3.141 \(\int \frac {1}{x (a+b x^3+c x^6)} \, dx\)

Optimal. Leaf size=69 \[ \frac {b \tanh ^{-1}\left (\frac {b+2 c x^3}{\sqrt {b^2-4 a c}}\right )}{3 a \sqrt {b^2-4 a c}}-\frac {\log \left (a+b x^3+c x^6\right )}{6 a}+\frac {\log (x)}{a} \]

[Out]

ln(x)/a-1/6*ln(c*x^6+b*x^3+a)/a+1/3*b*arctanh((2*c*x^3+b)/(-4*a*c+b^2)^(1/2))/a/(-4*a*c+b^2)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 69, normalized size of antiderivative = 1.00, 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} \[ \frac {b \tanh ^{-1}\left (\frac {b+2 c x^3}{\sqrt {b^2-4 a c}}\right )}{3 a \sqrt {b^2-4 a c}}-\frac {\log \left (a+b x^3+c x^6\right )}{6 a}+\frac {\log (x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^3 + c*x^6)),x]

[Out]

(b*ArcTanh[(b + 2*c*x^3)/Sqrt[b^2 - 4*a*c]])/(3*a*Sqrt[b^2 - 4*a*c]) + Log[x]/a - Log[a + b*x^3 + c*x^6]/(6*a)

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 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]]

Rubi steps

\begin {align*} \int \frac {1}{x \left (a+b x^3+c x^6\right )} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x \left (a+b x+c x^2\right )} \, dx,x,x^3\right )\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^3\right )}{3 a}+\frac {\operatorname {Subst}\left (\int \frac {-b-c x}{a+b x+c x^2} \, dx,x,x^3\right )}{3 a}\\ &=\frac {\log (x)}{a}-\frac {\operatorname {Subst}\left (\int \frac {b+2 c x}{a+b x+c x^2} \, dx,x,x^3\right )}{6 a}-\frac {b \operatorname {Subst}\left (\int \frac {1}{a+b x+c x^2} \, dx,x,x^3\right )}{6 a}\\ &=\frac {\log (x)}{a}-\frac {\log \left (a+b x^3+c x^6\right )}{6 a}+\frac {b \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^3\right )}{3 a}\\ &=\frac {b \tanh ^{-1}\left (\frac {b+2 c x^3}{\sqrt {b^2-4 a c}}\right )}{3 a \sqrt {b^2-4 a c}}+\frac {\log (x)}{a}-\frac {\log \left (a+b x^3+c x^6\right )}{6 a}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 66, normalized size = 0.96 \[ \frac {\log (x)}{a}-\frac {\text {RootSum}\left [\text {$\#$1}^6 c+\text {$\#$1}^3 b+a\& ,\frac {\text {$\#$1}^3 c \log (x-\text {$\#$1})+b \log (x-\text {$\#$1})}{2 \text {$\#$1}^3 c+b}\& \right ]}{3 a} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^3 + c*x^6)),x]

[Out]

Log[x]/a - RootSum[a + b*#1^3 + c*#1^6 & , (b*Log[x - #1] + c*Log[x - #1]*#1^3)/(b + 2*c*#1^3) & ]/(3*a)

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fricas [A]  time = 0.91, size = 223, normalized size = 3.23 \[ \left [\frac {\sqrt {b^{2} - 4 \, a c} b \log \left (\frac {2 \, c^{2} x^{6} + 2 \, b c x^{3} + b^{2} - 2 \, a c + {\left (2 \, c x^{3} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{6} + b x^{3} + a}\right ) - {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{6} + b x^{3} + a\right ) + 6 \, {\left (b^{2} - 4 \, a c\right )} \log \relax (x)}{6 \, {\left (a b^{2} - 4 \, a^{2} c\right )}}, \frac {2 \, \sqrt {-b^{2} + 4 \, a c} b \arctan \left (-\frac {{\left (2 \, c x^{3} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) - {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{6} + b x^{3} + a\right ) + 6 \, {\left (b^{2} - 4 \, a c\right )} \log \relax (x)}{6 \, {\left (a b^{2} - 4 \, a^{2} c\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(c*x^6+b*x^3+a),x, algorithm="fricas")

[Out]

