Integrand size = 23, antiderivative size = 191 \[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=\frac {a x}{e^2}-\frac {b n x}{e^2}-\frac {b \sqrt {d} n \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 e^{5/2}}+\frac {b x \log \left (c x^n\right )}{e^2}+\frac {d x \left (a+b \log \left (c x^n\right )\right )}{2 e^2 \left (d+e x^2\right )}-\frac {3 \sqrt {d} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^{5/2}}+\frac {3 i b \sqrt {d} n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{4 e^{5/2}}-\frac {3 i b \sqrt {d} n \operatorname {PolyLog}\left (2,\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{4 e^{5/2}} \]
a*x/e^2-b*n*x/e^2+b*x*ln(c*x^n)/e^2+1/2*d*x*(a+b*ln(c*x^n))/e^2/(e*x^2+d)- 1/2*b*n*arctan(x*e^(1/2)/d^(1/2))*d^(1/2)/e^(5/2)-3/2*arctan(x*e^(1/2)/d^( 1/2))*(a+b*ln(c*x^n))*d^(1/2)/e^(5/2)+3/4*I*b*n*polylog(2,-I*x*e^(1/2)/d^( 1/2))*d^(1/2)/e^(5/2)-3/4*I*b*n*polylog(2,I*x*e^(1/2)/d^(1/2))*d^(1/2)/e^( 5/2)
Time = 0.36 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.55 \[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=\frac {4 a \sqrt {e} x-4 b \sqrt {e} n x+4 b \sqrt {e} x \log \left (c x^n\right )-\frac {d \left (a+b \log \left (c x^n\right )\right )}{\sqrt {-d}-\sqrt {e} x}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{\sqrt {-d}+\sqrt {e} x}+\frac {b d n \left (\log (x)-\log \left (\sqrt {-d}-\sqrt {e} x\right )\right )}{\sqrt {-d}}+b \sqrt {-d} n \left (\log (x)-\log \left (\sqrt {-d}+\sqrt {e} x\right )\right )-3 \sqrt {-d} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )+3 \sqrt {-d} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )+3 b \sqrt {-d} n \operatorname {PolyLog}\left (2,\frac {\sqrt {e} x}{\sqrt {-d}}\right )-3 b \sqrt {-d} n \operatorname {PolyLog}\left (2,\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{4 e^{5/2}} \]
(4*a*Sqrt[e]*x - 4*b*Sqrt[e]*n*x + 4*b*Sqrt[e]*x*Log[c*x^n] - (d*(a + b*Lo g[c*x^n]))/(Sqrt[-d] - Sqrt[e]*x) + (d*(a + b*Log[c*x^n]))/(Sqrt[-d] + Sqr t[e]*x) + (b*d*n*(Log[x] - Log[Sqrt[-d] - Sqrt[e]*x]))/Sqrt[-d] + b*Sqrt[- d]*n*(Log[x] - Log[Sqrt[-d] + Sqrt[e]*x]) - 3*Sqrt[-d]*(a + b*Log[c*x^n])* Log[1 + (Sqrt[e]*x)/Sqrt[-d]] + 3*Sqrt[-d]*(a + b*Log[c*x^n])*Log[1 + (d*S qrt[e]*x)/(-d)^(3/2)] + 3*b*Sqrt[-d]*n*PolyLog[2, (Sqrt[e]*x)/Sqrt[-d]] - 3*b*Sqrt[-d]*n*PolyLog[2, (d*Sqrt[e]*x)/(-d)^(3/2)])/(4*e^(5/2))
Time = 0.45 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2793, 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.
