Integrand size = 19, antiderivative size = 207 \[ \int x^4 \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=-\frac {\left (b^4-4 a b^2 c+2 a^2 c^2\right ) n x}{5 c^4}+\frac {b \left (b^2-3 a c\right ) n x^2}{10 c^3}-\frac {\left (b^2-2 a c\right ) n x^3}{15 c^2}+\frac {b n x^4}{20 c}-\frac {2 n x^5}{25}+\frac {\sqrt {b^2-4 a c} \left (b^4-3 a b^2 c+a^2 c^2\right ) n \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{5 c^5}+\frac {b \left (b^4-5 a b^2 c+5 a^2 c^2\right ) n \log \left (a+b x+c x^2\right )}{10 c^5}+\frac {1}{5} x^5 \log \left (d \left (a+b x+c x^2\right )^n\right ) \]
-1/5*(2*a^2*c^2-4*a*b^2*c+b^4)*n*x/c^4+1/10*b*(-3*a*c+b^2)*n*x^2/c^3-1/15* (-2*a*c+b^2)*n*x^3/c^2+1/20*b*n*x^4/c-2/25*n*x^5+1/10*b*(5*a^2*c^2-5*a*b^2 *c+b^4)*n*ln(c*x^2+b*x+a)/c^5+1/5*x^5*ln(d*(c*x^2+b*x+a)^n)+1/5*(a^2*c^2-3 *a*b^2*c+b^4)*n*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))*(-4*a*c+b^2)^(1/2)/c ^5
Time = 0.14 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.92 \[ \int x^4 \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=\frac {c n x \left (-60 b^4+30 b^3 c x-20 b^2 c \left (-12 a+c x^2\right )+15 b c^2 x \left (-6 a+c x^2\right )-8 c^2 \left (15 a^2-5 a c x^2+3 c^2 x^4\right )\right )+60 \sqrt {b^2-4 a c} \left (b^4-3 a b^2 c+a^2 c^2\right ) n \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )+30 b \left (b^4-5 a b^2 c+5 a^2 c^2\right ) n \log (a+x (b+c x))+60 c^5 x^5 \log \left (d (a+x (b+c x))^n\right )}{300 c^5} \]
(c*n*x*(-60*b^4 + 30*b^3*c*x - 20*b^2*c*(-12*a + c*x^2) + 15*b*c^2*x*(-6*a + c*x^2) - 8*c^2*(15*a^2 - 5*a*c*x^2 + 3*c^2*x^4)) + 60*Sqrt[b^2 - 4*a*c] *(b^4 - 3*a*b^2*c + a^2*c^2)*n*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]] + 30 *b*(b^4 - 5*a*b^2*c + 5*a^2*c^2)*n*Log[a + x*(b + c*x)] + 60*c^5*x^5*Log[d *(a + x*(b + c*x))^n])/(300*c^5)
Time = 0.40 (sec) , antiderivative size = 201, 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.
| \(\displaystyle \int x^4 \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx\) |
| \(\Big \downarrow \) 3005 |
| \(\displaystyle \frac {1}{5} x^5 \log \left (d \left (a+b x+c x^2\right )^n\right )-\frac {1}{5} n \int \frac {x^5 (b+2 c x)}{c x^2+b x+a}dx\) |
| \(\Big \downarrow \) 1200 |
| \(\displaystyle \frac {1}{5} x^5 \log \left (d \left (a+b x+c x^2\right )^n\right )-\frac {1}{5} n \int \left (2 x^4-\frac {b x^3}{c}+\frac {\left (b^2-2 a c\right ) x^2}{c^2}-\frac {b \left (b^2-3 a c\right ) x}{c^3}+\frac {b^4-4 a c b^2+2 a^2 c^2}{c^4}-\frac {a \left (b^4-4 a c b^2+2 a^2 c^2\right )+b \left (b^4-5 a c b^2+5 a^2 c^2\right ) x}{c^4 \left (c x^2+b