Integrand size = 19, antiderivative size = 190 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x^5} \, dx=-\frac {b n}{12 a x^3}+\frac {\left (b^2-2 a c\right ) n}{8 a^2 x^2}-\frac {b \left (b^2-3 a c\right ) n}{4 a^3 x}-\frac {b \sqrt {b^2-4 a c} \left (b^2-2 a c\right ) n \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{4 a^4}-\frac {\left (b^4-4 a b^2 c+2 a^2 c^2\right ) n \log (x)}{4 a^4}+\frac {\left (b^4-4 a b^2 c+2 a^2 c^2\right ) n \log \left (a+b x+c x^2\right )}{8 a^4}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{4 x^4} \]
-1/12*b*n/a/x^3+1/8*(-2*a*c+b^2)*n/a^2/x^2-1/4*b*(-3*a*c+b^2)*n/a^3/x-1/4* (2*a^2*c^2-4*a*b^2*c+b^4)*n*ln(x)/a^4+1/8*(2*a^2*c^2-4*a*b^2*c+b^4)*n*ln(c *x^2+b*x+a)/a^4-1/4*ln(d*(c*x^2+b*x+a)^n)/x^4-1/4*b*(-2*a*c+b^2)*n*arctanh ((2*c*x+b)/(-4*a*c+b^2)^(1/2))*(-4*a*c+b^2)^(1/2)/a^4
Time = 0.29 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.91 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x^5} \, dx=-\frac {\frac {n x \left (2 a^3 b-3 a^2 \left (b^2-2 a c\right ) x+6 a b \left (b^2-3 a c\right ) x^2+6 b \sqrt {b^2-4 a c} \left (b^2-2 a c\right ) x^3 \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )+6 \left (b^4-4 a b^2 c+2 a^2 c^2\right ) x^3 \log (x)-3 \left (b^4-4 a b^2 c+2 a^2 c^2\right ) x^3 \log (a+x (b+c x))\right )}{a^4}+6 \log \left (d (a+x (b+c x))^n\right )}{24 x^4} \]
-1/24*((n*x*(2*a^3*b - 3*a^2*(b^2 - 2*a*c)*x + 6*a*b*(b^2 - 3*a*c)*x^2 + 6 *b*Sqrt[b^2 - 4*a*c]*(b^2 - 2*a*c)*x^3*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a* c]] + 6*(b^4 - 4*a*b^2*c + 2*a^2*c^2)*x^3*Log[x] - 3*(b^4 - 4*a*b^2*c + 2* a^2*c^2)*x^3*Log[a + x*(b + c*x)]))/a^4 + 6*Log[d*(a + x*(b + c*x))^n])/x^ 4
Time = 0.40 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3005, 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.
| \(\displaystyle \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x^5} \, dx\) |
| \(\Big \downarrow \) 3005 |
| \(\displaystyle \frac {1}{4} n \int \frac {b+2 c x}{x^4 \left (c x^2+b x+a\right )}dx-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{4 x^4}\) |
| \(\Big \downarrow \) 1200 |
| \(\displaystyle \frac {1}{4} n \int \left (\frac {b}{a x^4}+\frac {-b^4+4 a c b^2-2 a^2 c^2}{a^4 x}+\frac {b \left (b^4-5 a c b^2+5 a^2 c^2\right )+c \left (b^4-4 a c b^2+2 a^2 c^2\right ) x}{a^4 \left (c x^2+b x+a\right )}+\frac {b^3-3 a b c}{a^3 x^2}+\frac {2 a c-b^2}{a^2 x^3}\right )dx-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{4 x^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} n \left (-\frac {b \sqrt {b^2-4 a c} \left (b^2-2 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^4}-\frac {b \left (b^2-3 a c\right )}{a^3 x}+\frac {b^2-2 a c}{2 a^2 x^2}+\frac {\left (2 a^2 c^2-4 a b^2 c+b^4\right ) \log \left (a+b x+c x^2\right )}{2 a^4}-\frac {\log (x) \left (2 a^2 c^2-4 a b^2 c+b^4\right )}{a^4}-\frac {b}{3 a x^3}\right )-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{4 x^4}\) |
