Integrand size = 25, antiderivative size = 500 \[ \int \frac {\log ^3\left (c (a+b x)^n\right )}{d+e x+f x^2} \, dx=\frac {\log ^3\left (c (a+b x)^n\right ) \log \left (-\frac {b \left (e-\sqrt {e^2-4 d f}+2 f x\right )}{2 a f-b \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}-\frac {\log ^3\left (c (a+b x)^n\right ) \log \left (-\frac {b \left (e+\sqrt {e^2-4 d f}+2 f x\right )}{2 a f-b \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}+\frac {3 n \log ^2\left (c (a+b x)^n\right ) \operatorname {PolyLog}\left (2,\frac {2 f (a+b x)}{2 a f-b \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}-\frac {3 n \log ^2\left (c (a+b x)^n\right ) \operatorname {PolyLog}\left (2,\frac {2 f (a+b x)}{2 a f-b \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}-\frac {6 n^2 \log \left (c (a+b x)^n\right ) \operatorname {PolyLog}\left (3,\frac {2 f (a+b x)}{2 a f-b \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}+\frac {6 n^2 \log \left (c (a+b x)^n\right ) \operatorname {PolyLog}\left (3,\frac {2 f (a+b x)}{2 a f-b \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}+\frac {6 n^3 \operatorname {PolyLog}\left (4,\frac {2 f (a+b x)}{2 a f-b \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}-\frac {6 n^3 \operatorname {PolyLog}\left (4,\frac {2 f (a+b x)}{2 a f-b \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}} \]
ln(c*(b*x+a)^n)^3*ln(-b*(e+2*f*x-(-4*d*f+e^2)^(1/2))/(2*a*f-b*(e-(-4*d*f+e ^2)^(1/2))))/(-4*d*f+e^2)^(1/2)-ln(c*(b*x+a)^n)^3*ln(-b*(e+2*f*x+(-4*d*f+e ^2)^(1/2))/(2*a*f-b*(e+(-4*d*f+e^2)^(1/2))))/(-4*d*f+e^2)^(1/2)+3*n*ln(c*( b*x+a)^n)^2*polylog(2,2*f*(b*x+a)/(2*a*f-b*(e-(-4*d*f+e^2)^(1/2))))/(-4*d* f+e^2)^(1/2)-3*n*ln(c*(b*x+a)^n)^2*polylog(2,2*f*(b*x+a)/(2*a*f-b*(e+(-4*d *f+e^2)^(1/2))))/(-4*d*f+e^2)^(1/2)-6*n^2*ln(c*(b*x+a)^n)*polylog(3,2*f*(b *x+a)/(2*a*f-b*(e-(-4*d*f+e^2)^(1/2))))/(-4*d*f+e^2)^(1/2)+6*n^2*ln(c*(b*x +a)^n)*polylog(3,2*f*(b*x+a)/(2*a*f-b*(e+(-4*d*f+e^2)^(1/2))))/(-4*d*f+e^2 )^(1/2)+6*n^3*polylog(4,2*f*(b*x+a)/(2*a*f-b*(e-(-4*d*f+e^2)^(1/2))))/(-4* d*f+e^2)^(1/2)-6*n^3*polylog(4,2*f*(b*x+a)/(2*a*f-b*(e+(-4*d*f+e^2)^(1/2)) ))/(-4*d*f+e^2)^(1/2)
