Integrand size = 18, antiderivative size = 81 \[ \int \frac {1}{a x^2+b x^3+c x^4} \, dx=-\frac {1}{a x}-\frac {\left (b^2-2 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {b^2-4 a c}}-\frac {b \log (x)}{a^2}+\frac {b \log \left (a+b x+c x^2\right )}{2 a^2} \] Output:
-1/a/x-(-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)-b*ln(x)/a^2+1/2*b*ln(c*x^2+b*x+a)/a^2
Time = 0.09 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.95 \[ \int \frac {1}{a x^2+b x^3+c x^4} \, dx=\frac {-\frac {2 a}{x}+\frac {2 \left (b^2-2 a c\right ) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}-2 b \log (x)+b \log (a+x (b+c x))}{2 a^2} \] Input:
Integrate[(a*x^2 + b*x^3 + c*x^4)^(-1),x]
Output:
((-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)
Time = 0.30 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1949, 1145, 25, 1200, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{a x^2+b x^3+c x^4} \, dx\) |
\(\Big \downarrow \) 1949 |
\(\displaystyle \int \frac {1}{x^2 \left (a+b x+c x^2\right )}dx\) |
\(\Big \downarrow \) 1145 |
\(\displaystyle \frac {\int -\frac {b+c x}{x \left (c x^2+b x+a\right )}dx}{a}-\frac {1}{a x}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {b+c x}{x \left (c x^2+b x+a\right )}dx}{a}-\frac {1}{a x}\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle -\frac {\int \left (\frac {b}{a x}+\frac {-b^2-c x b+a c}{a \left (c x^2+b x+a\right )}\right )dx}{a}-\frac {1}{a x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {\left (b^2-2 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a \sqrt {b^2-4 a c}}-\frac {b \log \left (a+b x+c x^2\right )}{2 a}+\frac {b \log (x)}{a}}{a}-\frac {1}{a x}\) |
Input:
Int[(a*x^2 + b*x^3 + c*x^4)^(-1),x]
Output:
-(1/(a*x)) - (((b^2 - 2*a*c)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(a*Sq rt[b^2 - 4*a*c]) + (b*Log[x])/a - (b*Log[a + b*x + c*x^2])/(2*a))/a
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] + Simp [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}, x] && ILtQ[m, -1]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In tegersQ[n]
Int[((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol ] :> Int[x^(p*q)*(a + b*x^(n - q) + c*x^(2*(n - q)))^p, x] /; FreeQ[{a, b, c, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q] && IntegerQ[p]
Time = 0.10 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00
method | result | size |
default | \(\frac {\frac {b \ln \left (c \,x^{2}+b x +a \right )}{2}+\frac {2 \left (-a c +\frac {b^{2}}{2}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{a^{2}}-\frac {1}{a x}-\frac {b \ln \left (x \right )}{a^{2}}\) | \(81\) |
risch | \(\text {Expression too large to display}\) | \(1295\) |
Input:
int(1/(c*x^4+b*x^3+a*x^2),x,method=_RETURNVERBOSE)
Output:
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)))-1/a/x-b*ln(x)/a^2
Time = 0.09 (sec) , antiderivative size = 269, normalized size of antiderivative = 3.32 \[ \int \frac {1}{a x^2+b x^3+c x^4} \, dx=\left [-\frac {{\left (b^{2} - 2 \, a c\right )} \sqrt {b^{2} - 4 \, a c} x \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + 2 \, a b^{2} - 8 \, a^{2} c - {\left (b^{3} - 4 \, a b c\right )} x \log \left (c x^{2} + b x + a\right ) + 2 \, {\left (b^{3} - 4 \, a b c\right )} x \log \left (x\right )}{2 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x}, -\frac {2 \, {\left (b^{2} - 2 \, a c\right )} \sqrt {-b^{2} + 4 \, a c} x \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 2 \, a b^{2} - 8 \, a^{2} c - {\left (b^{3} - 4 \, a b c\right )} x \log \left (c x^{2} + b x + a\right ) + 2 \, {\left (b^{3} - 4 \, a b c\right )} x \log \left (x\right )}{2 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x}\right ] \] Input:
integrate(1/(c*x^4+b*x^3+a*x^2),x, algorithm="fricas")
Output:
[-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*arcta n(-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)]
Leaf count of result is larger than twice the leaf count of optimal. 862 vs. \(2 (75) = 150\).
