3.316 \(\int \frac{1}{x (a+b x^4+c x^8)} \, dx\)

Optimal. Leaf size=69 \[ \frac{b \tanh ^{-1}\left (\frac{b+2 c x^4}{\sqrt{b^2-4 a c}}\right )}{4 a \sqrt{b^2-4 a c}}-\frac{\log \left (a+b x^4+c x^8\right )}{8 a}+\frac{\log (x)}{a} \]

[Out]

(b*ArcTanh[(b + 2*c*x^4)/Sqrt[b^2 - 4*a*c]])/(4*a*Sqrt[b^2 - 4*a*c]) + Log[x]/a - Log[a + b*x^4 + c*x^8]/(8*a)

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Rubi [A]  time = 0.0683157, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {1357, 705, 29, 634, 618, 206, 628} \[ \frac{b \tanh ^{-1}\left (\frac{b+2 c x^4}{\sqrt{b^2-4 a c}}\right )}{4 a \sqrt{b^2-4 a c}}-\frac{\log \left (a+b x^4+c x^8\right )}{8 a}+\frac{\log (x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^4 + c*x^8)),x]

[Out]

(b*ArcTanh[(b + 2*c*x^4)/Sqrt[b^2 - 4*a*c]])/(4*a*Sqrt[b^2 - 4*a*c]) + Log[x]/a - Log[a + b*x^4 + c*x^8]/(8*a)

Rule 1357

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

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 29

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

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{1}{x \left (a+b x^4+c x^8\right )} \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{x \left (a+b x+c x^2\right )} \, dx,x,x^4\right )\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^4\right )}{4 a}+\frac{\operatorname{Subst}\left (\int \frac{-b-c x}{a+b x+c x^2} \, dx,x,x^4\right )}{4 a}\\ &=\frac{\log (x)}{a}-\frac{\operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,x^4\right )}{8 a}-\frac{b \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^4\right )}{8 a}\\ &=\frac{\log (x)}{a}-\frac{\log \left (a+b x^4+c x^8\right )}{8 a}+\frac{b \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^4\right )}{4 a}\\ &=\frac{b \tanh ^{-1}\left (\frac{b+2 c x^4}{\sqrt{b^2-4 a c}}\right )}{4 a \sqrt{b^2-4 a c}}+\frac{\log (x)}{a}-\frac{\log \left (a+b x^4+c x^8\right )}{8 a}\\ \end{align*}

Mathematica [C]  time = 0.0232458, size = 66, normalized size = 0.96 \[ \frac{\log (x)}{a}-\frac{\text{RootSum}\left [\text{$\#$1}^4 b+\text{$\#$1}^8 c+a\& ,\frac{\text{$\#$1}^4 c \log (x-\text{$\#$1})+b \log (x-\text{$\#$1})}{2 \text{$\#$1}^4 c+b}\& \right ]}{4 a} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^4 + c*x^8)),x]

[Out]

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

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Maple [A]  time = 0.007, size = 66, normalized size = 1. \begin{align*}{\frac{\ln \left ( x \right ) }{a}}-{\frac{\ln \left ( c{x}^{8}+b{x}^{4}+a \right ) }{8\,a}}-{\frac{b}{4\,a}\arctan \left ({(2\,c{x}^{4}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(c*x^8+b*x^4+a),x)

[Out]

ln(x)/a-1/8*ln(c*x^8+b*x^4+a)/a-1/4/a*b/(4*a*c-b^2)^(1/2)*arctan((2*c*x^4+b)/(4*a*c-b^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(c*x^8+b*x^4+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.80138, size = 510, normalized size = 7.39 \begin{align*} \left [\frac{\sqrt{b^{2} - 4 \, a c} b \log \left (\frac{2 \, c^{2} x^{8} + 2 \, b c x^{4} + b^{2} - 2 \, a c +{\left (2 \, c x^{4} + b\right )} \sqrt{b^{2} - 4 \, a c}}{c x^{8} + b x^{4} + a}\right ) -{\left (b^{2} - 4 \, a c\right )} \log \left (c x^{8} + b x^{4} + a\right ) + 8 \,{\left (b^{2} - 4 \, a c\right )} \log \left (x\right )}{8 \,{\left (a b^{2} - 4 \, a^{2} c\right )}}, \frac{2 \, \sqrt{-b^{2} + 4 \, a c} b \arctan \left (-\frac{{\left (2 \, c x^{4} + b\right )} \sqrt{-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) -{\left (b^{2} - 4 \, a c\right )} \log \left (c x^{8} + b x^{4} + a\right ) + 8 \,{\left (b^{2} - 4 \, a c\right )} \log \left (x\right )}{8 \,{\left (a b^{2} - 4 \, a^{2} c\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(c*x^8+b*x^4+a),x, algorithm="fricas")

