3.8.3 \(\int \frac {1}{x (a-b+2 a x^2+a x^4)} \, dx\)

Optimal. Leaf size=77 \[ \frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \left (x^2+1\right )}{\sqrt {b}}\right )}{2 \sqrt {b} (a-b)}-\frac {\log \left (a x^4+2 a x^2+a-b\right )}{4 (a-b)}+\frac {\log (x)}{a-b} \]

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Rubi [A]  time = 0.07, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1114, 705, 29, 634, 618, 206, 628} \begin {gather*} -\frac {\log \left (a x^4+2 a x^2+a-b\right )}{4 (a-b)}+\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \left (x^2+1\right )}{\sqrt {b}}\right )}{2 \sqrt {b} (a-b)}+\frac {\log (x)}{a-b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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 618

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

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

Rule 705

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

Rule 1114

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {1}{x \left (a-b+2 a x^2+a x^4\right )} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x \left (a-b+2 a x+a x^2\right )} \, dx,x,x^2\right )\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 (a-b)}+\frac {\operatorname {Subst}\left (\int \frac {-2 a-a x}{a-b+2 a x+a x^2} \, dx,x,x^2\right )}{2 (a-b)}\\ &=\frac {\log (x)}{a-b}-\frac {\operatorname {Subst}\left (\int \frac {2 a+2 a x}{a-b+2 a x+a x^2} \, dx,x,x^2\right )}{4 (a-b)}-\frac {a \operatorname {Subst}\left (\int \frac {1}{a-b+2 a x+a x^2} \, dx,x,x^2\right )}{2 (a-b)}\\ &=\frac {\log (x)}{a-b}-\frac {\log \left (a-b+2 a x^2+a x^4\right )}{4 (a-b)}+\frac {a \operatorname {Subst}\left (\int \frac {1}{4 a b-x^2} \, dx,x,2 a \left (1+x^2\right )\right )}{a-b}\\ &=\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \left (1+x^2\right )}{\sqrt {b}}\right )}{2 (a-b) \sqrt {b}}+\frac {\log (x)}{a-b}-\frac {\log \left (a-b+2 a x^2+a x^4\right )}{4 (a-b)}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 90, normalized size = 1.17 \begin {gather*} \frac {\left (\sqrt {a}+\sqrt {b}\right ) \log \left (\sqrt {a} \left (x^2+1\right )-\sqrt {b}\right )+\left (\sqrt {b}-\sqrt {a}\right ) \log \left (\sqrt {a} \left (x^2+1\right )+\sqrt {b}\right )-4 \sqrt {b} \log (x)}{4 \sqrt {b} (b-a)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.97, size = 151, normalized size = 1.96 \begin {gather*} \left [-\frac {\sqrt {\frac {a}{b}} \log \left (\frac {a x^{4} + 2 \, a x^{2} - 2 \, {\left (b x^{2} + b\right )} \sqrt {\frac {a}{b}} + a + b}{a x^{4} + 2 \, a x^{2} + a - b}\right ) + \log \left (a x^{4} + 2 \, a x^{2} + a - b\right ) - 4 \, \log \relax (x)}{4 \, {\left (a - b\right )}}, -\frac {2 \, \sqrt {-\frac {a}{b}} \arctan \left (\frac {b \sqrt {-\frac {a}{b}}}{a x^{2} + a}\right ) + \log \left (a x^{4} + 2 \, a x^{2} + a - b\right ) - 4 \, \log \relax (x)}{4 \, {\left (a - b\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.33, size = 71, normalized size = 0.92 \begin {gather*} -\frac {a \arctan \left (\frac {a x^{2} + a}{\sqrt {-a b}}\right )}{2 \, \sqrt {-a b} {\left (a - b\right )}} - \frac {\log \left (a x^{4} + 2 \, a x^{2} + a - b\right )}{4 \, {\left (a - b\right )}} + \frac {\log \left (x^{2}\right )}{2 \, {\left (a - b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 71, normalized size = 0.92 \begin {gather*} \frac {a \arctanh \left (\frac {2 a \,x^{2}+2 a}{2 \sqrt {a b}}\right )}{2 \left (a -b \right ) \sqrt {a b}}+\frac {\ln \relax (x )}{a -b}-\frac {\ln \left (a \,x^{4}+2 a \,x^{2}+a -b \right )}{4 \left (a -b \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.13, size = 85, normalized size = 1.10 \begin {gather*} -\frac {a \log \left (\frac {a x^{2} + a - \sqrt {a b}}{a x^{2} + a + \sqrt {a b}}\right )}{4 \, \sqrt {a b} {\left (a - b\right )}} - \frac {\log \left (a x^{4} + 2 \, a x^{2} + a - b\right )}{4 \, {\left (a - b\right )}} + \frac {\log \left (x^{2}\right )}{2 \, {\left (a - b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.56, size = 183, normalized size = 2.38 \begin {gather*} \frac {\ln \relax (x)}{a-b}-\frac {\ln \left (16\,a^4+20\,a^4\,x^2+\frac {\left (b-\sqrt {a\,b}\right )\,\left (x^2\,\left (16\,a^5+80\,b\,a^4\right )-16\,a^4\,b+16\,a^5\right )}{4\,\left (a\,b-b^2\right )}\right )\,\left (b-\sqrt {a\,b}\right )}{4\,\left (a\,b-b^2\right )}-\frac {\ln \left (16\,a^4+20\,a^4\,x^2+\frac {\left (b+\sqrt {a\,b}\right )\,\left (x^2\,\left (16\,a^5+80\,b\,a^4\right )-16\,a^4\,b+16\,a^5\right )}{4\,\left (a\,b-b^2\right )}\right )\,\left (b+\sqrt {a\,b}\right )}{4\,\left (a\,b-b^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)/(a - b) - (log(16*a^4 + 20*a^4*x^2 + ((b - (a*b)^(1/2))*(x^2*(80*a^4*b + 16*a^5) - 16*a^4*b + 16*a^5))/
(4*(a*b - b^2)))*(b - (a*b)^(1/2)))/(4*(a*b - b^2)) - (log(16*a^4 + 20*a^4*x^2 + ((b + (a*b)^(1/2))*(x^2*(80*a
^4*b + 16*a^5) - 16*a^4*b + 16*a^5))/(4*(a*b - b^2)))*(b + (a*b)^(1/2)))/(4*(a*b - b^2))

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sympy [B]  time = 5.31, size = 184, normalized size = 2.39 \begin {gather*} \left (- \frac {1}{4 \left (a - b\right )} - \frac {\sqrt {a b}}{4 b \left (a - b\right )}\right ) \log {\left (x^{2} + \frac {4 a b \left (- \frac {1}{4 \left (a - b\right )} - \frac {\sqrt {a b}}{4 b \left (a - b\right )}\right ) + a - 4 b^{2} \left (- \frac {1}{4 \left (a - b\right )} - \frac {\sqrt {a b}}{4 b \left (a - b\right )}\right ) + b}{a} \right )} + \left (- \frac {1}{4 \left (a - b\right )} + \frac {\sqrt {a b}}{4 b \left (a - b\right )}\right ) \log {\left (x^{2} + \frac {4 a b \left (- \frac {1}{4 \left (a - b\right )} + \frac {\sqrt {a b}}{4 b \left (a - b\right )}\right ) + a - 4 b^{2} \left (- \frac {1}{4 \left (a - b\right )} + \frac {\sqrt {a b}}{4 b \left (a - b\right )}\right ) + b}{a} \right )} + \frac {\log {\relax (x )}}{a - b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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