Integrand size = 26, antiderivative size = 267 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^6} \, dx=-\frac {16 b f m n}{225 e x^3}+\frac {12 b f^2 m n}{25 e^2 x}+\frac {2 b f^{5/2} m n \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{25 e^{5/2}}-\frac {2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}+\frac {2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}+\frac {2 f^{5/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac {b n \log \left (d \left (e+f x^2\right )^m\right )}{25 x^5}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}-\frac {i b f^{5/2} m n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{5 e^{5/2}}+\frac {i b f^{5/2} m n \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{5 e^{5/2}} \]
-16/225*b*f*m*n/e/x^3+12/25*b*f^2*m*n/e^2/x+2/25*b*f^(5/2)*m*n*arctan(x*f^ (1/2)/e^(1/2))/e^(5/2)-2/15*f*m*(a+b*ln(c*x^n))/e/x^3+2/5*f^2*m*(a+b*ln(c* x^n))/e^2/x+2/5*f^(5/2)*m*arctan(x*f^(1/2)/e^(1/2))*(a+b*ln(c*x^n))/e^(5/2 )-1/25*b*n*ln(d*(f*x^2+e)^m)/x^5-1/5*(a+b*ln(c*x^n))*ln(d*(f*x^2+e)^m)/x^5 -1/5*I*b*f^(5/2)*m*n*polylog(2,-I*x*f^(1/2)/e^(1/2))/e^(5/2)+1/5*I*b*f^(5/ 2)*m*n*polylog(2,I*x*f^(1/2)/e^(1/2))/e^(5/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.15 (sec) , antiderivative size = 399, normalized size of antiderivative = 1.49 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^6} \, dx=-\frac {16 b e^{3/2} f m n x^2-108 b \sqrt {e} f^2 m n x^4-18 b f^{5/2} m n x^5 \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )+30 a e^{3/2} f m x^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\frac {f x^2}{e}\right )+90 b f^{5/2} m n x^5 \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \log (x)+30 b e^{3/2} f m x^2 \log \left (c x^n\right )-90 b \sqrt {e} f^2 m x^4 \log \left (c x^n\right )-90 b f^{5/2} m x^5 \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \log \left (c x^n\right )-45 i b f^{5/2} m n x^5 \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+45 i b f^{5/2} m n x^5 \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )+45 a e^{5/2} \log \left (d \left (e+f x^2\right )^m\right )+9 b e^{5/2} n \log \left (d \left (e+f x^2\right )^m\right )+45 b e^{5/2} \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )+45 i b f^{5/2} m n x^5 \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-45 i b f^{5/2} m n x^5 \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{225 e^{5/2} x^5} \]
-1/225*(16*b*e^(3/2)*f*m*n*x^2 - 108*b*Sqrt[e]*f^2*m*n*x^4 - 18*b*f^(5/2)* m*n*x^5*ArcTan[(Sqrt[f]*x)/Sqrt[e]] + 30*a*e^(3/2)*f*m*x^2*Hypergeometric2 F1[-3/2, 1, -1/2, -((f*x^2)/e)] + 90*b*f^(5/2)*m*n*x^5*ArcTan[(Sqrt[f]*x)/ Sqrt[e]]*Log[x] + 30*b*e^(3/2)*f*m*x^2*Log[c*x^n] - 90*b*Sqrt[e]*f^2*m*x^4 *Log[c*x^n] - 90*b*f^(5/2)*m*x^5*ArcTan[(Sqrt[f]*x)/Sqrt[e]]*Log[c*x^n] - (45*I)*b*f^(5/2)*m*n*x^5*Log[x]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] + (45*I)*b* f^(5/2)*m*n*x^5*Log[x]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] + 45*a*e^(5/2)*Log[d *(e + f*x^2)^m] + 9*b*e^(5/2)*n*Log[d*(e + f*x^2)^m] + 45*b*e^(5/2)*Log[c* x^n]*Log[d*(e + f*x^2)^m] + (45*I)*b*f^(5/2)*m*n*x^5*PolyLog[2, ((-I)*Sqrt [f]*x)/Sqrt[e]] - (45*I)*b*f^(5/2)*m*n*x^5*PolyLog[2, (I*Sqrt[f]*x)/Sqrt[e ]])/(e^(5/2)*x^5)
Time = 0.41 (sec) , antiderivative size = 260, normalized size of antiderivative = 0.97, 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.