Time = 0.52 (sec) , antiderivative size = 993, normalized size of antiderivative = 1.99 \[ \int \frac {\log ^3\left (c (a+b x)^n\right )}{d+e x+f x^2} \, dx=\frac {-2 \sqrt {e^2-4 d f} n^3 \arctan \left (\frac {e+2 f x}{\sqrt {-e^2+4 d f}}\right ) \log ^3(a+b x)+6 \sqrt {e^2-4 d f} n^2 \arctan \left (\frac {e+2 f x}{\sqrt {-e^2+4 d f}}\right ) \log ^2(a+b x) \log \left (c (a+b x)^n\right )-6 \sqrt {e^2-4 d f} n \arctan \left (\frac {e+2 f x}{\sqrt {-e^2+4 d f}}\right ) \log (a+b x) \log ^2\left (c (a+b x)^n\right )+2 \sqrt {e^2-4 d f} \arctan \left (\frac {e+2 f x}{\sqrt {-e^2+4 d f}}\right ) \log ^3\left (c (a+b x)^n\right )+\sqrt {-e^2+4 d f} n^3 \log ^3(a+b x) \log \left (1-\frac {2 f (a+b x)}{-b e+2 a f+b \sqrt {e^2-4 d f}}\right )-3 \sqrt {-e^2+4 d f} n^2 \log ^2(a+b x) \log \left (c (a+b x)^n\right ) \log \left (1-\frac {2 f (a+b x)}{-b e+2 a f+b \sqrt {e^2-4 d f}}\right )+3 \sqrt {-e^2+4 d f} n \log (a+b x) \log ^2\left (c (a+b x)^n\right ) \log \left (1-\frac {2 f (a+b x)}{-b e+2 a f+b \sqrt {e^2-4 d f}}\right )-\sqrt {-e^2+4 d f} n^3 \log ^3(a+b x) \log \left (1+\frac {2 f (a+b x)}{-2 a f+b \left (e+\sqrt {e^2-4 d f}\right )}\right )+3 \sqrt {-e^2+4 d f} n^2 \log ^2(a+b x) \log \left (c (a+b x)^n\right ) \log \left (1+\frac {2 f (a+b x)}{-2 a f+b \left (e+\sqrt {e^2-4 d f}\right )}\right )-3 \sqrt {-e^2+4 d f} n \log (a+b x) \log ^2\left (c (a+b x)^n\right ) \log \left (1+\frac {2 f (a+b x)}{-2 a f+b \left (e+\sqrt {e^2-4 d f}\right )}\right )+3 \sqrt {-e^2+4 d f} n \log ^2\left (c (a+b x)^n\right ) \operatorname {PolyLog}\left (2,\frac {2 f (a+b x)}{2 a f+b \left (-e+\sqrt {e^2-4 d f}\right )}\right )-3 \sqrt {-e^2+4 d f} n \log ^2\left (c (a+b x)^n\right ) \operatorname {PolyLog}\left (2,\frac {2 f (a+b x)}{2 a f-b \left (e+\sqrt {e^2-4 d f}\right )}\right )-6 \sqrt {-e^2+4 d f} n^2 \log \left (c (a+b x)^n\right ) \operatorname {PolyLog}\left (3,\frac {2 f (a+b x)}{-b e+2 a f+b \sqrt {e^2-4 d f}}\right )+6 \sqrt {-e^2+4 d f} n^2 \log \left (c (a+b x)^n\right ) \operatorname {PolyLog}\left (3,\frac {2 f (a+b x)}{2 a f-b \left (e+\sqrt {e^2-4 d f}\right )}\right )+6 \sqrt {-e^2+4 d f} n^3 \operatorname {PolyLog}\left (4,\frac {2 f (a+b x)}{-b e+2 a f+b \sqrt {e^2-4 d f}}\right )-6 \sqrt {-e^2+4 d f} n^3 \operatorname {PolyLog}\left (4,\frac {2 f (a+b x)}{2 a f-b \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {-\left (e^2-4 d f\right )^2}} \]
