Integrand size = 18, antiderivative size = 104 \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right ) x^5} \, dx=-\frac {1}{2 a x^2}+\frac {b}{a^2 x}+\frac {b \left (b^2-3 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^3 \sqrt {b^2-4 a c}}+\frac {\left (b^2-a c\right ) \log (x)}{a^3}-\frac {\left (b^2-a c\right ) \log \left (a+b x+c x^2\right )}{2 a^3} \]
-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)
Time = 0.10 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right ) x^5} \, dx=\frac {-\frac {a^2}{x^2}+\frac {2 a b}{x}-\frac {2 b \left (b^2-3 a c\right ) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}+2 \left (b^2-a c\right ) \log (x)+\left (-b^2+a c\right ) \log (a+x (b+c x))}{2 a^3} \]
(-(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)
Time = 0.34 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1692, 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.
\(\displaystyle \int \frac {1}{x^5 \left (\frac {a}{x^2}+\frac {b}{x}+c\right )} \, dx\) |
\(\Big \downarrow \) 1692 |
\(\displaystyle \int \frac {1}{x^3 \left (a+b x+c x^2\right )}dx\) |
\(\Big \downarrow \) 1145 |
\(\displaystyle \frac {\int -\frac {b+c x}{x^2 \left (c x^2+b x+a\right )}dx}{a}-\frac {1}{2 a x^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {b+c x}{x^2 \left (c x^2+b x+a\right )}dx}{a}-\frac {1}{2 a x^2}\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle -\frac {\int \left (\frac {b}{a x^2}+\frac {a c-b^2}{a^2 x}+\frac {b \left (b^2-2 a c\right )+c \left (b^2-a c\right ) x}{a^2 \left (c x^2+b x+a\right )}\right )dx}{a}-\frac {1}{2 a x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {b \left (b^2-3 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {b^2-4 a c}}+\frac {\left (b^2-a c\right ) \log \left (a+b x+c x^2\right )}{2 a^2}-\frac {\log (x) \left (b^2-a c\right )}{a^2}-\frac {b}{a x}}{a}-\frac {1}{2 a x^2}\) |
-1/2*1/(a*x^2) - (-(b/(a*x)) - (b*(b^2 - 3*a*c)*ArcTanh[(b + 2*c*x)/Sqrt[b ^2 - 4*a*c]])/(a^2*Sqrt[b^2 - 4*a*c]) - ((b^2 - a*c)*Log[x])/a^2 + ((b^2 - a*c)*Log[a + b*x + c*x^2])/(2*a^2))/a
3.5.20.3.1 Defintions of rubi rules used
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[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + 2*n*p)*(c + b/x^n + a/x^(2*n))^p, x] /; FreeQ[{a, b, c, m, n }, x] && EqQ[n2, 2*n] && ILtQ[p, 0] && NegQ[n]
Time = 0.12 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.23
method | result | size |
default | \(-\frac {1}{2 a \,x^{2}}+\frac {\left (-a c +b^{2}\right ) \ln \left (x \right )}{a^{3}}+\frac {b}{a^{2} x}+\frac {\frac {\left (a \,c^{2}-b^{2} c \right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (2 a b c -b^{3}-\frac {\left (a \,c^{2}-b^{2} c \right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{a^{3}}\) | \(128\) |
risch | \(\text {Expression too large to display}\) | \(2265\) |
-1/2/a/x^2+(-a*c+b^2)*ln(x)/a^3+b/a^2/x+1/a^3*(1/2*(a*c^2-b^2*c)/c*ln(c*x^ 2+b*x+a)+2*(2*a*b*c-b^3-1/2*(a*c^2-b^2*c)*b/c)/(4*a*c-b^2)^(1/2)*arctan((2 *c*x+b)/(4*a*c-b^2)^(1/2)))