\(\displaystyle \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^6} \, dx\) |
\(\Big \downarrow \) 2823 |
\(\displaystyle -b n \int \left (\frac {2 m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) f^{5/2}}{5 e^{5/2} x}+\frac {2 m f^2}{5 e^2 x^2}-\frac {2 m f}{15 e x^4}-\frac {\log \left (d \left (f x^2+e\right )^m\right )}{5 x^6}\right )dx+\frac {2 f^{5/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}+\frac {2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}-\frac {2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 f^{5/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}+\frac {2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}-\frac {2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}-b n \left (-\frac {2 f^{5/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{25 e^{5/2}}+\frac {\log \left (d \left (e+f x^2\right )^m\right )}{25 x^5}+\frac {i f^{5/2} m \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{5 e^{5/2}}-\frac {i f^{5/2} m \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{5 e^{5/2}}-\frac {12 f^2 m}{25 e^2 x}+\frac {16 f m}{225 e x^3}\right )\) |
(-2*f*m*(a + b*Log[c*x^n]))/(15*e*x^3) + (2*f^2*m*(a + b*Log[c*x^n]))/(5*e ^2*x) + (2*f^(5/2)*m*ArcTan[(Sqrt[f]*x)/Sqrt[e]]*(a + b*Log[c*x^n]))/(5*e^ (5/2)) - ((a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m])/(5*x^5) - b*n*((16*f*m) /(225*e*x^3) - (12*f^2*m)/(25*e^2*x) - (2*f^(5/2)*m*ArcTan[(Sqrt[f]*x)/Sqr t[e]])/(25*e^(5/2)) + Log[d*(e + f*x^2)^m]/(25*x^5) + ((I/5)*f^(5/2)*m*Pol yLog[2, ((-I)*Sqrt[f]*x)/Sqrt[e]])/e^(5/2) - ((I/5)*f^(5/2)*m*PolyLog[2, ( I*Sqrt[f]*x)/Sqrt[e]])/e^(5/2))
3.1.99.3.1 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 = 158.25 (sec) , antiderivative size = 1146, normalized size of antiderivative = 4.29
-1/5*I*m*f^3/e^2/(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)-2/5*m*f^3*b/e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*n* ln(x)+1/5*m*f^3*b*n/e^2*ln(x)/(-e*f)^(1/2)*ln((-f*x+(-e*f)^(1/2))/(-e*f)^( 1/2))-1/5*m*f^3*b*n/e^2*ln(x)/(-e*f)^(1/2)*ln((f*x+(-e*f)^(1/2))/(-e*f)^(1 /2))+1/15*I*m*f/e/x^3*b*Pi*csgn(I*c*x^n)^3-1/5*I*m*f^2/e^2/x*b*Pi*csgn(I*c )*csgn(I*x^n)*csgn(I*c*x^n)+2/5*m*f^3*b/e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^( 1/2))*ln(x^n)+1/5*m*f^3*b*n/e^2/(-e*f)^(1/2)*dilog((-f*x+(-e*f)^(1/2))/(-e *f)^(1/2))-1/5*m*f^3*b*n/e^2/(-e*f)^(1/2)*dilog((f*x+(-e*f)^(1/2))/(-e*f)^ (1/2))+2/5*m*f^3/e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*ln(c)+2/25*m*f^ 3/e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*n-1/5*I*m*f^3/e^2/(e*f)^(1/2)* arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*c*x^n)^3+1/5*I*m*f^2/e^2/x*b*Pi*csgn(I *c)*csgn(I*c*x^n)^2+1/5*I*m*f^2/e^2/x*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/1 5*I*m*f/e/x^3*b*Pi*csgn(I*c)*csgn(I*c*x^n)^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*(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))*(-1/5*(-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^5-2/5*b/x^5*ln(x^n)-2/25*b/x^5*n)-1/5* I*m*f^2/e^2/x*b*Pi*csgn(I*c*x^n)^3+1/15*I*m*f/e/x^3*b*Pi*csgn(I*c)*csgn(I* x^n)*csgn(I*c*x^n)+1/5*I*m*f^3/e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*...
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^6} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{6}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^6} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^6} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^6} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{6}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^6} \, dx=\int \frac {\ln \left (d\,{\left (f\,x^2+e\right )}^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^6} \,d x \]