(-2*Sqrt[e^2 - 4*d*f]*n^3*ArcTan[(e + 2*f*x)/Sqrt[-e^2 + 4*d*f]]*Log[a + b *x]^3 + 6*Sqrt[e^2 - 4*d*f]*n^2*ArcTan[(e + 2*f*x)/Sqrt[-e^2 + 4*d*f]]*Log [a + b*x]^2*Log[c*(a + b*x)^n] - 6*Sqrt[e^2 - 4*d*f]*n*ArcTan[(e + 2*f*x)/ Sqrt[-e^2 + 4*d*f]]*Log[a + b*x]*Log[c*(a + b*x)^n]^2 + 2*Sqrt[e^2 - 4*d*f ]*ArcTan[(e + 2*f*x)/Sqrt[-e^2 + 4*d*f]]*Log[c*(a + b*x)^n]^3 + Sqrt[-e^2 + 4*d*f]*n^3*Log[a + b*x]^3*Log[1 - (2*f*(a + b*x))/(-(b*e) + 2*a*f + b*Sq rt[e^2 - 4*d*f])] - 3*Sqrt[-e^2 + 4*d*f]*n^2*Log[a + b*x]^2*Log[c*(a + b*x )^n]*Log[1 - (2*f*(a + b*x))/(-(b*e) + 2*a*f + b*Sqrt[e^2 - 4*d*f])] + 3*S qrt[-e^2 + 4*d*f]*n*Log[a + b*x]*Log[c*(a + b*x)^n]^2*Log[1 - (2*f*(a + b* x))/(-(b*e) + 2*a*f + b*Sqrt[e^2 - 4*d*f])] - Sqrt[-e^2 + 4*d*f]*n^3*Log[a + b*x]^3*Log[1 + (2*f*(a + b*x))/(-2*a*f + b*(e + Sqrt[e^2 - 4*d*f]))] + 3*Sqrt[-e^2 + 4*d*f]*n^2*Log[a + b*x]^2*Log[c*(a + b*x)^n]*Log[1 + (2*f*(a + b*x))/(-2*a*f + b*(e + Sqrt[e^2 - 4*d*f]))] - 3*Sqrt[-e^2 + 4*d*f]*n*Lo g[a + b*x]*Log[c*(a + b*x)^n]^2*Log[1 + (2*f*(a + b*x))/(-2*a*f + b*(e + S qrt[e^2 - 4*d*f]))] + 3*Sqrt[-e^2 + 4*d*f]*n*Log[c*(a + b*x)^n]^2*PolyLog[ 2, (2*f*(a + b*x))/(2*a*f + b*(-e + Sqrt[e^2 - 4*d*f]))] - 3*Sqrt[-e^2 + 4 *d*f]*n*Log[c*(a + b*x)^n]^2*PolyLog[2, (2*f*(a + b*x))/(2*a*f - b*(e + Sq rt[e^2 - 4*d*f]))] - 6*Sqrt[-e^2 + 4*d*f]*n^2*Log[c*(a + b*x)^n]*PolyLog[3 , (2*f*(a + b*x))/(-(b*e) + 2*a*f + b*Sqrt[e^2 - 4*d*f])] + 6*Sqrt[-e^2 + 4*d*f]*n^2*Log[c*(a + b*x)^n]*PolyLog[3, (2*f*(a + b*x))/(2*a*f - b*(e ...
Time = 0.93 (sec) , antiderivative size = 500, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2865, 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.
\(\displaystyle \int \frac {\log ^3\left (c (a+b x)^n\right )}{d+e x+f x^2} \, dx\) |
\(\Big \downarrow \) 2865 |
\(\displaystyle \int \left (\frac {2 f \log ^3\left (c (a+b x)^n\right )}{\sqrt {e^2-4 d f} \left (-\sqrt {e^2-4 d f}+e+2 f x\right )}-\frac {2 f \log ^3\left (c (a+b x)^n\right )}{\sqrt {e^2-4 d f} \left (\sqrt {e^2-4 d f}+e+2 f x\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {6 n^2 \log \left (c (a+b x)^n\right ) \operatorname {PolyLog}\left (3,\frac {2 f (a+b x)}{2 a f-b \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}+\frac {6 n^2 \log \left (c (a+b x)^n\right ) \operatorname {PolyLog}\left (3,\frac {2 f (a+b x)}{2 a f-b \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}+\frac {3 n \log ^2\left (c (a+b x)^n\right ) \operatorname {PolyLog}\left (2,\frac {2 f (a+b x)}{2 a f-b \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}-\frac {3 n \log ^2\left (c (a+b x)^n\right ) \operatorname {PolyLog}\left (2,\frac {2 f (a+b x)}{2 a f-b \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}+\frac {\log ^3\left (c (a+b x)^n\right ) \log \left (-\frac {b \left (-\sqrt {e^2-4 d f}+e+2 f x\right )}{2 