Integrand size = 26, antiderivative size = 248 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^5} \, dx=-\frac {3 b f m n}{16 e x^2}-\frac {b f^2 m n \log (x)}{8 e^2}+\frac {b f^2 m n \log ^2(x)}{4 e^2}-\frac {f m \left (a+b \log \left (c x^n\right )\right )}{4 e x^2}-\frac {f^2 m \log (x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {b f^2 m n \log \left (e+f x^2\right )}{16 e^2}-\frac {b f^2 m n \log \left (-\frac {f x^2}{e}\right ) \log \left (e+f x^2\right )}{8 e^2}+\frac {f^2 m \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^2\right )}{4 e^2}-\frac {b n \log \left (d \left (e+f x^2\right )^m\right )}{16 x^4}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{4 x^4}-\frac {b f^2 m n \operatorname {PolyLog}\left (2,1+\frac {f x^2}{e}\right )}{8 e^2} \]
-3/16*b*f*m*n/e/x^2-1/8*b*f^2*m*n*ln(x)/e^2+1/4*b*f^2*m*n*ln(x)^2/e^2-1/4* f*m*(a+b*ln(c*x^n))/e/x^2-1/2*f^2*m*ln(x)*(a+b*ln(c*x^n))/e^2+1/16*b*f^2*m *n*ln(f*x^2+e)/e^2-1/8*b*f^2*m*n*ln(-f*x^2/e)*ln(f*x^2+e)/e^2+1/4*f^2*m*(a +b*ln(c*x^n))*ln(f*x^2+e)/e^2-1/16*b*n*ln(d*(f*x^2+e)^m)/x^4-1/4*(a+b*ln(c *x^n))*ln(d*(f*x^2+e)^m)/x^4-1/8*b*f^2*m*n*polylog(2,1+f*x^2/e)/e^2
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 363, normalized size of antiderivative = 1.46 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^5} \, dx=-\frac {4 a e f m x^2+3 b e f m n x^2+8 a f^2 m x^4 \log (x)+2 b f^2 m n x^4 \log (x)-4 b f^2 m n x^4 \log ^2(x)+4 b e f m x^2 \log \left (c x^n\right )+8 b f^2 m x^4 \log (x) \log \left (c x^n\right )-4 b f^2 m n x^4 \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-4 b f^2 m n x^4 \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )-4 a f^2 m x^4 \log \left (e+f x^2\right )-b f^2 m n x^4 \log \left (e+f x^2\right )+4 b f^2 m n x^4 \log (x) \log \left (e+f x^2\right )-4 b f^2 m x^4 \log \left (c x^n\right ) \log \left (e+f x^2\right )+4 a e^2 \log \left (d \left (e+f x^2\right )^m\right )+b e^2 n \log \left (d \left (e+f x^2\right )^m\right )+4 b e^2 \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )-4 b f^2 m n x^4 \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-4 b f^2 m n x^4 \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{16 e^2 x^4} \]
-1/16*(4*a*e*f*m*x^2 + 3*b*e*f*m*n*x^2 + 8*a*f^2*m*x^4*Log[x] + 2*b*f^2*m* n*x^4*Log[x] - 4*b*f^2*m*n*x^4*Log[x]^2 + 4*b*e*f*m*x^2*Log[c*x^n] + 8*b*f ^2*m*x^4*Log[x]*Log[c*x^n] - 4*b*f^2*m*n*x^4*Log[x]*Log[1 - (I*Sqrt[f]*x)/ Sqrt[e]] - 4*b*f^2*m*n*x^4*Log[x]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] - 4*a*f^2 *m*x^4*Log[e + f*x^2] - b*f^2*m*n*x^4*Log[e + f*x^2] + 4*b*f^2*m*n*x^4*Log [x]*Log[e + f*x^2] - 4*b*f^2*m*x^4*Log[c*x^n]*Log[e + f*x^2] + 4*a*e^2*Log [d*(e + f*x^2)^m] + b*e^2*n*Log[d*(e + f*x^2)^m] + 4*b*e^2*Log[c*x^n]*Log[ d*(e + f*x^2)^m] - 4*b*f^2*m*n*x^4*PolyLog[2, ((-I)*Sqrt[f]*x)/Sqrt[e]] - 4*b*f^2*m*n*x^4*PolyLog[2, (I*Sqrt[f]*x)/Sqrt[e]])/(e^2*x^4)
Time = 0.47 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.96, 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.