(n*(-1/3*b/(a*x^3) + (b^2 - 2*a*c)/(2*a^2*x^2) - (b*(b^2 - 3*a*c))/(a^3*x) - (b*Sqrt[b^2 - 4*a*c]*(b^2 - 2*a*c)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c ]])/a^4 - ((b^4 - 4*a*b^2*c + 2*a^2*c^2)*Log[x])/a^4 + ((b^4 - 4*a*b^2*c + 2*a^2*c^2)*Log[a + b*x + c*x^2])/(2*a^4)))/4 - Log[d*(a + b*x + c*x^2)^n] /(4*x^4)
3.1.80.3.1 Defintions of rubi rules used
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[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_. ), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((a + b*Log[c*RFx^p])^n/(e*(m + 1))) , x] - Simp[b*n*(p/(e*(m + 1))) Int[SimplifyIntegrand[(d + e*x)^(m + 1)*( a + b*Log[c*RFx^p])^(n - 1)*(D[RFx, x]/RFx), x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunctionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]
Time = 1.05 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.18
| method | result | size |
| parts | \(-\frac {\ln \left (d \left (c \,x^{2}+b x +a \right )^{n}\right )}{4 x^{4}}+\frac {n \left (-\frac {b}{3 a \,x^{3}}-\frac {2 c a -b^{2}}{2 a^{2} x^{2}}+\frac {\left (-2 c^{2} a^{2}+4 a \,b^{2} c -b^{4}\right ) \ln \left (x \right )}{a^{4}}+\frac {b \left (3 c a -b^{2}\right )}{a^{3} x}+\frac {\frac {\left (2 c^{3} a^{2}-4 a \,b^{2} c^{2}+b^{4} c \right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (5 a^{2} b \,c^{2}-5 a \,b^{3} c +b^{5}-\frac {\left (2 c^{3} a^{2}-4 a \,b^{2} c^{2}+b^{4} c \right ) b}{2 c}\right ) \arctan \left (\frac {2 x c +b}{\sqrt {4 c a -b^{2}}}\right )}{\sqrt {4 c a -b^{2}}}}{a^{4}}\right )}{4}\) | \(225\) |
| risch | \(\text {Expression too large to display}\) | \(3583\) |
-1/4*ln(d*(c*x^2+b*x+a)^n)/x^4+1/4*n*(-1/3*b/a/x^3-1/2*(2*a*c-b^2)/a^2/x^2 +1/a^4*(-2*a^2*c^2+4*a*b^2*c-b^4)*ln(x)+b*(3*a*c-b^2)/a^3/x+1/a^4*(1/2*(2* a^2*c^3-4*a*b^2*c^2+b^4*c)/c*ln(c*x^2+b*x+a)+2*(5*a^2*b*c^2-5*a*b^3*c+b^5- 1/2*(2*a^2*c^3-4*a*b^2*c^2+b^4*c)*b/c)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/ (4*a*c-b^2)^(1/2))))
Time = 0.36 (sec) , antiderivative size = 404, normalized size of antiderivative = 2.13 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x^5} \, dx=\left [-\frac {3 \, {\left (b^{3} - 2 \, a b c\right )} \sqrt {b^{2} - 4 \, a c} n x^{4} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + 6 \, {\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2}\right )} n x^{4} \log \left (x\right ) + 2 \, a^{3} b n x + 6 \, {\left (a b^{3} - 3 \, a^{2} b c\right )} n x^{3} + 6 \, a^{4} \log \left (d\right ) - 3 \, {\left (a^{2} b^{2} - 2 \, a^{3} c\right )} n x^{2} - 3 \, {\left ({\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2}\right )} n x^{4} - 2 \, a^{4} n\right )} \log \left (c x^{2} + b x + a\right )}{24 \, a^{4} x^{4}}, -\frac {6 \, {\left (b^{3} - 2 \, a b c\right )} \sqrt {-b^{2} + 4 \, a c} n x^{4} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 6 \, {\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2}\right )} n x^{4} \log \left (x\right ) + 2 \, a^{3} b n x + 6 \, {\left (a b^{3} - 3 \, a^{2} b c\right )} n x^{3} + 6 \, a^{4} \log \left (d\right ) - 3 \, {\left (a^{2} b^{2} - 2 \, a^{3} c\right )} n x^{2} - 3 \, {\left ({\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2}\right )} n x^{4} - 2 \, a^{4} n\right )} \log \left (c x^{2} + b x + a\right )}{24 \, a^{4} x^{4}}\right ] \]
[-1/24*(3*(b^3 - 2*a*b*c)*sqrt(b^2 - 4*a*c)*n*x^4*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)) + 6*(b^ 4 - 4*a*b^2*c + 2*a^2*c^2)*n*x^4*log(x) + 2*a^3*b*n*x + 6*(a*b^3 - 3*a^2*b *c)*n*x^3 + 6*a^4*log(d) - 3*(a^2*b^2 - 2*a^3*c)*n*x^2 - 3*((b^4 - 4*a*b^2 *c + 2*a^2*c^2)*n*x^4 - 2*a^4*n)*log(c*x^2 + b*x + a))/(a^4*x^4), -1/24*(6 *(b^3 - 2*a*b*c)*sqrt(-b^2 + 4*a*c)*n*x^4*arctan(-sqrt(-b^2 + 4*a*c)*(2*c* x + b)/(b^2 - 4*a*c)) + 6*(b^4 - 4*a*b^2*c + 2*a^2*c^2)*n*x^4*log(x) + 2*a ^3*b*n*x + 6*(a*b^3 - 3*a^2*b*c)*n*x^3 + 6*a^4*log(d) - 3*(a^2*b^2 - 2*a^3 *c)*n*x^2 - 3*((b^4 - 4*a*b^2*c + 2*a^2*c^2)*n*x^4 - 2*a^4*n)*log(c*x^2 + b*x + a))/(a^4*x^4)]
Timed out. \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x^5} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x^5} \, dx=\text {Exception raised: ValueError} \]
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 = 0.34 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.11 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x^5} \, dx=\frac {{\left (b^{4} n - 4 \, a b^{2} c n + 2 \, a^{2} c^{2} n\right )} \log \left (c x^{2} + b x + a\right )}{8 \, a^{4}} - \frac {n \log \left (c x^{2} + b x + a\right )}{4 \, x^{4}} - \frac {{\left (b^{4} n - 4 \, a b^{2} c n + 2 \, a^{2} c^{2} n\right )} \log \left (x\right )}{4 \, a^{4}} + \frac {{\left (b^{5} n - 6 \, a b^{3} c n + 8 \, a^{2} b c^{2} n\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{4 \, \sqrt {-b^{2} + 4 \, a c} a^{4}} - \frac {6 \, b^{3} n x^{3} - 18 \, a b c n x^{3} - 3 \, a b^{2} n x^{2} + 6 \, a^{2} c n x^{2} + 2 \, a^{2} b n x + 6 \, a^{3} \log \left (d\right )}{24 \, a^{3} x^{4}} \]
1/8*(b^4*n - 4*a*b^2*c*n + 2*a^2*c^2*n)*log(c*x^2 + b*x + a)/a^4 - 1/4*n*l og(c*x^2 + b*x + a)/x^4 - 1/4*(b^4*n - 4*a*b^2*c*n + 2*a^2*c^2*n)*log(x)/a ^4 + 1/4*(b^5*n - 6*a*b^3*c*n + 8*a^2*b*c^2*n)*arctan((2*c*x + b)/sqrt(-b^ 2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*a^4) - 1/24*(6*b^3*n*x^3 - 18*a*b*c*n*x^3 - 3*a*b^2*n*x^2 + 6*a^2*c*n*x^2 + 2*a^2*b*n*x + 6*a^3*log(d))/(a^3*x^4)
