Integrand size = 26, antiderivative size = 179 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^2} \, dx=\frac {2 b \sqrt {f} m n \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e}}+\frac {2 \sqrt {f} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {e}}-\frac {b n \log \left (d \left (e+f x^2\right )^m\right )}{x}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x}-\frac {i b \sqrt {f} m n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e}}+\frac {i b \sqrt {f} m n \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e}} \] Output:
2*b*f^(1/2)*m*n*arctan(f^(1/2)*x/e^(1/2))/e^(1/2)+2*f^(1/2)*m*arctan(f^(1/ 2)*x/e^(1/2))*(a+b*ln(c*x^n))/e^(1/2)-b*n*ln(d*(f*x^2+e)^m)/x-(a+b*ln(c*x^ n))*ln(d*(f*x^2+e)^m)/x-I*b*f^(1/2)*m*n*polylog(2,-I*f^(1/2)*x/e^(1/2))/e^ (1/2)+I*b*f^(1/2)*m*n*polylog(2,I*f^(1/2)*x/e^(1/2))/e^(1/2)
Time = 0.17 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.70 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^2} \, dx=\frac {2 a \sqrt {f} m x \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )+2 b \sqrt {f} m n x \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )-2 b \sqrt {f} m n x \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \log (x)+2 b \sqrt {f} m x \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \log \left (c x^n\right )+i b \sqrt {f} m n x \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-i b \sqrt {f} m n x \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )-a \sqrt {e} \log \left (d \left (e+f x^2\right )^m\right )-b \sqrt {e} n \log \left (d \left (e+f x^2\right )^m\right )-b \sqrt {e} \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )-i b \sqrt {f} m n x \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+i b \sqrt {f} m n x \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e} x} \] Input:
Integrate[((a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m])/x^2,x]
Output:
(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 = 0.37 (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 |
| \(\displaystyle \frac {2 \sqrt {f} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {e}}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x}-b n \left (-\frac {2 \sqrt {f} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e}}+\frac {\log \left (d \left (e+f x^2\right )^m\right )}{x}+\frac {i \sqrt {f} m \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e}}-\frac {i \sqrt {f} m \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e}}\right )\) |
Input:
Int[((a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m])/x^2,x]
Output:
(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])
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 = 19.20 (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 x^{n}\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 ) \operatorname {csgn}\left (i c \right )-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+2 b \ln \left (c \right )+2 n b +2 a}{2 x}\right ) \ln \left (\left (f \,x^{2}+e \right )^{m}\right )+\left (-\frac {i \pi \,\operatorname {csgn}\left (i d \right ) \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 )}{4}+\frac {i \pi \,\operatorname {csgn}\left (i d \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 )}^{2}}{4}-\frac {i \pi {\operatorname {csgn}\left (i d \left (f \,x^{2}+e \right )^{m}\right )}^{3}}{4}+\frac {\ln \left (d \right )}{2}\right ) \left (-\frac {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 x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+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 ) \pi b \,\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 ) \pi b \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{\sqrt {e f}}-\frac {i m f \arctan \left (\frac {x f}{\sqrt {e f}}\right ) \pi b \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{\sqrt {e f}}+\frac {i m f \arctan \left (\frac {x f}{\sqrt {e f}}\right ) \pi b \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{\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 ) n b}{\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}}\) | \(733\) |
Input:
int((a+b*ln(c*x^n))*ln(d*(f*x^2+e)^m)/x^2,x,method=_RETURNVERBOSE)
Output:
(-b/x*ln(x^n)-1/2*(I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*csgn(I*x^n)*c sgn(I*c*x^n)*csgn(I*c)-I*Pi*b*csgn(I*c*x^n)^3+I*Pi*b*csgn(I*c*x^n)^2*csgn( I*c)+2*b*ln(c)+2*n*b+2*a)/x)*ln((f*x^2+e)^m)+(-1/4*I*Pi*csgn(I*d)*csgn(I*( f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)+1/4*I*Pi*csgn(I*d)*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)^2-1/4*I*Pi*csgn(I*d* (f*x^2+e)^m)^3+1/2*ln(d))*(-(I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*csg n(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-I*Pi*b*csgn(I*c*x^n)^3+I*Pi*b*csgn(I*c*x^ n)^2*csgn(I*c)+2*b*ln(c)+2*a)/x-2*b/x*ln(x^n)-2*b*n/x)+I*m*f/(e*f)^(1/2)*a rctan(x*f/(e*f)^(1/2))*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-I*m*f/(e*f)^(1/2)* arctan(x*f/(e*f)^(1/2))*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-I*m*f/(e* f)^(1/2)*arctan(x*f/(e*f)^(1/2))*Pi*b*csgn(I*c*x^n)^3+I*m*f/(e*f)^(1/2)*ar ctan(x*f/(e*f)^(1/2))*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)+2*m*f/(e*f)^(1/2)*arc tan(x*f/(e*f)^(1/2))*b*ln(c)+2*m*f/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*n*b +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*l n(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))
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{2}} \,d x } \] Input:
integrate((a+b*log(c*x^n))*log(d*(f*x^2+e)^m)/x^2,x, algorithm="fricas")
Output:
integral((b*log(c*x^n) + a)*log((f*x^2 + e)^m*d)/x^2, x)
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^2} \, dx=\text {Timed out} \] Input:
integrate((a+b*ln(c*x**n))*ln(d*(f*x**2+e)**m)/x**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((a+b*log(c*x^n))*log(d*(f*x^2+e)^m)/x^2,x, algorithm="maxima")
Output:
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
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{2}} \,d x } \] Input:
integrate((a+b*log(c*x^n))*log(d*(f*x^2+e)^m)/x^2,x, algorithm="giac")
Output:
integrate((b*log(c*x^n) + a)*log((f*x^2 + e)^m*d)/x^2, x)
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^2} \, dx=\int \frac {\ln \left (d\,{\left (f\,x^2+e\right )}^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^2} \,d x \] Input:
int((log(d*(e + f*x^2)^m)*(a + b*log(c*x^n)))/x^2,x)
Output:
int((log(d*(e + f*x^2)^m)*(a + b*log(c*x^n)))/x^2, x)
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^2} \, dx=\frac {2 \sqrt {f}\, \sqrt {e}\, \mathit {atan} \left (\frac {f x}{\sqrt {f}\, \sqrt {e}}\right ) a m x +2 \sqrt {f}\, \sqrt {e}\, \mathit {atan} \left (\frac {f x}{\sqrt {f}\, \sqrt {e}}\right ) b m n x -2 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{f \,x^{4}+e \,x^{2}}d x \right ) b \,e^{2} m x -\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) \mathrm {log}\left (x^{n} c \right ) b e -\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) a e -\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) b e n -2 \,\mathrm {log}\left (x^{n} c \right ) b e m -2 b e m n}{e x} \] Input:
int((a+b*log(c*x^n))*log(d*(f*x^2+e)^m)/x^2,x)
Output:
(2*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*m*x + 2*sqrt(f)*sqrt(e) *atan((f*x)/(sqrt(f)*sqrt(e)))*b*m*n*x - 2*int(log(x**n*c)/(e*x**2 + f*x** 4),x)*b*e**2*m*x - log((e + f*x**2)**m*d)*log(x**n*c)*b*e - log((e + f*x** 2)**m*d)*a*e - log((e + f*x**2)**m*d)*b*e*n - 2*log(x**n*c)*b*e*m - 2*b*e* m*n)/(e*x)