3.14.95 \(\int \frac {-b+a x^2}{(b+a x^2+x^4) \sqrt [4]{b x^2+a x^4}} \, dx\) [1395]

Optimal. Leaf size=100 \[ -\frac {1}{2} \text {RootSum}\left [b-a \text {$\#$1}^4+\text {$\#$1}^8\& ,\frac {2 a \log (x)-2 a \log \left (\sqrt [4]{b x^2+a x^4}-x \text {$\#$1}\right )-\log (x) \text {$\#$1}^4+\log \left (\sqrt [4]{b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}-2 \text {$\#$1}^5}\& \right ] \]

[Out]

Unintegrable

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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(627\) vs. \(2(100)=200\).
time = 1.39, antiderivative size = 627, normalized size of antiderivative = 6.27, number of steps used = 12, number of rules used = 7, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {2081, 6847, 6860, 385, 218, 214, 211} \begin {gather*} \frac {\sqrt {x} \left (a-\frac {a^2+2 b}{\sqrt {a^2-4 b}}\right ) \sqrt [4]{a x^2+b} \text {ArcTan}\left (\frac {\sqrt {x} \sqrt [4]{-a \sqrt {a^2-4 b}+a^2-2 b}}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt [4]{a x^2+b}}\right )}{\left (a-\sqrt {a^2-4 b}\right )^{3/4} \sqrt [4]{-a \sqrt {a^2-4 b}+a^2-2 b} \sqrt [4]{a x^4+b x^2}}+\frac {\sqrt {x} \left (\frac {a^2+2 b}{\sqrt {a^2-4 b}}+a\right ) \sqrt [4]{a x^2+b} \text {ArcTan}\left (\frac {\sqrt {x} \sqrt [4]{a \sqrt {a^2-4 b}+a^2-2 b}}{\sqrt [4]{\sqrt {a^2-4 b}+a} \sqrt [4]{a x^2+b}}\right )}{\left (\sqrt {a^2-4 b}+a\right )^{3/4} \sqrt [4]{a \sqrt {a^2-4 b}+a^2-2 b} \sqrt [4]{a x^4+b x^2}}+\frac {\sqrt {x} \left (a-\frac {a^2+2 b}{\sqrt {a^2-4 b}}\right ) \sqrt [4]{a x^2+b} \tanh ^{-1}\left (\frac {\sqrt {x} \sqrt [4]{-a \sqrt {a^2-4 b}+a^2-2 b}}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt [4]{a x^2+b}}\right )}{\left (a-\sqrt {a^2-4 b}\right )^{3/4} \sqrt [4]{-a \sqrt {a^2-4 b}+a^2-2 b} \sqrt [4]{a x^4+b x^2}}+\frac {\sqrt {x} \left (\frac {a^2+2 b}{\sqrt {a^2-4 b}}+a\right ) \sqrt [4]{a x^2+b} \tanh ^{-1}\left (\frac {\sqrt {x} \sqrt [4]{a \sqrt {a^2-4 b}+a^2-2 b}}{\sqrt [4]{\sqrt {a^2-4 b}+a} \sqrt [4]{a x^2+b}}\right )}{\left (\sqrt {a^2-4 b}+a\right )^{3/4} \sqrt [4]{a \sqrt {a^2-4 b}+a^2-2 b} \sqrt [4]{a x^4+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 211

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

Rule 214

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

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rule 6847

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [A]
time = 0.56, size = 132, normalized size = 1.32 \begin {gather*} -\frac {\sqrt {x} \sqrt [4]{b+a x^2} \text {RootSum}\left [b-a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {2 a \log (x)-4 a \log \left (\sqrt [4]{b+a x^2}-\sqrt {x} \text {$\#$1}\right )-\log (x) \text {$\#$1}^4+2 \log \left (\sqrt [4]{b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}-2 \text {$\#$1}^5}\&\right ]}{4 \sqrt [4]{x^2 \left (b+a x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(Sqrt[x]*(b + a*x^2)^(1/4)*RootSum[b - a*#1^4 + #1^8 & , (2*a*Log[x] - 4*a*Log[(b + a*x^2)^(1/4) - Sqrt[x
]*#1] - Log[x]*#1^4 + 2*Log[(b + a*x^2)^(1/4) - Sqrt[x]*#1]*#1^4)/(a*#1 - 2*#1^5) & ])/(x^2*(b + a*x^2))^(1/4)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {a \,x^{2}-b}{\left (x^{4}+a \,x^{2}+b \right ) \left (a \,x^{4}+b \,x^{2}\right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a*x^2-b)/(x^4+a*x^2+b)/(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*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x^2 - b)/((a*x^4 + b*x^2)^(1/4)*(x^4 + a*x^2 + b)), 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^2-b)/(x^4+a*x^2+b)/(a*x^4+b*x^2)^(1/4),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 {a x^{2} - b}{\sqrt [4]{x^{2} \left (a x^{2} + b\right )} \left (a x^{2} + b + x^{4}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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