Time = 2.20 (sec) , antiderivative size = 627, normalized size of antiderivative = 3.30 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x^5} \, dx=\frac {\ln \left (2\,a\,b^6+2\,b^7\,x-12\,a^4\,c^3+2\,a\,b^5\,\sqrt {b^2-4\,a\,c}+2\,b^6\,x\,\sqrt {b^2-4\,a\,c}-15\,a^2\,b^4\,c+31\,a^3\,b^2\,c^2+37\,a^2\,b^3\,c^2\,x-16\,a\,b^5\,c\,x-20\,a^3\,b\,c^3\,x-9\,a^2\,b^3\,c\,\sqrt {b^2-4\,a\,c}+7\,a^3\,b\,c^2\,\sqrt {b^2-4\,a\,c}-6\,a^3\,c^3\,x\,\sqrt {b^2-4\,a\,c}-12\,a\,b^4\,c\,x\,\sqrt {b^2-4\,a\,c}+19\,a^2\,b^2\,c^2\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (\frac {b^4\,n}{8}-a\,\left (\frac {b^2\,c\,n}{2}+\frac {b\,c\,n\,\sqrt {b^2-4\,a\,c}}{4}\right )+\frac {b^3\,n\,\sqrt {b^2-4\,a\,c}}{8}+\frac {a^2\,c^2\,n}{4}\right )}{a^4}-\frac {\ln \left (d\,{\left (c\,x^2+b\,x+a\right )}^n\right )}{4\,x^4}-\frac {\ln \left (x\right )\,\left (2\,n\,a^2\,c^2-4\,n\,a\,b^2\,c+n\,b^4\right )}{4\,a^4}-\frac {\ln \left (12\,a^4\,c^3-2\,b^7\,x-2\,a\,b^6+2\,a\,b^5\,\sqrt {b^2-4\,a\,c}+2\,b^6\,x\,\sqrt {b^2-4\,a\,c}+15\,a^2\,b^4\,c-31\,a^3\,b^2\,c^2-37\,a^2\,b^3\,c^2\,x+16\,a\,b^5\,c\,x+20\,a^3\,b\,c^3\,x-9\,a^2\,b^3\,c\,\sqrt {b^2-4\,a\,c}+7\,a^3\,b\,c^2\,\sqrt {b^2-4\,a\,c}-6\,a^3\,c^3\,x\,\sqrt {b^2-4\,a\,c}-12\,a\,b^4\,c\,x\,\sqrt {b^2-4\,a\,c}+19\,a^2\,b^2\,c^2\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (a\,\left (\frac {b^2\,c\,n}{2}-\frac {b\,c\,n\,\sqrt {b^2-4\,a\,c}}{4}\right )-\frac {b^4\,n}{8}+\frac {b^3\,n\,\sqrt {b^2-4\,a\,c}}{8}-\frac {a^2\,c^2\,n}{4}\right )}{a^4}-\frac {\frac {b\,n}{3\,a}+\frac {n\,x\,\left (2\,a\,c-b^2\right )}{2\,a^2}-\frac {b\,n\,x^2\,\left (3\,a\,c-b^2\right )}{a^3}}{4\,x^3} \]
(log(2*a*b^6 + 2*b^7*x - 12*a^4*c^3 + 2*a*b^5*(b^2 - 4*a*c)^(1/2) + 2*b^6* x*(b^2 - 4*a*c)^(1/2) - 15*a^2*b^4*c + 31*a^3*b^2*c^2 + 37*a^2*b^3*c^2*x - 16*a*b^5*c*x - 20*a^3*b*c^3*x - 9*a^2*b^3*c*(b^2 - 4*a*c)^(1/2) + 7*a^3*b *c^2*(b^2 - 4*a*c)^(1/2) - 6*a^3*c^3*x*(b^2 - 4*a*c)^(1/2) - 12*a*b^4*c*x* (b^2 - 4*a*c)^(1/2) + 19*a^2*b^2*c^2*x*(b^2 - 4*a*c)^(1/2))*((b^4*n)/8 - a *((b^2*c*n)/2 + (b*c*n*(b^2 - 4*a*c)^(1/2))/4) + (b^3*n*(b^2 - 4*a*c)^(1/2 ))/8 + (a^2*c^2*n)/4))/a^4 - log(d*(a + b*x + c*x^2)^n)/(4*x^4) - (log(x)* (b^4*n + 2*a^2*c^2*n - 4*a*b^2*c*n))/(4*a^4) - (log(12*a^4*c^3 - 2*b^7*x - 2*a*b^6 + 2*a*b^5*(b^2 - 4*a*c)^(1/2) + 2*b^6*x*(b^2 - 4*a*c)^(1/2) + 15* a^2*b^4*c - 31*a^3*b^2*c^2 - 37*a^2*b^3*c^2*x + 16*a*b^5*c*x + 20*a^3*b*c^ 3*x - 9*a^2*b^3*c*(b^2 - 4*a*c)^(1/2) + 7*a^3*b*c^2*(b^2 - 4*a*c)^(1/2) - 6*a^3*c^3*x*(b^2 - 4*a*c)^(1/2) - 12*a*b^4*c*x*(b^2 - 4*a*c)^(1/2) + 19*a^ 2*b^2*c^2*x*(b^2 - 4*a*c)^(1/2))*(a*((b^2*c*n)/2 - (b*c*n*(b^2 - 4*a*c)^(1 /2))/4) - (b^4*n)/8 + (b^3*n*(b^2 - 4*a*c)^(1/2))/8 - (a^2*c^2*n)/4))/a^4 - ((b*n)/(3*a) + (n*x*(2*a*c - b^2))/(2*a^2) - (b*n*x^2*(3*a*c - b^2))/a^3 )/(4*x^3)