Integrand size = 26, antiderivative size = 179 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^2} \, dx=\frac {2 b \sqrt {f} m n \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e}}+\frac {2 \sqrt {f} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {e}}-\frac {b n \log \left (d \left (e+f x^2\right )^m\right )}{x}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x}-\frac {i b \sqrt {f} m n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e}}+\frac {i b \sqrt {f} m n \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e}} \]
-b*n*ln(d*(f*x^2+e)^m)/x-(a+b*ln(c*x^n))*ln(d*(f*x^2+e)^m)/x+2*b*m*n*arcta n(x*f^(1/2)/e^(1/2))*f^(1/2)/e^(1/2)+2*m*arctan(x*f^(1/2)/e^(1/2))*(a+b*ln (c*x^n))*f^(1/2)/e^(1/2)-I*b*m*n*polylog(2,-I*x*f^(1/2)/e^(1/2))*f^(1/2)/e ^(1/2)+I*b*m*n*polylog(2,I*x*f^(1/2)/e^(1/2))*f^(1/2)/e^(1/2)
Time = 0.09 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.70 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^2} \, dx=\frac {2 a \sqrt {f} m x \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )+2 b \sqrt {f} m n x \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )-2 b \sqrt {f} m n x \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \log (x)+2 b \sqrt {f} m x \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \log \left (c x^n\right )+i b \sqrt {f} m n x \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-i b \sqrt {f} m n x \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )-a \sqrt {e} \log \left (d \left (e+f x^2\right )^m\right )-b \sqrt {e} n \log \left (d \left (e+f x^2\right )^m\right )-b \sqrt {e} \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )-i b \sqrt {f} m n x \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+i b \sqrt {f} m n x \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e} x} \]
(2*a*Sqrt[f]*m*x*ArcTan[(Sqrt[f]*x)/Sqrt[e]] + 2*b*Sqrt[f]*m*n*x*ArcTan[(S qrt[f]*x)/Sqrt[e]] - 2*b*Sqrt[f]*m*n*x*ArcTan[(Sqrt[f]*x)/Sqrt[e]]*Log[x] + 2*b*Sqrt[f]*m*x*ArcTan[(Sqrt[f]*x)/Sqrt[e]]*Log[c*x^n] + I*b*Sqrt[f]*m*n *x*Log[x]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] - I*b*Sqrt[f]*m*n*x*Log[x]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] - a*Sqrt[e]*Log[d*(e + f*x^2)^m] - b*Sqrt[e]*n*Lo g[d*(e + f*x^2)^m] - b*Sqrt[e]*Log[c*x^n]*Log[d*(e + f*x^2)^m] - I*b*Sqrt[ f]*m*n*x*PolyLog[2, ((-I)*Sqrt[f]*x)/Sqrt[e]] + I*b*Sqrt[f]*m*n*x*PolyLog[ 2, (I*Sqrt[f]*x)/Sqrt[e]])/(Sqrt[e]*x)
Time = 0.33 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2823, 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.
| \(\displaystyle \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^2} \, dx\) |
| \(\Big \downarrow \) 2823 |
| \(\displaystyle -b n \int \left (\frac {2 \sqrt {f} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e} x}-\frac {\log \left (d \left (f x^2+e\right )^m\right )}{x^2}\right )dx+\frac {2 \sqrt {f} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {e}}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \sqrt {f} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {e}}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x}-b n \left (-\frac {2 \sqrt {f} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e}}+\frac {\log \left (d \left (e+f x^2\right )^m\right )}{x}+\frac {i \sqrt {f} m \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e}}-\frac {i \sqrt {f} m \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e}}\right )\) |
(2*Sqrt[f]*m*ArcTan[(Sqrt[f]*x)/Sqrt[e]]*(a + b*Log[c*x^n]))/Sqrt[e] - ((a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m])/x - b*n*((-2*Sqrt[f]*m*ArcTan[(Sqrt [f]*x)/Sqrt[e]])/Sqrt[e] + Log[d*(e + f*x^2)^m]/x + (I*Sqrt[f]*m*PolyLog[2 , ((-I)*Sqrt[f]*x)/Sqrt[e]])/Sqrt[e] - (I*Sqrt[f]*m*PolyLog[2, (I*Sqrt[f]* x)/Sqrt[e]])/Sqrt[e])
3.1.97.3.1 Defintions of rubi rules used
Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_. )]*(b_.))*((g_.)*(x_))^(q_.), x_Symbol] :> With[{u = IntHide[(g*x)^q*Log[d* (e + f*x^m)^r], x]}, Simp[(a + b*Log[c*x^n]) u, x] - Simp[b*n Int[1/x u, x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m, n, q}, x] && (IntegerQ[(q + 1)/m] || (RationalQ[m] && RationalQ[q])) && NeQ[q, -1]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 16.98 (sec) , antiderivative size = 733, normalized size of antiderivative = 4.09
