\(\int \frac {(a+b \log (c x^n)) \log (d (e+f x^2)^m)}{x^6} \, dx\) [105]

Optimal result

Integrand size = 26, antiderivative size = 267 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^6} \, dx=-\frac {16 b f m n}{225 e x^3}+\frac {12 b f^2 m n}{25 e^2 x}+\frac {2 b f^{5/2} m n \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{25 e^{5/2}}-\frac {2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}+\frac {2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}+\frac {2 f^{5/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac {b n \log \left (d \left (e+f x^2\right )^m\right )}{25 x^5}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}-\frac {i b f^{5/2} m n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{5 e^{5/2}}+\frac {i b f^{5/2} m n \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{5 e^{5/2}} \] Output:

-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(f^(1 
/2)*x/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(f^(1/2)*x/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*f^(1/2)*x/e^(1/2))/e^(5/2)+1/5*I*b*f^(5/ 
2)*m*n*polylog(2,I*f^(1/2)*x/e^(1/2))/e^(5/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.28 (sec) , antiderivative size = 399, normalized size of antiderivative = 1.49 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^6} \, dx=-\frac {16 b e^{3/2} f m n x^2-108 b \sqrt {e} f^2 m n x^4-18 b f^{5/2} m n x^5 \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )+30 a e^{3/2} f m x^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\frac {f x^2}{e}\right )+90 b f^{5/2} m n x^5 \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \log (x)+30 b e^{3/2} f m x^2 \log \left (c x^n\right )-90 b \sqrt {e} f^2 m x^4 \log \left (c x^n\right )-90 b f^{5/2} m x^5 \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \log \left (c x^n\right )-45 i b f^{5/2} m n x^5 \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+45 i b f^{5/2} m n x^5 \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )+45 a e^{5/2} \log \left (d \left (e+f x^2\right )^m\right )+9 b e^{5/2} n \log \left (d \left (e+f x^2\right )^m\right )+45 b e^{5/2} \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )+45 i b f^{5/2} m n x^5 \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-45 i b f^{5/2} m n x^5 \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{225 e^{5/2} x^5} \] Input:

Integrate[((a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m])/x^6,x]
 

Output:

-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)
 

Rubi [A] (verified)

Time = 0.46 (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

\(\displaystyle \frac {2 f^{5/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}+\frac {2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}-\frac {2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}-b n \left (-\frac {2 f^{5/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{25 e^{5/2}}+\frac {\log \left (d \left (e+f x^2\right )^m\right )}{25 x^5}+\frac {i f^{5/2} m \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{5 e^{5/2}}-\frac {i f^{5/2} m \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{5 e^{5/2}}-\frac {12 f^2 m}{25 e^2 x}+\frac {16 f m}{225 e x^3}\right )\)

Input:

Int[((a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m])/x^6,x]
 

Output:

(-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))
 

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 = 189.00 (sec) , antiderivative size = 1146, normalized size of antiderivative = 4.29

method	result	size
risch	\(\text {Expression too large to display}\)	\(1146\)

Input:

int((a+b*ln(c*x^n))*ln(d*(f*x^2+e)^m)/x^6,x,method=_RETURNVERBOSE)
 

Output:

-2/15*m*f/e/x^3*a+(-1/5*b/x^5*ln(x^n)-1/50*(5*I*b*Pi*csgn(I*x^n)*csgn(I*c* 
x^n)^2-5*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-5*I*b*Pi*csgn(I*c*x^n) 
^3+5*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)+10*b*ln(c)+2*n*b+10*a)/x^5)*ln((f*x^ 
2+e)^m)-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))-1/5*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*x^n)^3-1/5* 
I*m*f^3/e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*x^n)*csgn(I*c* 
x^n)*csgn(I*c)-2/15*m*f/e/x^3*b*ln(c)+(-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))*(-1/5*(I*Pi*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^5-2/5*b/x^5*ln(x^n)-2/25*b/x^5*n)+2/5*m*f^3/e 
^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*a+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) 
)*n*b+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^2/e^2/x 
*a+12/25*b*f^2*m*n/e^2/x+2/5*m*f^2/e^2/x*b*ln(c)+2/5*m*f^2*b*ln(x^n)/e^...
 

Fricas [F]

\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^6} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{6}} \,d x } \] Input:

integrate((a+b*log(c*x^n))*log(d*(f*x^2+e)^m)/x^6,x, algorithm="fricas")
 

Output:

integral((b*log(c*x^n) + a)*log((f*x^2 + e)^m*d)/x^6, x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^6} \, dx=\text {Timed out} \] Input:

integrate((a+b*ln(c*x**n))*ln(d*(f*x**2+e)**m)/x**6,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^6} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((a+b*log(c*x^n))*log(d*(f*x^2+e)^m)/x^6,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
 

Giac [F]

\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^6} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{6}} \,d x } \] Input:

integrate((a+b*log(c*x^n))*log(d*(f*x^2+e)^m)/x^6,x, algorithm="giac")
 

Output:

integrate((b*log(c*x^n) + a)*log((f*x^2 + e)^m*d)/x^6, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^6} \, dx=\int \frac {\ln \left (d\,{\left (f\,x^2+e\right )}^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^6} \,d x \] Input:

int((log(d*(e + f*x^2)^m)*(a + b*log(c*x^n)))/x^6,x)
 

Output:

int((log(d*(e + f*x^2)^m)*(a + b*log(c*x^n)))/x^6, x)
 

Reduce [F]

\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^6} \, dx=\frac {150 \sqrt {f}\, \sqrt {e}\, \mathit {atan} \left (\frac {f x}{\sqrt {f}\, \sqrt {e}}\right ) a \,f^{2} m \,x^{5}+30 \sqrt {f}\, \sqrt {e}\, \mathit {atan} \left (\frac {f x}{\sqrt {f}\, \sqrt {e}}\right ) b \,f^{2} m n \,x^{5}-150 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{f \,x^{8}+e \,x^{6}}d x \right ) b \,e^{4} m \,x^{5}-75 \,\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) \mathrm {log}\left (x^{n} c \right ) b \,e^{3}-75 \,\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) a \,e^{3}-15 \,\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) b \,e^{3} n -30 \,\mathrm {log}\left (x^{n} c \right ) b \,e^{3} m -50 a \,e^{2} f m \,x^{2}+150 a e \,f^{2} m \,x^{4}-6 b \,e^{3} m n -10 b \,e^{2} f m n \,x^{2}+30 b e \,f^{2} m n \,x^{4}}{375 e^{3} x^{5}} \] Input:

int((a+b*log(c*x^n))*log(d*(f*x^2+e)^m)/x^6,x)
 

Output:

(150*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*f**2*m*x**5 + 30*sqrt 
(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*f**2*m*n*x**5 - 150*int(log(x* 
*n*c)/(e*x**6 + f*x**8),x)*b*e**4*m*x**5 - 75*log((e + f*x**2)**m*d)*log(x 
**n*c)*b*e**3 - 75*log((e + f*x**2)**m*d)*a*e**3 - 15*log((e + f*x**2)**m* 
d)*b*e**3*n - 30*log(x**n*c)*b*e**3*m - 50*a*e**2*f*m*x**2 + 150*a*e*f**2* 
m*x**4 - 6*b*e**3*m*n - 10*b*e**2*f*m*n*x**2 + 30*b*e*f**2*m*n*x**4)/(375* 
e**3*x**5)