\(\int \frac {(a+b \log (c x^n)) \log (d (e+f x^2)^m)}{x^5} \, dx\) [100]

Optimal result

Integrand size = 26, antiderivative size = 248 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^5} \, dx=-\frac {3 b f m n}{16 e x^2}-\frac {b f^2 m n \log (x)}{8 e^2}+\frac {b f^2 m n \log ^2(x)}{4 e^2}-\frac {f m \left (a+b \log \left (c x^n\right )\right )}{4 e x^2}-\frac {f^2 m \log (x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {b f^2 m n \log \left (e+f x^2\right )}{16 e^2}-\frac {b f^2 m n \log \left (-\frac {f x^2}{e}\right ) \log \left (e+f x^2\right )}{8 e^2}+\frac {f^2 m \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^2\right )}{4 e^2}-\frac {b n \log \left (d \left (e+f x^2\right )^m\right )}{16 x^4}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{4 x^4}-\frac {b f^2 m n \operatorname {PolyLog}\left (2,1+\frac {f x^2}{e}\right )}{8 e^2} \] Output:

-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
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.21 (sec) , antiderivative size = 363, normalized size of antiderivative = 1.46 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^5} \, dx=-\frac {4 a e f m x^2+3 b e f m n x^2+8 a f^2 m x^4 \log (x)+2 b f^2 m n x^4 \log (x)-4 b f^2 m n x^4 \log ^2(x)+4 b e f m x^2 \log \left (c x^n\right )+8 b f^2 m x^4 \log (x) \log \left (c x^n\right )-4 b f^2 m n x^4 \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-4 b f^2 m n x^4 \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )-4 a f^2 m x^4 \log \left (e+f x^2\right )-b f^2 m n x^4 \log \left (e+f x^2\right )+4 b f^2 m n x^4 \log (x) \log \left (e+f x^2\right )-4 b f^2 m x^4 \log \left (c x^n\right ) \log \left (e+f x^2\right )+4 a e^2 \log \left (d \left (e+f x^2\right )^m\right )+b e^2 n \log \left (d \left (e+f x^2\right )^m\right )+4 b e^2 \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )-4 b f^2 m n x^4 \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-4 b f^2 m n x^4 \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{16 e^2 x^4} \] Input:

Integrate[((a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m])/x^5,x]
 

Output:

-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)
 

Rubi [A] (verified)

Time = 0.53 (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

\(\displaystyle -\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{4 x^4}-\frac {f^2 m \log (x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {f^2 m \log \left (e+f x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{4 e^2}-\frac {f m \left (a+b \log \left (c x^n\right )\right )}{4 e x^2}-b n \left (\frac {\log \left (d \left (e+f x^2\right )^m\right )}{16 x^4}+\frac {f^2 m \operatorname {PolyLog}\left (2,\frac {f x^2}{e}+1\right )}{8 e^2}-\frac {f^2 m \log \left (e+f x^2\right )}{16 e^2}+\frac {f^2 m \log \left (-\frac {f x^2}{e}\right ) \log \left (e+f x^2\right )}{8 e^2}-\frac {f^2 m \log ^2(x)}{4 e^2}+\frac {f^2 m \log (x)}{8 e^2}+\frac {3 f m}{16 e x^2}\right )\)

Input:

Int[((a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m])/x^5,x]
 

Output:

-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))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2823

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]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 106.10 (sec) , antiderivative size = 1074, normalized size of antiderivative = 4.33

method	result	size
risch	\(\text {Expression too large to display}\)	\(1074\)

Input:

int((a+b*ln(c*x^n))*ln(d*(f*x^2+e)^m)/x^5,x,method=_RETURNVERBOSE)
 

Output:

1/4*m*f^2*b*n/e^2*ln(x)*ln((-f*x+(-e*f)^(1/2))/(-e*f)^(1/2))+1/4*m*f^2*b*n 
/e^2*ln(x)*ln((f*x+(-e*f)^(1/2))/(-e*f)^(1/2))-1/2*m*f^2/e^2*ln(x)*a+(-1/4 
*b/x^4*ln(x^n)-1/16*(2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-2*I*b*Pi*csgn(I* 
x^n)*csgn(I*c*x^n)*csgn(I*c)-2*I*b*Pi*csgn(I*c*x^n)^3+2*I*b*Pi*csgn(I*c*x^ 
n)^2*csgn(I*c)+4*b*ln(c)+n*b+4*a)/x^4)*ln((f*x^2+e)^m)+1/4*m*f^2/e^2*ln(f* 
x^2+e)*a+(-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))*(-1/4*(I*P 
i*b*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*csgn(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^ 
4-1/2*b/x^4*ln(x^n)-1/8*b/x^4*n)-1/4*m*f/e/x^2*a-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^2-1/4*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*m*f*b*ln(x^n)/e/x^2+1/4*m*f^2/e^2*ln(f*x^2 
+e)*b*ln(c)-1/2*m*f^2*b*ln(x^n)/e^2*ln(x)+1/4*m*f^2*b*ln(x^n)/e^2*ln(f*x^2 
+e)+1/4*m*f^2*b*n/e^2*dilog((-f*x+(-e*f)^(1/2))/(-e*f)^(1/2))+1/4*m*f^2*b* 
n/e^2*dilog((f*x+(-e*f)^(1/2))/(-e*f)^(1/2))-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*c*x^n)^3+1/8*I*m*f/e/x^2*b*Pi*csgn( 
I*c*x^n)^3+1/4*I*m*f^2/e^2*ln(x)*b*Pi*csgn(I*c*x^n)^3-1/8*I*m*f/e/x^2*b*Pi 
*csgn(I*x^n)*csgn(I*c*x^n)^2-1/8*I*m*f/e/x^2*b*Pi*csgn(I*c*x^n)^2*csgn(I*c 
)-1/4*I*m*f^2/e^2*ln(x)*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/4*I*m*f^2/e^...
 

Fricas [F]

\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^5} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{5}} \,d x } \] Input:

integrate((a+b*log(c*x^n))*log(d*(f*x^2+e)^m)/x^5,x, algorithm="fricas")
 

Output:

integral((b*log(c*x^n) + a)*log((f*x^2 + e)^m*d)/x^5, x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^5} \, dx=\text {Timed out} \] Input:

integrate((a+b*ln(c*x**n))*ln(d*(f*x**2+e)**m)/x**5,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^5} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{5}} \,d x } \] Input:

integrate((a+b*log(c*x^n))*log(d*(f*x^2+e)^m)/x^5,x, algorithm="maxima")
 

Output:

-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)
 

Giac [F]

\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^5} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{5}} \,d x } \] Input:

integrate((a+b*log(c*x^n))*log(d*(f*x^2+e)^m)/x^5,x, algorithm="giac")
 

Output:

integrate((b*log(c*x^n) + a)*log((f*x^2 + e)^m*d)/x^5, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^5} \, dx=\int \frac {\ln \left (d\,{\left (f\,x^2+e\right )}^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^5} \,d x \] Input:

int((log(d*(e + f*x^2)^m)*(a + b*log(c*x^n)))/x^5,x)
 

Output:

int((log(d*(e + f*x^2)^m)*(a + b*log(c*x^n)))/x^5, x)
 

Reduce [F]

\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^5} \, dx=\frac {-16 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{f \,x^{7}+e \,x^{5}}d x \right ) b \,e^{3} m \,x^{4}-8 \,\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) \mathrm {log}\left (x^{n} c \right ) b \,e^{2}-8 \,\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) a \,e^{2}+8 \,\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) a \,f^{2} x^{4}-2 \,\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) b \,e^{2} n +2 \,\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) b \,f^{2} n \,x^{4}-4 \,\mathrm {log}\left (x^{n} c \right ) b \,e^{2} m -16 \,\mathrm {log}\left (x \right ) a \,f^{2} m \,x^{4}-4 \,\mathrm {log}\left (x \right ) b \,f^{2} m n \,x^{4}-8 a e f m \,x^{2}-b \,e^{2} m n -2 b e f m n \,x^{2}}{32 e^{2} x^{4}} \] Input:

int((a+b*log(c*x^n))*log(d*(f*x^2+e)^m)/x^5,x)
 

Output:

( - 16*int(log(x**n*c)/(e*x**5 + f*x**7),x)*b*e**3*m*x**4 - 8*log((e + f*x 
**2)**m*d)*log(x**n*c)*b*e**2 - 8*log((e + f*x**2)**m*d)*a*e**2 + 8*log((e 
 + f*x**2)**m*d)*a*f**2*x**4 - 2*log((e + f*x**2)**m*d)*b*e**2*n + 2*log(( 
e + f*x**2)**m*d)*b*f**2*n*x**4 - 4*log(x**n*c)*b*e**2*m - 16*log(x)*a*f** 
2*m*x**4 - 4*log(x)*b*f**2*m*n*x**4 - 8*a*e*f*m*x**2 - b*e**2*m*n - 2*b*e* 
f*m*n*x**2)/(32*e**2*x**4)