[1/6*(sqrt(b^2 - 4*a*c)*b*log((2*c^2*x^6 + 2*b*c*x^3 + b^2 - 2*a*c + (2*c*x^3 + b)*sqrt(b^2 - 4*a*c))/(c*x^6 +
b*x^3 + a)) - (b^2 - 4*a*c)*log(c*x^6 + b*x^3 + a) + 6*(b^2 - 4*a*c)*log(x))/(a*b^2 - 4*a^2*c), 1/6*(2*sqrt(-
b^2 + 4*a*c)*b*arctan(-(2*c*x^3 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) - (b^2 - 4*a*c)*log(c*x^6 + b*x^3 + a)
+ 6*(b^2 - 4*a*c)*log(x))/(a*b^2 - 4*a^2*c)]

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giac [A]  time = 1.00, size = 66, normalized size = 0.96 \[ -\frac {b \arctan \left (\frac {2 \, c x^{3} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{3 \, \sqrt {-b^{2} + 4 \, a c} a} - \frac {\log \left (c x^{6} + b x^{3} + a\right )}{6 \, a} + \frac {\log \left ({\left | x \right |}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(c*x^6+b*x^3+a),x, algorithm="giac")

[Out]

-1/3*b*arctan((2*c*x^3 + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*a) - 1/6*log(c*x^6 + b*x^3 + a)/a + log(ab
s(x))/a

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maple [A]  time = 0.01, size = 66, normalized size = 0.96 \[ -\frac {b \arctan \left (\frac {2 c \,x^{3}+b}{\sqrt {4 a c -b^{2}}}\right )}{3 \sqrt {4 a c -b^{2}}\, a}+\frac {\ln \relax (x )}{a}-\frac {\ln \left (c \,x^{6}+b \,x^{3}+a \right )}{6 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(c*x^6+b*x^3+a),x)

[Out]

ln(x)/a-1/6*ln(c*x^6+b*x^3+a)/a-1/3/a*b/(4*a*c-b^2)^(1/2)*arctan((2*c*x^3+b)/(4*a*c-b^2)^(1/2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(c*x^6+b*x^3+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 1.92, size = 1362, normalized size = 19.74 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*x^3 + c*x^6)),x)

[Out]

log(x)/a + (log(a + b*x^3 + c*x^6)*(12*a*c - 3*b^2))/(2*(9*a*b^2 - 36*a^2*c)) - (b*atan((3*(4*a*c - b^2)^2*(4*
b^4 + 7*a^2*c^2 - 15*a*b^2*c)*((b^3*(27*b^3*c^3 - (27*a*b^3*c^3*(12*a*c - 3*b^2))/(2*(9*a*b^2 - 36*a^2*c))))/(
216*a^3*(4*a*c - b^2)^(3/2)) + (9*b^4*c^3*(12*a*c - 3*b^2)^3)/(16*(9*a*b^2 - 36*a^2*c)^3*(4*a*c - b^2)^(1/2))
- (3*b^6*c^3*(12*a*c - 3*b^2))/(16*a^2*(9*a*b^2 - 36*a^2*c)*(4*a*c - b^2)^(3/2)) - (b*(12*a*c - 3*b^2)^2*(27*b
^3*c^3 - (27*a*b^3*c^3*(12*a*c - 3*b^2))/(2*(9*a*b^2 - 36*a^2*c))))/(8*a*(9*a*b^2 - 36*a^2*c)^2*(4*a*c - b^2)^
(1/2))))/(b^3*c^6*(49*a*c - 12*b^2)) - (3*(4*a*c - b^2)^(3/2)*(4*b^5 + 29*a^2*b*c^2 - 23*a*b^3*c)*(((12*a*c -
3*b^2)^3*(27*b^3*c^3 - (27*a*b^3*c^3*(12*a*c - 3*b^2))/(2*(9*a*b^2 - 36*a^2*c))))/(8*(9*a*b^2 - 36*a^2*c)^3) -
(b^7*c^3)/(48*a^3*(4*a*c - b^2)^2) - (b^2*(12*a*c - 3*b^2)*(27*b^3*c^3 - (27*a*b^3*c^3*(12*a*c - 3*b^2))/(2*(
9*a*b^2 - 36*a^2*c))))/(24*a^2*(9*a*b^2 - 36*a^2*c)*(4*a*c - b^2)) + (9*b^5*c^3*(12*a*c - 3*b^2)^2)/(16*a*(9*a
*b^2 - 36*a^2*c)^2*(4*a*c - b^2))))/(b^3*c^6*(49*a*c - 12*b^2)) + (48*a^4*x^3*(((4*b^4 + 7*a^2*c^2 - 15*a*b^2*
c)*((b^3*(63*b^2*c^4 - ((108*b^4*c^3 - 378*a*b^2*c^4)*(12*a*c - 3*b^2))/(2*(9*a*b^2 - 36*a^2*c))))/(216*a^3*(4
*a*c - b^2)^(3/2)) + (b*(108*b^4*c^3 - 378*a*b^2*c^4)*(12*a*c - 3*b^2)^3)/(48*a*(9*a*b^2 - 36*a^2*c)^3*(4*a*c
- b^2)^(1/2)) - (b^3*(108*b^4*c^3 - 378*a*b^2*c^4)*(12*a*c - 3*b^2))/(144*a^3*(9*a*b^2 - 36*a^2*c)*(4*a*c - b^
2)^(3/2)) - (b*(63*b^2*c^4 - ((108*b^4*c^3 - 378*a*b^2*c^4)*(12*a*c - 3*b^2))/(2*(9*a*b^2 - 36*a^2*c)))*(12*a*
c - 3*b^2)^2)/(8*a*(9*a*b^2 - 36*a^2*c)^2*(4*a*c - b^2)^(1/2))))/(16*a^4*c^3*(49*a*c - 12*b^2)) - ((4*b^5 + 29
*a^2*b*c^2 - 23*a*b^3*c)*(((63*b^2*c^4 - ((108*b^4*c^3 - 378*a*b^2*c^4)*(12*a*c - 3*b^2))/(2*(9*a*b^2 - 36*a^2
*c)))*(12*a*c - 3*b^2)^3)/(8*(9*a*b^2 - 36*a^2*c)^3) - (b^4*(108*b^4*c^3 - 378*a*b^2*c^4))/(1296*a^4*(4*a*c -
b^2)^2) + (b^2*(108*b^4*c^3 - 378*a*b^2*c^4)*(12*a*c - 3*b^2)^2)/(48*a^2*(9*a*b^2 - 36*a^2*c)^2*(4*a*c - b^2))
- (b^2*(63*b^2*c^4 - ((108*b^4*c^3 - 378*a*b^2*c^4)*(12*a*c - 3*b^2))/(2*(9*a*b^2 - 36*a^2*c)))*(12*a*c - 3*b
^2))/(24*a^2*(9*a*b^2 - 36*a^2*c)*(4*a*c - b^2))))/(16*a^4*c^3*(4*a*c - b^2)^(1/2)*(49*a*c - 12*b^2)))*(4*a*c
- b^2)^2)/(b^3*c^3)))/(3*a*(4*a*c - b^2)^(1/2))