Time = 156.71 (sec) , antiderivative size = 862, normalized size of antiderivative = 10.64 \[ \int \frac {1}{a x^2+b x^3+c x^4} \, dx =\text {Too large to display} \] Input:
integrate(1/(c*x**4+b*x**3+a*x**2),x)
Output:
(b/(2*a**2) - sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*a**2*(4*a*c - b**2)))* log(x + (-28*a**6*b*c**2*(b/(2*a**2) - sqrt(-4*a*c + b**2)*(2*a*c - b**2)/ (2*a**2*(4*a*c - b**2)))**2 + 15*a**5*b**3*c*(b/(2*a**2) - sqrt(-4*a*c + b **2)*(2*a*c - b**2)/(2*a**2*(4*a*c - b**2)))**2 - 4*a**5*c**3*(b/(2*a**2) - sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*a**2*(4*a*c - b**2))) - 2*a**4*b** 5*(b/(2*a**2) - sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*a**2*(4*a*c - b**2)) )**2 - 3*a**4*b**2*c**2*(b/(2*a**2) - sqrt(-4*a*c + b**2)*(2*a*c - b**2)/( 2*a**2*(4*a*c - b**2))) + a**3*b**4*c*(b/(2*a**2) - sqrt(-4*a*c + b**2)*(2 *a*c - b**2)/(2*a**2*(4*a*c - b**2))) - 4*a**3*b*c**3 + 25*a**2*b**3*c**2 - 14*a*b**5*c + 2*b**7)/(2*a**3*c**4 + 15*a**2*b**2*c**3 - 12*a*b**4*c**2 + 2*b**6*c)) + (b/(2*a**2) + sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*a**2*(4 *a*c - b**2)))*log(x + (-28*a**6*b*c**2*(b/(2*a**2) + sqrt(-4*a*c + b**2)* (2*a*c - b**2)/(2*a**2*(4*a*c - b**2)))**2 + 15*a**5*b**3*c*(b/(2*a**2) + sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*a**2*(4*a*c - b**2)))**2 - 4*a**5*c* *3*(b/(2*a**2) + sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*a**2*(4*a*c - b**2) )) - 2*a**4*b**5*(b/(2*a**2) + sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*a**2* (4*a*c - b**2)))**2 - 3*a**4*b**2*c**2*(b/(2*a**2) + sqrt(-4*a*c + b**2)*( 2*a*c - b**2)/(2*a**2*(4*a*c - b**2))) + a**3*b**4*c*(b/(2*a**2) + sqrt(-4 *a*c + b**2)*(2*a*c - b**2)/(2*a**2*(4*a*c - b**2))) - 4*a**3*b*c**3 + 25* a**2*b**3*c**2 - 14*a*b**5*c + 2*b**7)/(2*a**3*c**4 + 15*a**2*b**2*c**3...
Exception generated. \[ \int \frac {1}{a x^2+b x^3+c x^4} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(c*x^4+b*x^3+a*x^2),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.98 \[ \int \frac {1}{a x^2+b x^3+c x^4} \, dx=\frac {b \log \left (c x^{2} + b x + a\right )}{2 \, a^{2}} - \frac {b \log \left ({\left | x \right |}\right )}{a^{2}} + \frac {{\left (b^{2} - 2 \, a c\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} a^{2}} - \frac {1}{a x} \] Input:
integrate(1/(c*x^4+b*x^3+a*x^2),x, algorithm="giac")
Output:
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))/(sqrt(-b^2 + 4*a*c)*a^2) - 1/(a*x)
Time = 22.27 (sec) , antiderivative size = 339, normalized size of antiderivative = 4.19 \[ \int \frac {1}{a x^2+b x^3+c x^4} \, dx=\frac {\ln \left (2\,a\,b^3+2\,b^4\,x-2\,a\,b^2\,\sqrt {b^2-4\,a\,c}+a^2\,c\,\sqrt {b^2-4\,a\,c}-2\,b^3\,x\,\sqrt {b^2-4\,a\,c}+2\,a^2\,c^2\,x-7\,a^2\,b\,c-8\,a\,b^2\,c\,x+4\,a\,b\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (a\,\left (2\,b\,c-c\,\sqrt {b^2-4\,a\,c}\right )-\frac {b^3}{2}+\frac {b^2\,\sqrt {b^2-4\,a\,c}}{2}\right )}{4\,a^3\,c-a^2\,b^2}-\frac {1}{a\,x}-\frac {\ln \left (2\,a\,b^3+2\,b^4\,x+2\,a\,b^2\,\sqrt {b^2-4\,a\,c}-a^2\,c\,\sqrt {b^2-4\,a\,c}+2\,b^3\,x\,\sqrt {b^2-4\,a\,c}+2\,a^2\,c^2\,x-7\,a^2\,b\,c-8\,a\,b^2\,c\,x-4\,a\,b\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (\frac {b^3}{2}-a\,\left (2\,b\,c+c\,\sqrt {b^2-4\,a\,c}\right )+\frac {b^2\,\sqrt {b^2-4\,a\,c}}{2}\right )}{4\,a^3\,c-a^2\,b^2}-\frac {b\,\ln \left (x\right )}{a^2} \] Input:
int(1/(a*x^2 + b*x^3 + c*x^4),x)
Output:
(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
Time = 0.20 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.95 \[ \int \frac {1}{a x^2+b x^3+c x^4} \, dx=\frac {-4 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) a c x +2 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) b^{2} x +4 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) a b c x -\mathrm {log}\left (c \,x^{2}+b x +a \right ) b^{3} x -8 \,\mathrm {log}\left (x \right ) a b c x +2 \,\mathrm {log}\left (x \right ) b^{3} x -8 a^{2} c +2 a \,b^{2}}{2 a^{2} x \left (4 a c -b^{2}\right )} \] Input:
int(1/(c*x^4+b*x^3+a*x^2),x)
Output:
( - 4*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*c*x + 2*sq rt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b**2*x + 4*log(a + b *x + c*x**2)*a*b*c*x - log(a + b*x + c*x**2)*b**3*x - 8*log(x)*a*b*c*x + 2 *log(x)*b**3*x - 8*a**2*c + 2*a*b**2)/(2*a**2*x*(4*a*c - b**2))