x+a\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} x^5 \log \left (d \left (a+b x+c x^2\right )^n\right )-\frac {1}{5} n \left (-\frac {\sqrt {b^2-4 a c} \left (a^2 c^2-3 a b^2 c+b^4\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^5}-\frac {b \left (5 a^2 c^2-5 a b^2 c+b^4\right ) \log \left (a+b x+c x^2\right )}{2 c^5}+\frac {x \left (2 a^2 c^2-4 a b^2 c+b^4\right )}{c^4}-\frac {b x^2 \left (b^2-3 a c\right )}{2 c^3}+\frac {x^3 \left (b^2-2 a c\right )}{3 c^2}-\frac {b x^4}{4 c}+\frac {2 x^5}{5}\right )\) |
-1/5*(n*(((b^4 - 4*a*b^2*c + 2*a^2*c^2)*x)/c^4 - (b*(b^2 - 3*a*c)*x^2)/(2* c^3) + ((b^2 - 2*a*c)*x^3)/(3*c^2) - (b*x^4)/(4*c) + (2*x^5)/5 - (Sqrt[b^2 - 4*a*c]*(b^4 - 3*a*b^2*c + a^2*c^2)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c ]])/c^5 - (b*(b^4 - 5*a*b^2*c + 5*a^2*c^2)*Log[a + b*x + c*x^2])/(2*c^5))) + (x^5*Log[d*(a + b*x + c*x^2)^n])/5
3.1.71.3.1 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 = 1.06 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.15
| method | result | size |
| parts | \(\frac {x^{5} \ln \left (d \left (c \,x^{2}+b x +a \right )^{n}\right )}{5}-\frac {n \left (\frac {\frac {2}{5} c^{4} x^{5}-\frac {1}{4} b \,x^{4} c^{3}-\frac {2}{3} a \,c^{3} x^{3}+\frac {1}{3} b^{2} c^{2} x^{3}+\frac {3}{2} a b \,c^{2} x^{2}-\frac {1}{2} b^{3} c \,x^{2}+2 a^{2} x \,c^{2}-4 a \,b^{2} c x +b^{4} x}{c^{4}}+\frac {\frac {\left (-5 a^{2} b \,c^{2}+5 a \,b^{3} c -b^{5}\right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (-2 c^{2} a^{3}+4 a^{2} b^{2} c -b^{4} a -\frac {\left (-5 a^{2} b \,c^{2}+5 a \,b^{3} c -b^{5}\right ) b}{2 c}\right ) \arctan \left (\frac {2 x c +b}{\sqrt {4 c a -b^{2}}}\right )}{\sqrt {4 c a -b^{2}}}}{c^{4}}\right )}{5}\) | \(238\) |
| risch | \(\text {Expression too large to display}\) | \(1621\) |
1/5*x^5*ln(d*(c*x^2+b*x+a)^n)-1/5*n*(1/c^4*(2/5*c^4*x^5-1/4*b*x^4*c^3-2/3* a*c^3*x^3+1/3*b^2*c^2*x^3+3/2*a*b*c^2*x^2-1/2*b^3*c*x^2+2*a^2*x*c^2-4*a*b^ 2*c*x+b^4*x)+1/c^4*(1/2*(-5*a^2*b*c^2+5*a*b^3*c-b^5)/c*ln(c*x^2+b*x+a)+2*( -2*c^2*a^3+4*a^2*b^2*c-b^4*a-1/2*(-5*a^2*b*c^2+5*a*b^3*c-b^5)*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 = 444, normalized size of antiderivative = 2.14 \[ \int x^4 \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=\left [-\frac {24 \, c^{5} n x^{5} - 60 \, c^{5} x^{5} \log \left (d\right ) - 15 \, b c^{4} n x^{4} + 20 \, {\left (b^{2} c^{3} - 2 \, a c^{4}\right )} n x^{3} - 30 \, {\left (b^{3} c^{2} - 3 \, a b c^{3}\right )} n x^{2} - 30 \, {\left (b^{4} - 3 \, a b^{2} c + a^{2} c^{2}\right )} \sqrt {b^{2} - 4 \, a c} n \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + 60 \, {\left (b^{4} c - 4 \, a b^{2} c^{2} + 2 \, a^{2} c^{3}\right )} n x - 30 \, {\left (2 \, c^{5} n x^{5} + {\left (b^{5} - 5 \, a b^{3} c + 5 \, a^{2} b c^{2}\right )} n\right )} \log \left (c x^{2} + b x + a\right )}{300 \, c^{5}}, -\frac {24 \, c^{5} n x^{5} - 60 \, c^{5} x^{5} \log \left (d\right ) - 15 \, b c^{4} n x^{4} + 20 \, {\left (b^{2} c^{3} - 2 \, a c^{4}\right )} n x^{3} - 30 \, {\left (b^{3} c^{2} - 3 \, a b c^{3}\right )} n x^{2} - 60 \, {\left (b^{4} - 3 \, a b^{2} c + a^{2} c^{2}\right )} \sqrt {-b^{2} + 4 \, a c} n \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 60 \, {\left (b^{4} c - 4 \, a b^{2} c^{2} + 2 \, a^{2} c^{3}\right )} n x - 30 \, {\left (2 \, c^{5} n x^{5} + {\left (b^{5} - 5 \, a b^{3} c + 5 \, a^{2} b c^{2}\right )} n\right )} \log \left (c x^{2} + b x + a\right )}{300 \, c^{5}}\right ] \]
[-1/300*(24*c^5*n*x^5 - 60*c^5*x^5*log(d) - 15*b*c^4*n*x^4 + 20*(b^2*c^3 - 2*a*c^4)*n*x^3 - 30*(b^3*c^2 - 3*a*b*c^3)*n*x^2 - 30*(b^4 - 3*a*b^2*c + a ^2*c^2)*sqrt(b^2 - 4*a*c)*n*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)) + 60*(b^4*c - 4*a*b^2*c^2 + 2 *a^2*c^3)*n*x - 30*(2*c^5*n*x^5 + (b^5 - 5*a*b^3*c + 5*a^2*b*c^2)*n)*log(c *x^2 + b*x + a))/c^5, -1/300*(24*c^5*n*x^5 - 60*c^5*x^5*log(d) - 15*b*c^4* n*x^4 + 20*(b^2*c^3 - 2*a*c^4)*n*x^3 - 30*(b^3*c^2 - 3*a*b*c^3)*n*x^2 - 60 *(b^4 - 3*a*b^2*c + a^2*c^2)*sqrt(-b^2 + 4*a*c)*n*arctan(-sqrt(-b^2 + 4*a* c)*(2*c*x + b)/(b^2 - 4*a*c)) + 60*(b^4*c - 4*a*b^2*c^2 + 2*a^2*c^3)*n*x - 30*(2*c^5*n*x^5 + (b^5 - 5*a*b^3*c + 5*a^2*b*c^2)*n)*log(c*x^2 + b*x + a) )/c^5]
Timed out. \[ \int x^4 \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=\text {Timed out} \]
Exception generated. \[ \int x^4 \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=\text {Exception raised: ValueError} \]
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 = 0.42 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.07 \[ \int x^4 \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=\frac {1}{5} \, n x^{5} \log \left (c x^{2} + b x + a\right ) - \frac {1}{25} \, {\left (2 \, n - 5 \, \log \left (d\right )\right )} x^{5} + \frac {b n x^{4}}{20 \, c} - \frac {{\left (b^{2} n - 2 \, a c n\right )} x^{3}}{15 \, c^{2}} + \frac {{\left (b^{3} n - 3 \, a b c n\right )} x^{2}}{10 \, c^{3}} - \frac {{\left (b^{4} n - 4 \, a b^{2} c n + 2 \, a^{2} c^{2} n\right )} x}{5 \, c^{4}} + \frac {{\left (b^{5} n - 5 \, a b^{3} c n + 5 \, a^{2} b c^{2} n\right )} \log \left (c x^{2} + b x + a\right )}{10 \, c^{5}} - \frac {{\left (b^{6} n - 7 \, a b^{4} c n + 13 \, a^{2} b^{2} c^{2} n - 4 \, a^{3} c^{3} n\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{5 \, \sqrt {-b^{2} + 4 \, a c} c^{5}} \]