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sympy [B]  time = 6.73, size = 253, normalized size = 3.67 \[ \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{6 a \left (4 a c - b^{2}\right )} - \frac {1}{6 a}\right ) \log {\left (x^{3} + \frac {- 12 a^{2} c \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{6 a \left (4 a c - b^{2}\right )} - \frac {1}{6 a}\right ) + 3 a b^{2} \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{6 a \left (4 a c - b^{2}\right )} - \frac {1}{6 a}\right ) - 2 a c + b^{2}}{b c} \right )} + \left (\frac {b \sqrt {- 4 a c + b^{2}}}{6 a \left (4 a c - b^{2}\right )} - \frac {1}{6 a}\right ) \log {\left (x^{3} + \frac {- 12 a^{2} c \left (\frac {b \sqrt {- 4 a c + b^{2}}}{6 a \left (4 a c - b^{2}\right )} - \frac {1}{6 a}\right ) + 3 a b^{2} \left (\frac {b \sqrt {- 4 a c + b^{2}}}{6 a \left (4 a c - b^{2}\right )} - \frac {1}{6 a}\right ) - 2 a c + b^{2}}{b c} \right )} + \frac {\log {\relax (x )}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(c*x**6+b*x**3+a),x)

[Out]

(-b*sqrt(-4*a*c + b**2)/(6*a*(4*a*c - b**2)) - 1/(6*a))*log(x**3 + (-12*a**2*c*(-b*sqrt(-4*a*c + b**2)/(6*a*(4
*a*c - b**2)) - 1/(6*a)) + 3*a*b**2*(-b*sqrt(-4*a*c + b**2)/(6*a*(4*a*c - b**2)) - 1/(6*a)) - 2*a*c + b**2)/(b
*c)) + (b*sqrt(-4*a*c + b**2)/(6*a*(4*a*c - b**2)) - 1/(6*a))*log(x**3 + (-12*a**2*c*(b*sqrt(-4*a*c + b**2)/(6
*a*(4*a*c - b**2)) - 1/(6*a)) + 3*a*b**2*(b*sqrt(-4*a*c + b**2)/(6*a*(4*a*c - b**2)) - 1/(6*a)) - 2*a*c + b**2
)/(b*c)) + log(x)/a

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