\(\int \frac {(a+b \log (c x^n)) \log (d (e+f x^2)^m)}{x^4} \, dx\) [104]

Optimal result

Integrand size = 26, antiderivative size = 227 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^4} \, dx=-\frac {8 b f m n}{9 e x}-\frac {2 b f^{3/2} m n \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{9 e^{3/2}}-\frac {2 f m \left (a+b \log \left (c x^n\right )\right )}{3 e x}-\frac {2 f^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 e^{3/2}}-\frac {b n \log \left (d \left (e+f x^2\right )^m\right )}{9 x^3}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{3 x^3}+\frac {i b f^{3/2} m n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 e^{3/2}}-\frac {i b f^{3/2} m n \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 e^{3/2}} \] Output:

-8/9*b*f*m*n/e/x-2/9*b*f^(3/2)*m*n*arctan(f^(1/2)*x/e^(1/2))/e^(3/2)-2/3*f 
*m*(a+b*ln(c*x^n))/e/x-2/3*f^(3/2)*m*arctan(f^(1/2)*x/e^(1/2))*(a+b*ln(c*x 
^n))/e^(3/2)-1/9*b*n*ln(d*(f*x^2+e)^m)/x^3-1/3*(a+b*ln(c*x^n))*ln(d*(f*x^2 
+e)^m)/x^3+1/3*I*b*f^(3/2)*m*n*polylog(2,-I*f^(1/2)*x/e^(1/2))/e^(3/2)-1/3 
*I*b*f^(3/2)*m*n*polylog(2,I*f^(1/2)*x/e^(1/2))/e^(3/2)
 

Mathematica [C] (verified)

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

Time = 0.20 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.59 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^4} \, dx=\frac {-8 b \sqrt {e} f m n x^2-2 b f^{3/2} m n x^3 \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )-6 a \sqrt {e} f m x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\frac {f x^2}{e}\right )+6 b f^{3/2} m n x^3 \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \log (x)-6 b \sqrt {e} f m x^2 \log \left (c x^n\right )-6 b f^{3/2} m x^3 \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \log \left (c x^n\right )-3 i b f^{3/2} m n x^3 \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+3 i b f^{3/2} m n x^3 \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )-3 a e^{3/2} \log \left (d \left (e+f x^2\right )^m\right )-b e^{3/2} n \log \left (d \left (e+f x^2\right )^m\right )-3 b e^{3/2} \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )+3 i b f^{3/2} m n x^3 \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-3 i b f^{3/2} m n x^3 \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{9 e^{3/2} x^3} \] Input:

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

Output:

(-8*b*Sqrt[e]*f*m*n*x^2 - 2*b*f^(3/2)*m*n*x^3*ArcTan[(Sqrt[f]*x)/Sqrt[e]] 
- 6*a*Sqrt[e]*f*m*x^2*Hypergeometric2F1[-1/2, 1, 1/2, -((f*x^2)/e)] + 6*b* 
f^(3/2)*m*n*x^3*ArcTan[(Sqrt[f]*x)/Sqrt[e]]*Log[x] - 6*b*Sqrt[e]*f*m*x^2*L 
og[c*x^n] - 6*b*f^(3/2)*m*x^3*ArcTan[(Sqrt[f]*x)/Sqrt[e]]*Log[c*x^n] - (3* 
I)*b*f^(3/2)*m*n*x^3*Log[x]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] + (3*I)*b*f^(3/ 
2)*m*n*x^3*Log[x]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] - 3*a*e^(3/2)*Log[d*(e + 
f*x^2)^m] - b*e^(3/2)*n*Log[d*(e + f*x^2)^m] - 3*b*e^(3/2)*Log[c*x^n]*Log[ 
d*(e + f*x^2)^m] + (3*I)*b*f^(3/2)*m*n*x^3*PolyLog[2, ((-I)*Sqrt[f]*x)/Sqr 
t[e]] - (3*I)*b*f^(3/2)*m*n*x^3*PolyLog[2, (I*Sqrt[f]*x)/Sqrt[e]])/(9*e^(3 
/2)*x^3)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.98, 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.

\(\displaystyle \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^4} \, dx\)

\(\Big \downarrow \) 2823

\(\displaystyle -b n \int \left (-\frac {2 m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) f^{3/2}}{3 e^{3/2} x}-\frac {2 m f}{3 e x^2}-\frac {\log \left (d \left (f x^2+e\right )^m\right )}{3 x^4}\right )dx-\frac {2 f^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 e^{3/2}}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{3 x^3}-\frac {2 f m \left (a+b \log \left (c x^n\right )\right )}{3 e x}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 f^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 e^{3/2}}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{3 x^3}-\frac {2 f m \left (a+b \log \left (c x^n\right )\right )}{3 e x}-b n \left (\frac {2 f^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{9 e^{3/2}}+\frac {\log \left (d \left (e+f x^2\right )^m\right )}{9 x^3}-\frac {i f^{3/2} m \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 e^{3/2}}+\frac {i f^{3/2} m \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 e^{3/2}}+\frac {8 f m}{9 e x}\right )\)

Input:

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

Output:

(-2*f*m*(a + b*Log[c*x^n]))/(3*e*x) - (2*f^(3/2)*m*ArcTan[(Sqrt[f]*x)/Sqrt 
[e]]*(a + b*Log[c*x^n]))/(3*e^(3/2)) - ((a + b*Log[c*x^n])*Log[d*(e + f*x^ 
2)^m])/(3*x^3) - b*n*((8*f*m)/(9*e*x) + (2*f^(3/2)*m*ArcTan[(Sqrt[f]*x)/Sq 
rt[e]])/(9*e^(3/2)) + Log[d*(e + f*x^2)^m]/(9*x^3) - ((I/3)*f^(3/2)*m*Poly 
Log[2, ((-I)*Sqrt[f]*x)/Sqrt[e]])/e^(3/2) + ((I/3)*f^(3/2)*m*PolyLog[2, (I 
*Sqrt[f]*x)/Sqrt[e]])/e^(3/2))
 

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 = 41.11 (sec) , antiderivative size = 965, normalized size of antiderivative = 4.25

method	result	size
risch	\(\text {Expression too large to display}\)	\(965\)

Input:

int((a+b*ln(c*x^n))*ln(d*(f*x^2+e)^m)/x^4,x,method=_RETURNVERBOSE)
 

Output:

(-1/3*b/x^3*ln(x^n)-1/18*(3*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-3*I*b*Pi*cs 
gn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-3*I*b*Pi*csgn(I*c*x^n)^3+3*I*b*Pi*csgn(I 
*c*x^n)^2*csgn(I*c)+6*b*ln(c)+2*n*b+6*a)/x^3)*ln((f*x^2+e)^m)+(-1/4*I*Pi*c 
sgn(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))*(-1/3*(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^3-2/3*b/x^3*ln(x^n)-2 
/9*b/x^3*n)+1/3*I*m*f/e/x*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/3*I*m 
*f/e/x*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)-1/3*I*m*f^2/e/(e*f)^(1/2)*arctan(x*f 
/(e*f)^(1/2))*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/3*I*m*f/e/x*b*Pi*csgn(I*x 
^n)*csgn(I*c*x^n)^2-2/3*m*f/e/x*b*ln(c)-8/9*b*f*m*n/e/x-2/3*m*f/e/x*a+1/3* 
I*m*f/e/x*b*Pi*csgn(I*c*x^n)^3-1/3*I*m*f^2/e/(e*f)^(1/2)*arctan(x*f/(e*f)^ 
(1/2))*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)+1/3*I*m*f^2/e/(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)+1/3*I*m*f^2/e/(e*f) 
^(1/2)*arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*c*x^n)^3-2/3*m*f^2/e/(e*f)^(1/2 
)*arctan(x*f/(e*f)^(1/2))*b*ln(c)-2/9*m*f^2/e/(e*f)^(1/2)*arctan(x*f/(e*f) 
^(1/2))*n*b-2/3*m*f^2/e/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*a-2/3*m*f*b*ln 
(x^n)/e/x+2/3*m*f^2*b/e/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*n*ln(x)-2/3*m* 
f^2*b/e/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*ln(x^n)-1/3*m*f^2*b*n/e*ln(...
 

Fricas [F]

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

integrate((a+b*log(c*x^n))*log(d*(f*x^2+e)^m)/x^4,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

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

integrate((a+b*ln(c*x**n))*ln(d*(f*x**2+e)**m)/x**4,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^4} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((a+b*log(c*x^n))*log(d*(f*x^2+e)^m)/x^4,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]

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

integrate((a+b*log(c*x^n))*log(d*(f*x^2+e)^m)/x^4,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^4} \, dx=\frac {-18 \sqrt {f}\, \sqrt {e}\, \mathit {atan} \left (\frac {f x}{\sqrt {f}\, \sqrt {e}}\right ) a f m \,x^{3}-6 \sqrt {f}\, \sqrt {e}\, \mathit {atan} \left (\frac {f x}{\sqrt {f}\, \sqrt {e}}\right ) b f m n \,x^{3}-18 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{f \,x^{6}+e \,x^{4}}d x \right ) b \,e^{3} m \,x^{3}-9 \,\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) \mathrm {log}\left (x^{n} c \right ) b \,e^{2}-9 \,\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) a \,e^{2}-3 \,\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) b \,e^{2} n -6 \,\mathrm {log}\left (x^{n} c \right ) b \,e^{2} m -18 a e f m \,x^{2}-2 b \,e^{2} m n -6 b e f m n \,x^{2}}{27 e^{2} x^{3}} \] Input:

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

Output:

( - 18*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*f*m*x**3 - 6*sqrt(f 
)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*f*m*n*x**3 - 18*int(log(x**n*c)/ 
(e*x**4 + f*x**6),x)*b*e**3*m*x**3 - 9*log((e + f*x**2)**m*d)*log(x**n*c)* 
b*e**2 - 9*log((e + f*x**2)**m*d)*a*e**2 - 3*log((e + f*x**2)**m*d)*b*e**2 
*n - 6*log(x**n*c)*b*e**2*m - 18*a*e*f*m*x**2 - 2*b*e**2*m*n - 6*b*e*f*m*n 
*x**2)/(27*e**2*x**3)