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

Optimal. Leaf size=130 \[ \frac {\tan ^{-1}\left (\frac {\frac {x^2}{\sqrt [4]{2} \sqrt [4]{n}}-\frac {\sqrt [4]{n} \sqrt {a x^2+b}}{2^{3/4}}}{x \sqrt [4]{a x^2+b}}\right )}{2^{3/4} n^{3/4}}+\frac {\tanh ^{-1}\left (\frac {\frac {\sqrt [4]{n} \sqrt {a x^2+b}}{2^{3/4}}+\frac {x^2}{\sqrt [4]{2} \sqrt [4]{n}}}{x \sqrt [4]{a x^2+b}}\right )}{2^{3/4} n^{3/4}} \]

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Rubi [C]  time = 0.84, antiderivative size = 571, normalized size of antiderivative = 4.39, number of steps used = 10, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1692, 399, 490, 1218} \begin {gather*} \frac {\sqrt [4]{b} \left (\frac {\sqrt {a^2 n-8 b}}{\sqrt {n}}+a\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (-\frac {2 \sqrt {b}}{\sqrt {-n a^2-\sqrt {n} \sqrt {a^2 n-8 b} a+4 b}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 x \sqrt {-a \sqrt {n} \sqrt {a^2 n-8 b}+a^2 (-n)+4 b}}-\frac {\sqrt [4]{b} \left (\frac {\sqrt {a^2 n-8 b}}{\sqrt {n}}+a\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (\frac {2 \sqrt {b}}{\sqrt {-n a^2-\sqrt {n} \sqrt {a^2 n-8 b} a+4 b}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 x \sqrt {-a \sqrt {n} \sqrt {a^2 n-8 b}+a^2 (-n)+4 b}}+\frac {\sqrt [4]{b} \left (a-\frac {\sqrt {a^2 n-8 b}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (-\frac {2 \sqrt {b}}{\sqrt {-n a^2+\sqrt {n} \sqrt {a^2 n-8 b} a+4 b}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 x \sqrt {a \sqrt {n} \sqrt {a^2 n-8 b}+a^2 (-n)+4 b}}-\frac {\sqrt [4]{b} \left (a-\frac {\sqrt {a^2 n-8 b}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (\frac {2 \sqrt {b}}{\sqrt {-n a^2+\sqrt {n} \sqrt {a^2 n-8 b} a+4 b}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 x \sqrt {a \sqrt {n} \sqrt {a^2 n-8 b}+a^2 (-n)+4 b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^(1/4)*(a + Sqrt[-8*b + a^2*n]/Sqrt[n])*Sqrt[-((a*x^2)/b)]*EllipticPi[(-2*Sqrt[b])/Sqrt[4*b - a^2*n - a*Sqrt
[n]*Sqrt[-8*b + a^2*n]], ArcSin[(b + a*x^2)^(1/4)/b^(1/4)], -1])/(2*Sqrt[4*b - a^2*n - a*Sqrt[n]*Sqrt[-8*b + a
^2*n]]*x) - (b^(1/4)*(a + Sqrt[-8*b + a^2*n]/Sqrt[n])*Sqrt[-((a*x^2)/b)]*EllipticPi[(2*Sqrt[b])/Sqrt[4*b - a^2
*n - a*Sqrt[n]*Sqrt[-8*b + a^2*n]], ArcSin[(b + a*x^2)^(1/4)/b^(1/4)], -1])/(2*Sqrt[4*b - a^2*n - a*Sqrt[n]*Sq
rt[-8*b + a^2*n]]*x) + (b^(1/4)*(a - Sqrt[-8*b + a^2*n]/Sqrt[n])*Sqrt[-((a*x^2)/b)]*EllipticPi[(-2*Sqrt[b])/Sq
rt[4*b - a^2*n + a*Sqrt[n]*Sqrt[-8*b + a^2*n]], ArcSin[(b + a*x^2)^(1/4)/b^(1/4)], -1])/(2*Sqrt[4*b - a^2*n +
a*Sqrt[n]*Sqrt[-8*b + a^2*n]]*x) - (b^(1/4)*(a - Sqrt[-8*b + a^2*n]/Sqrt[n])*Sqrt[-((a*x^2)/b)]*EllipticPi[(2*
Sqrt[b])/Sqrt[4*b - a^2*n + a*Sqrt[n]*Sqrt[-8*b + a^2*n]], ArcSin[(b + a*x^2)^(1/4)/b^(1/4)], -1])/(2*Sqrt[4*b
- a^2*n + a*Sqrt[n]*Sqrt[-8*b + a^2*n]]*x)