[Out]

[1/8*(sqrt(b^2 - 4*a*c)*b*log((2*c^2*x^8 + 2*b*c*x^4 + b^2 - 2*a*c + (2*c*x^4 + b)*sqrt(b^2 - 4*a*c))/(c*x^8 +
b*x^4 + a)) - (b^2 - 4*a*c)*log(c*x^8 + b*x^4 + a) + 8*(b^2 - 4*a*c)*log(x))/(a*b^2 - 4*a^2*c), 1/8*(2*sqrt(-
b^2 + 4*a*c)*b*arctan(-(2*c*x^4 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) - (b^2 - 4*a*c)*log(c*x^8 + b*x^4 + a)
+ 8*(b^2 - 4*a*c)*log(x))/(a*b^2 - 4*a^2*c)]

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Sympy [B]  time = 5.46836, size = 253, normalized size = 3.67 \begin{align*} \left (- \frac{b \sqrt{- 4 a c + b^{2}}}{8 a \left (4 a c - b^{2}\right )} - \frac{1}{8 a}\right ) \log{\left (x^{4} + \frac{- 16 a^{2} c \left (- \frac{b \sqrt{- 4 a c + b^{2}}}{8 a \left (4 a c - b^{2}\right )} - \frac{1}{8 a}\right ) + 4 a b^{2} \left (- \frac{b \sqrt{- 4 a c + b^{2}}}{8 a \left (4 a c - b^{2}\right )} - \frac{1}{8 a}\right ) - 2 a c + b^{2}}{b c} \right )} + \left (\frac{b \sqrt{- 4 a c + b^{2}}}{8 a \left (4 a c - b^{2}\right )} - \frac{1}{8 a}\right ) \log{\left (x^{4} + \frac{- 16 a^{2} c \left (\frac{b \sqrt{- 4 a c + b^{2}}}{8 a \left (4 a c - b^{2}\right )} - \frac{1}{8 a}\right ) + 4 a b^{2} \left (\frac{b \sqrt{- 4 a c + b^{2}}}{8 a \left (4 a c - b^{2}\right )} - \frac{1}{8 a}\right ) - 2 a c + b^{2}}{b c} \right )} + \frac{\log{\left (x \right )}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(c*x**8+b*x**4+a),x)

[Out]

(-b*sqrt(-4*a*c + b**2)/(8*a*(4*a*c - b**2)) - 1/(8*a))*log(x**4 + (-16*a**2*c*(-b*sqrt(-4*a*c + b**2)/(8*a*(4
*a*c - b**2)) - 1/(8*a)) + 4*a*b**2*(-b*sqrt(-4*a*c + b**2)/(8*a*(4*a*c - b**2)) - 1/(8*a)) - 2*a*c + b**2)/(b
*c)) + (b*sqrt(-4*a*c + b**2)/(8*a*(4*a*c - b**2)) - 1/(8*a))*log(x**4 + (-16*a**2*c*(b*sqrt(-4*a*c + b**2)/(8
*a*(4*a*c - b**2)) - 1/(8*a)) + 4*a*b**2*(b*sqrt(-4*a*c + b**2)/(8*a*(4*a*c - b**2)) - 1/(8*a)) - 2*a*c + b**2
)/(b*c)) + log(x)/a

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Giac [A]  time = 7.70529, size = 92, normalized size = 1.33 \begin{align*} -\frac{b \arctan \left (\frac{2 \, c x^{4} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{4 \, \sqrt{-b^{2} + 4 \, a c} a} - \frac{\log \left (c x^{8} + b x^{4} + a\right )}{8 \, a} + \frac{\log \left (x^{4}\right )}{4 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(c*x^8+b*x^4+a),x, algorithm="giac")

[Out]

-1/4*b*arctan((2*c*x^4 + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*a) - 1/8*log(c*x^8 + b*x^4 + a)/a + 1/4*lo
g(x^4)/a