\(\int \frac {(a+b \log (c x^n)) \log (d (e+f x^2)^m)}{x} \, dx\) [98]

Optimal result

Integrand size = 26, antiderivative size = 113 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x} \, dx=\frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (d \left (e+f x^2\right )^m\right )}{2 b n}-\frac {m \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {f x^2}{e}\right )}{2 b n}-\frac {1}{2} m \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {f x^2}{e}\right )+\frac {1}{4} b m n \operatorname {PolyLog}\left (3,-\frac {f x^2}{e}\right ) \] Output:

1/2*(a+b*ln(c*x^n))^2*ln(d*(f*x^2+e)^m)/b/n-1/2*m*(a+b*ln(c*x^n))^2*ln(1+f 
*x^2/e)/b/n-1/2*m*(a+b*ln(c*x^n))*polylog(2,-f*x^2/e)+1/4*b*m*n*polylog(3, 
-f*x^2/e)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.63 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x} \, dx=\frac {1}{2} \left (b m n \log ^2(x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-2 b m \log (x) \log \left (c x^n\right ) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+b m n \log ^2(x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )-2 b m \log (x) \log \left (c x^n\right ) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )-b n \log ^2(x) \log \left (d \left (e+f x^2\right )^m\right )+a \log \left (-\frac {f x^2}{e}\right ) \log \left (d \left (e+f x^2\right )^m\right )+2 b \log (x) \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )-2 b m \log \left (c x^n\right ) \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-2 b m \log \left (c x^n\right ) \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )+a m \operatorname {PolyLog}\left (2,1+\frac {f x^2}{e}\right )+2 b m n \operatorname {PolyLog}\left (3,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+2 b m n \operatorname {PolyLog}\left (3,\frac {i \sqrt {f} x}{\sqrt {e}}\right )\right ) \] Input:

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

Output:

(b*m*n*Log[x]^2*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] - 2*b*m*Log[x]*Log[c*x^n]*L 
og[1 - (I*Sqrt[f]*x)/Sqrt[e]] + b*m*n*Log[x]^2*Log[1 + (I*Sqrt[f]*x)/Sqrt[ 
e]] - 2*b*m*Log[x]*Log[c*x^n]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] - b*n*Log[x]^ 
2*Log[d*(e + f*x^2)^m] + a*Log[-((f*x^2)/e)]*Log[d*(e + f*x^2)^m] + 2*b*Lo 
g[x]*Log[c*x^n]*Log[d*(e + f*x^2)^m] - 2*b*m*Log[c*x^n]*PolyLog[2, ((-I)*S 
qrt[f]*x)/Sqrt[e]] - 2*b*m*Log[c*x^n]*PolyLog[2, (I*Sqrt[f]*x)/Sqrt[e]] + 
a*m*PolyLog[2, 1 + (f*x^2)/e] + 2*b*m*n*PolyLog[3, ((-I)*Sqrt[f]*x)/Sqrt[e 
]] + 2*b*m*n*PolyLog[3, (I*Sqrt[f]*x)/Sqrt[e]])/2
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2822, 2775, 2821, 7143}

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.

\(\displaystyle \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x} \, dx\)

\(\Big \downarrow \) 2822

\(\displaystyle \frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (d \left (e+f x^2\right )^m\right )}{2 b n}-\frac {f m \int \frac {x \left (a+b \log \left (c x^n\right )\right )^2}{f x^2+e}dx}{b n}\)

\(\Big \downarrow \) 2775

\(\displaystyle \frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (d \left (e+f x^2\right )^m\right )}{2 b n}-\frac {f m \left (\frac {\log \left (\frac {f x^2}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{2 f}-\frac {b n \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (\frac {f x^2}{e}+1\right )}{x}dx}{f}\right )}{b n}\)

\(\Big \downarrow \) 2821

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

((a + b*Log[c*x^n])^2*Log[d*(e + f*x^2)^m])/(2*b*n) - (f*m*(((a + b*Log[c* 
x^n])^2*Log[1 + (f*x^2)/e])/(2*f) - (b*n*(-1/2*((a + b*Log[c*x^n])*PolyLog 
[2, -((f*x^2)/e)]) + (b*n*PolyLog[3, -((f*x^2)/e)])/4))/f))/(b*n)
 