a f-b \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}-\frac {\log ^3\left (c (a+b x)^n\right ) \log \left (-\frac {b \left (\sqrt {e^2-4 d f}+e+2 f x\right )}{2 a f-b \left (\sqrt {e^2-4 d f}+e\right )}\right )}{\sqrt {e^2-4 d f}}+\frac {6 n^3 \operatorname {PolyLog}\left (4,\frac {2 f (a+b x)}{2 a f-b \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}-\frac {6 n^3 \operatorname {PolyLog}\left (4,\frac {2 f (a+b x)}{2 a f-b \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}\) |
(Log[c*(a + b*x)^n]^3*Log[-((b*(e - Sqrt[e^2 - 4*d*f] + 2*f*x))/(2*a*f - b *(e - Sqrt[e^2 - 4*d*f])))])/Sqrt[e^2 - 4*d*f] - (Log[c*(a + b*x)^n]^3*Log [-((b*(e + Sqrt[e^2 - 4*d*f] + 2*f*x))/(2*a*f - b*(e + Sqrt[e^2 - 4*d*f])) )])/Sqrt[e^2 - 4*d*f] + (3*n*Log[c*(a + b*x)^n]^2*PolyLog[2, (2*f*(a + b*x ))/(2*a*f - b*(e - Sqrt[e^2 - 4*d*f]))])/Sqrt[e^2 - 4*d*f] - (3*n*Log[c*(a + b*x)^n]^2*PolyLog[2, (2*f*(a + b*x))/(2*a*f - b*(e + Sqrt[e^2 - 4*d*f]) )])/Sqrt[e^2 - 4*d*f] - (6*n^2*Log[c*(a + b*x)^n]*PolyLog[3, (2*f*(a + b*x ))/(2*a*f - b*(e - Sqrt[e^2 - 4*d*f]))])/Sqrt[e^2 - 4*d*f] + (6*n^2*Log[c* (a + b*x)^n]*PolyLog[3, (2*f*(a + b*x))/(2*a*f - b*(e + Sqrt[e^2 - 4*d*f]) )])/Sqrt[e^2 - 4*d*f] + (6*n^3*PolyLog[4, (2*f*(a + b*x))/(2*a*f - b*(e - Sqrt[e^2 - 4*d*f]))])/Sqrt[e^2 - 4*d*f] - (6*n^3*PolyLog[4, (2*f*(a + b*x) )/(2*a*f - b*(e + Sqrt[e^2 - 4*d*f]))])/Sqrt[e^2 - 4*d*f]
3.4.47.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Sy mbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunctionQ[ RFx, x] && IntegerQ[p]
\[\int \frac {\ln \left (c \left (b x +a \right )^{n}\right )^{3}}{f \,x^{2}+e x +d}d x\]
\[ \int \frac {\log ^3\left (c (a+b x)^n\right )}{d+e x+f x^2} \, dx=\int { \frac {\log \left ({\left (b x + a\right )}^{n} c\right )^{3}}{f x^{2} + e x + d} \,d x } \]
Timed out. \[ \int \frac {\log ^3\left (c (a+b x)^n\right )}{d+e x+f x^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\log ^3\left (c (a+b x)^n\right )}{d+e x+f x^2} \, dx=\text {Exception raised: ValueError} \]
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*d*f-e^2>0)', see `assume?` for more deta
\[ \int \frac {\log ^3\left (c (a+b x)^n\right )}{d+e x+f x^2} \, dx=\int { \frac {\log \left ({\left (b x + a\right )}^{n} c\right )^{3}}{f x^{2} + e x + d} \,d x } \]
Timed out. \[ \int \frac {\log ^3\left (c (a+b x)^n\right )}{d+e x+f x^2} \, dx=\int \frac {{\ln \left (c\,{\left (a+b\,x\right )}^n\right )}^3}{f\,x^2+e\,x+d} \,d x \]