Rule 399

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[(2*Sqrt[-((b*x^2)/a)])/x, Subst[I
nt[x^2/(Sqrt[1 - x^4/a]*(b*c - a*d + d*x^4)), x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0]

Rule 490

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(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)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), In
t[1/((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1218

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(c/a), 4]}, Simp[(1*Ellipt
icPi[-(e/(d*q^2)), ArcSin[q*x], -1])/(d*Sqrt[a]*q), x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]

Rule 1692

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandInteg
rand[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, q}, x] && PolyQ[Px, x^2] && NeQ[b
^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {2 b+a x^2}{\sqrt [4]{b+a x^2} \left (b n+a n x^2+2 x^4\right )} \, dx &=\int \left (\frac {a-\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}}{\left (a n-\sqrt {n} \sqrt {-8 b+a^2 n}+4 x^2\right ) \sqrt [4]{b+a x^2}}+\frac {a+\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}}{\left (a n+\sqrt {n} \sqrt {-8 b+a^2 n}+4 x^2\right ) \sqrt [4]{b+a x^2}}\right ) \, dx\\ &=\left (a-\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \int \frac {1}{\left (a n-\sqrt {n} \sqrt {-8 b+a^2 n}+4 x^2\right ) \sqrt [4]{b+a x^2}} \, dx+\left (a+\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \int \frac {1}{\left (a n+\sqrt {n} \sqrt {-8 b+a^2 n}+4 x^2\right ) \sqrt [4]{b+a x^2}} \, dx\\ &=\frac {\left (2 \left (a-\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (-4 b+a \left (a n-\sqrt {n} \sqrt {-8 b+a^2 n}\right )+4 x^4\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{x}+\frac {\left (2 \left (a+\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (-4 b+a \left (a n+\sqrt {n} \sqrt {-8 b+a^2 n}\right )+4 x^4\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{x}\\ &=-\frac {\left (\left (a-\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {4 b-a^2 n+a \sqrt {n} \sqrt {-8 b+a^2 n}}-2 x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{2 x}+\frac {\left (\left (a-\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {4 b-a^2 n+a \sqrt {n} \sqrt {-8 b+a^2 n}}+2 x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{2 x}-\frac {\left (\left (a+\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {4 b-a^2 n-a \sqrt {n} \sqrt {-8 b+a^2 n}}-2 x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{2 x}+\frac {\left (\left (a+\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {4 b-a^2 n-a \sqrt {n} \sqrt {-8 b+a^2 n}}+2 x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{2 x}\\ &=\frac {\sqrt [4]{b} \left (a+\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (-\frac {2 \sqrt {b}}{\sqrt {4 b-a^2 n-a \sqrt {n} \sqrt {-8 b+a^2 n}}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 \sqrt {4 b-a^2 n-a \sqrt {n} \sqrt {-8 b+a^2 n}} x}-\frac {\sqrt [4]{b} \left (a+\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (\frac {2 \sqrt {b}}{\sqrt {4 b-a^2 n-a \sqrt {n} \sqrt {-8 b+a^2 n}}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 \sqrt {4 b-a^2 n-a \sqrt {n} \sqrt {-8 b+a^2 n}} x}+\frac {\sqrt [4]{b} \left (a-\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (-\frac {2 \sqrt {b}}{\sqrt {4 b-a^2 n+a \sqrt {n} \sqrt {-8 b+a^2 n}}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 \sqrt {4 b-a^2 n+a \sqrt {n} \sqrt {-8 b+a^2 n}} x}-\frac {\sqrt [4]{b} \left (a-\frac {\sqrt {-8 b+a^2 n}}{\sqrt {n}}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (\frac {2 \sqrt {b}}{\sqrt {4 b-a^2 n+a \sqrt {n} \sqrt {-8 b+a^2 n}}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 \sqrt {4 b-a^2 n+a \sqrt {n} \sqrt {-8 b+a^2 n}} x}\\ \end {align*}

