3.318 \(\int \frac{1}{x^5 (a+b x^4+c x^8)} \, dx\)

Optimal. Leaf size=89 \[ -\frac{\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^4}{\sqrt{b^2-4 a c}}\right )}{4 a^2 \sqrt{b^2-4 a c}}+\frac{b \log \left (a+b x^4+c x^8\right )}{8 a^2}-\frac{b \log (x)}{a^2}-\frac{1}{4 a x^4} \]

[Out]

-1/(4*a*x^4) - ((b^2 - 2*a*c)*ArcTanh[(b + 2*c*x^4)/Sqrt[b^2 - 4*a*c]])/(4*a^2*Sqrt[b^2 - 4*a*c]) - (b*Log[x])
/a^2 + (b*Log[a + b*x^4 + c*x^8])/(8*a^2)

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Rubi [A]  time = 0.123629, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {1357, 709, 800, 634, 618, 206, 628} \[ -\frac{\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^4}{\sqrt{b^2-4 a c}}\right )}{4 a^2 \sqrt{b^2-4 a c}}+\frac{b \log \left (a+b x^4+c x^8\right )}{8 a^2}-\frac{b \log (x)}{a^2}-\frac{1}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a + b*x^4 + c*x^8)),x]

[Out]

-1/(4*a*x^4) - ((b^2 - 2*a*c)*ArcTanh[(b + 2*c*x^4)/Sqrt[b^2 - 4*a*c]])/(4*a^2*Sqrt[b^2 - 4*a*c]) - (b*Log[x])
/a^2 + (b*Log[a + b*x^4 + c*x^8])/(8*a^2)

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 709

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

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

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^5 \left (a+b x^4+c x^8\right )} \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x+c x^2\right )} \, dx,x,x^4\right )\\ &=-\frac{1}{4 a x^4}+\frac{\operatorname{Subst}\left (\int \frac{-b-c x}{x \left (a+b x+c x^2\right )} \, dx,x,x^4\right )}{4 a}\\ &=-\frac{1}{4 a x^4}+\frac{\operatorname{Subst}\left (\int \left (-\frac{b}{a x}+\frac{b^2-a c+b c x}{a \left (a+b x+c x^2\right )}\right ) \, dx,x,x^4\right )}{4 a}\\ &=-\frac{1}{4 a x^4}-\frac{b \log (x)}{a^2}+\frac{\operatorname{Subst}\left (\int \frac{b^2-a c+b c x}{a+b x+c x^2} \, dx,x,x^4\right )}{4 a^2}\\ &=-\frac{1}{4 a x^4}-\frac{b \log (x)}{a^2}+\frac{b \operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,x^4\right )}{8 a^2}+\frac{\left (b^2-2 a c\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^4\right )}{8 a^2}\\ &=-\frac{1}{4 a x^4}-\frac{b \log (x)}{a^2}+\frac{b \log \left (a+b x^4+c x^8\right )}{8 a^2}-\frac{\left (b^2-2 a c\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^4\right )}{4 a^2}\\ &=-\frac{1}{4 a x^4}-\frac{\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^4}{\sqrt{b^2-4 a c}}\right )}{4 a^2 \sqrt{b^2-4 a c}}-\frac{b \log (x)}{a^2}+\frac{b \log \left (a+b x^4+c x^8\right )}{8 a^2}\\ \end{align*}

Mathematica [C]  time = 0.0310873, size = 92, normalized size = 1.03 \[ \frac{\text{RootSum}\left [\text{$\#$1}^4 b+\text{$\#$1}^8 c+a\& ,\frac{\text{$\#$1}^4 b c \log (x-\text{$\#$1})-a c \log (x-\text{$\#$1})+b^2 \log (x-\text{$\#$1})}{2 \text{$\#$1}^4 c+b}\& \right ]}{4 a^2}-\frac{b \log (x)}{a^2}-\frac{1}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(a + b*x^4 + c*x^8)),x]

[Out]

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

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Maple [A]  time = 0.009, size = 119, normalized size = 1.3 \begin{align*} -{\frac{1}{4\,a{x}^{4}}}-{\frac{b\ln \left ( x \right ) }{{a}^{2}}}+{\frac{b\ln \left ( c{x}^{8}+b{x}^{4}+a \right ) }{8\,{a}^{2}}}-{\frac{c}{2\,a}\arctan \left ({(2\,c{x}^{4}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+{\frac{{b}^{2}}{4\,{a}^{2}}\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^5/(c*x^8+b*x^4+a),x)

[Out]

-1/4/a/x^4-b*ln(x)/a^2+1/8*b*ln(c*x^8+b*x^4+a)/a^2-1/2/a/(4*a*c-b^2)^(1/2)*arctan((2*c*x^4+b)/(4*a*c-b^2)^(1/2
))*c+1/4/a^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x^4+b)/(4*a*c-b^2)^(1/2))*b^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^5/(c*x^8+b*x^4+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.5792, size = 664, normalized size = 7.46 \begin{align*} \left [-\frac{{\left (b^{2} - 2 \, a c\right )} \sqrt{b^{2} - 4 \, a c} x^{4} \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^{3} - 4 \, a b c\right )} x^{4} \log \left (c x^{8} + b x^{4} + a\right ) + 8 \,{\left (b^{3} - 4 \, a b c\right )} x^{4} \log \left (x\right ) + 2 \, a b^{2} - 8 \, a^{2} c}{8 \,{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x^{4}}, -\frac{2 \,{\left (b^{2} - 2 \, a c\right )} \sqrt{-b^{2} + 4 \, a c} x^{4} \arctan \left (-\frac{{\left (2 \, c x^{4} + b\right )} \sqrt{-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) -{\left (b^{3} - 4 \, a b c\right )} x^{4} \log \left (c x^{8} + b x^{4} + a\right ) + 8 \,{\left (b^{3} - 4 \, a b c\right )} x^{4} \log \left (x\right ) + 2 \, a b^{2} - 8 \, a^{2} c}{8 \,{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(c*x^8+b*x^4+a),x, algorithm="fricas")

[Out]

[-1/8*((b^2 - 2*a*c)*sqrt(b^2 - 4*a*c)*x^4*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^3 - 4*a*b*c)*x^4*log(c*x^8 + b*x^4 + a) + 8*(b^3 - 4*a*b*c)*x^4*log(x) + 2*
a*b^2 - 8*a^2*c)/((a^2*b^2 - 4*a^3*c)*x^4), -1/8*(2*(b^2 - 2*a*c)*sqrt(-b^2 + 4*a*c)*x^4*arctan(-(2*c*x^4 + b)
*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) - (b^3 - 4*a*b*c)*x^4*log(c*x^8 + b*x^4 + a) + 8*(b^3 - 4*a*b*c)*x^4*log(x)
+ 2*a*b^2 - 8*a^2*c)/((a^2*b^2 - 4*a^3*c)*x^4)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(c*x**8+b*x**4+a),x)

[Out]

Timed out

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Giac [A]  time = 7.50556, size = 127, normalized size = 1.43 \begin{align*} \frac{b \log \left (c x^{8} + b x^{4} + a\right )}{8 \, a^{2}} - \frac{b \log \left (x^{4}\right )}{4 \, a^{2}} + \frac{{\left (b^{2} - 2 \, a c\right )} \arctan \left (\frac{2 \, c x^{4} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{4 \, \sqrt{-b^{2} + 4 \, a c} a^{2}} + \frac{b x^{4} - a}{4 \, a^{2} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(c*x^8+b*x^4+a),x, algorithm="giac")

[Out]

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