3.17 \(\int \frac {1}{x (a x^2+b x^3+c x^4)} \, dx\)

Optimal. Leaf size=104 \[ -\frac {\left (b^2-a c\right ) \log \left (a+b x+c x^2\right )}{2 a^3}+\frac {\log (x) \left (b^2-a c\right )}{a^3}+\frac {b \left (b^2-3 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^3 \sqrt {b^2-4 a c}}+\frac {b}{a^2 x}-\frac {1}{2 a x^2} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1585, 709, 800, 634, 618, 206, 628} \[ -\frac {\left (b^2-a c\right ) \log \left (a+b x+c x^2\right )}{2 a^3}+\frac {\log (x) \left (b^2-a c\right )}{a^3}+\frac {b \left (b^2-3 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^3 \sqrt {b^2-4 a c}}+\frac {b}{a^2 x}-\frac {1}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a*x^2 + b*x^3 + c*x^4)),x]

[Out]

-1/(2*a*x^2) + b/(a^2*x) + (b*(b^2 - 3*a*c)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(a^3*Sqrt[b^2 - 4*a*c]) +
((b^2 - a*c)*Log[x])/a^3 - ((b^2 - a*c)*Log[a + b*x + c*x^2])/(2*a^3)

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 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 1585

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(m +
n*p)*(a + b*x^(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, m, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] &
& PosQ[r - p]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 102, normalized size = 0.98 \[ \frac {-\frac {a^2}{x^2}+2 \log (x) \left (b^2-a c\right )+\left (a c-b^2\right ) \log (a+x (b+c x))-\frac {2 b \left (b^2-3 a c\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+\frac {2 a b}{x}}{2 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a*x^2 + b*x^3 + c*x^4)),x]

[Out]

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

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fricas [A]  time = 0.98, size = 358, normalized size = 3.44 \[ \left [-\frac {{\left (b^{3} - 3 \, a b c\right )} \sqrt {b^{2} - 4 \, a c} x^{2} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c - \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + a^{2} b^{2} - 4 \, a^{3} c + {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} x^{2} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} x^{2} \log \relax (x) - 2 \, {\left (a b^{3} - 4 \, a^{2} b c\right )} x}{2 \, {\left (a^{3} b^{2} - 4 \, a^{4} c\right )} x^{2}}, \frac {2 \, {\left (b^{3} - 3 \, a b c\right )} \sqrt {-b^{2} + 4 \, a c} x^{2} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) - a^{2} b^{2} + 4 \, a^{3} c - {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} x^{2} \log \left (c x^{2} + b x + a\right ) + 2 \, {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} x^{2} \log \relax (x) + 2 \, {\left (a b^{3} - 4 \, a^{2} b c\right )} x}{2 \, {\left (a^{3} b^{2} - 4 \, a^{4} c\right )} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(c*x^4+b*x^3+a*x^2),x, algorithm="fricas")

[Out]

[-1/2*((b^3 - 3*a*b*c)*sqrt(b^2 - 4*a*c)*x^2*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c - sqrt(b^2 - 4*a*c)*(2*c*x
+ b))/(c*x^2 + b*x + a)) + a^2*b^2 - 4*a^3*c + (b^4 - 5*a*b^2*c + 4*a^2*c^2)*x^2*log(c*x^2 + b*x + a) - 2*(b^
4 - 5*a*b^2*c + 4*a^2*c^2)*x^2*log(x) - 2*(a*b^3 - 4*a^2*b*c)*x)/((a^3*b^2 - 4*a^4*c)*x^2), 1/2*(2*(b^3 - 3*a*
b*c)*sqrt(-b^2 + 4*a*c)*x^2*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) - a^2*b^2 + 4*a^3*c - (b^4 -
5*a*b^2*c + 4*a^2*c^2)*x^2*log(c*x^2 + b*x + a) + 2*(b^4 - 5*a*b^2*c + 4*a^2*c^2)*x^2*log(x) + 2*(a*b^3 - 4*a
^2*b*c)*x)/((a^3*b^2 - 4*a^4*c)*x^2)]

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giac [A]  time = 0.51, size = 105, normalized size = 1.01 \[ -\frac {{\left (b^{2} - a c\right )} \log \left (c x^{2} + b x + a\right )}{2 \, a^{3}} + \frac {{\left (b^{2} - a c\right )} \log \left ({\left | x \right |}\right )}{a^{3}} - \frac {{\left (b^{3} - 3 \, a b c\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} a^{3}} + \frac {2 \, a b x - a^{2}}{2 \, a^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(c*x^4+b*x^3+a*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 150, normalized size = 1.44 \[ \frac {3 b c \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a^{2}}-\frac {b^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a^{3}}-\frac {c \ln \relax (x )}{a^{2}}+\frac {c \ln \left (c \,x^{2}+b x +a \right )}{2 a^{2}}+\frac {b^{2} \ln \relax (x )}{a^{3}}-\frac {b^{2} \ln \left (c \,x^{2}+b x +a \right )}{2 a^{3}}+\frac {b}{a^{2} x}-\frac {1}{2 a \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(c*x^4+b*x^3+a*x^2),x)