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Mathematica [F]  time = 0.30, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 b+a x^2}{\sqrt [4]{b+a x^2} \left (b n+a n x^2+2 x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.42, size = 130, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\frac {x^2}{\sqrt [4]{2} \sqrt [4]{n}}-\frac {\sqrt [4]{n} \sqrt {b+a x^2}}{2^{3/4}}}{x \sqrt [4]{b+a x^2}}\right )}{2^{3/4} n^{3/4}}+\frac {\tanh ^{-1}\left (\frac {\frac {x^2}{\sqrt [4]{2} \sqrt [4]{n}}+\frac {\sqrt [4]{n} \sqrt {b+a x^2}}{2^{3/4}}}{x \sqrt [4]{b+a x^2}}\right )}{2^{3/4} n^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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^2+2*b)/(a*x^2+b)^(1/4)/(2*x^4+1881*a*x^2+1881*b),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{2} + 2 \, b}{{\left (2 \, x^{4} + 1881 \, a x^{2} + 1881 \, b\right )} {\left (a x^{2} + b\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^2+2*b)/(a*x^2+b)^(1/4)/(2*x^4+1881*a*x^2+1881*b),x, algorithm="giac")

[Out]

integrate((a*x^2 + 2*b)/((2*x^4 + 1881*a*x^2 + 1881*b)*(a*x^2 + b)^(1/4)), x)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {a \,x^{2}+2 b}{\left (a \,x^{2}+b \right )^{\frac {1}{4}} \left (2 x^{4}+1881 a \,x^{2}+1881 b \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2+2*b)/(a*x^2+b)^(1/4)/(2*x^4+1881*a*x^2+1881*b),x)

[Out]

int((a*x^2+2*b)/(a*x^2+b)^(1/4)/(2*x^4+1881*a*x^2+1881*b),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{2} + 2 \, b}{{\left (2 \, x^{4} + 1881 \, a x^{2} + 1881 \, b\right )} {\left (a x^{2} + b\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^2+2*b)/(a*x^2+b)^(1/4)/(2*x^4+1881*a*x^2+1881*b),x, algorithm="maxima")

[Out]

integrate((a*x^2 + 2*b)/((2*x^4 + 1881*a*x^2 + 1881*b)*(a*x^2 + b)^(1/4)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a\,x^2+2\,b}{{\left (a\,x^2+b\right )}^{1/4}\,\left (2\,x^4+1881\,a\,x^2+1881\,b\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*b + a*x^2)/((b + a*x^2)^(1/4)*(1881*b + 1881*a*x^2 + 2*x^4)),x)

[Out]

int((2*b + a*x^2)/((b + a*x^2)^(1/4)*(1881*b + 1881*a*x^2 + 2*x^4)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{2} + 2 b}{\sqrt [4]{a x^{2} + b} \left (1881 a x^{2} + 1881 b + 2 x^{4}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**2+2*b)/(a*x**2+b)**(1/4)/(2*x**4+1881*a*x**2+1881*b),x)

[Out]

Integral((a*x**2 + 2*b)/((a*x**2 + b)**(1/4)*(1881*a*x**2 + 1881*b + 2*x**4)), x)

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