3.10.9 \(\int \frac {1}{x (a+b+2 a x^2+a x^4)} \, dx\) [909]

Optimal. Leaf size=69 \[ -\frac {\sqrt {a} \tan ^{-1}\left (\frac {\sqrt {a} \left (1+x^2\right )}{\sqrt {b}}\right )}{2 \sqrt {b} (a+b)}+\frac {\log (x)}{a+b}-\frac {\log \left (a+b+2 a x^2+a x^4\right )}{4 (a+b)} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1128, 719, 29, 648, 632, 210, 642} \begin {gather*} -\frac {\sqrt {a} \text {ArcTan}\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Sqrt[a]*ArcTan[(Sqrt[a]*(1 + x^2))/Sqrt[b]])/(Sqrt[b]*(a + 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 210

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

Rule 632

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 642

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 648

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 719

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 1128

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} \text {Subst}\left (\int \frac {1}{x \left (a+b+2 a x+a x^2\right )} \, dx,x,x^2\right )\\ &=\frac {\text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 (a+b)}+\frac {\text {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 {\text {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 \text {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 \text {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} \tan ^{-1}\left (\frac {\sqrt {a} \left (1+x^2\right )}{\sqrt {b}}\right )}{2 \sqrt {b} (a+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 [C] Result contains complex when optimal does not.
time = 0.04, size = 105, normalized size = 1.52 \begin {gather*} \frac {4 \sqrt {b} \log (x)+i \left (\sqrt {a}+i \sqrt {b}\right ) \log \left (-i \sqrt {b}+\sqrt {a} \left (1+x^2\right )\right )+\left (-i \sqrt {a}-\sqrt {b}\right ) \log \left (i \sqrt {b}+\sqrt {a} \left (1+x^2\right )\right )}{4 \sqrt {b} (a+b)} \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] + I*(Sqrt[a] + I*Sqrt[b])*Log[(-I)*Sqrt[b] + Sqrt[a]*(1 + x^2)] + ((-I)*Sqrt[a] - Sqrt[b])*L
og[I*Sqrt[b] + Sqrt[a]*(1 + x^2)])/(4*Sqrt[b]*(a + b))

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Maple [A]
time = 0.02, size = 63, normalized size = 0.91

method	result	size
risch	\(\frac {\ln \left (x \right )}{a +b}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (1+\left (a b +b^{2}\right ) \textit {\_Z}^{2}+2 b \textit {\_Z} \right )}{\sum }\textit {\_R} \ln \left (\left (\left (-a +5 b \right ) \textit {\_R} +5\right ) x^{2}+\left (-a -b \right ) \textit {\_R} +4\right )\right )}{4}\)	\(62\)
default	\(-\frac {a \left (\frac {\ln \left (a \,x^{4}+2 a \,x^{2}+a +b \right )}{2 a}+\frac {\arctan \left (\frac {2 a \,x^{2}+2 a}{2 \sqrt {a b}}\right )}{\sqrt {a b}}\right )}{2 \left (a +b \right )}+\frac {\ln \left (x \right )}{a +b}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 61, normalized size = 0.88 \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="maxima")

[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*log(
x^2)/(a + b)

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Fricas [A]
time = 0.36, size = 147, normalized size = 2.13 \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 \left (x\right )}{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 \left (x\right )}{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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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (61) = 122\).
time = 2.52, size = 194, normalized size = 2.81 \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 {\left (x \right )}}{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)))*lo
g(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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Giac [A]
time = 2.55, size = 61, normalized size = 0.88 \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*log(
x^2)/(a + b)

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Mupad [B]
time = 4.64, size = 71, normalized size = 1.03 \begin {gather*} \frac {\ln \left (x\right )}{a+b}-\frac {4\,b\,\ln \left (a\,x^4+2\,a\,x^2+a+b\right )}{16\,b^2+16\,a\,b}-\frac {\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {a}\,x^2}{\sqrt {b}}\right )}{2\,\sqrt {b}\,\left (a+b\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) - (4*b*log(a + b + 2*a*x^2 + a*x^4))/(16*a*b + 16*b^2) - (a^(1/2)*atan(a^(1/2)/b^(1/2) + (a^(1/
2)*x^2)/b^(1/2)))/(2*b^(1/2)*(a + b))

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