Integrand size = 23, antiderivative size = 165 \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )} \, dx=-\frac {b n}{9 d x^3}+\frac {b e n}{d^2 x}-\frac {a+b \log \left (c x^n\right )}{3 d x^3}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}+\frac {e^{3/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^{5/2}}-\frac {i b e^{3/2} n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2}}+\frac {i b e^{3/2} n \operatorname {PolyLog}\left (2,\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2}} \]
-1/9*b*n/d/x^3+b*e*n/d^2/x+1/3*(-a-b*ln(c*x^n))/d/x^3+e*(a+b*ln(c*x^n))/d^ 2/x+e^(3/2)*arctan(x*e^(1/2)/d^(1/2))*(a+b*ln(c*x^n))/d^(5/2)-1/2*I*b*e^(3 /2)*n*polylog(2,-I*x*e^(1/2)/d^(1/2))/d^(5/2)+1/2*I*b*e^(3/2)*n*polylog(2, I*x*e^(1/2)/d^(1/2))/d^(5/2)
Time = 0.13 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.28 \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )} \, dx=\frac {1}{18} \left (-\frac {2 b n}{d x^3}+\frac {18 b e n}{d^2 x}-\frac {6 \left (a+b \log \left (c x^n\right )\right )}{d x^3}+\frac {18 e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}-\frac {9 e^{3/2} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )}{(-d)^{5/2}}+\frac {9 e^{3/2} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{(-d)^{5/2}}+\frac {9 b e^{3/2} n \operatorname {PolyLog}\left (2,\frac {\sqrt {e} x}{\sqrt {-d}}\right )}{(-d)^{5/2}}-\frac {9 b e^{3/2} n \operatorname {PolyLog}\left (2,\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{(-d)^{5/2}}\right ) \]
((-2*b*n)/(d*x^3) + (18*b*e*n)/(d^2*x) - (6*(a + b*Log[c*x^n]))/(d*x^3) + (18*e*(a + b*Log[c*x^n]))/(d^2*x) - (9*e^(3/2)*(a + b*Log[c*x^n])*Log[1 + (Sqrt[e]*x)/Sqrt[-d]])/(-d)^(5/2) + (9*e^(3/2)*(a + b*Log[c*x^n])*Log[1 + (d*Sqrt[e]*x)/(-d)^(3/2)])/(-d)^(5/2) + (9*b*e^(3/2)*n*PolyLog[2, (Sqrt[e] *x)/Sqrt[-d]])/(-d)^(5/2) - (9*b*e^(3/2)*n*PolyLog[2, (d*Sqrt[e]*x)/(-d)^( 3/2)])/(-d)^(5/2))/18
Time = 0.66 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {2780, 2741, 2780, 2741, 2761, 27, 5355, 2838}
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.
| \(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )} \, dx\) |
| \(\Big \downarrow \) 2780 |
| \(\displaystyle \frac {\int \frac {a+b \log \left (c x^n\right )}{x^4}dx}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (e x^2+d\right )}dx}{d}\) |
| \(\Big \downarrow \) 2741 |
| \(\displaystyle \frac {-\frac {a+b \log \left (c x^n\right )}{3 x^3}-\frac {b n}{9 x^3}}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (e x^2+d\right )}dx}{d}\) |
| \(\Big \downarrow \) 2780 |
| \(\displaystyle \frac {-\frac {a+b \log \left (c x^n\right )}{3 x^3}-\frac {b n}{9 x^3}}{d}-\frac {e \left (\frac {\int \frac {a+b \log \left (c x^n\right )}{x^2}dx}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{e x^2+d}dx}{d}\right )}{d}\) |
| \(\Big \downarrow \) 2741 |
| \(\displaystyle \frac {-\frac {a+b \log \left (c x^n\right )}{3 x^3}-\frac {b n}{9 x^3}}{d}-\frac {e \left (\frac {-\frac {a+b \log \left (c x^n\right )}{x}-\frac {b n}{x}}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{e x^2+d}dx}{d}\right )}{d}\) |
| \(\Big \downarrow \) 2761 |
| \(\displaystyle \frac {-\frac {a+b \log \left (c x^n\right )}{3 x^3}-\frac {b n}{9 x^3}}{d}-\frac {e \left (\frac {-\frac {a+b \log \left (c x^n\right )}{x}-\frac {b n}{x}}{d}-\frac {e \left (\frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d} \sqrt {e}}-b n \int \frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e} x}dx\right )}{d}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {a+b \log \left (c x^n\right )}{3 x^3}-\frac {b n}{9 x^3}}{d}-\frac {e \left (\frac {-\frac {a+b \log \left (c x^n\right )}{x}-\frac {b n}{x}}{d}-\frac {e \left (\frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d} \sqrt {e}}-\frac {b n \int \frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{x}dx}{\sqrt {d} \sqrt {e}}\right )}{d}\right )}{d}\) |
| \(\Big \downarrow \) 5355 |
