\(\int \frac {1}{a x^2+b x^3+c x^4} \, dx\) [16]

Optimal result

Integrand size = 18, antiderivative size = 81 \[ \int \frac {1}{a x^2+b x^3+c x^4} \, dx=-\frac {1}{a x}-\frac {\left (b^2-2 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {b^2-4 a c}}-\frac {b \log (x)}{a^2}+\frac {b \log \left (a+b x+c x^2\right )}{2 a^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {1608, 723, 814, 648, 632, 212, 642} \[ \int \frac {1}{a x^2+b x^3+c x^4} \, dx=-\frac {\left (b^2-2 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {b^2-4 a c}}+\frac {b \log \left (a+b x+c x^2\right )}{2 a^2}-\frac {b \log (x)}{a^2}-\frac {1}{a x} \]

[In]

Int[(a*x^2 + b*x^3 + c*x^4)^(-1),x]

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[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 723

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 814

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 1608

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

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.95 \[ \int \frac {1}{a x^2+b x^3+c x^4} \, dx=\frac {-\frac {2 a}{x}+\frac {2 \left (b^2-2 a c\right ) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}-2 b \log (x)+b \log (a+x (b+c x))}{2 a^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00

method	result	size
default	\(-\frac {1}{a x}-\frac {b \ln \left (x \right )}{a^{2}}+\frac {\frac {b \ln \left (c \,x^{2}+b x +a \right )}{2}+\frac {2 \left (-a c +\frac {b^{2}}{2}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{a^{2}}\)	\(81\)
risch	\(-\frac {1}{a x}-\frac {b \ln \left (x \right )}{a^{2}}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (4 a^{3} c -a^{2} b^{2}\right ) \textit {\_Z}^{2}+\left (-4 a b c +b^{3}\right ) \textit {\_Z} +c^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (6 a^{3} c -2 a^{2} b^{2}\right ) \textit {\_R}^{2}-2 \textit {\_R} a b c +c^{2}\right ) x -a^{3} b \,\textit {\_R}^{2}+\left (c \,a^{2}-b^{2} a \right ) \textit {\_R} +b c \right )\right )\)	\(117\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 269, normalized size of antiderivative = 3.32 \[ \int \frac {1}{a x^2+b x^3+c x^4} \, dx=\left [-\frac {{\left (b^{2} - 2 \, a c\right )} \sqrt {b^{2} - 4 \, a c} x \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + 2 \, a b^{2} - 8 \, a^{2} c - {\left (b^{3} - 4 \, a b c\right )} x \log \left (c x^{2} + b x + a\right ) + 2 \, {\left (b^{3} - 4 \, a b c\right )} x \log \left (x\right )}{2 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x}, -\frac {2 \, {\left (b^{2} - 2 \, a c\right )} \sqrt {-b^{2} + 4 \, a c} x \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 2 \, a b^{2} - 8 \, a^{2} c - {\left (b^{3} - 4 \, a b c\right )} x \log \left (c x^{2} + b x + a\right ) + 2 \, {\left (b^{3} - 4 \, a b c\right )} x \log \left (x\right )}{2 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x}\right ] \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{a x^2+b x^3+c x^4} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{a x^2+b x^3+c x^4} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(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 deta

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.98 \[ \int \frac {1}{a x^2+b x^3+c x^4} \, dx=\frac {b \log \left (c x^{2} + b x + a\right )}{2 \, a^{2}} - \frac {b \log \left ({\left | x \right |}\right )}{a^{2}} + \frac {{\left (b^{2} - 2 \, a c\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} a^{2}} - \frac {1}{a x} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.88 (sec) , antiderivative size = 339, normalized size of antiderivative = 4.19 \[ \int \frac {1}{a x^2+b x^3+c x^4} \, dx=\frac {\ln \left (2\,a\,b^3+2\,b^4\,x-2\,a\,b^2\,\sqrt {b^2-4\,a\,c}+a^2\,c\,\sqrt {b^2-4\,a\,c}-2\,b^3\,x\,\sqrt {b^2-4\,a\,c}+2\,a^2\,c^2\,x-7\,a^2\,b\,c-8\,a\,b^2\,c\,x+4\,a\,b\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (a\,\left (2\,b\,c-c\,\sqrt {b^2-4\,a\,c}\right )-\frac {b^3}{2}+\frac {b^2\,\sqrt {b^2-4\,a\,c}}{2}\right )}{4\,a^3\,c-a^2\,b^2}-\frac {1}{a\,x}-\frac {\ln \left (2\,a\,b^3+2\,b^4\,x+2\,a\,b^2\,\sqrt {b^2-4\,a\,c}-a^2\,c\,\sqrt {b^2-4\,a\,c}+2\,b^3\,x\,\sqrt {b^2-4\,a\,c}+2\,a^2\,c^2\,x-7\,a^2\,b\,c-8\,a\,b^2\,c\,x-4\,a\,b\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (\frac {b^3}{2}-a\,\left (2\,b\,c+c\,\sqrt {b^2-4\,a\,c}\right )+\frac {b^2\,\sqrt {b^2-4\,a\,c}}{2}\right )}{4\,a^3\,c-a^2\,b^2}-\frac {b\,\ln \left (x\right )}{a^2} \]

[In]

int(1/(a*x^2 + b*x^3 + c*x^4),x)

[Out]

(log(2*a*b^3 + 2*b^4*x - 2*a*b^2*(b^2 - 4*a*c)^(1/2) + a^2*c*(b^2 - 4*a*c)^(1/2) - 2*b^3*x*(b^2 - 4*a*c)^(1/2)
+ 2*a^2*c^2*x - 7*a^2*b*c - 8*a*b^2*c*x + 4*a*b*c*x*(b^2 - 4*a*c)^(1/2))*(a*(2*b*c - c*(b^2 - 4*a*c)^(1/2)) -
b^3/2 + (b^2*(b^2 - 4*a*c)^(1/2))/2))/(4*a^3*c - a^2*b^2) - 1/(a*x) - (log(2*a*b^3 + 2*b^4*x + 2*a*b^2*(b^2 -
4*a*c)^(1/2) - a^2*c*(b^2 - 4*a*c)^(1/2) + 2*b^3*x*(b^2 - 4*a*c)^(1/2) + 2*a^2*c^2*x - 7*a^2*b*c - 8*a*b^2*c*
x - 4*a*b*c*x*(b^2 - 4*a*c)^(1/2))*(b^3/2 - a*(2*b*c + c*(b^2 - 4*a*c)^(1/2)) + (b^2*(b^2 - 4*a*c)^(1/2))/2))/
(4*a^3*c - a^2*b^2) - (b*log(x))/a^2