1/5*n*x^5*log(c*x^2 + b*x + a) - 1/25*(2*n - 5*log(d))*x^5 + 1/20*b*n*x^4/ c - 1/15*(b^2*n - 2*a*c*n)*x^3/c^2 + 1/10*(b^3*n - 3*a*b*c*n)*x^2/c^3 - 1/ 5*(b^4*n - 4*a*b^2*c*n + 2*a^2*c^2*n)*x/c^4 + 1/10*(b^5*n - 5*a*b^3*c*n + 5*a^2*b*c^2*n)*log(c*x^2 + b*x + a)/c^5 - 1/5*(b^6*n - 7*a*b^4*c*n + 13*a^ 2*b^2*c^2*n - 4*a^3*c^3*n)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b ^2 + 4*a*c)*c^5)
Time = 1.65 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.91 \[ \int x^4 \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=x^2\,\left (\frac {b\,\left (\frac {b^2\,n}{5\,c^2}-\frac {2\,a\,n}{5\,c}\right )}{2\,c}-\frac {a\,b\,n}{10\,c^2}\right )-\frac {2\,n\,x^5}{25}+x\,\left (\frac {a\,\left (\frac {b^2\,n}{5\,c^2}-\frac {2\,a\,n}{5\,c}\right )}{c}-\frac {b\,\left (\frac {b\,\left (\frac {b^2\,n}{5\,c^2}-\frac {2\,a\,n}{5\,c}\right )}{c}-\frac {a\,b\,n}{5\,c^2}\right )}{c}\right )+\frac {x^5\,\ln \left (d\,{\left (c\,x^2+b\,x+a\right )}^n\right )}{5}-x^3\,\left (\frac {b^2\,n}{15\,c^2}-\frac {2\,a\,n}{15\,c}\right )+\frac {\ln \left (b\,\sqrt {b^2-4\,a\,c}-4\,a\,c+b^2+2\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (\frac {b^5\,n}{10}+c^2\,\left (\frac {a^2\,n\,\sqrt {b^2-4\,a\,c}}{10}+\frac {a^2\,b\,n}{2}\right )-c\,\left (\frac {a\,b^3\,n}{2}+\frac {3\,a\,b^2\,n\,\sqrt {b^2-4\,a\,c}}{10}\right )+\frac {b^4\,n\,\sqrt {b^2-4\,a\,c}}{10}\right )}{c^5}-\frac {\ln \left (4\,a\,c+b\,\sqrt {b^2-4\,a\,c}-b^2+2\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (c^2\,\left (\frac {a^2\,n\,\sqrt {b^2-4\,a\,c}}{10}-\frac {a^2\,b\,n}{2}\right )-\frac {b^5\,n}{10}+c\,\left (\frac {a\,b^3\,n}{2}-\frac {3\,a\,b^2\,n\,\sqrt {b^2-4\,a\,c}}{10}\right )+\frac {b^4\,n\,\sqrt {b^2-4\,a\,c}}{10}\right )}{c^5}+\frac {b\,n\,x^4}{20\,c} \]
x^2*((b*((b^2*n)/(5*c^2) - (2*a*n)/(5*c)))/(2*c) - (a*b*n)/(10*c^2)) - (2* n*x^5)/25 + x*((a*((b^2*n)/(5*c^2) - (2*a*n)/(5*c)))/c - (b*((b*((b^2*n)/( 5*c^2) - (2*a*n)/(5*c)))/c - (a*b*n)/(5*c^2)))/c) + (x^5*log(d*(a + b*x + c*x^2)^n))/5 - x^3*((b^2*n)/(15*c^2) - (2*a*n)/(15*c)) + (log(b*(b^2 - 4*a *c)^(1/2) - 4*a*c + b^2 + 2*c*x*(b^2 - 4*a*c)^(1/2))*((b^5*n)/10 + c^2*((a ^2*n*(b^2 - 4*a*c)^(1/2))/10 + (a^2*b*n)/2) - c*((a*b^3*n)/2 + (3*a*b^2*n* (b^2 - 4*a*c)^(1/2))/10) + (b^4*n*(b^2 - 4*a*c)^(1/2))/10))/c^5 - (log(4*a *c + b*(b^2 - 4*a*c)^(1/2) - b^2 + 2*c*x*(b^2 - 4*a*c)^(1/2))*(c^2*((a^2*n *(b^2 - 4*a*c)^(1/2))/10 - (a^2*b*n)/2) - (b^5*n)/10 + c*((a*b^3*n)/2 - (3 *a*b^2*n*(b^2 - 4*a*c)^(1/2))/10) + (b^4*n*(b^2 - 4*a*c)^(1/2))/10))/c^5 + (b*n*x^4)/(20*c)