| \(\displaystyle \frac {-\frac {a+b \log \left (c x^n\right )}{3 x^3}-\frac {b n}{9 x^3}}{d}-\frac {e \left (\frac {-\frac {a+b \log \left (c x^n\right )}{x}-\frac {b n}{x}}{d}-\frac {e \left (\frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d} \sqrt {e}}-\frac {b n \left (\frac {1}{2} i \int \frac {\log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{x}dx-\frac {1}{2} i \int \frac {\log \left (\frac {i \sqrt {e} x}{\sqrt {d}}+1\right )}{x}dx\right )}{\sqrt {d} \sqrt {e}}\right )}{d}\right )}{d}\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle \frac {-\frac {a+b \log \left (c x^n\right )}{3 x^3}-\frac {b n}{9 x^3}}{d}-\frac {e \left (\frac {-\frac {a+b \log \left (c x^n\right )}{x}-\frac {b n}{x}}{d}-\frac {e \left (\frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d} \sqrt {e}}-\frac {b n \left (\frac {1}{2} i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {e} x}{\sqrt {d}}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,\frac {i \sqrt {e} x}{\sqrt {d}}\right )\right )}{\sqrt {d} \sqrt {e}}\right )}{d}\right )}{d}\) |
(-1/9*(b*n)/x^3 - (a + b*Log[c*x^n])/(3*x^3))/d - (e*((-((b*n)/x) - (a + b *Log[c*x^n])/x)/d - (e*((ArcTan[(Sqrt[e]*x)/Sqrt[d]]*(a + b*Log[c*x^n]))/( Sqrt[d]*Sqrt[e]) - (b*n*((I/2)*PolyLog[2, ((-I)*Sqrt[e]*x)/Sqrt[d]] - (I/2 )*PolyLog[2, (I*Sqrt[e]*x)/Sqrt[d]]))/(Sqrt[d]*Sqrt[e])))/d))/d
3.3.20.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^( m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> With[{u = IntHide[1/(d + e*x^2), x]}, Simp[u*(a + b*Log[c*x^n]), x] - Si mp[b*n Int[u/x, x], x]] /; FreeQ[{a, b, c, d, e, n}, x]
Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.))/((d_) + (e_.)* (x_)^(r_.)), x_Symbol] :> Simp[1/d Int[x^m*(a + b*Log[c*x^n])^p, x], x] - Simp[e/d Int[(x^(m + r)*(a + b*Log[c*x^n])^p)/(d + e*x^r), x], x] /; Fre eQ[{a, b, c, d, e, m, n, r}, x] && IGtQ[p, 0] && IGtQ[r, 0] && ILtQ[m, -1]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Simp[I*(b/2) Int[Log[1 - I*c*x]/x, x], x] - Simp[I*(b/2) Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.56 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.24
| method | result | size |
| risch | \(-\frac {b \,e^{2} \arctan \left (\frac {x e}{\sqrt {d e}}\right ) n \ln \left (x \right )}{d^{2} \sqrt {d e}}+\frac {b \,e^{2} \arctan \left (\frac {x e}{\sqrt {d e}}\right ) \ln \left (x^{n}\right )}{d^{2} \sqrt {d e}}-\frac {b \ln \left (x^{n}\right )}{3 d \,x^{3}}+\frac {b \ln \left (x^{n}\right ) e}{d^{2} x}+\frac {b n \,e^{2} \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{2} \sqrt {-d e}}-\frac {b n \,e^{2} \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{2} \sqrt {-d e}}+\frac {b n \,e^{2} \operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{2} \sqrt {-d e}}-\frac {b n \,e^{2} \operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{2} \sqrt {-d e}}-\frac {b n}{9 d \,x^{3}}+\frac {b e n}{d^{2} x}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {e^{2} \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{d^{2} \sqrt {d e}}-\frac {1}{3 d \,x^{3}}+\frac {e}{d^{2} x}\right )\) | \(369\) |
-b*e^2/d^2/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*n*ln(x)+b*e^2/d^2/(d*e)^(1/ 2)*arctan(x*e/(d*e)^(1/2))*ln(x^n)-1/3*b*ln(x^n)/d/x^3+b*ln(x^n)*e/d^2/x+1 /2*b*n*e^2/d^2*ln(x)/(-d*e)^(1/2)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/2 *b*n*e^2/d^2*ln(x)/(-d*e)^(1/2)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1/2*b* n*e^2/d^2/(-d*e)^(1/2)*dilog((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/2*b*n*e^2 /d^2/(-d*e)^(1/2)*dilog((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/9*b*n/d/x^3+b*e *n/d^2/x+(-1/2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/2*I*b*Pi*csgn( I*c)*csgn(I*c*x^n)^2+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/2*I*b*Pi*csg n(I*c*x^n)^3+b*ln(c)+a)*(e^2/d^2/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))-1/3/d /x^3+e/d^2/x)
\[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )} x^{4}} \,d x } \]
\[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )} \, dx=\int \frac {a + b \log {\left (c x^{n} \right )}}{x^{4} \left (d + e x^{2}\right )}\, dx \]
Exception generated. \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )} x^{4}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^4\,\left (e\,x^2+d\right )} \,d x \]