Defintions of rubi rules used

rule 2775

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) 
+ (e_.)*(x_)^(r_)), x_Symbol] :> Simp[f^m*Log[1 + e*(x^r/d)]*((a + b*Log[c* 
x^n])^p/(e*r)), x] - Simp[b*f^m*n*(p/(e*r))   Int[Log[1 + e*(x^r/d)]*((a + 
b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, r}, x] & 
& EqQ[m, r - 1] && IGtQ[p, 0] && (IntegerQ[m] || GtQ[f, 0]) && NeQ[r, n]
 

rule 2821

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b 
_.))^(p_.))/(x_), x_Symbol] :> Simp[(-PolyLog[2, (-d)*f*x^m])*((a + b*Log[c 
*x^n])^p/m), x] + Simp[b*n*(p/m)   Int[PolyLog[2, (-d)*f*x^m]*((a + b*Log[c 
*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 
0] && EqQ[d*e, 1]
 

rule 2822

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_ 
.)]*(b_.))^(p_.))/(x_), x_Symbol] :> Simp[Log[d*(e + f*x^m)^r]*((a + b*Log[ 
c*x^n])^(p + 1)/(b*n*(p + 1))), x] - Simp[f*m*(r/(b*n*(p + 1)))   Int[x^(m 
- 1)*((a + b*Log[c*x^n])^(p + 1)/(e + f*x^m)), x], x] /; FreeQ[{a, b, c, d, 
 e, f, r, m, n}, x] && IGtQ[p, 0] && NeQ[d*e, 1]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

Maple [C] (warning: unable to verify)

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

Time = 18.55 (sec) , antiderivative size = 649, normalized size of antiderivative = 5.74

method	result	size
risch	\(\frac {n b \ln \left (x \right )^{2} \ln \left (\left (f \,x^{2}+e \right )^{m}\right )}{2}-\frac {n b m \ln \left (x \right )^{2} \ln \left (1+\frac {f \,x^{2}}{e}\right )}{2}-\frac {n b m \ln \left (x \right ) \operatorname {polylog}\left (2, -\frac {f \,x^{2}}{e}\right )}{2}+\frac {b m n \operatorname {polylog}\left (3, -\frac {f \,x^{2}}{e}\right )}{4}+\frac {\left (\ln \left (\left (f \,x^{2}+e \right )^{m}\right )-m \ln \left (f \,x^{2}+e \right )\right ) \left (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 b \left (\ln \left (x^{n}\right )-n \ln \left (x \right )\right )+2 a \right ) \ln \left (x \right )}{2}+\frac {m \left (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 b \left (\ln \left (x^{n}\right )-n \ln \left (x \right )\right )+2 a \right ) \left (\ln \left (x \right ) \ln \left (f \,x^{2}+e \right )-2 f \left (\frac {\ln \left (x \right ) \left (\ln \left (\frac {-f x +\sqrt {-e f}}{\sqrt {-e f}}\right )+\ln \left (\frac {f x +\sqrt {-e f}}{\sqrt {-e f}}\right )\right )}{2 f}+\frac {\operatorname {dilog}\left (\frac {-f x +\sqrt {-e f}}{\sqrt {-e f}}\right )+\operatorname {dilog}\left (\frac {f x +\sqrt {-e f}}{\sqrt {-e f}}\right )}{2 f}\right )\right )}{2}+\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 (i \pi \ln \left (x \right ) b \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i \pi \ln \left (x \right ) b \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+2 \ln \left (x \right ) a +2 \ln \left (c \right ) b \ln \left (x \right )+\frac {b \ln \left (x^{n}\right )^{2}}{n}-i \pi \ln \left (x \right ) b \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-i \pi \ln \left (x \right ) b \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )\right )\)	\(649\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

-1/2*(b*n*log(x)^2 - 2*b*log(x)*log(x^n) - 2*(b*log(c) + a)*log(x))*log((f 
*x^2 + e)^m) - integrate(-(b*f*m*n*x^2*log(x)^2 + b*e*log(c)*log(d) - 2*(b 
*f*m*log(c) + a*f*m)*x^2*log(x) + (b*f*log(c)*log(d) + a*f*log(d))*x^2 + a 
*e*log(d) - (2*b*f*m*x^2*log(x) - b*f*x^2*log(d) - b*e*log(d))*log(x^n))/( 
f*x^3 + e*x), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(4*int(log((e + f*x**2)**m*d)/(e*x + f*x**3),x)*a*e*m + 4*int((log((e + f* 
x**2)**m*d)*log(x**n*c))/x,x)*b*m + log((e + f*x**2)**m*d)**2*a)/(4*m)