{4}-2 b}{x^{4} \left (a \,x^{4}+b \,x^{2}\right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{4} - 2 \, b}{{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.25, size = 75, normalized size = 0.80 \begin {gather*} \frac {4\,{\left (a\,x^4+b\,x^2\right )}^{3/4}\,\left (3\,b-4\,a\,x^2\right )}{21\,b\,x^5}+\frac {2\,a\,x\,{\left (\frac {a\,x^2}{b}+1\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{4};\ \frac {5}{4};\ -\frac {a\,x^2}{b}\right )}{{\left (a\,x^4+b\,x^2\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*(a*x^4 + b*x^2)^(3/4)*(3*b - 4*a*x^2))/(21*b*x^5) + (2*a*x*((a*x^2)/b + 1)^(1/4)*hypergeom([1/4, 1/4], 5/4,
-(a*x^2)/b))/(a*x^4 + b*x^2)^(1/4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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