| \(\displaystyle \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 2793 |
| \(\displaystyle \int \left (\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{e^2 \left (d+e x^2\right )^2}-\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^2 \left (d+e x^2\right )}+\frac {a+b \log \left (c x^n\right )}{e^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 \sqrt {d} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^{5/2}}+\frac {d x \left (a+b \log \left (c x^n\right )\right )}{2 e^2 \left (d+e x^2\right )}+\frac {a x}{e^2}-\frac {b \sqrt {d} n \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 e^{5/2}}+\frac {b x \log \left (c x^n\right )}{e^2}+\frac {3 i b \sqrt {d} n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{4 e^{5/2}}-\frac {3 i b \sqrt {d} n \operatorname {PolyLog}\left (2,\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{4 e^{5/2}}-\frac {b n x}{e^2}\) |
(a*x)/e^2 - (b*n*x)/e^2 - (b*Sqrt[d]*n*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*e^( 5/2)) + (b*x*Log[c*x^n])/e^2 + (d*x*(a + b*Log[c*x^n]))/(2*e^2*(d + e*x^2) ) - (3*Sqrt[d]*ArcTan[(Sqrt[e]*x)/Sqrt[d]]*(a + b*Log[c*x^n]))/(2*e^(5/2)) + (((3*I)/4)*b*Sqrt[d]*n*PolyLog[2, ((-I)*Sqrt[e]*x)/Sqrt[d]])/e^(5/2) - (((3*I)/4)*b*Sqrt[d]*n*PolyLog[2, (I*Sqrt[e]*x)/Sqrt[d]])/e^(5/2)
3.3.26.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* (x_)^(r_.))^(q_.), x_Symbol] :> With[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && Integer Q[r]))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.52 (sec) , antiderivative size = 559, normalized size of antiderivative = 2.93
| method | result | size |
| risch | \(\frac {b \ln \left (x^{n}\right ) d x}{2 e^{2} \left (e \,x^{2}+d \right )}+\frac {3 b d \arctan \left (\frac {x e}{\sqrt {d e}}\right ) n \ln \left (x \right )}{2 e^{2} \sqrt {d e}}-\frac {3 b d \arctan \left (\frac {x e}{\sqrt {d e}}\right ) \ln \left (x^{n}\right )}{2 e^{2} \sqrt {d e}}+\frac {b \ln \left (x^{n}\right ) x}{e^{2}}-\frac {b n x}{e^{2}}-\frac {b n d \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{e^{2} \sqrt {-d e}}+\frac {b n d \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{e^{2} \sqrt {-d e}}-\frac {3 b n d \operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{4 e^{2} \sqrt {-d e}}+\frac {3 b n d \operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{4 e^{2} \sqrt {-d e}}-\frac {b n d \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{2 e^{2} \sqrt {d e}}+\frac {b n d \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right ) x^{2}}{4 e \left (e \,x^{2}+d \right ) \sqrt {-d e}}-\frac {b n d \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right ) x^{2}}{4 e \left (e \,x^{2}+d \right ) \sqrt {-d e}}+\frac {b n \,d^{2} \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{4 e^{2} \left (e \,x^{2}+d \right ) \sqrt {-d e}}-\frac {b n \,d^{2} \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{4 e^{2} \left (e \,x^{2}+d \right ) \sqrt {-d e}}+\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 {d \left (-\frac {x}{2 \left (e \,x^{2}+d \right )}+\frac {3 \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{2 \sqrt {d e}}\right )}{e^{2}}+\frac {x}{e^{2}}\right )\) | \(559\) |
1/2*b*ln(x^n)*d/e^2*x/(e*x^2+d)+3/2*b*d/e^2/(d*e)^(1/2)*arctan(x*e/(d*e)^( 1/2))*n*ln(x)-3/2*b*d/e^2/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*ln(x^n)+b*ln (x^n)/e^2*x-b*n*x/e^2-b*n*d/e^2*ln(x)/(-d*e)^(1/2)*ln((-e*x+(-d*e)^(1/2))/ (-d*e)^(1/2))+b*n*d/e^2*ln(x)/(-d*e)^(1/2)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1 /2))-3/4*b*n*d/e^2/(-d*e)^(1/2)*dilog((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+3/ 4*b*n*d/e^2/(-d*e)^(1/2)*dilog((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/2*b*n*d/ e^2/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))+1/4*b*n*d/e*ln(x)/(e*x^2+d)/(-d*e) ^(1/2)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))*x^2-1/4*b*n*d/e*ln(x)/(e*x^2+d )/(-d*e)^(1/2)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))*x^2+1/4*b*n*d^2/e^2*ln( x)/(e*x^2+d)/(-d*e)^(1/2)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/4*b*n*d^2 /e^2*ln(x)/(e*x^2+d)/(-d*e)^(1/2)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+(-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*csgn(I*c*x^n)^3+ b*ln(c)+a)*(-d/e^2*(-1/2*x/(e*x^2+d)+3/2/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2 )))+x/e^2)
\[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{4}}{{\left (e x^{2} + d\right )}^{2}} \,d x } \]
\[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=\int \frac {x^{4} \left (a + b \log {\left (c x^{n} \right )}\right )}{\left (d + e x^{2}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{4}}{{\left (e x^{2} + d\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=\int \frac {x^4\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^2} \,d x \]