Time = 0.32 (sec) , antiderivative size = 358, normalized size of antiderivative = 3.44 \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right ) x^5} \, dx=\left [-\frac {{\left (b^{3} - 3 \, a b c\right )} \sqrt {b^{2} - 4 \, a c} x^{2} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c - \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + a^{2} b^{2} - 4 \, a^{3} c + {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} x^{2} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} x^{2} \log \left (x\right ) - 2 \, {\left (a b^{3} - 4 \, a^{2} b c\right )} x}{2 \, {\left (a^{3} b^{2} - 4 \, a^{4} c\right )} x^{2}}, \frac {2 \, {\left (b^{3} - 3 \, a b c\right )} \sqrt {-b^{2} + 4 \, a c} x^{2} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) - a^{2} b^{2} + 4 \, a^{3} c - {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} x^{2} \log \left (c x^{2} + b x + a\right ) + 2 \, {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} x^{2} \log \left (x\right ) + 2 \, {\left (a b^{3} - 4 \, a^{2} b c\right )} x}{2 \, {\left (a^{3} b^{2} - 4 \, a^{4} c\right )} x^{2}}\right ] \]
[-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)]
Timed out. \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right ) x^5} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right ) x^5} \, dx=\text {Exception raised: ValueError} \]
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 = 0.31 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.01 \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right ) x^5} \, dx=-\frac {{\left (b^{2} - a c\right )} \log \left (c x^{2} + b x + a\right )}{2 \, a^{3}} + \frac {{\left (b^{2} - a c\right )} \log \left ({\left | x \right |}\right )}{a^{3}} - \frac {{\left (b^{3} - 3 \, a b c\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} a^{3}} + \frac {2 \, a b x - a^{2}}{2 \, a^{3} x^{2}} \]
-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)/sqrt(-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)
Time = 8.59 (sec) , antiderivative size = 447, normalized size of antiderivative = 4.30 \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right ) x^5} \, dx=\frac {\ln \left (2\,a\,b^4+2\,b^5\,x+6\,a^3\,c^2+2\,a\,b^3\,\sqrt {b^2-4\,a\,c}+2\,b^4\,x\,\sqrt {b^2-4\,a\,c}-9\,a^2\,b^2\,c-10\,a\,b^3\,c\,x-3\,a^2\,b\,c\,\sqrt {b^2-4\,a\,c}+9\,a^2\,b\,c^2\,x+3\,a^2\,c^2\,x\,\sqrt {b^2-4\,a\,c}-6\,a\,b^2\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (\frac {b^4}{2}-a\,\left (\frac {5\,b^2\,c}{2}+\frac {3\,b\,c\,\sqrt {b^2-4\,a\,c}}{2}\right )+\frac {b^3\,\sqrt {b^2-4\,a\,c}}{2}+2\,a^2\,c^2\right )}{4\,a^4\,c-a^3\,b^2}-\frac {\ln \left (2\,a\,b^4+2\,b^5\,x+6\,a^3\,c^2-2\,a\,b^3\,\sqrt {b^2-4\,a\,c}-2\,b^4\,x\,\sqrt {b^2-4\,a\,c}-9\,a^2\,b^2\,c-10\,a\,b^3\,c\,x+3\,a^2\,b\,c\,\sqrt {b^2-4\,a\,c}+9\,a^2\,b\,c^2\,x-3\,a^2\,c^2\,x\,\sqrt {b^2-4\,a\,c}+6\,a\,b^2\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (a\,\left (\frac {5\,b^2\,c}{2}-\frac {3\,b\,c\,\sqrt {b^2-4\,a\,c}}{2}\right )-\frac {b^4}{2}+\frac {b^3\,\sqrt {b^2-4\,a\,c}}{2}-2\,a^2\,c^2\right )}{4\,a^4\,c-a^3\,b^2}-\frac {\frac {1}{2\,a}-\frac {b\,x}{a^2}}{x^2}-\frac {\ln \left (x\right )\,\left (a\,c-b^2\right )}{a^3} \]
(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