[Out]

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

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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^4+b*x^3+a*x^2),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 = 0.59, size = 447, normalized size = 4.30 \[ \frac {\ln \left (2\,a\,b^4+2\,b^5\,x+6\,a^3\,c^2+2\,a\,b^3\,\sqrt {b^2-4\,a\,c}+2\,b^4\,x\,\sqrt {b^2-4\,a\,c}-9\,a^2\,b^2\,c-10\,a\,b^3\,c\,x-3\,a^2\,b\,c\,\sqrt {b^2-4\,a\,c}+9\,a^2\,b\,c^2\,x+3\,a^2\,c^2\,x\,\sqrt {b^2-4\,a\,c}-6\,a\,b^2\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (\frac {b^4}{2}-a\,\left (\frac {5\,b^2\,c}{2}+\frac {3\,b\,c\,\sqrt {b^2-4\,a\,c}}{2}\right )+\frac {b^3\,\sqrt {b^2-4\,a\,c}}{2}+2\,a^2\,c^2\right )}{4\,a^4\,c-a^3\,b^2}-\frac {\ln \left (2\,a\,b^4+2\,b^5\,x+6\,a^3\,c^2-2\,a\,b^3\,\sqrt {b^2-4\,a\,c}-2\,b^4\,x\,\sqrt {b^2-4\,a\,c}-9\,a^2\,b^2\,c-10\,a\,b^3\,c\,x+3\,a^2\,b\,c\,\sqrt {b^2-4\,a\,c}+9\,a^2\,b\,c^2\,x-3\,a^2\,c^2\,x\,\sqrt {b^2-4\,a\,c}+6\,a\,b^2\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (a\,\left (\frac {5\,b^2\,c}{2}-\frac {3\,b\,c\,\sqrt {b^2-4\,a\,c}}{2}\right )-\frac {b^4}{2}+\frac {b^3\,\sqrt {b^2-4\,a\,c}}{2}-2\,a^2\,c^2\right )}{4\,a^4\,c-a^3\,b^2}-\frac {\frac {1}{2\,a}-\frac {b\,x}{a^2}}{x^2}-\frac {\ln \relax (x)\,\left (a\,c-b^2\right )}{a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a*x^2 + b*x^3 + c*x^4)),x)

[Out]

(log(2*a*b^4 + 2*b^5*x + 6*a^3*c^2 + 2*a*b^3*(b^2 - 4*a*c)^(1/2) + 2*b^4*x*(b^2 - 4*a*c)^(1/2) - 9*a^2*b^2*c -
10*a*b^3*c*x - 3*a^2*b*c*(b^2 - 4*a*c)^(1/2) + 9*a^2*b*c^2*x + 3*a^2*c^2*x*(b^2 - 4*a*c)^(1/2) - 6*a*b^2*c*x*
(b^2 - 4*a*c)^(1/2))*(b^4/2 - a*((5*b^2*c)/2 + (3*b*c*(b^2 - 4*a*c)^(1/2))/2) + (b^3*(b^2 - 4*a*c)^(1/2))/2 +
2*a^2*c^2))/(4*a^4*c - a^3*b^2) - (log(2*a*b^4 + 2*b^5*x + 6*a^3*c^2 - 2*a*b^3*(b^2 - 4*a*c)^(1/2) - 2*b^4*x*(
b^2 - 4*a*c)^(1/2) - 9*a^2*b^2*c - 10*a*b^3*c*x + 3*a^2*b*c*(b^2 - 4*a*c)^(1/2) + 9*a^2*b*c^2*x - 3*a^2*c^2*x*
(b^2 - 4*a*c)^(1/2) + 6*a*b^2*c*x*(b^2 - 4*a*c)^(1/2))*(a*((5*b^2*c)/2 - (3*b*c*(b^2 - 4*a*c)^(1/2))/2) - b^4/
2 + (b^3*(b^2 - 4*a*c)^(1/2))/2 - 2*a^2*c^2))/(4*a^4*c - a^3*b^2) - (1/(2*a) - (b*x)/a^2)/x^2 - (log(x)*(a*c -
b^2))/a^3

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(c*x**4+b*x**3+a*x**2),x)

[Out]

Timed out

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