| method | result | size |
| risch | \(\left (-\frac {b \ln \left (x^{n}\right )}{x}-\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 )+i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 b \ln \left (c \right )+2 b n +2 a}{2 x}\right ) \ln \left (\left (f \,x^{2}+e \right )^{m}\right )+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (f \,x^{2}+e \right )^{m}\right ) {\operatorname {csgn}\left (i d \left (f \,x^{2}+e \right )^{m}\right )}^{2}}{4}-\frac {i \pi \,\operatorname {csgn}\left (i \left (f \,x^{2}+e \right )^{m}\right ) \operatorname {csgn}\left (i d \left (f \,x^{2}+e \right )^{m}\right ) \operatorname {csgn}\left (i d \right )}{4}-\frac {i \pi {\operatorname {csgn}\left (i d \left (f \,x^{2}+e \right )^{m}\right )}^{3}}{4}+\frac {i \pi {\operatorname {csgn}\left (i d \left (f \,x^{2}+e \right )^{m}\right )}^{2} \operatorname {csgn}\left (i d \right )}{4}+\frac {\ln \left (d \right )}{2}\right ) \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 )+i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 b \ln \left (c \right )+2 a}{x}-\frac {2 b \ln \left (x^{n}\right )}{x}-\frac {2 b n}{x}\right )-\frac {i m f \arctan \left (\frac {x f}{\sqrt {e f}}\right ) b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{\sqrt {e f}}+\frac {i m f \arctan \left (\frac {x f}{\sqrt {e f}}\right ) b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{\sqrt {e f}}+\frac {i m f \arctan \left (\frac {x f}{\sqrt {e f}}\right ) b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{\sqrt {e f}}-\frac {i m f \arctan \left (\frac {x f}{\sqrt {e f}}\right ) b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{\sqrt {e f}}+\frac {2 m f \arctan \left (\frac {x f}{\sqrt {e f}}\right ) b \ln \left (c \right )}{\sqrt {e f}}+\frac {2 m f \arctan \left (\frac {x f}{\sqrt {e f}}\right ) b n}{\sqrt {e f}}+\frac {2 m f \arctan \left (\frac {x f}{\sqrt {e f}}\right ) a}{\sqrt {e f}}-\frac {2 m f b \arctan \left (\frac {x f}{\sqrt {e f}}\right ) n \ln \left (x \right )}{\sqrt {e f}}+\frac {2 m f b \arctan \left (\frac {x f}{\sqrt {e f}}\right ) \ln \left (x^{n}\right )}{\sqrt {e f}}+\frac {m f b n \ln \left (x \right ) \ln \left (\frac {-f x +\sqrt {-e f}}{\sqrt {-e f}}\right )}{\sqrt {-e f}}-\frac {m f b n \ln \left (x \right ) \ln \left (\frac {f x +\sqrt {-e f}}{\sqrt {-e f}}\right )}{\sqrt {-e f}}+\frac {m f b n \operatorname {dilog}\left (\frac {-f x +\sqrt {-e f}}{\sqrt {-e f}}\right )}{\sqrt {-e f}}-\frac {m f b n \operatorname {dilog}\left (\frac {f x +\sqrt {-e f}}{\sqrt {-e f}}\right )}{\sqrt {-e f}}\) | \(733\) |
(-b/x*ln(x^n)-1/2*(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I*b*Pi*csgn (I*c)*csgn(I*c*x^n)^2+I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-I*b*Pi*csgn(I*c*x ^n)^3+2*b*ln(c)+2*b*n+2*a)/x)*ln((f*x^2+e)^m)+(1/4*I*Pi*csgn(I*(f*x^2+e)^m )*csgn(I*d*(f*x^2+e)^m)^2-1/4*I*Pi*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^ m)*csgn(I*d)-1/4*I*Pi*csgn(I*d*(f*x^2+e)^m)^3+1/4*I*Pi*csgn(I*d*(f*x^2+e)^ m)^2*csgn(I*d)+1/2*ln(d))*(-(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I *b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-I*b*Pi* csgn(I*c*x^n)^3+2*b*ln(c)+2*a)/x-2*b/x*ln(x^n)-2*b*n/x)-I*m*f/(e*f)^(1/2)* arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I*m*f/(e* f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+I*m*f/(e*f )^(1/2)*arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-I*m*f/(e* f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*c*x^n)^3+2*m*f/(e*f)^(1/2)*ar ctan(x*f/(e*f)^(1/2))*b*ln(c)+2*m*f/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b* n+2*m*f/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*a-2*m*f*b/(e*f)^(1/2)*arctan(x *f/(e*f)^(1/2))*n*ln(x)+2*m*f*b/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*ln(x^n )+m*f*b*n*ln(x)/(-e*f)^(1/2)*ln((-f*x+(-e*f)^(1/2))/(-e*f)^(1/2))-m*f*b*n* ln(x)/(-e*f)^(1/2)*ln((f*x+(-e*f)^(1/2))/(-e*f)^(1/2))+m*f*b*n/(-e*f)^(1/2 )*dilog((-f*x+(-e*f)^(1/2))/(-e*f)^(1/2))-m*f*b*n/(-e*f)^(1/2)*dilog((f*x+ (-e*f)^(1/2))/(-e*f)^(1/2))
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^2} \, dx=\int \frac {\ln \left (d\,{\left (f\,x^2+e\right )}^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^2} \,d x \]