| \(\displaystyle \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^5} \, dx\) |
| \(\Big \downarrow \) 2823 |
| \(\displaystyle -b n \int \left (-\frac {m \log (x) f^2}{2 e^2 x}+\frac {m \log \left (f x^2+e\right ) f^2}{4 e^2 x}-\frac {m f}{4 e x^3}-\frac {\log \left (d \left (f x^2+e\right )^m\right )}{4 x^5}\right )dx-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{4 x^4}-\frac {f^2 m \log (x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {f^2 m \log \left (e+f x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{4 e^2}-\frac {f m \left (a+b \log \left (c x^n\right )\right )}{4 e x^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{4 x^4}-\frac {f^2 m \log (x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {f^2 m \log \left (e+f x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{4 e^2}-\frac {f m \left (a+b \log \left (c x^n\right )\right )}{4 e x^2}-b n \left (\frac {\log \left (d \left (e+f x^2\right )^m\right )}{16 x^4}+\frac {f^2 m \operatorname {PolyLog}\left (2,\frac {f x^2}{e}+1\right )}{8 e^2}-\frac {f^2 m \log \left (e+f x^2\right )}{16 e^2}+\frac {f^2 m \log \left (-\frac {f x^2}{e}\right ) \log \left (e+f x^2\right )}{8 e^2}-\frac {f^2 m \log ^2(x)}{4 e^2}+\frac {f^2 m \log (x)}{8 e^2}+\frac {3 f m}{16 e x^2}\right )\) |
-1/4*(f*m*(a + b*Log[c*x^n]))/(e*x^2) - (f^2*m*Log[x]*(a + b*Log[c*x^n]))/ (2*e^2) + (f^2*m*(a + b*Log[c*x^n])*Log[e + f*x^2])/(4*e^2) - ((a + b*Log[ c*x^n])*Log[d*(e + f*x^2)^m])/(4*x^4) - b*n*((3*f*m)/(16*e*x^2) + (f^2*m*L og[x])/(8*e^2) - (f^2*m*Log[x]^2)/(4*e^2) - (f^2*m*Log[e + f*x^2])/(16*e^2 ) + (f^2*m*Log[-((f*x^2)/e)]*Log[e + f*x^2])/(8*e^2) + Log[d*(e + f*x^2)^m ]/(16*x^4) + (f^2*m*PolyLog[2, 1 + (f*x^2)/e])/(8*e^2))
3.1.94.3.1 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 = 98.26 (sec) , antiderivative size = 1074, normalized size of antiderivative = 4.33
1/4*I*m*f^2/e^2*ln(x)*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+(-1/4*b/x^4 *ln(x^n)-1/16*(-2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+2*I*b*Pi*csgn (I*c)*csgn(I*c*x^n)^2+2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-2*I*b*Pi*csgn(I *c*x^n)^3+4*b*ln(c)+b*n+4*a)/x^4)*ln((f*x^2+e)^m)-1/4*m*f*b*ln(x^n)/e/x^2- 1/4*b*f^2*m*n*ln(x)/e^2*ln(f*x^2+e)-1/4*m*f/e/x^2*b*ln(c)+1/4*I*m*f^2/e^2* ln(x)*b*Pi*csgn(I*c*x^n)^3-1/2*m*f^2/e^2*ln(x)*a-1/4*m*f/e/x^2*a+(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))*(-1/4*(-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^4-1/2*b/x^4*ln(x^n )-1/8*b/x^4*n)-1/8*b*f^2*m*n*ln(x)/e^2+1/4*b*f^2*m*n*ln(x)^2/e^2-3/16*b*f* m*n/e/x^2+1/8*I*m*f/e/x^2*b*Pi*csgn(I*c*x^n)^3-1/8*I*m*f^2/e^2*ln(f*x^2+e) *b*Pi*csgn(I*c*x^n)^3-1/2*m*f^2*b*ln(x^n)/e^2*ln(x)-1/2*m*f^2/e^2*ln(x)*b* ln(c)+1/8*I*m*f^2/e^2*ln(f*x^2+e)*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2+1/8*I*m *f^2/e^2*ln(f*x^2+e)*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+1/8*I*m*f/e/x^2*b*Pi*c sgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/8*I*m*f^2/e^2*ln(f*x^2+e)*b*Pi*csgn(I *c)*csgn(I*x^n)*csgn(I*c*x^n)-1/8*I*m*f/e/x^2*b*Pi*csgn(I*c)*csgn(I*c*x^n) ^2-1/8*I*m*f/e/x^2*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/4*I*m*f^2/e^2*ln(x)* b*Pi*csgn(I*c)*csgn(I*c*x^n)^2-1/4*I*m*f^2/e^2*ln(x)*b*Pi*csgn(I*x^n)*c...
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^5} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{5}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^5} \, dx=\text {Timed out} \]
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^5} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{5}} \,d x } \]
-1/16*(b*(n + 4*log(c)) + 4*b*log(x^n) + 4*a)*log((f*x^2 + e)^m)/x^4 + int egrate(1/8*(8*b*e*log(c)*log(d) + (4*(f*m + 2*f*log(d))*a + (f*m*n + 4*(f* m + 2*f*log(d))*log(c))*b)*x^2 + 8*a*e*log(d) + 4*((f*m + 2*f*log(d))*b*x^ 2 + 2*b*e*log(d))*log(x^n))/(f*x^7 + e*x^5), x)
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^5} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{5}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^5} \, dx=\int \frac {\ln \left (d\,{\left (f\,x^2+e\right )}^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^5} \,d x \]